MATH592

Introduction to Algebraic Topology

Pingbang Hu

September 19, 2024

 Abstract

This is a graduate-level course on algebraic topology taught by Jennifer Wilson at University of Michigan.
It is self-contained enough that only requires a background in abstract algebra and some point set
topology. We will use Hatcher [HPM02] as the main text, but the order may differ here and there. Enjoy
this fun course! In particular, I added some extra content that is not covered in lectures, things like
groupoid, fibered coproduct, feel free to skip these contents.

This course is taken in Winter 2022, and the date on the cover page is the last updated time.
Contents

1 Foundation of Algebraic Topology 3

1.1 Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Homotopy Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.3 CW Complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.4 Operations on CW Complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 Category Theory 16
2.1 Category Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.2 Free Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3 The Fundamental Group 23

3.1 Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.2 Fundamental Group and Groupoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.3 Calculations of Fundamental Group of Spheres . . . . . . . . . . . . . . . . . . . . . . . . 31
3.4 Fundamental Group and Groupoid Define Functors . . . . . . . . . . . . . . . . . . . . . . 35
3.5 Free Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.6 Seifert-Van Kampen Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.7 Group Presentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4 Covering Spaces 55
4.1 Lifting Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.2 Deck Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5 Homology 74
5.1 Motivation for Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
5.2 Simplicial Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
5.3 Singular Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
5.4 Functoriality and Homotopy Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
5.5 Relative Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
5.6 Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
5.7 Cellular Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
5.8 Eilenberg-Steenrod Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

6 Lefschetz Fixed Point Theorem 111

6.1 Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
6.2 Lefschetz Fixed Point Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
6.3 Simplicial Approximation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
6.4 Proof of Lefschetz Fixed Point Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

7 Epilogue 117
7.1 Proof of Seifert-Van-Kampen Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
7.2 Review Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

A Additional Proofs 124

A.1 Seifert-Van Kampen Theorem on Groupoid . . . . . . . . . . . . . . . . . . . . . . . . . . 124
A.2 An alternative proof of Seifert Van-Kampen Theorem . . . . . . . . . . . . . . . . . . . . 127
A.3 Cellular Boundary Formula in Definition 5.7.1 . . . . . . . . . . . . . . . . . . . . . . . . . 128

1
B Algebra 129
B.1 Abelian Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
B.2 Free Abelian Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
B.3 Finitely Generated Abelian Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
B.4 Homological Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

CONTENTS 2
Chapter 1

Foundation of Algebraic Topology

Lecture 1: Homotopies of Maps

In this chapter, we’ll give a few, but important definitions which will be used throughout the discussion 05 Jan. 10:00
of this course. This will help us build up a solid understanding about spaces we’re working on both
mathematically and visually.

1.1 Homotopy
We start with the most important and fundamental concept, homotopy.

Definition. Let X, Y be topological spaces, and f , g : X → Y being two continuous maps.

Definition 1.1.1 (Homotopy). A homotopy from f to g is a 1-parameter family of maps that

continuously deforms f to g, i.e., it’s a continuous function F : X × I → Y , where I = [0, 1],
such that
F (x, 0) = f (x), F (x, 1) = g(x).
We often write Ft (x) for F (x, t).

Ft

Figure 1.1: The continuous deforming from f to g described by Ft

Definition 1.1.2 (Homotopic). If a homotopy exists between f and g, we say they are homotopic
and write
f ≃ g.

Definition 1.1.3 (Nullhomotopic). If f is homotopic to a constant map, we call it nullhomotopic.

Remark. Later, we’ll not state that a map is continuous explicitly since we almost always assume
this in this context.

3
 Lecture 1: Homotopies of Maps

Example (Straight line homotopy). Any two (continuous) maps with specification

f, g : X → Rn

are homotopic by considering

Ft (x) = (1 − t)f (x) + tg(x).
We call it the straight line homotopy.

Example. Let S 1 denotes the unit circle in R2 , and D2 denotes the unit disk in R2 . Then the
inclusion f : S 1 ,→ D2 is nullhomotopic.

Proof. We see this by considering

Ft (x) = (1 − t)f (x) + (t · 0).

1
t=0 t= 2
t=1

Figure 1.2: The illustration of Ft (x)

We see that there is a homotopy from f (x) to 0 (the zero map which maps everything to 0),
and since 0 is a constant map, hence it’s actually a nullhomotopy. ⊛

Example. The maps

S1 → S1 S1 → S1
and
Θ 7 → S1 Θ 7 → −Θ
are not homotopy.

Remark. It will essentially flip the orientation, hence we can’t deform one to another contin-
uously.

Exercise. A subset S ⊆ Rn is star-shaped if ∃x0 ∈ S s.t. ∀x ∈ S, the line from x0 to x lies in S.

Show that id : S → S is nullhomotopic.

x0

Figure 1.3: Star-shaped illustration

Answer. Consider
Ft (x) := (1 − t)x + tx0 ,

CHAPTER 1. FOUNDATION OF ALGEBRAIC TOPOLOGY 4

 Lecture 2: Homotopy Equivalence

which essentially just concentrates all points x to x0 . ⊛

Exercise. Suppose
f1 g1
X Y Z
f0 g0

where
f0 ≃ f1 , g0 ≃ g1 .
Ft Gt

Show
g0 ◦ f0 ≃ g1 ◦ f1 .

Answer. Consider I × X → Z, where

X ×I → Y ×I → Z
(x, t) 7→ (Ft (x), t) 7→ Gt (Ft (x)).

Remark. Noting that if one wants to be precise, you need to check the continuity of this
construction.

Exercise. How could you show 2 maps are not homotopic?

Answer. We’ll see! ⊛

Lecture 2: Homotopy Equivalence

As previously seen. Two maps f, g : X → Y is homotopy if there exists a map 07 Jan. 10:00

Ft (x) : X × I → Y

with the properties

(a) Continuous
(b) F0 (x) = f (x)

(c) F1 (x) = g(x)

Remark. The continuity of Ft is an even stronger condition for the continuity of Ft for a fixed
t.

We now introduce another concept.

Definition 1.1.4 (Homotopy relative). Given two spaces X, Y , and let B ⊆ X. Then a homotopy
Ft (x) : X → Y is called homotopy relative B (denotes rel B) if Ft (b) is independent of t for all b ∈ B.

Example. Given X and B = {0, 1}. Then the homotopy of paths from [0, 1] → X is rel{0, 1}.

CHAPTER 1. FOUNDATION OF ALGEBRAIC TOPOLOGY 5

 Lecture 2: Homotopy Equivalence

F0 (t)

F0 (0) F0 (1)

1.2 Homotopy Equivalence

With this, we can introduce the concept of homotopy equivalence.

Definition. Let X, Y be topological spaces, and f : X → Y being a continuous map.

Definition 1.2.1 (Homotopy equivalence). A map f : X → Y is a homotopy equivalence if

∃g : Y → X such that
f ◦ g ≃ idY , g ◦ f ≃ idX .
f
X Y
g ◦ f ≃ id f ◦ g ≃ id
g

Definition 1.2.2 (Homotopy equivalent). If f : X → Y is a homotopy equivalence, we then say

that X, Y are homotopy equivalent.

Definition 1.2.3 (Homotopy inverse). If f : X → Y is a homotopy equivalence realize with g,

then g is called the homotopy inverse of f .

Definition 1.2.4 (Homotopy type). If X, Y are homotopy equivalent, then we say that they
have the same homotopy type.

Notation. We denote a closed n-disk as Dn .

Example. Dn is homotopy equivalent to a point.

∗
n
g
D

Proof. We see that f ◦ g = id∗ and

g ◦ f = constant map at |{z}

0 ,
g(∗)

which is homotopic to idDn by straight line homotopy Ft (x) = tx. Specifically, we see that this
holds for any convex set. ⊛

Definition 1.2.5 (Contractible). We say that a space X is contractible if X is homotopy equivalent

to a point.
The following proposition is added much after, which may uses some concepts not yet covered.

CHAPTER 1. FOUNDATION OF ALGEBRAIC TOPOLOGY 6

 Lecture 2: Homotopy Equivalence

Proposition 1.2.1. The followings are equivalent.

(a) X is contractible.
(b) ∀x ∈ X, idX ≃ cx .
(c) ∃x ∈ X, idX ≃ cx .
Proof. We see that 2. ⇒ 3. is obvious. We consider 3. ⇒ 2. This follows from the following general
lemma.

Lemma 1.2.1. Given a topological space X such that ∃x ∈ X, idX ≃ cp , with f, g : Y → X,

then f ≃ g.
Proof. Let x ∈ X such that idX ≃ cx . Then

f = idX ◦f ≃ cx ◦ f = cx ◦ g ≃ idX ◦g = g.

■
Then, from this Lemma 1.2.1, we see that assuming x0 ∈ X such that idX ≃ cx0 , then consider cx
for all x ∈ X, then from Lemma 1.2.1, we see that cx ≃ idX .
To show 3. ⇒ 1., we let x0 ∈ X such that idX ≃ cx0 .

f
X {∗}
g

Since g(∗) = x0 , and

g◦f: X →X
x 7→ x0 ,

which is just cx0 , from the assumption we’re done.

Now, we show 1. ⇒ 3. Let
f
X {∗}
g

be a homotopy equivalent, let g(∗) = x0 . We see that cx0 ≃ idX since

g ◦ f = cx0 ≃ idX .

Remark. Note that the above notation cx is introduced here.

Definition. Before doing exercises, we introduce the following new concepts.

Definition 1.2.6 (Retraction). Given B ⊆ X, a retraction from X to B is a map f : X → X

(or X → B) such that ∀b ∈ B f (b) = b, namely r|B = idB . Or one can see this from

i r
B X B
r◦i

where r is a retraction if and only if r ◦ i = idB , where i is an inclusion identity.

Definition 1.2.7 (Retract). If the above r exists, we say that B is a retract of X.

CHAPTER 1. FOUNDATION OF ALGEBRAIC TOPOLOGY 7

 Lecture 2: Homotopy Equivalence

Definition 1.2.8 (Deformation retraction). Given X and B ⊆ X, a (strong) deformation re-

traction Ft : X → X onto B is a homotopy rel B from the idX to a retraction from X to B.
i.e.,
F0 (x) =x ∀x ∈ X
F1 (x) ∈ B ∀x ∈ X
Ft (b) =b ∀t ∀b ∈ B.

Exercise. Let X ≃ Y . Show X is path-connected if and only if Y is.

Answer. Suppose X is path-connected. Then we see that given two points x1 and x2 in X, there
exists a path γ(t) with
γ : [0, 1] → X, γ(0) = x1 , γ(1) = x2 .
Since X ≃ Y , then there exists a pair of f and g such that f : X → Y and g : Y → X with

f ◦ g ≃ idY , g ◦ f ≃ idX .
F G

(Notice the abuse of notation)

For any two y1 and y2 ∈ Y , we want to construct a path γ ′ (t) such that

γ ′ : [0, 1] → Y, γ ′ (0) = y1 , γ ′ (1) = y2 .

Firstly, we let g(y1 ) =: x1 and g(y2 ) =: x2 . From the argument above, we know there exists such
a γ starting at x1 = g(y1 ) ending at x2 = g(y2 ). Now, consider f (γ(t)) = (f ◦ γ)(t) such that

f ◦ γ : I → Y, f ◦ γ(0) = y1′ , f ◦ γ(1) = y2′ ,

we immediately see that y1′ and y2′ is path connected. Now, we claim that y1 and y1′ are path
connected in Y , hence so are y2 and y2′ . To see this, note that

f ◦ g ≃ idY ,
F

which means that there exists F : Y × I → Y such that

(
F (y1 , 0) = f ◦ g(y1 ) = f (x1 ) = f (γ(0)) = (f ◦ γ)(0) = y1′
F (y1 , 1) = idY (y1 ) = y1 .

Since F is continuous in I, we see that there must exist a path connects y1 and y1′ . The same
argument applies to y2 and y2′ . Now, we see that the path

y1 → y1′ → y2′ → y2

is a path in Y for any two y1 and y2 , which shows Y is path-connected.

CHAPTER 1. FOUNDATION OF ALGEBRAIC TOPOLOGY 8

 Lecture 3: Deformation Retraction

I I
0 1 0 1

f
γ y1

Ft (y1 ) γ′
g(y1 ) = γ(0) =: x1 y2
y1′

f Ft (y2 )

y2′
g(y2 ) = γ(1) =: x2
g
X Y

Figure 1.4: Demonstration of the proof.

⊛
One can further show that the connectedness is also preserved by any homotopy equivalence.

Corollary 1.2.1. A contractible space is path-connected.

Exercise. Show that if there exists deformation retraction from X to B ⊆ X, then X ≃ B.

Lecture 3: Deformation Retraction

As previously seen. A deformation retraction is a homotopy of maps rel B X → X from idX to a 10 Jan. 10:00
retraction from X to B. Then B is a deformation retract.

Example. S 1 is a deformation retraction of D2 \ {0}.

Proof. Indeed, since

x
Ft (x) = t · + (1 − t)x.
∥x∥

Figure 1.5: The deformation retraction of D2 \ {0} is just to enlarge that hole and push all the
interior of D2 to the boundary, which is S 1 .

CHAPTER 1. FOUNDATION OF ALGEBRAIC TOPOLOGY 9

 Lecture 4: Cell Complex (CW Complex)

Example. Rn deformation retracts to 0.

Proof. Indeed, since

Ft (x). = (1 − t)x.

Remark. This implies that Rn ≃ ∗, hence we see that

• dimension
• compactness
• etc.

are not homotopy invariants.

Example. S 1 is a deformation retract of a cylinder and a Möbius band.

Proof. For a cylinder, consider X × I → X. Define homotopy on a closed rectangle, then verify it
induces map on quotient.
For a Möbius band, we define a homotopy on a closed rectangle, then verify that it respect the
equivalence relation.
Finally, we use the universal property of quotient topology to argue that we get a homotopy on
Möbius band.

Figure 1.6: The deformation retraction for Cylinder and Möbius band

Remark. We see that Möbius band ≃ S 1 ≃ cylinder, hence the orientability is not homotopy
invariant.

Lecture 4: Cell Complex (CW Complex)

As previously seen. We saw that 12 Jan. 10:00

• homotopy equivalence
• homotopy invariants

CHAPTER 1. FOUNDATION OF ALGEBRAIC TOPOLOGY 10

 Lecture 4: Cell Complex (CW Complex)

– path-connectedness
• not invariant
– dimension
– orientability
– compactness

1.3 CW Complexes
Example (Constructing spheres). We now see how to construct S 1 and S 2 from ground up.
• S 1 (up to homeomorphisma )

• S2
– glue boundary of 2-disk to a point
– glue 2 disks onto a circle

Figure 1.7: Left: Glue a 2-disk to a point along its boundary. Right: Glue 2 disks to S 1 .

The gluing instruction to construct S 2 in the right-hand side can be demonstrated as follows.

CHAPTER 1. FOUNDATION OF ALGEBRAIC TOPOLOGY 11

 Lecture 4: Cell Complex (CW Complex)

top hemi-sphere

bottom hemi-sphere

a This is just the term for isomorphism in topology.

Example (Constructing torus). A torus T is just T = S 1 × S 1 .

view as gluing instructions

vertex + 2 edges + 2-disks.
Soon, we’ll see that above is just

0-skeleton 1-skeleton 2-skeleton

Gluing edges Gluing disk

Notation. Let Dn denotes a closed n-disk (or n-ball) Dn ≃ {x ∈ Rn : ∥x∥ ≤ 1}, and let S n denotes
an n-sphere S n ≃ x ∈ Rn+1 : ∥x∥ = 1 .

Then, formally, we have the following definition.

Definition 1.3.1 (CW Complex). A CW Complex X is a topological space constructed inductively

as follows. First, we have the basic elements called cells.

Definition 1.3.2 (Cell). We call a point as a 0-cell, and the interior of Dn Int(Dn ) for n ≥ 1
as an n-cell.

CHAPTER 1. FOUNDATION OF ALGEBRAIC TOPOLOGY 12

 Lecture 4: Cell Complex (CW Complex)

Then, we use cells to build our skeletons.

Definition 1.3.3 (Skeleton). The 0-skeleton X 0 is a set of discrete points, i.e, 0-cell.
We inductively construct the n-skeleton X n from X n−1 by attaching n-cells enα , where α is the
index. Specifically, we follow the gluing instructions called attaching maps defined below and
obtain X n from X n−1 by
!
a
n−1 n .
Xn = X Dα
x ∼ φ (x)
α α

with identification x ∼ φα (x) for all x ∈ ∂Dαn with quotient topology.

Remark. We write X (n) for n-skeleton if we need to distinguish from the Cartesian prod-
uct.

Definition 1.3.4 (Attaching map). The attaching map for every cell enα is a continuous map φα
such that
φα : ∂Dαn → X n−1 .
Finally, we let X be defined as [
X= X n,
n=0

equipped with the so-called weak topology defined below.

Definition 1.3.5 (Weak topology). The weak topology, denoted as w, contains open set u such
that
u ⊆ X is open ⇔ ∀n u ∩ X n is open .
If all cells have dimension less than N and there exists an N -cell, then X = X N and we call it
N -dimensional CW complex.

Example. Let’s look at some examples.

(a) 0-dim CW complex is a discrete space.

(b) 1-dim CW complex is a graph.
(c) A CW complex X is finite if it has finitely many cells.

Definition 1.3.6 (CW subcomplex). A CW subcomplex A ⊆ X is a closed subset equal to a union of

cells
enα = Int (Dαn ) .

Remark. This inherits a CW complex structure. Check the

images of
attaching
Exercise. Given the following gluing instruction: maps.

CHAPTER 1. FOUNDATION OF ALGEBRAIC TOPOLOGY 13

 Lecture 5: Operation on Spaces

identify Torus, Klein bottle, Cylinder, Möbius band, 2-sphere, RP .

Notation. Notice that we call the real projection space as RP , and we also have so-called
complex projection space, denote as CP .

Answer. We see that

1. Torus 2. Cylinder 3. 2-sphere

4. Klein bottle 5. Möbius band 6. RP

Lecture 5: Operation on Spaces

14 Jan. 10:00
1.4 Operations on CW Complexes
1.4.1 Products
We can consider the product of two CW complex given by a CW complex structure. Namely, given X
and Y two CW complexes, we can take two cells enα from X and em β from Y and form the product space
enα × em
β , which is homeomorphic to an (n + m)-cell. We then take these products as the cells for X × Y .
Specifically, given X, Y are CW complexes, then X × Y has a cells structure

α × eα : eα is an m-cell on X, eα is an n-cell on Y } .
{em n m n

Remark. The product topology may not agree with the weak topology on the X × Y . However,
they do agree if X or Y is locally compact or if X and Y both have at most countably many cells.

1.4.2 Wedge Sum

Given X, Y are CW complexes, and x0 ∈ X 0 , y0 ∈ Y 0 (only points). Then we define X ∨ Y = X
`
Y
with quotient topology.

Remark. X ∨ Y is a CW complex.

CHAPTER 1. FOUNDATION OF ALGEBRAIC TOPOLOGY 14

 Lecture 5: Operation on Spaces

1.4.3 Quotients
Let X be a CW complex, and A ⊆ X subcomplex (closed union of cells), then X / A is a quotient space
collapse A to one point and inherits a CW complex structure.

Remark. X / A is a CW complex.

Proof. With the 0-skeleton being a

(X 0 − A0 ) ∗
n
where ∗ is a point for A, then each cell of X − A is attached to (X / A) by attaching map

ϕα quotient
Sn Xn X n / An

Example. We can take the sphere and squish the equator down to form a wedge of two spheres.

X X /A

A
A

Example. We can take the torus and squish down a ring around the hole.
X X /A

≃ ≃

A A
We see that X / A is homotopy equivalent to a 2-sphere wedged with a 1-sphere via extending the
orange point into a line, and then sliding the left point to the line along the 2-sphere towards the
other points, forming a circle.

CHAPTER 1. FOUNDATION OF ALGEBRAIC TOPOLOGY 15

Chapter 2

Category Theory

Lecture 6: A Foray into Category Theory

Category theory is a general theory of mathematical structures and their relations introduced in the 19 Jan. 10:00
middle of the 20th century along with the development of algebraic topology. Nowadays, category
theory is used in almost all areas of mathematics, and in some areas of computer science. In particular,
many constructions of new mathematical objects from previous ones, that appear similarly in several
contexts are conveniently expressed and unified in terms of categories. Hence, we now study this in depth
and will use it heavily afterwards.

2.1 Category Theory

When discussing algebraic-related topics, category theory is an essential tool doing this. It exploits some
internal structure of objects we care about, such as fundamental group, homology group and so on, as
we’ll see.
Now, we see start with the definition.

Definition 2.1.1 (Category). A category C consists of 3 pieces of data

• A class of objects Ob(C ).

Definition 2.1.2 (Object). An object in a category can be realized as a node in a directed

diagram.

• ∀X, Y ∈ Ob(C ) a class of morphisms or arrows, HomC (X, Y ).

Definition 2.1.3 (Morphism). A morphism in a category is just an arrow between objects.

• ∀X, Y, Z ∈ Ob(C ), there exists a composition law

Hom(X, Y ) × Hom(Y, Z) → Hom(X, Z), (f, g) 7→ g ◦ f.

In addition, it also contains 2 axioms

• Associativity. (f ◦ g) ◦ h = f ◦ (g ◦ h) for all morphisms f, g, h where composites are defined.
• Identity. ∀X ∈ Ob(C ) ∃ idX ∈ HomC (X, X) such that

f ◦ idX = f, idX ◦g = g

for all f, g where this makes sense.

16
 Lecture 6: A Foray into Category Theory

Example. We introduce some common category.

C Ob(C) Mor(C)
set Sets X All maps of sets
fset Finite sets All maps
Gp Groups Group Homomorphisms
Ab Abelian groups Group Homomorphisms
k−vect Vector spaces over k k-linear maps
Rng Rings Ring Homomorphisms
Top Topological spaces Continuous maps
Haus Hausdorff Spaces Continuous maps
hTop Topological spaces Homotopy classes of continuous maps
Top∗ Based topological spacesa Based mapsb
a Topological spaces with a distinguished base point x0 ∈ X
b Continuous maps that presence base point f : (X, x0 ) → (Y, y0 ) such that f : X → Y, f (x0 ) = y0 is continuous.

Remark. From Definition 2.1.1, we see that a category is just any directed diagram plus compo-
sition law and identities.
idA A B idB .

Definition. Given a morphism f : M → N , then we have the following definitions.

Definition 2.1.4 (Monic). f is monic if

∀g1 , g2 f ◦ g1 = f ◦ g2 ⇒ g1 = g2 .
g1
f
A M N
g2

Definition 2.1.5 (Epic). Dually, f is epic if

∀g1 , g2 g1 ◦ f = g2 ◦ f ⇒ g1 = g2 .
g1
f
M N B
g2

Lemma 2.1.1. In set, Ab, Top, Gp, a map is monic if and only if f is injective, and epic if and only
if f is surjective.
Proof. In set, we prove that f is monic if and only if f is injective. Suppose f ◦ g1 = f ◦ g2 and f
is injective, then for any a,

f (g1 (a)) = f (g2 (a)) ⇒ g1 (a) = g2 (a),

hence g1 = g2 .
Now we prove another direction, with contrapositive. Namely, we assume that f is not injective
and show that f is not monic. Suppose f (a) = f (b) and a ̸= b, we want to show such gi exists.
This is easy by considering
g1 : ∗ 7→ a, g2 : ∗ 7→ b.
■

CHAPTER 2. CATEGORY THEORY 17

 Lecture 7: Functors

2.1.1 Functor
After introducing the category, we then see the most important concept we’ll use, a functor. Again, we
start with the definition.

Definition 2.1.6 (Functor). Given C , D be two categories. A (covariant) functor F : C → D is

(a) a map on objects

F : Ob(C ) → Ob(D)
X 7→ F (X);

(b) maps of morphisms

HomC (X, Y ) → HomD (F (X), F (Y ))

[f : X → Y ] 7→ [F (f ) : F (X) → F (Y )]

such that
• F (idX ) = idF (x) ;
• F (f ◦ g) = F (f ) ◦ F (g).

Lecture 7: Functors
Example (Applying a covariant functor). Assume that we initially have a commutative diagram in C 21 Jan. 10:00
as
f
X Y
g
g◦f
Z
and a functor F : C → D. After applying F , we’ll have

F (f )
F (X) F (Y )
F (g)
F (g◦f )=F (g)◦F (f )
F (Z)

which is a commutative diagram in D.

We can also have a so-called contravariant functor.

Definition 2.1.7 (Contravariant functor). Given C , D be two categories. A contravariant functor

F : C → D is
(a) a map on objects

F : Ob(C ) → Ob(D)
X 7→ F (X).

(b) maps of morphisms

HomC (X, Y ) → HomD (F (Y ), F (X))

[f : X → Y ] 7→ [F (f ) : F (Y ) → F (X)]

such that

CHAPTER 2. CATEGORY THEORY 18

 Lecture 7: Functors

• F (idX ) = idF (x)

• F (f ◦ g) = F (g) ◦ F (f )

Example (Applying a contravariant functor). Then, we see that in this case, when we apply a con-
travariant functor F , the diagram becomes

F (f )
F (X) F (Y )
F (g)
F (g◦f )=F (f )◦F (g)
F (Z)

which is a commutative diagram in D.

We now see some common functors as examples.

Example (Identity functor). Define I as I : C → C such that it just send an object C ∈ C to itself.

Example (Forgetful functor). We see two examples.

• Define F as F : Gp → set such that G 7→ G.a Specifically,

[f : G → H] 7→ [f : G → H] .

• Define F as F : Top → set such that X 7→ X.b Specifically,

[f : X → Y ] 7→ [f : X → Y ] .
aG is now just the underlying set of the group G.
bX is now just the underlying set of the topological space X.

Example (Free functor). Define a functor as

set → k−vect
s 7→ "free" k-vector space on s

i.e., vector space with basis s such that

[f : A → B] 7→ [unique k-linear map extending f ]

Example. Let T be defined as

T : k−vect → k−vect
V 7→ V ∗ = Homk (V, k).

If we are working on a basis, we can then represent T as a matrix A, and we further have

A 7→ A⊤ .

Remark. Specifically, we care about two functors.

(a) π1 : Top∗ → Gp such that

π1 : Top∗ → Gp
(X, x0 ) 7→ π1 (X, x0 ),

CHAPTER 2. CATEGORY THEORY 19

 Lecture 7: Functors

where π1 is the so-called fundamental group.

(b) Hp : Top → Ab such that

Hp : Top → Ab
X 7→ Hp (X),

where Hp is the so-called pth homology group.

We are not building toward fundamental groups.
Let’s see some building blocks we need.

2.2 Free Groups

Definition 2.2.1 (Free group). Given a set S, the free group is a group FS on S with a map S → FS
satisfying the following universal property: If G is any group, f : S → G is any map of sets, f
extends uniquely to group homomorphism f : FS → G.

S FS
∃!f : group homorphism
f
G

Note. This defines a natural bijection

Homset (S, U (G)) ∼

= HomGp (FS , G),

where U (G) is the forgetful functor from the category of groups to the category of sets. This is
the statement that the free functor and the forgetful functor are adjoint; specifically that the free
functor is the left adjoint (appears on the left in the Hom above).

Definition 2.2.2 (Adjoints functor). A free and forgetful functor is adjoints.

Remark. Whenever we state a universal property for an object (plus a map), an object (plus a map)
may or may not exist. If such object exists, then it defines the object uniquely up to unique
isomorphism, so we can use the universal property as the definition of the object (plus a map).

Lemma 2.2.1. Universal property defines FS (plus a map S → F (S) ) uniquely up to unique
isomorphism.
Proof. Fix S. Suppose
S → FS , S → FeS
both satisfy the unique property. By universal property, there exist maps such that

S FeS S FS
∃!φ ∃!ψ
f f

FS FeS

We’ll show φ and ψ are inverses (and the unique isomorphism making above commute). Since we

CHAPTER 2. CATEGORY THEORY 20

 Lecture 8: The Fundamental Group π1

must have the following two commutative graphs.

FS FeS
f f

S idFS S idFe
S

f f
FS FeS

Hence, we see that

FS FeS
f ψ f
φ

φ◦ψ=idFS S FS ψ◦φ=idFe
S FeS S

φ ψ
f f
FS FeS

where the identity makes these outer triangles commute, then by the uniqueness in universal prop-
erty, we must have
φ ◦ ψ = idFS , ψ ◦ φ = idFeS ,
so φ and ψ are inverses (thus group isomorphism). ■

Lecture 8: The Fundamental Group π1

Example. In category Ab free Abelian group on a set S is 24 Jan. 10:00
M
Z.
S

In category of fields, no such thing as free field on S.

2.2.1 Constructing the Free Groups FS

Proposition 2.2.1. The free group defined by the universal property exists.
Proof. We’ll just give a construction below. First, we see the definition.

Definition 2.2.3 (Word). Fix a set S, and we define a word as a finite sequence (possibly ∅)
in the formal symbols  −1
s, s | s ∈ S .

Then we see that elements in FS are equivalence classes of words with the equivalence relation
being
• deleted ss−1 or s−1 s. i.e.,

vs−1 sw ∼ vw
vss−1 w ∼ vw

for every word v, w, s ∈ S,

with the group operation being concatenation.

CHAPTER 2. CATEGORY THEORY 21

 Lecture 8: The Fundamental Group π1

Example. Given words ab−1 , bba, their product is

ab−1 · bba = ab−1 bba = aba.

Exercise. There are something left to check.

(a) This product is well-defined on equivalence classes.
(b) Every equivalence class of words has a unique reduced form, namely the representation.

(c) Check that FS satisfies the universal property with respect to the map

S → FS , s 7→ s.

CHAPTER 2. CATEGORY THEORY 22

Chapter 3

The Fundamental Group

The fundamental group is the first and the simplest homotopy group, which records information about
the basic shape, or holes, of the topological space. Not surprisingly, this is already interesting enough
to study and can help us distinguish between different shapes just by finding the spaces’ corresponding
fundamental groups.

3.1 Path
We start with the definition.

Definition 3.1.1 (Path). A path in a space X is a continuous map

γ: I → X

where I = [0, 1].

Definition 3.1.2 (Homotopy path). A homotopy of paths γ0 , γ1 is a homotopy from γ0 to γ1 rel{0, 1}.

γ0 (0) = γ1 (0) γ0 (1) = γ1 (1)

γt

Example. Fix x1 , x0 ∈ X, then ∃ homotopy of paths is an equivalence relation on paths from x0 to

x1 (i.e., γ with γ(0) = x0 , γ(1) = x1 ).

Definition 3.1.3 (Path composition). For paths α, β in X with α(1) = β(0), the composition a α · β
is  
1


 α(2t), if t ∈ 0,
 2
(α · β)(t) :=  
1
β(2t − 1), if t ∈

 ,1 .
2

α α(1) = β(0) β(1)

α(0) β

a Also named product, concatenation.

23
 Lecture 8: The Fundamental Group π1

Remark. By the pasting lemma, this is continuous, hence α · β is actually a path from α(0) to β(1).

Definition 3.1.4 (Reparameterization). Let γ : I → X be a path, then a reparameterization of γ is a

path
φ γ
γ ′ : I −→ I −→ X
where φ is continuous and
φ(0) = 0, φ(1) = 1.

Exercise. A path γ is homotopic rel{0, 1} to all of its reparameterizations.

Answer. We show that γ and γ ◦ ϕ are homotopic rel{0, 1} by showing that there exists a continuous
Ft such that
F0 = γ, F1 = γ ◦ ϕ.
Notice that since ϕ is continuous, so we define

Ft (x) = (1 − t)γ(x) + t · γ ◦ ϕ(x).

We see that
F0 (x) = γ(x), F1 (x) = γ ◦ ϕ(x),
and also, we have
Ft (x) ∈ X
for all x, t ∈ I.
Now, we check that Ft really gives us a homotopic rel{0, 1}. We have

Ft (0) = (1 − t)γ(0) + t · γ ◦ ϕ(0) = (1 − t)γ(0) + t · γ(ϕ(0)) = γ(0),

|{z}
0
Ft (1) = (1 − t)γ(1) + t · γ ◦ ϕ(1) = (1 − t)γ(1) + t · γ(ϕ(1)) = γ(1),
|{z}
1

which shows that 0 and 1 are independent of t, hence γ and γ ◦ ϕ are homotopic rel{0, 1}. ⊛

Exercise. Fix x1 , x1 ∈ X. Then homotopy of paths (relative {0, 1}) is an equivalence relation on
paths from x0 to x1 .

3.2 Fundamental Group and Groupoid

In class, we only discuss fundamental groups. But there is an also important concept called fundamental
groupoid, which is kind of similar but more general and power to use. I include this just for completeness,
feel free to skip the additional content. Now, let’s start with fundamental group.

3.2.1 Fundamental Group

Definition 3.2.1 (Fundamental group). Let X denotes the space and let x0 ∈ X be the base point.
The fundamental group of X based at x0 , denoted by π1 (X, x0 ), is a group such that
• Elements: Homotopy classes rel{0, 1} of paths [γ] where γ is a loop with γ(0) = γ(1) = x0 a

CHAPTER 3. THE FUNDAMENTAL GROUP 24

 Lecture 8: The Fundamental Group π1

x0
γ

• Operation: Composition of paths.

• Identity: Constant loop γ based at x0 such that

γ : I → X, t 7→ x0

• Inverses: The inverse [γ]−1 of [γ] is represented by the loop γ such that

γ(t) = γ(1 − t).

x0 x0
γ(t) γ(1 − t)

a We say γ is based at x0 .

Proof. We actually need to prove that the defined π1 actually is a group, hence, we prove that

Associativity. [γ1 · (γ2 · γ3 )] = [(γ1 · γ2 ) · γ3 ]. We break this down into

  
1
  
 γ 1 (2t), t ∈ 0, ;
1 2
 

γ1 (2t), t ∈ 0, ; 
 

  
 2 1 3
γ1 · (γ2 · γ3 )(t) =   = γ2 (4t − 2), t ∈ , ;
 1  2 4
(γ2 · γ3 )(2t − 1), t ∈
 ,1 
  
2 3


γ3 (4t − 3), t ∈ ,1 ,


4

and   
1
  
 γ1 (4t), t ∈ 0, ;
1 4
 

(γ · γ )(2t), t ∈ 0, ;
 

 1 2

2

 
1 1

(γ1 · γ2 ) · γ3 (t) =   = γ2 (4t − 1), t ∈ , ;
 1  4 2
γ3 (2t − 1),
 t∈ ,1 
  
2 1


γ3 (2t − 1), t ∈ ,1 .


2
Then, we define ϕ : I → I such that
    
1 1

 2t ∈ 0, , t ∈ 0, ;
2 4




    
 1 1 3 1 1
ϕ(t) = t + ∈ , , t∈ , ;

 4 2 4 4 2
    
t+1 3 1


∈ ,1 , t ∈ ,1 .



2 4 2

We easily see that

γ1 · (γ2 · γ3 )(t) = (γ1 · γ2 ) · γ3 ◦ ϕ(t)
and ϕ(t) is continuous and satisfied ϕ(0) = 0 and ϕ(1) = 1, which implies that the associativity
holds.

CHAPTER 3. THE FUNDAMENTAL GROUP 25

 Lecture 8: The Fundamental Group π1

Identity. We want to show that [γ · c] = [γ]. Again, we consider

 
1

γ(2t),

 t ∈ 0, ;
2
(γ · c)(t) =  
 1
c(2t − 1) = c = x0 = γ(0), t ∈
 ,1 .
2

Now, consider ϕ : I → I such that

 
1

2t,

 t ∈ 0, ;
2
ϕ(t) =  
 1
1,
 t∈ ,1 .
2

We easily see that

(γ · c)(t) = (γ ◦ ϕ)(t)
and ϕ(t) is continuous and satisfied ϕ(0) = 0 and ϕ(1) = 1.

Inverses. We want to show that γ · γ ≃ c, where γ(t) = γ(1 − t). Firstly, we have
 
1


 γ(2t), t ∈ 0, ;
 2
(γ · γ)(t) =  
 1
γ(1 − 2t), t ∈
 ,1 .
2

We consider Ft given by
 
1

γ(2xt),

 x ∈ 0, ;
2
Ft (x) =  
 1
γ(1 − 2xt), x ∈
 ,1 .
2

If t = 0, we have  
1

γ(0),

 x ∈ 0, ;
2
F0 (x) =   = x0
 1
γ(1), x ∈
 ,1
2
for all x ∈ I, namely F0 = c, while when t = 1, we have
 
1

γ(2x),
 x ∈ 0, ;
 2
F1 (x) =   = (γ · γ)(x),
 1
γ(1 − 2x), x ∈
 ,1
2

and we see that Ft is continuous since at x = 12 , we have

γ(2x) = γ(1) = γ(0) = γ(1 − 2x),

hence we see that Ft is the homotopy between γ · γ and c.

CHAPTER 3. THE FUNDAMENTAL GROUP 26

 Lecture 8: The Fundamental Group π1

γ
x0 x0 x0
γ
γ γ

t
γ

γ γ
Figure 3.1: Illustration of Ft . Intuitively, the path γ · γ is x0 → x0 → x0 . But now, Ft is
γ γ
x0 → t → x0 . We can think of this homotopy is pulling back the turning point along the original
path.

Theorem 3.2.1. If X is path-connected, then

∀x0 , x1 ∈ X π1 (X, x0 ) ∼
= π1 (X, x1 ).
Proof. To show that the change-of-basepoint map is isomorphism, we show that it’s one-to-one and
onto.
• One-to-one. Consider that if [h · γ · h] = [h · γ ′ · h], then since we know that h−1 = h, hence
in the fundamental group π1 (X, x0 ), we see that

h · h · γ · h · h = h · h · γ ′ · h · h. ⇒ γ = γ ′

as we desired.
• Onto. We see that for every α ∈ π1 (X, x0 ), there exists a γ ∈ π1 (X, x0 ) such that

γ = h · α · h ∈ π1 (X, x1 )

since h · γ · h = α.a
We then see that the fundamental group of X does not depend on the choice of basepoint, only
on the choice of the path component of the basepoint. If X is path-connected, it now makes sense to
refer to the fundamental group of X and write π1 (X) for the abstract group (up to isomorphism).
■
a Notice that this is indeed the case, one can verify this by the fact that h : x0 → x1 and h : x1 → x0 .

Remark. We see that we can write π1 (X) up to isomorphism if X is path-connected from Theo-
rem 3.2.1.

Exercise. Composition of paths is well-defined on homotopy classes rel{0, 1}.

Exercise. If X is a contractible space, then X is path-connected and π1 (X) is trivial.

The followings are the properties about homotopy path. They are useful when we introduce funda-
mental groupoid.

Lemma 3.2.1. Given x0 , x1 , x2 ∈ X, α, α′ are two paths from x0 to x1 , and β, β ′ are two paths
from x1 to x2 . If ⟨α⟩ = ⟨α′ ⟩, ⟨β⟩ = ⟨β ′ ⟩, then ⟨α · β⟩ = ⟨α′ · β ′ ⟩.

CHAPTER 3. THE FUNDAMENTAL GROUP 27

 Lecture 8: The Fundamental Group π1

Proof. Given α ≃ α′ rel{0, 1}, β ≃ β ′ rel{0, 1}, then we want to prove

F G

α · β ≃ α′ · β ′ rel{0, 1}.

This is done by using homotopy H : I × I → X such that it combines F (2s, t) and G(2s − 1, t).

α β
x0 x1 x2
α′ β′

α′ β′

F (2s, t) G(2s − 1, t)
α β

Lemma 3.2.2. Let x0 , x1 , x2 , x3 ∈ X, α is a path from x0 to x1 , β is a path from x1 to x2 , γ is a

path from x2 to x3 . Then
⟨(α · β) · γ⟩ = ⟨α · (β · γ)⟩.
Proof. We can write out the homotopy by the following diagram.

α′ β′ γ′

α β γ

Lemma 3.2.3. Let X be a topological space, and x0 ∈ X. Then for every path homotopy ⟨α⟩ from
x1 to x2 , we have
⟨cx1 · α⟩ = ⟨α⟩ = ⟨α · cx2 ⟩.
Proof. We only need to prove cx1 · α ≃ α rel{0, 1}. The homotopy can be written out explicitly by
the following diagram.

CHAPTER 3. THE FUNDAMENTAL GROUP 28

 Lecture 8: The Fundamental Group π1

α
1

0
cx1 α

Lemma 3.2.4. For every path homotopy ⟨α⟩ from x1 to x2 , then

⟨α · α−1 ⟩ = ⟨cx1 ⟩, ⟨α−1 · α⟩ = ⟨cx2 ⟩.

Proof. For the first case, we have the following diagram.

cx1
1
cα−1 (1−t)
t From α−1 (0) to α−1 (1 − t)

From α(t) to α(1)

0
α−1 α

The second case follows similarly. ■

3.2.2 Fundamental Groupoid

This section is not covered in class, but it’s a useful concept. The idea is that after giving Definition 3.2.1,
we see that we actually create a fundamental group at every point in X, furthermore, when we use
Theorem 3.2.1 if X is path-connected, we actually lose some information about this space. Here is how
we can store all the information.

Notation (Constant loop). We denote cx , where x ∈ X such that

cx : [0, 1] → X
t 7→ x

as a constant loop.

Definition 3.2.2 (Groupoid). A category C is a groupoid if any morphisms in C is and isomorphism.

Remark. We’ll soon see that for any topological space x, Definition 3.2.1 defines a groupoid, denoted
by Π(X).

Definition 3.2.3 (Fundamental groupoid). Let X denotes the space, then the category Π(X) is a
fundamental groupoid of X such that

• Ob(Π(X)) := X

CHAPTER 3. THE FUNDAMENTAL GROUP 29

 Lecture 8: The Fundamental Group π1

• Hom(Π(X)) : ∀p, q ∈ Ob(Π(X)) = X,

HomΠ(X) (p, q) := {Paths from p to q} ∼.

.

• Composition: For every p, q, r ∈ Ob(Π(X)) = X,

◦ : HomΠ(X) (p, q) × HomΠ(X) (q, r) → HomΠ(X) (p, r)

(⟨α⟩, ⟨β⟩) 7→ ⟨β⟩ ◦ ⟨α⟩ := ⟨α · β⟩.

• Identity: For every p ∈ Ob(Π(X)) = X, we define 1p := ⟨cp ⟩ ∈ HomΠ(X) (p, p) be the constant
loop based at p such that for every ⟨α⟩ ∈ HomΠ(X) (p, q),

⟨α⟩ ◦ idp = idq ◦⟨α⟩ = ⟨α⟩.

• Associativity: Given p, q, r, s ∈ Ob(Π(X)) = X, with the paths

⟨α⟩ ⟨γ⟩
p q ⟨β⟩ r s

Then
⟨γ⟩ ◦ (⟨β⟩ ◦ ⟨α⟩) = (⟨γ⟩ ◦ ⟨β⟩) ◦ ⟨α⟩.
Proof. Note that in Definition 3.2.3, we need to show some definitions are indeed well-defined, and
we also need to show that Π(X) is actually a groupoid.
• Composition: Since if α ≃ α′ , β ≃ β ′ , we have

α · β ≃ α′ · β ′

from Lemma 3.2.1.

• Identity: It follows that
⟨α⟩ ◦ idp = ⟨cp · α⟩ = ⟨α⟩
from Lemma 3.2.3. The left identity can be shown similarly.
• Associativity: It’s trivial in the sense that all the homotopy can be easily derived from
Lemma 3.2.2.

Additionally, from Lemma 3.2.4, we see that given α is a path from p to q, then
( −1
⟨α · α⟩ = ⟨cq ⟩ =: idq
⟨α · α−1 ⟩ = ⟨cp ⟩ =: idp .

Furthermore, since ⟨α−1 · α⟩ = ⟨α⟩ ◦ ⟨α−1 ⟩ and ⟨α · α−1 ⟩ = ⟨α−1 ⟩ ◦ ⟨α⟩, hence this means Π(X) is
indeed a groupoid. ■

Remark. Assume C is a groupoid, then for every x ∈ Ob(C ), we can define

· : HomC (x, x) × HomC (x, x) → HomC (x, x)

such that
(f, g) 7→ f · g := g ◦ f.
We can prove that
(HomC (x, x), ·)
defines a group AutC (x) called the isotropy group of C at x.

CHAPTER 3. THE FUNDAMENTAL GROUP 30

 Lecture 9: Calculate Fundamental Group

Exercise. For every x, y ∈ Ob(C ), if there exists f ∈ HomC (x, y), then f induces
≃
f∗ : AutC (x) → AutC (y),

where f∗ is a group homomorphism.

Remark. For every p ∈ X = Ob(Π(X)), we have

AutΠ(X) (p) = π1 (X, p).

Firstly, since they’re the same in the sense of set:

AutΠ(X) (p) = HomΠ(X) (p, p) = {Loops in X based at p} ∼ = π1 (X, p).

.

Hence, we only need to verify their group composition agrees. But this is trivial, since for every
two ⟨α⟩, ⟨β⟩ ∈ AutΠ(X) (p),

⟨α⟩ · ⟨β⟩ = ⟨β⟩ ◦ ⟨α⟩ = ⟨α · β⟩ .

| {z } | {z }
Composition from AutΠ(X) Composition from π1

This implies that Theorem 3.2.1 is just a particular example as a groupoid.

Lecture 9: Calculate Fundamental Group

As previously seen. Intuitively, fundamental group is like a hole detector. 26 Jan. 10:00

3.3 Calculations of Fundamental Group of Spheres

Let’s start with a basic but important theorem.

Theorem 3.3.1 (The fundamental group of S 1 ). The fundamental group of S 1 is

π1 (S 1 ) ∼
= Z,

and this identification is given by the paths

n ↔ [ωn (t) = (cos(2πnt), sin(2πnt))].

Proof. With the help of covering spaces and the theorems build around which, we can define

p : R → S1 φ : Z → π1 (S 1 , 1)
and
x 7→ e2πix n 7→ ⟨p ◦ γn ⟩

where p defined above is a covering map. We need to show that this is well-defined.
From the definition of φ, we see that it’s a homomorphism. But we also need to show

• φ is a surjection. This is shown by Corollary 4.1.1, specifically in the case of path.

• φ is an injection. This is shown by Corollary 4.1.1, specifically in the case of homotopy of
paths.

Remark. Intuitively, this winds around S 1 n times. The key to this proof was to understand S 1 via

CHAPTER 3. THE FUNDAMENTAL GROUP 31

 Lecture 9: Calculate Fundamental Group

the covering space R → S 1 . We will talk about covering spaces much later.

Theorem 3.3.2. Given (X, x0 ) and (Y, y0 ), then

π(X × Y, (x0 , y0 )) ∼
= π1 (X, x0 ) × π1 (Y, y0 )

such that " #

r: I → X × Y
7→ (rX , rY ).
r(t) = (rX (t), rY (t))

f f
Proof. Let Z → X × Y with z 7→ (fX (z), fY (z)). Then we have

f continuous ⇔ fX , fY are continuous.

Now, apply above to

• Paths I → X × Y .
• Homotopies of paths I × I → X × Y .

Corollary 3.3.1 (The fundamental group of S k ). The torus T ∼

= S 1 × S 1 has fundamental group
∼
π1 (T ) = Z . Additionally, for a k-torus
2

S 1 × S 1 × · · · × S 1 = (S 1 )k ,
| {z }
k times

the fundamental group is then Zk , i.e.

π1 (S 1 )k ∼= Zk .

Proof idea. Since

π1 ∼
= Z2 ∼
= Za ⊕ Zb .

b a
∼
=
b

Remark. One way to think of the k-torus is as a k-dimensional cube with opposite (k−1)-dimensional
faces identified by translation.

CHAPTER 3. THE FUNDAMENTAL GROUP 32

 Lecture 9: Calculate Fundamental Group

Figure 3.2: 3-torus with cube identified with parallel sides.

Lemma 3.3.1. Let f, g : X → Y such that f ≃ g. Let x0 ∈ X, then given

F

f∗ : π1 (X, x0 ) → π1 (Y, f (x0 ))

g∗ : π1 (X, x0 ) → π1 (Y, g(x0 ))

with γ : [0, 1] → Y , t 7→ F (x0 , t),

γ∗ : π1 (Y, f (x0 )) → π1 (Y, g(x0 ))

⟨α⟩ 7→ ⟨γ −1 · α · γ⟩,

the following diagram commutes.

π1 (Y, f (x0 ))
f∗

π1 (X, x0 ) γ∗
∼
=

g∗

π1 (Y, g(x0 ))

Proof. We want to prove that for any ⟨α⟩ ∈ π1 (X, x0 ), we have

γ∗ ◦ f∗ (⟨α⟩) = g∗ (⟨α⟩).

The left-hand side is just

γ∗ ◦ f∗ (⟨α⟩) = γ∗ (⟨f ◦ α⟩) = ⟨γ −1 · (f ◦ α) · γ⟩,

while the right-hand side is just

g∗ (⟨α⟩) = ⟨g ◦ α⟩.
That is, we now want to show
⟨γ −1 · (f ◦ α) · γ⟩ = ⟨g ◦ α⟩.

CHAPTER 3. THE FUNDAMENTAL GROUP 33

 Lecture 9: Calculate Fundamental Group

F1 = g
x0
X ×I α
g(x0 ) Y

g◦α
(x0 , t) F F (x0 , t)

f ◦α
x0 α f (x0 )

F0 = f
We see that we can obtain a homotopy G : I × I → Y such that

G := F ◦ (α × id),

where we define α × id by

α × id : I × I → X × I, (s, t) 7→ (α(s), t).

Figure 3.3: α × id’s image.

We see that by defining such G, we have the following.

G1 g(x0 ) Y
1
g◦α
G
t γ

0 f ◦α
G0 f (x0 )

g(x0 )
g(x0 )
g(x0 )
g(x0 )

To write out this homotopy explicitly, we see the following diagram.

CHAPTER 3. THE FUNDAMENTAL GROUP 34

 Lecture 9: Calculate Fundamental Group

g◦α
1
γ −1
G g(x0 ) → γ(t)
t
γ
γ(t) → g(x0 )
0
γ −1 f ◦ α γ

Theorem 3.3.3 (Fundamental group is a homotopy invariant). If X, Y are homotopy equivalent, then
their fundamental groups are isomorphic.
Proof. ■ HW.

Remark. This gives us a powerful tool to calculate π1 .

Example. π1 (S ∞ × S 1 ) ∼
= Z.

Example. π1 (R2 \ {0}) ∼

= 0 × Z = Z since

R2 \ {0} ∼
= S 1 × R,

which means that the generators are just loops around the hole intuitively.

3.4 Fundamental Group and Groupoid Define Functors

Theorem 3.4.1 (Fundamental group defines a functor). π1 is a functor such that

π1 : Top∗ → Gp
(X, x0 ) 7→ π1 (X, x0 ).

While on a map f : X → Y taking base point x0 to y0 , π1 induces a map

f∗ : π1 (X, x0 ) → π1 (Y, y0 )
[γ] 7→ [f ◦ γ]

i.e.,
[f : X → Y ] 7→ [f∗ : π1 (X, x0 ) → π1 (Y, f (x0 ))] .
Proof. We need to check
• well-defined on path homotopy classes.
• f∗ is a group homomorphism.
 
1

f (α(2s)),

 if s ∈ 0,
2
f∗ (α · β) = f∗ (α) · f∗ (β) =  
1
if s ∈

f (β(1 − 2s)),
 ,1 .
2

• id(X,x0 )


∗
= idπ1 (X,x0 )

CHAPTER 3. THE FUNDAMENTAL GROUP 35

 Lecture 9: Calculate Fundamental Group

• (f∗ ◦ g∗ ) = (f ◦ g)∗

(f ◦ g)∗ [γ] = [f ◦ g ◦ γ] = [f ◦ (g ◦ γ)] ⇒ f∗ (γ∗ (γ)).

DIY

(X, x0 ) π1 (X, x0 )
f f∗

(Y, y0 ) π1 (Y, y0 )
■

Remark. We usually write f∗ if it’s a covarant functor, while writing f ∗ if it’s a contravariant
functor.

Remark. We see that the construction of fundamental group is actually constructing a functor.
Specifically,
π1 : Top∗ → Gp
such that
• on objects:

∀(X, x0 ) ∈ Ob(Top∗ ), π1 (X, x0 ) = fundamental group based at x0 .

• on morphisms:

∀f : (X, x0 ) → (Y, y0 ), π1 (f ) = f∗ : π1 (X, x0 ) → π1 (Y, y0 ).

Our initial motivation is to construct a topological invariant, but we see that using π1 , we need an
additional base point. But as you already imagined, the fundamental groupoid actually is a functor as
well.
Before we proceed further, we need to see the category of groupoid, denoted by Gpd.

Definition 3.4.1 (Category of groupoid). The category of groupoid, denoted as Gpd, contains the
following data.

• Ob(Gpd): groupoids.
• Hom(Gpd): functors between groupoids.
• Composition: For every X, Y, Z ∈ Ob(Gpd),

F G
X Y Z

then G ◦ F : X → Z is a functor defined as

– on objects: ∀X ∈ Ob(X),
G ◦ F (X) := G(F (X)).
– on morphisms: ∀X, Y ∈ Ob(X) and f : X → Y ,

G ◦ F (f ) := G(F (f )).

• Identity. For every groupoid X, we define idX : X → X, where

– ∀X ∈ Ob(X), idX (X) = X

– ∀f ∈ Hom(X), idX (f ) = f .

CHAPTER 3. THE FUNDAMENTAL GROUP 36

 Lecture 9: Calculate Fundamental Group

• Associativity. Since the composition is defined based on two functors,a this holds trivially.
a For F G
example, given X → Y → Z.

Proof. We need to show that the composition is well-defined. Specifically, we need to check

• G ◦ F (idX ) = idG◦F (X) , since

G ◦ F (idX ) = G(F (idX )) = G(idF (X) ) = idG(F (X)) = idG◦F (X) .

• Given X1 , X2 , X3 ∈ Ob(X) and

f g
X1 X2 X3

we want to show G ◦ F (g ◦ f ) = G ◦ F (g) ◦ G ◦ F (f ). Firstly, since G is a functor, hence

G ◦ F (g) ◦ G ◦ F (f ) = G(F (g)) ◦ G(F (f )) = G (F (g) ◦ F (f )) .

Again, since F is a functor, so we further have

G ◦ F (g) ◦ G ◦ F (f ) = G(F (g ◦ f )) = G ◦ F (g ◦ f ).

Theorem 3.4.2 (Fundamental groupoid defines a functor). Π is a functor such that

Π : Top → Gpd,

where
• on objects: For every X ∈ Ob(Top),

X 7→ Π(X);

• on morphisms: for every X, Y ∈ Ob(Top), f : X → Y , define a functor

Π(f ) : Π(X) → Π(Y )

such that
– on objects: For every p ∈ Ob(Π(X)) = X, Π(f )(p) = f (p). i.e.,

Π(f ) : Ob(Π(X)) → Ob(Π(Y )) .

| {z } | {z }
X Y

– on morphisms: For every ⟨α⟩ ∈ HomΠ(X) (p, q), define

Π(f )(⟨α⟩) := ⟨f ◦ α⟩ ∈ HomΠ(Y ) (f (p), f (q)).

Proof. We need to check that the defined functor Π(f ) satisfies

• Π(f )(idp ) = idf (p) . Indeed, since

Π(f )(idp ) = Π(f )(⟨cp ⟩) = ⟨f ◦ dp ⟩ = ⟨cf (p) ⟩ = idf (p) .

• For every p, q, r ∈ X = Ob(Π(X)),

⟨α⟩ ⟨β⟩
p q r

CHAPTER 3. THE FUNDAMENTAL GROUP 37

 Lecture 10: Seifert-Van Kampen Theorem

we want to show Π(f ) (⟨β⟩ ◦ ⟨α⟩) = Π(f )(⟨β⟩) ◦ Π(f )(⟨α⟩). Indeed, since

Π(f ) (⟨β⟩ ◦ ⟨α⟩) = Π(f )(⟨α · β⟩) = ⟨f ◦ (α · β)⟩,

and
Π(f )(⟨β⟩) ◦ Π(f )(⟨α⟩) = ⟨f ◦ β⟩ ◦ ⟨f ◦ α⟩ = ⟨(f ◦ α) · (f ◦ β)⟩.
Since ⟨f ◦ (α · β)⟩ = ⟨(f ◦ α) · (f ◦ β)⟩, hence Π(f ) is well-defined.

Now, we need to prove the same thing for Π, namely Π satisfies

• Π(idX ) = idΠ(X) for all X ∈ Ob(Top). This is trivial since

Π(idX ) : Π(X) → Π(X),

– on objects: p 7→ idX (p) = p.

⟨α⟩
– on morphisms: p q 7→ ⟨idX ◦α⟩ = ⟨α⟩.

• For all X, Y, Z ∈ Ob(Top),

f g
X Y Z
then Π(g ◦ f ) = Π(g) ◦ Π(f ). The diagrams are as follows.

Π(g ◦ f ) : Π(X) → Π(Z)

and
Π(f ) Π(g)
Π(X) Π(Y ) Π(Z)

We see that this equality is in the sense of functor, hence we consider

– on objects: For every p ∈ Ob(Π(X)) = X, Π(g ◦ f )(p) = g ◦ f (p) and

Π(g) ◦ Π(f )(p) = Π(g)(Π(f )(p)) = Π(g)(f (p) = g(f (p))),

hence they’re the same;

– on morphisms: For all ⟨α⟩ ∈ HomΠ(X) (p, q),
∗ Π(g ◦ f )(⟨α⟩) = ⟨(g ◦ f ) ◦ α⟩.
∗ Π(g) ◦ Π(f )(⟨α⟩) = Π(g)(Π(f )(⟨α⟩)) = ⟨g ◦ (f ◦ α)⟩.
| {z }
⟨f ◦α⟩

We see that they’re the same.

Lecture 10: Seifert-Van Kampen Theorem

The goal is to compute π1 (X) where X = A ∪ B using the data 26 Jan. 10:00

π1 (A), π1 (B), π1 (A ∩ B).

3.5 Free Product

We first introduce a definition.

Definition 3.5.1 (Free product). Given some collections of groups {Gα }α , the free product, denoted
by ∗Gα is a group such that
α

CHAPTER 3. THE FUNDAMENTAL GROUP 38

 Lecture 10: Seifert-Van Kampen Theorem

• Elements: Words in {g : g ∈ Gα for any α} modulo by the equivalence relation generated by

wgi gj v ∼ w(gi gj )v

when both gi , gj ∈ Gα . Also, for the identity element id = eα ∈ Gα for any α such that

weα v ∼ wv.

Specifically,
∗α Gα := {words in {Gα }α } ∼.
.

• Operation: Concatenation of words.

Remark. In particular, we have the following universal property of ∗α Gα . For every α, there is a
ια such that
ια : Gα → ∗α Gα , g 7→ g,
where ια is a group homomorphism, obviously. Further, (∗α Gα , ια ) satisfies the following property:
For every group H and a group homomorphism φα : Gα → G for all α, there exists a unique group
homomorphism φ : ∗α Gα → H such that φ ◦ ια = φα , i.e., the following diagram commutes.

Gα1
ια1

Gα2 ια2 ∗α Gα
φα1

.. φα2
∃!φ
.

Proof. The proof is straightforward. Firstly, we define w = g1 g2 . . . gn ∈ ∗α Gα , gi ∈ Gαi ,

φ(w) := φα1 (g1 ) . . . φαn (gn ).

Now, we just need to check

• It’s well-defined, since φα is a group homomorphism.

• φ is a group homomorphism.
• φ ◦ ια = φα .
• Such φ is unique. Suppose there exists another ψ : ∗α Gα → H, then

ψ ◦ ι α = φα ⇒ ∀ ψ(g) = ψα (g),
g∈Gα

But then for every w = g1 g2 . . . gn ∈ ∗α Gα , gi ∈ Gαi , we have

ψ(w) = ψ(g1 . . . gn ) = ψ(g1 ) . . . ψ(gn ) = ψα1 (g1 ) . . . ψαn (gn ),

which is just φ.

Remark. We further claim that this universal property determines such free product uniquely. i.e.,
assume there are another group Ge and eια : Gα → G.e Assume (G, ια ) also satisfies the following
e e

CHAPTER 3. THE FUNDAMENTAL GROUP 39

 Lecture 10: Seifert-Van Kampen Theorem

property: For every group H and group homomorphism φα : Gα → H, then there exists a unique
group homomorphism φ : G e → H such that the following diagram commutes.

Gα1
ια1
e

Gα2 ια2
e
∗α G
eα
φα1

.. φα2
∃!φ
.

e∼
Then, G = ∗α Gα .

Proof. Assume (G, ια ) satisfies the universal property mentioned above. Then from the universal
e e
property and viewing Ge and ∗α Gα as H separately, we obtain the following diagram.

∃!f
G
e ∗α Gα
ια1
e ∃!g ια1

ια2
e ια2
Gα1

Gα2

..
.

We claim that
g ◦ f = id, f ◦ g = id .
To see this, we simply apply the same observation, for example,

Gα1
ια1
e

Gα2 ια2
e
G
e

ια1
e
∃!g◦f
..
. ια2
e

G
e

where g ◦ f comes from the previous diagram. But notice that id let the diagram commutes also,
and since it’s unique, hence g ◦ f = id. Similarly, we have f ◦ g = id. ⊛
If you’re careful enough, you may find out that all we’re doing is just writing out a specific example
of Lemma 2.2.1! Indeed, this is exactly the construction of a free group.

Definition 3.5.2 (Fibered coproduct). Given a category C , let f : Z → X, g : Z → Y . The fibered

coproduct between f and g is the data (W, p1 , p2 ), where W ∈ Ob(C ), ;1 : X → W , p2 : Y → W
satisfy the following.

CHAPTER 3. THE FUNDAMENTAL GROUP 40

 Lecture 10: Seifert-Van Kampen Theorem

• The diagram commutes.

f
Z X
g p1

Y p2 W

• For every u : X → U , v : Y → U such that the following diagram commutes

f
Z X
g p1
u
Y p2 W
∃!h
v
U

there exists a unique h : W → U such that h ◦ p = u, h ◦ p2 = v.

We say
Z X

Y W
is a Cocartesian diagram.

Exercise. Prove that in a category C , if the fibered coproduct of f and g exists

f
Z X
g

then such fibered coproduct is unique up to isomorphism.

Remark. If we reverse all the directions of morphism, then we have so-called fibered product.

Example. Let C = Top, and let X ∈ Ob(Top). Given X0 , X1 ∈ X, and Int(X0 ) ∪ Int(X1 ) = X, if
we have
i0 : X0 ,→ X, i1 : X1 ,→ X
j0 : X0 ∩ X1 ,→ X0 , j1 : X0 ∩ X1 ,→ X1 ,
then
j0
X0 ∩ X1 X0
j1 i0

X1 i1
X

is a cocartesian diagram.

Proof. All we need to show is that given a topological space Y ∈ Top and f : X0 → Y , g : X1 → Y
in Top, we have
f ◦ j0 = g ◦ j1 .

CHAPTER 3. THE FUNDAMENTAL GROUP 41

 Lecture 10: Seifert-Van Kampen Theorem

j0
X0 ∩ X1 X0
j1 i0
f
X1 i1
X
∃!h
g
Y
We simply define h : X → Y , x 7→ h(x) such that
(
f (x), if x ∈ X0 ;
h(x) =
g(x), if x ∈ X1 .

h is clearly well-defined since the diagram commutes, so if x ∈ X0 ∩ X1 , then f (x) = g(x). The only
thing we need to show is that h is continuous. But this is obvious too since X = Int(X0 ) ∪ Int(X1 ),
and
h|Int(X0 ) = f |Int(X0 ) , h|Int(X1 ) = g|Int(X1 ) .
The uniqueness is trivial, hence this is indeed a cocartesian diagram. ⊛

Example. Let C = Top∗ . Given p ∈ X0 ∩ X1 , where all other data are the same with the above
example, we see that
j0
(X0 ∩ X1 , p) (X0 , p)
j1 i0

(X1 , p) i1
(X, p)

is a cocartesian diagram.

Example. Let C = Gp. Given P, G, H ∈ Ob(Gp), we claim that the fibered coproduct of i and j
exists.
i
P G
j

H
Consider G ∗ H be the free product between G and H, with two inclusions

ι1 : G ,→ G ∗ H, ι2 : H ,→ G ∗ H.
i
P G
j ι1

H ι2 G∗H
Let
N := ⟨ ι1 ◦ i(x) · (ι2 ◦ j(x))−1 | x ∈ P ⟩,


we define
G ∗p H = G ∗ H N .
.

i
P G
j ι1
τ
H ι2 G∗H
π
ν
G ∗p H

CHAPTER 3. THE FUNDAMENTAL GROUP 42

 Lecture 10: Seifert-Van Kampen Theorem

We claim that
i
P G
j τ

H ν G ∗p H

is a cocartesian diagram in Gp.

Proof. Firstly, since it’s just an outer diagram from above, hence it commutes. So we only need
to prove this diagram satisfies the second diagram. Given any group K, for every f : G → K,
g : H → K such that the following diagram commutes.
i
P G
j τ
f
H ν G ∗p H
h
g
K

We want to prove that there exists a unique h : G ∗p H → K such that this diagram still commutes.
The idea is simple, from the universal property of G∗H, we see that there exists a unique e
h : G∗H →
K such that
h ◦ ι1 = f, e
e h ◦ ι2 = g.
i
P G
ι1

j G∗H τ

ι2 π f
∃!e
h
H ν G∗H

h
g

G ∗p H

We see that we can actually factor e

h through π, as long as ker(e h) ⊃ ker(π). Now, since

ker(π) = ⟨ ι1 ◦ i(x) · (ι2 ◦ j(x))−1 | x ∈ p ⟩,



we see that the kernel of π is indeed in the kernel of e

h since for every x ∈ P ,

h ι1 ◦ i(x) · (ι2 ◦ j(x))−1 = e h ◦ ι2 ◦j(x−1 ) = 1,

e h ◦ τ1 ◦i(x) · e
| {z } | {z }
f g

which implies ker(e

h) ⊃ ker(π).
π
G∗H K
h
e

G ∗p H

We then see that there exists a unique h : G ∗p H → K such that the above diagram commutes. ⊛

3.5.1 Free Product with Amalgamation

After seeing the above examples, the following definition should make sense.

CHAPTER 3. THE FUNDAMENTAL GROUP 43

 Lecture 10: Seifert-Van Kampen Theorem

Definition 3.5.3 (Free product with amalgamation). If two groups Gα and Gβ have a common sub-
group S{α,β} ,a given two inclusion mapsb iαβ : S{α,β} → Gα and iβα : S{α,β} → Gβ , the free product
with amalgamation α ∗S Gα is defined as ∗Gα modulo the normal subgroup generated by
α

iαβ (s{α,β} )iβα (s{α,β} )−1 | s{α,β} ∈ S{α,β} ,



Namely,c
∗Gα
.
α∗S Gα = α ⟨iαβ (s{α,β} )iβα (s{α,β} )−1 ⟩
and satisfies the universal property described by the following commutative diagram.
iαβ
S Gα
iβα

Gβ Gα ∗S Gβ
∃!

X
a In general, we don’t need S{α,β} to be a subgroup.
b We don’t actually need iαβ , iβα to be inclusive as well.
c i.e., i
αβ (s) and iβα (s) will be identified in the quotient.

Remark. We see that

• We can then write out words such as gα · s · gβ for s ∈ S, and view s as an element of Gα or
Gβ . In fact, we can do this construction even when iα and iβ are not injective, though this
means we are not working with a subgroup.
• Aside, in Top, the same universal property defines union

iα
A∩B A
iβ

B A∪B
∃!

for A, B are open subsets and the inclusion of intersection.

3.6 Seifert-Van Kampen Theorem

With Definition 3.5.3, we can now see the important theorem.

Theorem 3.6.1 (Seifert-Van Kampen Theorem). Given (X, x0 ) such that X = Aα with
S
α

• Aα are open and path-connected and ∀α x0 ∈ Aα

• Aα ∩ Aβ is path-connected for all α, β.

Then there exists a surjective group homomorphism

∗ : π1 (Aα , x0 ) → π1 (X, x0 ).
α

CHAPTER 3. THE FUNDAMENTAL GROUP 44

 Lecture 10: Seifert-Van Kampen Theorem

If we additionally have Aα ∩ Aβ ∩ Aγ where they are all path-connected for every α, β, γ, then

π1 (X, x0 ) ∼
=α∗π1 (Aα ∩Aβ ,x0 ) π1 (Aα , x0 )

associated to all maps πa (Aα ∩Aβ ) → π1 (Aα ), π1 (Aβ ) induced by inclusions of spaces. i.e., π1 (X, x0 )
is a quotient of the free product ∗α π1 (Aα ) where we have

(iαβ )∗ : π1 (Aα ∩ Aβ ) → π1 (Aα )

which is induced by the inclusion iαβ : Aα ∩ Aβ → Aα . We then take the quotient by the normal
subgroup generated by
{(iαβ )∗ (γ)(iβα )∗ | γ ∈ π1 (Aα ∩ Aβ )} .
We’ll defer the proof of Theorem 3.6.1 until we get familiar with this theorem.

Example. We first see a great visualization of the Theorem 3.6.1.

A∩B X =A∪B
γ

x0
α
β
A
B

Intuitively we see the fundamental group of X, which is built by gluing A and B along their
intersection. As the fundamental group of A and B glued along the fundamental group of their
intersection. In essence, π1 (X, x0 ) is the quotient of π1 (A) ∗ π1 (B) by relations to impose the
condition that loops like γ lying in A ∩ B can be viewed as elements of either π1 (A) or π1 (B).

Remark. We canSuse a more abstract way to describe Theorem 3.6.1. Specifically, in the case that
2
n = 2, i.e., X = i=1 Ai , we let Ai =: Xi , then we have the following. The functor π1 : Top∗ → Gp
maps the cocartesian diagram in Top∗ to a cocartesian diagram in Gp as follows.

j0 (j0 )∗
(X0 ∩ X1 , x0 ) (X0 , x0 ) π1 (X0 ∩ X1 , x0 ) π1 (X0 , x0 )
π1
j1 i0 7−→ (j1 )∗ (i0 )∗

(X1 , x0 ) i1
(X, x0 ) π1 (X1 , x0 ) π1 (X, x0 )
(i1 )∗

Then, simply from the property of cocartesian diagram, we see that

π1 (X, x0 ) ∼
= π1 (X0 , x0 ) ∗π1 (X0 ∩X1 ,x0 ) π1 (X1 , x0 ).
Additionally, there is a more general version of Theorem 3.6.1, which is defined on groupoid. The
theorem is stated in Appendix A.1 with the proof.
With this more general version and the proof of which, we can apply it to Theorem 3.6.1. But one
question is that, the above proof works in Gpd rather than in Gp. We now see how to generalize a group
to a groupoid.
For any group G, we can define a groupoid, denoted as G also, as follows.
• Ob(G) = {∗}, a one point set.
• Hom(G) = {g ∈ G}.

• Composition: We define g ◦ h := h · g.

CHAPTER 3. THE FUNDAMENTAL GROUP 45

 Lecture 11: Group Presentations

We see that the associativity of group elements implies the associativity of composition defined above,
and since there is an identity element in G, hence we also have an identity morphism, these two facts
ensure that G is a category.
Furthermore, since for every g ∈ G, there is a g −1 ∈ G, hence every morphism is an isomorphism,
which implies G is a groupoid.
With this, we see that we can view the following diagram in the category of groupoid Gpd.

(j0 )∗
π1 (X0 ∩ X1 , x0 ) π1 (X0 , x0 )
(j1 )∗ (i0 )∗

π1 (X1 , x0 ) π1 (X, x0 )
(i1 )∗

And to prove Theorem 3.6.1, we only need to show this diagram is cocartesian. This version of proof
is given in Appendix A.2.

Lecture 11: Group Presentations

Example (Fundamental group of S 2 ). We can use Seifert Van Kampen Theorem to compute the 31 Jan. 10:00
fundamental group of S 2 .

Proof. We have the following CW complex structure on S 2 as follows.

We see that π1 (S 2 ) must be a quotient of π1 (A) ∗ π1 (B), but since A, B ≃ D2 , we know that
π1 (A) and π1 (B) are both zero groups, thus π1 (A) ∗ π1 (B) is the zero group, and π1 (S 2 ) is also the
zero group.

Remark. Note that the inclusion of A ∩ B ,→ A induces the zero map π1 (A ∩ B) → π1 (A),
which cannot be an injection. In fact, we know that π1 (A ∩ B) ∼
= Z since A ∩ B ≃ S 1 .

Example (Fundamental group of T ). We can use Seifert Van Kampen Theorem to compute the
fundamental group of a torus T .

Proof. In the case of torus, consider the following CW complex structure.

CHAPTER 3. THE FUNDAMENTAL GROUP 46

 Lecture 11: Group Presentations

Figure 3.4: A is the interior, while B is the neighborhood of the boundary.

Now note that A ≃ D2 and B ≃ S 1 ∨ S 1 , and since it’s a thickening of the two loops around the
torus in both ways, this suggests the question of how do we find π1 (B)? We grab a bit of knowledge
from Seifert Van Kampen Theorem before we continue.

Exercise. Suppose we have path-connected spaces (Xα , xα ), and we take their wedge sum
by identifying the points xα to a single point x. We also suppose a mild condition for
W
α X α
all α, the point xα is a deformation retract of some neighborhood of xα .
For example, this doesn’t work if we choose the bad point on the Hawaiian earring. Then
we can use Seifert Van Kampen Theorem to show that
_ 
π1 Xα , x ∼= ∗π1 (Xα , xα ) .
α α

Answer. If we denote the wedge of circles as Cn ,a then π1 (Cn ) ∼

= Fn .

X1 U3

U2

X1 ∪ U2 ∪ U3 ≃ X1

We can then apply Theorem 3.6.1 to Aα = Xα ∪β Uβ . Specifically, take Aα = Xα ∪β Uβ ≃

Xα , where Uβ is a neighborhood of xβ which deformation retracts to xβ . This makes Aα open
as desired. ⊛
a In this case, n = 3.

Corollary 3.6.1. The wedge sum of circles π1 ( α∈A S 1 ) = ∗α Z is a free group on A. In

W
particular, when A is finite, the fundamental group of a bouquet of circles is the free group on
|A|.
Returning to the current example, we see that
• π1 (A) = 0

• π1 (B) = π1 (S 1 ∨ S 1 ) = Z ∗ Z = F2
• π1 (A ∩ B) = π1 (S 1 ) = Z
Further, we know that π1 (A ∩ B) → π1 (A) is the zero map. We need to understand π1 (A ∩ B) →
π1 (B). To do so we need to understand how we’re able to identify π1 (S 1 ∨ S1 ) with F2 and how we
identify π1 (S 1 ) with Z. We update our Figure 3.4 to talk about this.

CHAPTER 3. THE FUNDAMENTAL GROUP 47

 Lecture 11: Group Presentations

B b
γ
a A a−1

b−1

From this, we have

π1 (A ∩ B) → π1 (B) ∼
= Fa,b
γ 7→ aba−1 b−1 .

By Seifert Van Kampen Theorem, we identify the image of γ in π1 (B) as [aba−1 b−1 ] with its image
in π1 (A), which is just trivial. Therefore, we have
.
π1 (T 2 ) = Fa,b ⟨aba−1 b−1 ⟩ ∼
= Z2 .

Example. Let’s see the last example which illustrate the power of Seifert Van Kampen Theorem.
Start with a torus, and we glue in two disks into the hollow inside.

Glue in 2 disks

We’ll call this space X, and our goal is to find π1 (X).

Proof. We can place a CW complex structure on this space so that each disk is a subcomplex.
Then, we take quotient of each disk to a point without changing the homotopy type, hence X is
homotopy to

By the same property, we can expand one of those points into an interval, and then contract the
red path as follows.

CHAPTER 3. THE FUNDAMENTAL GROUP 48

 Lecture 11: Group Presentations

This is exactly S 2 ∨ S 2 ∨ S 1 . With Seifert Van Kampen Theorem, we have

π1 (X) = π1 (S 2 ∨ S 2 ∨ S 1 ) = 0 ∗ 0 ∗ Z ∼
= Z.

Exercise. Consider R2 \ {x1 , . . . , xn }, that is the plane punctured at n points. Show that

π1 (X) ≃ Fn .

Answer. Observe that X ≃ n S 1 , so one way to do this is to convince yourself that you can do a
W
deformation retract the plane onto the following wedge.

Figure 3.5: Deformation retract X onto wedge.

3.7 Group Presentation

In order to go further, we introduce the concept of group presentation.

Definition 3.7.1 (Group presentation). A presentation ⟨S | R⟩ of a group G such that

.
G∼ = FS ⟨R⟩,

where ⟨R⟩ is a subgroup normally generated by the elements of R. We further call

Definition 3.7.2 (Generator). S is the set of generators.

Definition 3.7.3 (Relater). R is the set of relaters (words in a generator and inverses).

Definition 3.7.4 (Finite presentation). If S and R are both finite, then G = ⟨S | R⟩ is a finite
presentation if S, R are, and we say that G is finitely presented.

CHAPTER 3. THE FUNDAMENTAL GROUP 49

 Lecture 11: Group Presentations

Note. One way to think about whether G is finitely presented is that if r is a word in R then r = 1,
where 1 is the identity of G.

Example. We see that

(a) F2 = ⟨a, b | ⟩

(b) Z2 = ⟨a, b | aba−1 b−1 ⟩ = ⟨a, b⟩ / {aba−1 b−1 }

(c) Z / 3Z = ⟨a | a3 ⟩
(d) S3 = ⟨a, b | a2 , b2 , (ab)3 ⟩

Theorem 3.7.1. Any group G has a presentation.

Proof. We first choose a generating set S for G. Notice that we can even choose S = G directly.
From the universal property of free group, we see that there exists a surjective map φ : FS → G, s 7→
s. Now, let R be the generating set for ker(φ), by the first isomorphism theorema , G ∼ = FS / ker φ.
In fact, we have G = ⟨S | R⟩.
Specifically, i : S → G with ι : S → FS , we have φ ◦ ι = i.
ι
S FS
∃!φ
i
G

■
a https://en.wikipedia.org/wiki/Isomorphism_theorems

Remark. The advantages of using group presentation are that given G = ⟨S | R⟩, it’s now easy to
define a homomorphism ψ : G → H given a map φ : S → H, ψ extends to a group homomorphism
G → H if and only if ψ vanishes on R, i.e., ψ(r) = 0 for all r ∈ R. We see an example to illustrate
this.

Example. If we have G = ⟨a, b | aba⟩, a map φ : {a, b} → H gives a group homomorphism if

and only if
φ(aba) = φ(a)φ(b)φ(a) = 1H .
This essentially uses the universal property of quotients.

Remark. It’s sometimes easy to calculate GAb

GAb = ⟨S | R, commutators in S⟩.

Example. Suppose all relations in R are commutators, so R ⊆ [G, G]. Then,

M
GAb = (FS )Ab = Z.
S

Remark. The disadvantages are that this is computationally very difficult. Let’s see an example
to illustrate this.

CHAPTER 3. THE FUNDAMENTAL GROUP 50

 Lecture 12: Presentations for Fundamental Group of CW Complexes

Example. Given Z2 = ⟨a, b | aba−1 b−1 ⟩, let

ψ : {a, b} → H

extends to a homomorphism if and only if

ψ(a)ψ(b)ψ(a)−1 ψ(b)−1 = 1H ∈ H.

Namely, this is a presentation of the trivial group, but this is entirely unclear.

Lecture 12: Presentations for Fundamental Group of CW Com-

plexes
Let’s first see an exercise. 2 Feb. 10:00

Exercise. Consider G1 = ⟨S1 | R1 ⟩ and G2 = ⟨S2 | R2 ⟩. Then

• G1 ∗ G2 = ⟨S1 ∪ S2 | R1 ∪ R2 ⟩
• G1 ⊕ G2 = ⟨S1 ∪ S2 | R1 ∪ R2 ∪ {[g1 , g2 ] | g1 ∈ G1 , g2 ∈ G2 }⟩
• G1 ∗H G2 where f1 : H → G1 and f2 : H → G2 . Then we have

G1 ∗H G2 = ⟨S1 ∪ S2 | R1 ∪ R2 ∪ f1 (h)f2 (h)−1 | h ∈ H ⟩.



3.7.1 Presentations for Fundamental Group of CW Complexes

For X a CW complex, we have
(a) A 1-dimensional CW complex has free π1 (call its generators as a1 , . . . , an ).

(b) Gluing a 2-disk by its boundary along a word w in the generators kills w in π1 . We then get a
presentation for π1 (X 2 ) given by

⟨a1 , . . . , an | w for each 2-cell in X2 ⟩.

(c) Gluing in any higher dimensional cells along their boundary will not change π1 . That is, in a CW
complex, we have π1 (X) = π1 (X 2 ).

Remark. We can write the above more precise.

(a) Find free generators {ai }i∈I for π1 (X 1 ).

(b) For each 2-disk Dα2 , write attaching map as word wα in ai . i.e., π1 (X 2 ) = ⟨ai | wα ⟩.
(c) π1 (X) = π1 (X 2 ).

Remark. Every group is π1 of some space. Specifically, given a group G, we work with its presen-
tation ⟨S | R⟩.

Proof. We first see a simple example to grab some intuition.

Example (Fundamental group as Z / nZ). Given G = Z / nZ = ⟨a | an ⟩, find a space with its

fundamental group being G.

CHAPTER 3. THE FUNDAMENTAL GROUP 51

 Lecture 13: Graph Structure of CW Complex

Proof. We see that we can simply take a loop and then wind a 2-disk around the loop a for n
times.

a
a

Figure 3.6: For G = Z / nZ = ⟨a | an ⟩, we wind the boundary around a for n times.

⊛
We then see that given a group G with presentation ⟨S | R⟩, one can construct a 2-dimensional
CW complex with π1 = G by
• Set X 1 = s∈S S 1
W

• For each relation r ∈ R, glue in a 2-disk along loops specified by the word r.

Theorem 3.7.2. If X is a CW complex and ι1 : X 1 ,→ X and ι2 : X 2 ,→ X, then (ι1 )∗ surjects onto

π1 and (ι2 )∗ is an isomorphism on π1 .

Proof. ■ HW

Lecture 13: Graph Structure of CW Complex

We can now put some kind of graph structure on a space X and relate it to the calculation of π1 . 4 Feb. 10:00

Definition. We import some topological definitions of graph theoretic concepts.

Definition 3.7.5 (Graph). A graph is a 1-dimensional CW complex.

Definition 3.7.6 (Subgraph). A subgraph is a subcomplex.

Definition 3.7.7 (Tree). A tree is a contractible graph.

Definition 3.7.8 (Maximal tree). A tree in a graph X (necessarily a subgraph) is maximal or

spanning if it contains all the vertices.

Theorem 3.7.3. Every connected graph has a maximal tree. Every tree is contained in a maximal
tree.
Proof. ■ DIY

CHAPTER 3. THE FUNDAMENTAL GROUP 52

 Lecture 13: Graph Structure of CW Complex

Corollary 3.7.1. Suppose X is a connected graph with basepoint x0 . Then π1 (X, x0 ) is a free group.
Furthermore, we can give a presentation for π1 (X, x0 ) by finding a spanning tree T in X. The
generators of π1 will be indexed by cells eα ∈ X − T , and eα will correspond to a loop that passes
through T , traverses eα once, then returns to the basepoint x0 through T .
Proof. The idea is simple. X is homotopy equivalent to X / T via previous work on the homework,
T contains all the vertices, so the quotient has a single vertex. Thus, it is a wedge of circles, and
each eα projects to a loop in X / T .

T X
T
X

To be formal, we calculate the fundamental group of X by considering its CW complex structures.

For now, we need to see that the fundamental group of a 1-skeleton (a graph) can be found by taking
a maximal tree, and then quotienting out the space by the tree to get a wedge of circles.

T a
b q

x0
X X /T

Free generators for π1

(lifts of loops in X / T )

We now prove that the maximal trees exist. Recall that X is a quotient of X 0 α Iα . Since
`
each subset U is open if and only if it intersects each edge eα in an open subset. A map X → Y if
and only if its restriction to each edge eα is continuous. Now, take X0 to be a subgraph. Our goal
is to construct a subgraph Y with

• X0 ⊂ Y ⊂ X
• Y deformation retracts to X0
• Y contains all vertices of X.

So if we take X0 to be a vertex, then Y is our tree, and the result follows.

Hence, we now build a sequence X0 ⊂ X1 ⊂ . . . and correspondingly, Y0 ⊂ Y1 ⊂ . . .. We start
with X0 and inductively define
[
Xi := Xi−1 all edges eα with one or both vertices in Xi−1 .

We then see that X = Xi .a Now, let Y0 = X0 . By induction, we’ll assume that Yi is a subgraph
S
i Check.
of Xi such that
• Yi contains all vertices of Xi .
• Yi deformation retracts to Yi−1 .

CHAPTER 3. THE FUNDAMENTAL GROUP 53

 Lecture 13: Graph Structure of CW Complex

We can then construct Yi+1 by taking Yi and adding to it one edge to adjoin every vertex of Xi+1 ,
namely [
Yi+1 := Yi one edge to adjoint every vertex of Xi ,
which is possible if we assume Axiom of Choice.
We then see that Yi+1 deformation retracts to Yi by just smashing down each edge. Now, we
can show that Y deformation retracts to Y0 = X0 by performing the deformation retraction from
Yi to Yi−1 during the time interval [1/2i , 1/2i−1 ]. ■
a Hatcher [HPM02] do this by arguing the union on the right is both open and closed.

Example (Fundamental group of S n ). Let

• S n : decompose into 2 open disks

• A1 : neighborhood of top hemisphere

• A2 : neighborhood of lower hemisphere
We can then apply Theorem 3.6.1 and conclude that π1 (S n ) = 0.

Proof. We see that A1 ∩ A2 ≃ S n−1 , where we need n ≥ 2 to let S n−1 be connected. We then have

π1 (S n ) ∼
=0 ∗ 0 = 0.
π1 (A1 ∩A2 )

On the other hand, if n ≥ 3, then we see that

n
S n = D ∪ ∗ ∼.
.

Since 2-skeleton is a point, thus π1 (S n ) = 0. ⊛

CHAPTER 3. THE FUNDAMENTAL GROUP 54

Chapter 4

Covering Spaces

Lecture 14: Covering Spaces Theory

7 Feb. 10:00
4.1 Lifting Properties
Lack of
As always, we start with a definition. content...
Things are
Definition 4.1.1 (Covering space). A covering space X
e of X is a space X e and a map p : X
e →X in Hw.
such that ∀x ∈ X ∃ neighborhood ux with p−1 (ux ) the disjoint union of open sets α uα such that
`

p|uα : ux → ux

is a homeomorphism for every α.

..
.

X
e
u1
u2
u3

X
p

ux
x

Definition 4.1.2 (Covering map). We sometimes call p as the covering map.

Although we already investigate into covering spaces quite a lot in homework, but some terminologies
are still worth mentioning.

Definition 4.1.3 (Evenly covered). Let p : X

e → X be a continuous map of spaces. Then an open
subset U ⊆ X is called evenly covered by p if

p|Vi : Vi → U

is a homeomorphism.

55
 Lecture 14: Covering Spaces Theory

` (Slice).−1Given a covering space X and the relating map p, we call the parts Vi of
Definition 4.1.4 e
the partition i Vi of p (U ) slices.

Remark. We see that p is a covering map if and only if every point x ∈ X has a neighborhood which
is evenly covered.
We immediately have the following proposition.

Proposition 4.1.1 (Homotopy lifting property). The covering spaces satisfy the homotopy lifting
property such that the following diagram commutes.

F
e0
X × {0} Ye
∃!F
et
p

X ×I Ft
Y

Proof. We already proved this in homework! ■

Definition 4.1.5 (Lift). We call Fet the lift of Ft in Proposition 4.1.1.

Corollary 4.1.1 (Path lifting property). For each path γ : I → X in X, x

e0 ∈ p−1 (γ(0)) such that
there exists a unique lift γ
e starting at x
e0 .
And for each path homotopy I × I → X, there exists a unique path homotopy γ e: I × I → Xe
starting at x
e0 .

γ
e
p
γ
x0 x
e0

Proof. We prove them separately.

Note. Though we can directly use Proposition 4.1.1 to prove this, but we can see some insight
by directly proving this.

Lifting a path. Assume that we have the following lift.

CHAPTER 4. COVERING SPACES 56

 Lecture 14: Covering Spaces Theory

X
e

γ
e
x
e0

∃!e
γ
p
I

0 1 γ X
γ(0)
γ

We first prove that a path will be lifted uniquely to a path γ

e from x
e0 . For every x ∈ X, there
exists an open neighborhood Ux such that
a
p−1 (Ux ) = Uxα ,
α

where for every α,

p|Ux : Uxα → Ux
α

is a homeomorphism. We see that {Ux | x ∈ X} is an open cover of X, hence

 −1
γ (Ux ) | x ∈ X

is an open cover of [0, 1]. Note that since [0, 1] is a compact metric space, from Lebesgue Lemmaa ,
there exists a partition of [0, 1] such that

0 = t0 < t1 < · · · < tk = 1

such that for every i, [ti , ti+1 ] ⊂ γ −1 (Ux ) for some x. Without loss of generality, we assume that
[ti , ti+1 ] ⊂ γ −1 (Uxi ), i.e.,
γ([ti , ti+1 ]) ⊂ Uxi .

..
.
Ux2 α3
Ux2 α2
Ux2 α1
.. ..
. .
Ux1 α3 Ux3 α3
Ux1 α2 Ux3 α2
Ux1 α1 p Ux3 α1
p
p
I X
γ(t1 )
γ(0)
0 t1 t2 1 γ γ
Ux1 γ(t2 )
Ux2 Ux3

Now, since p(e x0 ) = γ(0) for γ0 ∈ Ux1 and x

e0 ∈ p−1 (Ux1 ), we may assume x
e0 ∈ Ux1 α1 . Consider
lifting the first segment, namely γ([0, t1 ]).

CHAPTER 4. COVERING SPACES 57

 Lecture 14: Covering Spaces Theory

x
e0
Ux1 α1
p|Ux
1 α1

I X
γ(t1 )
γ(0)
0 t1 t2 1 γ|[0,t1 ]
Ux1

 −1
Specifically, let γ
e1 (t) = p|Ux ◦ γ(t) for 0 ≤ t ≤ t1 , we see that
1 α1

e1 : [0, t1 ] → X
γ e

is a lift of γ|[0,t1 ] from x

e0 . We claim that this lift is unique. Consider there exists another lift from
x
e0 γe : [0, t1 ] → X,
e 1
e then since

• γ
e1 (0) = x
e e0

• γ
e1 is continuous
e

• x
e0 ∈ Ux1 α1 ,

we see that γ
e1 (0, t1 ) ⊂ Ux1 α1 , which implies
e

γ
e
e1
[0, t1 ] Ux1 α1  −1
p|Ux ⇒γ
e1 = p|Ux
e ◦ γ|[0,t1 ] = γ
e1 ,
1 α1 1 α1
γ|[0,t
1]
Ux1

hence this lift is unique. Now, we see that we can simply repeat this argument, namely replacing ti
by ti+1 , γ
ei (ti ) by γ
ei+1 (ti+1 ) and so on. Since this partition is finite, hence in finitely many steps,
we obtain a unique path homotopy γ e by concatenating all γ
ei starting at xe0 .

Lifting a path homotopy. We now consider lifting a path homotopy. Consider

γ1 ≃ γ2 rel{0, 1}
F

we’ll show that γ e2 rel{0, 1} where p ◦ Fe = F . Firstly, we denote x0 := γ1 (0) = γ2 (0), such
e1 ≃ γ
F
e
that

CHAPTER 4. COVERING SPACES 58

 Lecture 14: Covering Spaces Theory

?
γ
e1 (1) = γ
e2 (1)
γ
e1
X
e
γ
e2
x
e0

Fe
p
F1 = γ2
[0, 1] × [0, 1] γ2 (1)
X
x0
Ft F (s, t) γ1
t
γ2
γ1 (1) = γ2 (1)
x0 γ1 (1)
F0 = γ1

We claim that it’s sufficient to show that there exists a continuous Fe : I × I → X such that
p ◦ Fe = F , and Fe({0} × I) = x0 . It’s because

p ◦ Fe0 = F0 = γ1 , p ◦ Fe1 = F1 = γ2

where Fe0 , Fe1 is γ1 , γ2 ’s lifting starting at x

e0 , respectively. And since p ◦ Fe = F , we have
 
p Fe({1} × I) = x1 ⇒ Fe({1} × I) ⊂ p−1 ({x1 }),

which implies ∃ex1 ∈ p−1 ({x1 }) such that Fe({1}×I) = x

e1 since we know that p−1 ({x1 }) is a discrete
points-set and Fe is assumed to be continuous, and {1} × I is connected. We now show Fe exists.
We define

Fe : I × I → X
(s, t) 7→ Fet (s),

where Fet : [0, 1] → X

e is a lift starting at x
e0 of Ft : [0, 1] → X, s 7→ F (s, t). Obviously, p ◦ Fe = F
from the uniqueness of the lift of a path, and also, Fe({0} × I) = x e0 holds trivially, hence we only
need to show F is continuous.
e

(a) We show that ∃ϵ0 > 0 such that Fe is continuous.

[0,ϵ0 ]×I

CHAPTER 4. COVERING SPACES 59

 Lecture 14: Covering Spaces Theory

..
.
Ux0 α3
Ux0 α2
Ux0 α1
Fe x
e0
p

[0, 1] × [0, 1] γ2 (1)

X
x0
F (s, t)
Ux0

x0 γ1 (1) F |[0,ϵ0 ]×I

0 ϵ0

Since F is continuous, we see that there exists an open neighborhood Ux0 of x0 such that
p−1 (Ux0 ) = α Ux0 α , where
`
∼
=
p|Ux : Ux0 α → Ux0 .
0α

Since F −1 (Ux0 ) is an open set contain {0} × I, there exists a ϵ0 > 0 such that [0, ϵ0 ] × I ⊂
F −1 (Ux0 ),b which implies
F ([0, ϵ0 ] × I) ⊂ Ux0 .
 −1
Note that x0 ∈ Ux0 and p(e x0 ) = x0 , we may assume x e0 ∈ Ux0 α1 . Consider p|Ux α ◦
0 1
F |[0,ϵ0 ]×I , which is a lift of F |[0,ϵ0 ]×I . We claim that
 −1
p|Ux ◦ F |[0,ϵ0 ]×I = Fe .
0 α1 [0,ϵ0 ]×I

This is because for every t ∈ I,

 −1
s 7→ p|Ux ◦ F |[0,ϵ0 ]×I (s, t)
0 α1

is a lift starting at x
e0 ; also, for every t ∈ I,

s 7→ Fe (s, t)
[0,ϵ0 ]×I

is a lift of Ft starting at x
e0 . From the uniqueness of the lift of paths, we see that they’re equal.
Note that this implies Fe is now continuous at [0, ϵ0 ] × I, since F is continuous and p|Ux α is
0 1
a homeomorphism, hence continuous, then from
 −1
Fe = p|Ux α ◦ F |[0,ϵ0 ]×I ,
[0,ϵ0 ]×I 0 1
| {z } | {z }
continuous continuous

we see that Fe is indeed continuous at [0, ϵ0 ] × I.

CHAPTER 4. COVERING SPACES 60

 Lecture 14: Covering Spaces Theory

(b) We now prove that Fe : I × I → X e is continuous. Assume there exists (s0 , t0 ) ∈ I × I such
that Fe is discontinuous at (s0 , t0 ). Then consider
n o
0 < ϵ0 ≤ inf s | Fe is discontinuous at s, t0 =: s1 ,
| {z }
∋s0 ⇏=∅

where the first inequality is from the first step.

X
e

x
e1

Fe
p
[0, 1] × [0, 1]
γ2 (1)
X

F (s, t)
(s1 , t0 ) x1
Ux1
x0 γ1 (1)
0 ϵ0

Let x1 := F (s1 , t` e1 := Fe(s1 , t0 ), then there exists an open neighborhood Ux1 in X such
0 ), x
that x1 ∈ Ux1 = α Ux1 α , where
∼
=
p|Ux : Ux1 α → Ux1 .
1α

Since F is continuous, there exists an ϵ1 > 0, δ1 > 0 such that

F ((s1 − ϵ1 , s1 + ϵ1 ) × (t0 − δ1 , t0 + δ1 )) ⊂ Ux1 .

Notice that here we’re considering open box.

CHAPTER 4. COVERING SPACES 61

 Lecture 14: Covering Spaces Theory

Ux1 α3

Ux1 α2

Ux1 α1

Fe
p
[0, 1] × [0, 1]

X
t0 + δ1 x1
F (s, t)
(s1 , t0 )
t0 − δ1
Ux1

s1 − ϵ1 s1 + ϵ1

We may assume x e1 ∈ Ux1 α1 . Then, we see that Fet0 is a lift of Ft0 , which means Fet0 is
continuous, hence there exists an s2 such that s1 − ϵ1 < s2 < s1 such that

Fe(s2 , t0 ) ∈ Ux1 α1 .

Fe(s2 , t0 ) =: x
e2
x
e1 Ux1 α1

Fe
p|Ux
1 α1

[0, 1] × [0, 1]

X
t0 + δ1
(s2 , t0 ) F (s, t)
t0
(s1 , t0 ) x1
t0 − δ1
Ux1

s1 − ϵ1 s1 + ϵ1

We see that Fe is continuous at (s2 , t0 ), hence there exists a δ2 > 0 such that

Fe ({s2 } × (t0 − δ2 , t0 + δ2 )) ⊂ Ux1 α1 .

Note that here we can also consider a closed interval, which matches what we’re going to do.
Namely, we’re going to construct a closed box B. But this is just a technical detail.

CHAPTER 4. COVERING SPACES 62

 Lecture 14: Covering Spaces Theory

Fe(L)

x
e2 x
e1 Ux1 α1

Fe
B p|Ux
1 α1

[0, 1] × [0, 1] (s2 , t0 )

(s1 , t0 ) X
t0 + δ2
F |B
t0
x1
t0 − δ2 B
Ux1
L

s1 − ϵ1 s1 + ϵ1

Now, observe that Fe(B) ⊂ Ux1 α1 . To see this, consider a fixed t ∈ (t0 + δ2 , t0 − δ2 ), then the
map Fe is
[s1 − ϵ1 , s1 + ϵ1 ] → X,
e s 7→ Fe(s, t) = Fet (s).
Specifically, a
Fet ([s1 − ϵ1 , s1 + ϵ1 ]) ⊂ p−1 (Ux1 ) = Ux1 α ,
α

with the fact that Fet ([s1 − ϵ1 , s1 + ϵ1 ]) is connected, and Fet (s2 ) ∈ Ux1 α1 with Fet is a lift of Ft ,
hence continuous, so
Fet ([s1 − ϵ1 , s1 + ϵ1 ]) ⊂ Ux1 α1 .

This is true for every t ∈ [t0 − δ2 , t0 + δ2 ], hence Fe ⊂ Ux1 α1 . Now, since

B

p|Ux ◦ Fe = F |B ,
1 α1 B

and  −1
p|Ux ◦ F |B : B → Ux1 α1 ,
1 α1

so  −1 
p|Ux ◦ p|Ux ◦ F |B = F |B
1 α1 1 α1

obviously. Since p|Ux is a homeomorphism, we have

1 α1

 −1
Fe = p|Ux α ◦ F |B ,
B 1 1
| {z } | {z }
continuous continuous

hence we have Fe is continuous, which leads to a contradiction since

B
n o
s1 = inf s | Fe is discontinuous at s, t0 ,

while Fe is continuous for all B, hence we see that Fe : I × I → X

e is continuous.c

CHAPTER 4. COVERING SPACES 63

 Lecture 14: Covering Spaces Theory

■
a https://en.wikipedia.org/wiki/Lebesgue%27s_number_lemma
b Notice that we’re working on product topology here.
c There is a tricky situation, namely while s1 = 1. But this can be considered also.

Example (Covers of S 1 ∨ S 1 ). We have the following covers of S 1 ∨ S 1 .

X
e1 X
e2 X
e3

a b b b
b a
b
a b
b b a a a
a

p1 p2 p3

x0

a X = S1 ∨ S1
b

Note that in each cover (those three on the top), the black dot is the preimage of {x0 }, namely
p−1
i ({x0 }).

Remark. We see that for each p−1

i ({x0 }), there are exactly

• one a edge goes out

• one b edge goes out

• one a edge goes in

• one b edge goes in
It turns out that there are much more covers of S 1 ∨ S 1 , as long as this main property is
satisfied.

Proposition 4.1.2. Let p : (X,

e xe0 ) → (X, x0 ) be a covering map. Then

(a) p∗ : π1 (X,
e xe0 ) → π1 (X, x0 ) is injective.

(b) p∗ (π1 (X,

e xe0 )) ⊆ π1 (X, x0 ), which picks out the subset

{[γ] | lift γ
e starting at x
e0 is a loop.} .
Proof. We prove this one by one.
(a) Suppose γ
e ∈ π1 (X,
e xe0 ) is in ker(p∗ ). Then

γ ]) = [p ◦ γ
[γ] = p∗ ([e e] .

Let γt be a nullhomotopy from γ to the constant loop cx0 rel{0, 1}. We can then lift γt to γ
et
where γ e. Now, we claim that
e0 = γ

CHAPTER 4. COVERING SPACES 64

 Lecture 15: Lifting

• γ
e is a homotopy rel{0, 1}.
• γ
e1 is the constant loop cxe0 .

X
e X
e
γ γ
et
e p p

I γ X I ×I γt X

We see that the above diagrams prove the first claim, since we know that the left and right
edge of I × I maps to x0 under γt , and cxe0 lifts this, so by uniqueness t 7→ γ
et (0) and t 7→ γ
et (1)
must be constant paths at xe0 as desired.
Then the lift γ
et is a homotopy of paths to the constant loop, so [e
γ ] = 1.

(b) Let see an example to show the idea of the proof.

Example. Given

max tree

X
e b
b x
e0 a

x0
X a b

Then
p∗ π1 = ⟨b, a2 , aba⟩ ⊆ π1 (X) = ⟨a, b | ⟩.

Proposition 4.1.3 (Lifting criterion). Let p : (Ye , ye0 ) → (Y, y0 ) be a covering map. Given
• f : (X, x0 ) → (Y, y0 );
• X is path-connected, locally path-connected,

then a lift
fe: (X, x0 ) → (Ye , y0 )
exists if and only if
f∗ (π1 (X, x0 )) ⊆ p∗ (π1 (Ye , ye0 )).
In diagram, we have

(Ye , ye0 ) π1 (Ye , ye0 )

∃fe fe∗
p p∗

(X, x0 ) f
(Y, y0 ) π1 (X, x0 ) f∗
π1 (Y, y0 )

Lecture 15: Lifting

Before proving Proposition 4.1.3, we first see an application. 9 Feb. 10:00

CHAPTER 4. COVERING SPACES 65

 Lecture 15: Lifting

Example. Every continuous map f : RP 2 → S 1 is nullhomotopic.

Proof. If we can show that there is a lift fe: RP 2 → R of f , then we’re done since we can apply the
straight line nullhomotopy on R, i.e.,

R
fe
p

RP 2 f
S1

and consider f = p ◦ fe compose nullhomotopy with p, so f ≃ constant map. Specifically, since

π1 (RP 2 ) = Z / 2Z and π1 (S 1 ) = Z, hence

f∗ (π1 (RP 2 )) = 0

since Z has no (nonzero) torsion. So it lifts by Proposition 4.1.3. ⊛

Now we can proof Proposition 4.1.3.
Proof of Proposition 4.1.3. We prove two directions as follows.

Necessary. We see that we can factorize f∗ as

f∗ = p∗ ◦ fe∗

follows from the functoriality of π1 .

Sufficient. Let x ∈ X. Choose a path γ from x0 to x by the assumption that X is path-connected.

Then, f γ has a unique lift starting at ye0 , denote by ffγ. Now, define

fe(x) = ffγ(1).

Then, we need to check

(a) fe is well-defined. Suppose γ, γ ′ are paths in X from x0 to x. We want to show

γ ′ (1) = ffγ(1).
fg

Since γ · γ ′ is a loop in X at x0 , we know that [(f γ) · (f γ ′ )] is a class of loops in Y in Im(f∗ ).

By hypothesis, this class of loops is in Im(p∗ ). It lifts to a loop which is based at ye0 . By
uniqueness of lifts, this loop lifting (f γ) · (f γ ′ ) to Ye must be equal to the lifts ffγ · fg
γ ′ with a
common value at t = 1/2. Hence, ffγ(1) = fg γ (1) as desired, namely the endpoints agree.
′

CHAPTER 4. COVERING SPACES 66

 Lecture 16: Proving Proposition 4.1.3

Ye
fe(x)

◦γ
f]

ye0
fe

y0
x0
γ
f ◦γ
x
γ ′ f
f ◦ γ′
f (x)

X Y

Lecture 16: Proving Proposition 4.1.3

We now continue our proof of Proposition 4.1.3. 11 Feb. 10:00
2. fe is continuous. Choose x ∈ X and a neighborhood U
e of fe(x) in Ye . Note that we can choose
e small enough to p| e is homeomorphism to U in Y . Now, there exists a neighborhood V of
U U
x in X with f (V ) ⊆ U .

V
α
x′
γ x f fα

x0 f (V )
fγ

The goal is fe(V ) ⊆ U

e . Without loss of generality, we can assume that V is path-connected.
Then,
α = [f^
ffγ · ff γ · f α].
Hence,
α = ( p|Ue )−1 ◦ f ◦ α,
ff

where ( p|Ue )−1 ’s image is in U

e , so

fe(x′ ) = f γ · f α(1) ∈ U
e,

which implies
fe(V ) ⊆ U
e.

CHAPTER 4. COVERING SPACES 67

 Lecture 16: Proving Proposition 4.1.3

Proposition 4.1.4 (Uniqueness of lifts). Let p : Ye → Y be a covering map with X is a connected

space. If two lifts fe1 , fe2 of the same map f agree at a single point, then they agree everywhere.

Ye
fe1
p

fe2
X f
Y

Proof. Let S being n o

S := x ∈ X | fe1 (x) = fe2 (x) .

We want to show that S is both closed and open, so if S is nonempty, S = X.

fe1 (x) U
e1

fe2 (x) U
e2

fe1
p
fe2

U
f (x)
x
Y

We see that U
e1 and U
e2 are slices of p−1 (U ), where U is an evenly covered neighborhood of f (x).

(a) If fe1 (x) ̸= fe2 (x). Then U e2 are disjoint. Since fe1 , fe2 are continuous, there exists a neigh-
e1 , U
borhood N of x with
fe1 (N ) ⊆ U
e1 , fe2 (N ) ⊆ U e2 ,

with the fact that they’re disjoint, so x is an interior point of S c .

(b) If fe1 (x) = fe2 (x). Then U e2 . Choose N as before, then we have
e1 = U

fe1 (n) = ( p|ue1 )−1 (f (n)) = fe2 (n),

hence x ∈ Int(S).

4.2 Deck Transformation

We now want to introduce a special kind of transformation.

Definition 4.2.1 (Isomorphism of covers). Given covering maps

e1 → X,
p1 : X e2 → X,
p2 : X

CHAPTER 4. COVERING SPACES 68

 Lecture 17: Deck Transformation

an isomorphism of covers is a homeomorphism f : X e2 such that p1 = p2 ◦ f .

e1 → X

f
X
e1 X
e2

p1 p2
X

Exercise. This defines equivalent relation on covers of X.

Definition 4.2.2 (Deck transformation). Given a covering map p : X

e → X, the isomorphisms of
covers X → X are called deck transformation.
e e

Definition 4.2.3 (Set of deck transformation). Also, we let G(X)

e denotes the set of deck trans-
formations.

Note. Note that we’ve suppressed the data of p in the notation, but this data is essential to what
a deck transformation is, when this is unclear we write G(X,
e p).

Lecture 17: Deck Transformation

14 Feb. 10:00
Example. Deck transformations G(X)
e are a subgroup of the group of homeomorphisms of X.
e

Example. Given the cover p : R → S 1 .

• Deck maps: translation by n ∈ Z units.

• G(R) ∼
=Z

Example. Given the cover pn : S 1 → S 1 be an n-sheeted cover.

• Deck maps: rotation by 2π/n.

• G(S 1 , pn ) ∼
= Z / nZ

∼
= S1

S1

Figure 4.1: pn : S 1 → S 1 be an n-sheeted cover, here n = 3.

Exercise (Deck Transformation is determined by the image of one point). Given X, X e are path-
connected, locally path-connected, deck map is determined by the image of any one point.

CHAPTER 4. COVERING SPACES 69

 Lecture 17: Deck Transformation

Answer.
X
e
f
p

X
e
p X
⊛

Corollary 4.2.1. If a deck transformation has a fixed point, it is the identity transformation.

Exercise. Let X be connected. Given a deck transformation τ : X e → X,e τ defines a permutation of

p−1 ({x0 }). If this permutation has a fixed point, then it is the identity.

Definition 4.2.4 (Regular (normal) cover). A covering space p : X e → X is regular or normal if

∀x0 ∈ X, ∀e e1 ∈ p ({x0 }), there exists a deck transformation such that
x0 , x −1

e0 7→ x
x e1 .

Example (Regular and non-regular cover of S 1 ∨ S 1 ). Given the following covers of S 1 ∨ S 1 , determine
which cover is regular.

x
e1

x
e2
x
e0

Proof. The left one is regular, while the right one is not since there is no automorphism from x
e0 to
e1 or x
x e2 . ⊛

Remark. A regular cover is as symmetric as possible.

Exercise. Regular means that the group G(X)

e acts transitively on p−1 ({x0 }). Explain why we
cannot ask for more than this:
e cannot induce the full symmetric group on p−1 ({x0 }) provided that p−1 ({x0 }) > 2.
G(X)

Answer. The key is uniqueness. ⊛

Since we’re talking about symmetric, it’s natural to introduce the following concept for groups.

Definition 4.2.5 (Normal subgroup). A subgroup N of G is called a normal subgroup if it’s invariant
under conjugation, and we denote this relation as N ◁ G.

Definition 4.2.6 (Normalizer). Given G as a group, H ⊆ G is a subgroup of G. Then the normalizer

of H, denoted by N (H), is defined as

N (H) := {g ∈ G | gH = Hg} .

Exercise. We can prove the followings.

(a) N (H) is a subgroup.

CHAPTER 4. COVERING SPACES 70

 Lecture 18: Proving Proposition 4.2.1

(b) H ≤ N (H).
(c) H is normal in N (H).
(d) If H ≤ G is normal, N (H) = G.

(e) N (H) is the largest subgroup (under containment) of G containing H as normal subgroup.

Proposition 4.2.1. Given p : (X,

e xe0 ) → (X, x0 ) be a cover, and X, e X are path-connected, locally
path-connected. Let
H = p∗ (π1 (X,
e xe0 )) ⊆ π1 (X, x0 ).
Then
(a) p is normal if and only if H ⊆ π1 (X, x0 ) is normal.
(b) We have .
e ∼
G(X) = N (H) H ,

where G(X)
e are deck maps, and N (H) is the normalizer of H in π1 (X, x0 ).

Remark. A fact is worth noting is the following. Let γ

e be a path x
e1 to x
e0 . Then

p∗ (π1 (X,
e xe0 )) = [γ]p∗ (π1 (X,
e xe1 ))[γ −1 ].

γ
e
x
e0 x
e1

γ
γ =p◦γ
x0
e

Lecture 18: Proving Proposition 4.2.1

Now let’s prove Proposition 4.2.1. 16 Feb. 10:00
Proof of Proposition 4.2.1. Let X, x0 be the base space and x e1 ∈ p−1 ({x0 }) where p : X
e0 , x e →X
is a covering map. Further, let H := p∗ (π1 (X, x
e e0 )).
In homework, given (X, x0 ), x e1 ∈ p−1 ({x0 }) if we change the basepoint from π1 (X,
e0 , x e0 ) to
e x
π1 (X,
e xe1 ), then we have the induced subgroups of the base spaces fundamental group are conjugate
by some loop [γ] ∈ π1 (X, x0 ), i.e.,

p∗ (π1 (X,
e xe1 )) = [γ] · p∗ (π1 (X,
e xe0 )) · [γ]−1

where γ is lifted to a path from x

e0 to x
e1 .
Therefore, [γ] ∈ N (H) if and only if p∗ (π1 (X,
e xe1 )) = p∗ (π1 (X,
e xe0 )), and this holds if and only if
there is a deck transformation taking xe0 to xe1 by the classification of based covering spaces in the
homework.

Note. Alternatively, we can use the lifting criterion.

This shows that p is a normal cover if and only if H is normal, which proves the first claim.

CHAPTER 4. COVERING SPACES 71

 Lecture 18: Proving Proposition 4.2.1

We then define a map Φ such that

Φ : N (H) → G(X)[γ],
e · 7→ τ

where τ lifts to a path from x

e0 to x
e1 and τ is a deck transformation mapping x e0 to x
e1 , which will
be uniquely defined by the uniqueness of lifts with specified base points. We now need to check

(a) Φ is surjective.
(b) ker(Φ) = H.
(c) Φ is a group homomorphism.

If we can prove all the above, then the result follows directly from the first isomorphism theorem.

(a) We’ve proved that Φ is surjective before in our work above.

(b) Φ([γ]) is the identity if and only if τ sends x

e0 to x
e0 , meaning that [γ] lifts to a loop. Then by
our characterization of the fundamental group downstairs:

ker(Φ) = {[γ] | [γ] lifts to a loop} = H.

Φ Φ
(c) Suppose we have loops [γ1 ] 7→ τ1 and [γ2 ] 7→ τ2 . We claim that γ1 · γ2 lifts to γ
e1 · τ (e
γ2 ).

· γ′
γ]
γ
e x
x
e0 x
e2 e3

p
lifts of γ ′ starting at x
e2

γ γ′
x0

Exercise. Check that the lift of γ2 starting at x

e1 is exactly Φ1 (e
γ2 ), where γ
e2 is a lift
starting at x
e0 .

x
e0 x
e2
τ γ′)
τ (e
e′
γ

Figure 4.2: Must be lift of γ ′ starting at x

e2
Answer. The key is the uniqueness of lifts. ⊛

We then just observe that this path γ γ2 ) is a path from x

e1 · τ1 (e e0 to γ1 (e
γ2 (1)) = τ1 (τ2 (e
x0 )),
so the image must be a deck transformation sending x e0 to τ1 (τ2 (e
x0 )). But then τ1 ◦ τ2 maps
e0 to this same point, and from this exercise, we know that the deck transformations are
x
determined by where they send a single point, hence we’re done.

CHAPTER 4. COVERING SPACES 72

 Lecture 18: Proving Proposition 4.2.1

e ∼
Corollary 4.2.2. If p is a normal covering, then G(X) = π1 (X, x0 ) / H.

Definition 4.2.7 (Universal covering). A cover p : X

e → X is called a universal covering if X
e is simply
connected.

Corollary 4.2.3. If X e ∼
e is the universal cover, then G(X) = π1 (X, x0 ).

Exercise. Whether Im(p∗ ) is normal is independent of the basepoint in X

e and X.

So, p is normal if and only if G(X)

e is transitive on p−1 (x0 ) for at least one x0 ∈ X.

Exercise. Let Σg be the genus g surface. Prove that Σg has a normal n-sheeted path-connected
cover for every n.

CHAPTER 4. COVERING SPACES 73

Chapter 5

Homology

Lecture 19: Simplex and Homology

Homology is a general way of associating a sequence of algebraic objects, such as Abelian groups or 18 Feb. 10:00
modules, with other mathematical objects such as topological spaces. In this section, we’ll develop this
notion rigorously, and make some connection with fundamental group to see why this is much more
powerful.

5.1 Motivation for Homology

Informally, the higher homotopy groups is defined as

πn (X, x0 ) : I∗n → (X, x0 ), ∂I n 7→ x0 .

X X
γ1
γ

x0 x0
γ2

We see that it’s extremely hard to compute higher fundamental group. Hence, instead, we will study
the higher dimensional structure of X via homology.

• Cons.
– The definition is more opaque at first encounter.
• Pros.

– Lots of computational tools

– Functional
– Abelian Groups

74
 Lecture 19: Simplex and Homology

Remark. More like πn for n > 1.

– No basepoints
– Can compute using CW structure.
– Good properties. For example, Hn = 0 if n > dim X

5.2 Simplicial Homology

5.2.1 ∆-Simplex
This is a stricter version of a CW complex which allows us to decompose our spaces into cells. In terms
of how things fit together, we have the following diagram.

Simplicial Complex structure

(triangulation)

△-Complex structure

CW Complex structure

Now we try to give the definition.

Definition 5.2.1 (Simplex). We see that

• 0-simplex. A point.
• 1-simplex. Interval.

• 2-simplex. Triangle.
• 3-simplex. Tetrahedron.
• n-simplex. The convex hull of (n + 1)-points position in Rn .

v2 v3

v2

v0 v0 v1 v0 v1 v0 v1

Remark. We see that

• The top of which is the 2-disk and remember cell structure (edges and vertices) and remember
orientation (ordering on vertices).

CHAPTER 5. HOMOLOGY 75
 Lecture 20: Simplicial Complex

• The top of which is the 3-disk and cells and the orientation.
• We can view simplices as both combinatorial and topological objects.
An alternative definition can be made.

Definition 5.2.2 (Standard simplex). We say that an n-dimensional standard simplex, denoted by
∆n is ( )
X
n n+1
∆ = (t0 , . . . , tn ) ∈ R | ti ≥ 0, ti = 1 .
i

We’ll call such a simplex as standard n-simplex.

v1 v2

v0 v0 v1
v0
∆0 ∆1 ∆2

Remark. The standard simplices will implicitly come with a choice of ordering of the vertices as

∆n = [v0 , v1 , . . . , vn ]

such that the convex hull of these points is taken with this ordering.

Lecture 20: Simplicial Complex

We now give some definitions about standard simplex. 21 Feb. 10:00

Definition. With Definition 5.2.2, we have the followings.

Definition 5.2.3 (Subsimplex). Combinatorially, a subsimplex of a simplex ∆n is a subset of

the vertices; while topologically, it’s the convex hull of the subset of vertices.

v2

[v0 , v2 ] [v1 , v2 ]
[v0 , v1 , v2 ]

v0 [v0 , v1 ] v1

Definition 5.2.4 (Face). A face of a simplex ∆n is a subsimplex of 1 dimensional lower than

∆n (codimension 1).

Definition 5.2.5 (Boundary). The boundary ∂σ of a simplex σ is the union of its faces.

Definition 5.2.6 (Open simplex). The open simplex of ∆ is defined as

˚ n := ∆n − ∂∆n .
∆

CHAPTER 5. HOMOLOGY 76
 Lecture 20: Simplicial Complex

Definition 5.2.7 (∆-Complex). A ∆-complex structure on X is a collection of maps

σα : ∆ n → X

such that
(a) σα |∆
˚ n injective, each point of X is in the image of exactly one such map.

(b) Each restriction of σα to a face coincides with a map

σβ : ∆n−1 → X.

(c) A set A ⊆ X is open if and only if σα−1 (A) is open in ∆

˚ n for all σα , i.e., X is a quotient
`
σα
∆nα
`
X.
n,α

Exercise. A ∆-complex X is a CW complex W with characteristic maps σα with extra constraints

on the attaching maps.

Note. We see that the second condition of Definition 5.2.7 implies that attaching maps injective on
interior of faces.

Definition 5.2.8 (Simplicial complex). A simplicial complex is a ∆-complex such that

• σα must map every face to a different (n − 1)-simplex.

• Every simplex is uniquely determined by its vertex set.

• Any (n + 1) vertices in X 0 is the vertex set of at most 1 n-simplex.

Example (Difference between simplicial and ∆-complex structure of S 1 ). With Definition 5.2.7 and
Definition 5.2.8, we see the followings.

∆-simplex ∆-simplex ∆-simplex

not simplicial not simplicial is simplicial

Example (Difference between simplicial and ∆-complex structure of a torus). The torus with the
following edges, a, b, c and the gluing in triangles A and B can be seen as follows.

CHAPTER 5. HOMOLOGY 77
 Lecture 21: Simplicial Homology

A
a c a

This structure is only valid as a ∆-complex.

Proof. For this ∆-complex, notice that we’ve glued down a triangle whose vertices are all identified.
This is not allowed in a simplicial complex/triangulation.

Remark. The minimum number of triangles in a simplicial complex structure is 14.

Exercise. Try to come up with the corresponding simplicial complex.

Lecture 21: Simplicial Homology

To demonstrate how the definition of homology arise, we first see the idea behind it. Fix a space X 23 Feb. 10:00
which equips with the ∆-complex structure. Then, we define Cn (X) to be the free Abelian group on the
n-simplices of X. That is,
( )
X
Cn (X) = finite sums n
mα σα | mα ∈ Z, σα : ∆ → X .

σ2

σ1 σ3

Figure 5.1: C2 (X) = Zσ1 ⊕ Zσ2 ⊕ Zσ3 .

Then, the n-th homology group will be a subquotient of Cn (X), where the heuristic/imprecise idea
is
• Take subgroup of Cn of cycles. These are sums of simplices satisfying a combinatorial condition
on the boundary gluing maps to ensure that they close up, i.e., they have no boundary.

CHAPTER 5. HOMOLOGY 78
 Lecture 21: Simplicial Homology

σ1

σ
σ
σ2

σ1 + σ2 cycles σ cycle σ not a cycle

σ1
σ2
σ2
σ1 σ3
σ4

σ3
σ1 + σ2 + σ3 + σ4 cycles σ1 + σ2 + σ3 not a cycles

• To take the quotient, we consider two cycles to be equivalent if their difference is a boundary.
For example, in the case of torus, a is homologous to b since a − b is the boundary of the shaded
subsurface S on of the torus below.

b S

In fact, a and b are homotopic (which will imply they’re homologous essentially), but two loops do
not need to be homotopic to be homologous. For example, in the figure below, a + b is homologous
to c, since a + b − c is the boundary of S (a + b1 and c are not homotopic).

S a

b
c

Let’s now see the formal definition.

Definition 5.2.9 (Simplicial chain group). We define the simplicial chain group Cn (X) of order n to
be the free Abelian group on the n-simplices of X such that
( )
X
Cn (X) := finite sums n
mα σα | mα ∈ Z, σα : ∆ → X .

Definition 5.2.10 (Cycle). Given any chain group Cn (X), a cycle of Cn (X) is those chains
P
mα σα
with no boundaries.

1 Which isn’t even a loop

CHAPTER 5. HOMOLOGY 79
 Lecture 21: Simplicial Homology

Definition 5.2.11 (Boundary homomorphism). A map ∂n : Cn (X) → Cn−1 (X) is called a boundary
homomorphism such that

∂n : Cn (X) → Cn−1 (X)

n
X
[σα ] 7→ (−1)i σα |[v0 ,...,v̂i ,...,vn ] ,
i=1

which defines the map on the basis, and we extend it linearly.

Remark. We see that the definition of boundary homomorphism indeed coincides with the definition
of boundary when considering either ∆-complex or simplicial complex structure.

Example. We give some lower dimensions examples of Definition 5.2.11 to motivate the general
definition.
• For n = 1, ∂1 : C1 (X) → C0 (X) such that

[σα : [v0 , v1 ] → X] 7→ σα |[v1 ] − σα |[v0 ] .

• For n = 2, ∂2 : C2 (X) → C1 (X) such that

[σα : [v0 , v1 , v2 ] → X] 7→ σα |[v1 ,v2 ] − σα |[v0 ,v2 ] + σα |[v0 ,v1 ] .

Lemma 5.2.1. Let Cn being the simplicial chain group and ∂n being the boundary homomorphism,
for any n ≥ 2, we have
∂n ∂n−1
Cn (X) Cn−1 (X) Cn−2 (X)
∂n−1 ◦∂n =0

Proof. Since all Ci are free Abelian group, hence we only need to consider ∂n−1 ◦ ∂n (σ) = 0 for a
generator σ. Given a generator σ, the result follows from directly applying the definition and with
some calculation. ■

Definition 5.2.12 (Chain complex). A chain complex (C∗ , d∗ ) is a collection of maps dn between
groups Cn such that
dn+1 dn dn−1
... Cn+1 Cn Cn−1 ...

of Abelian groups Cn and group homomorphism dn such that

dn−1 ◦ dn = 0.

Additionally, we have the followings.

Definition 5.2.13 (Chain group). We call Cn the n-th chain group.

Definition 5.2.14 (Differential). We call dn the n-th differential.

We see that we can certainly think Cn and dn as the simplicial chain group and the boundary
homomorphism as defined before since we already showed this combination satisfies Definition 5.2.12 in
Lemma 5.2.1. But we note that the definition of Cn and dn can be further abstract as what we have
defined.

CHAPTER 5. HOMOLOGY 80
 Lecture 22: Calculation of Homology

Note. As just discussed, we can put different chain group structure on Cn . We’ll see what this
means later.a But for now, we think Cn be equipped with the definition we gave for simplicial chain
group.
a Spoiler: It just means we can give different definition about the map σ.

Remark. We see that

• Lemma 5.2.1 guarantees that our simplicial chain groups form a chain complex.

• Definition 5.2.12 means that ker(dn ) contains Im(dn+1 ), since dn ◦ dn+1 = 0.

Definition 5.2.15 (Exact). We say that the sequence is exact at Cn provided that ker(dn ) =
Im(dn+1 ). A chain complex is exact if it is exact at each point.

Definition 5.2.16 (Homology group). The nth homology group of a chain complex (C∗ , d∗ ), denoted
as Hn or Hn (C∗ ), is the quotient
.
Hn := ker(dn ) Im(d ). n+1

Remark. The homology group measures how far the chain complex is from being exact at Cn .
With what we have just defined, it’s natural to define homology groups of space X with a ∆-complex
structure.

Definition 5.2.17 (Homology class). We say ker(∂n ) is the subgroup of cycles is Cn (X), and Im(∂n+1 )
is the subgroup of boundaries in Cn (X). We then set

{cycles}
. .
Hn (X) := ker(∂n ) Im(∂ ) = {boundaries}.
n+1

In other words, it’s the homology of the chain complex

∂n+1 ∂n ∂n−1
... Cn+1 Cn Cn−1 ...

where we take it to be 0 in all negative indices, namely

∂3 ∂2 ∂1 ∂0
... Cn+1 Cn Cn−1 0

We then call the elements of Hn (X) as homology classes.

Remark. Definition 5.2.17 is saying that we should call the element of a homology group whose
chain group is some kinds of geometric subjects. In this case, we’re just considering ∆-complex
structure, but the definition of homology class is in fact more general.

Definition 5.2.18 (Simplicial homology group). By considering the chain complex being the simplicial
chain groups and boundary homomorphisms, we have so-called simplicial homology groups induced
by Definition 5.2.16.

Lecture 22: Calculation of Homology

5.2.2 Calculation of Homology 25 Feb. 10:00

We start from some calculation about homology group of some spaces.

CHAPTER 5. HOMOLOGY 81
 Lecture 22: Calculation of Homology

Example (Homology group of RP 2 ). Calculate the homology group by Definition 5.2.16 with the
chain complex being the simplicial chain complex.

Proof. Let X = RP 2 .

w b v v1 v2 u2

A A
a c a
B
B

v b w v0 u0 u1

We see that we have

• C0 = Z⟨v, w⟩
• C1 = Z⟨a, b, c⟩

• C2 = Z⟨A, B⟩ = ZA ⊕ ZB
The chain complex is then
∂3 ∂2 ∂1 ∂0
0 C2 C1 C0 0

Where we let A = [v0 , v1 , v2 ] and B = [u0 , u1 , u2 ], then


a →7 w−v
(
7→ b − c + a

A
∂2 : , ∂1 : b → 7 v−w
B 7→ −a − c − b 
c → 7 v−v =0


We can also calculate the image and the kernel at Ci , i.e.,

C2 : Im ∂3 = 0, ker ∂2 = 0,
C1 : Im ∂2 = ⟨2c, b − c + a⟩, ker ∂1 = ⟨b + a, c⟩,
C0 : Im ∂1 = ⟨v − w⟩, ker ∂0 = ⟨v, w⟩.

Hence,
.
= Z⟨v, w⟩ Z⟨v − w⟩ ∼
H0 ∼ =Z

= Z⟨b + a − c, c⟩ Z⟨2c, b + a − c⟩ ∼
. . .
= Z⟨b + a, c⟩ Z⟨2c, b + a − c⟩ ∼
H1 ∼ = Z 2Z
H2 = 0

Remark. Given a basis for a free Abelian group ⟨b1 , . . . , bn ⟩ we can replace bi with

bi ± m 1 b1 ± · · · ± m
di bi ± · · · ± mn bn .

Remark. Care is needed when doing change of bases over Z. For example, if b1 , b2 is a basis for
A ⊆ Zn , then b1 − b2 , b1 + b2 is not a basis, it is an index-2 subgroup. The key to this is that

CHAPTER 5. HOMOLOGY 82
 Lecture 23: Singular Homology

 
1 1
has determinant −2 (not unit in Z).
1 −1
We can transform a basis for a free group into a different basis by applying a matrix of determinant
±1. If we apply a matrix of determinant D we will obtain generators for a subgroup of index |D|.
 
1 0 ··· 0 ±m1 0 ··· 0
0 1 · · · 0 ±m2 0 ··· 0 
 .. .. .. . . . .. .
 
. . . .. .. .. . .. 

 
0 0 · · · 1 ±m i−1 0 · · · 0 
 
0 0 · · · 0 1 0 · · · 0 
 
0 0 · · · 0 ±m i+1 1 · · · 0 
. . . . . . . .
 
 .. .. .. .. .. .. . . .. 


0 0 ··· 0 ±mn 0 ··· 1

As a summary, we have the following procedures to compute Hn (X).

(a) Choose ∆-complex structure on X. (We will prove H∗ (X) is independent of the choice of ∆-complex
structure)
(b) Choose orientations on each simplex (Any choice is okay, but you must commit to a choice, or you
will make a sign error!)
(c) For each n-simplex σ compute ∂n (σ) (careful with signs!)
(d) Im ∂n = ⟨∂n (σ) | σ an n-simplex⟩. Use linear algebra to compute ker(∂n ).
(e) For each n compute Hn (X) = ker ∂n / Im ∂n+1 . Be careful that any change-of-variables map you
apply is invertible over Z.

Lecture 23: Singular Homology

07 Mar. 10:00
5.3 Singular Homology
As we noted before, we can give a different structure of chain complex, which shall induces a different
homology group compare to simplicial homology group.
We now see one abstract way to define σ, which will give us so-called singular homology group.

Definition 5.3.1 (Singular simplex). A singular n-simplex in a space X is a continuous map

σ : ∆n → X.

Definition 5.3.2 (Singular chain complex). The chain complex defined with singular chain group and
singular boundary map defined as follows is called singular chain complex.

Definition 5.3.3 (Singular chain group). Let Cn (X) be the free group on singular n-simplices
in X, which we call it the singular n-chain group.

Definition 5.3.4 (Singular boundary map). With C∗ being the singular chain group, we defined
so-called singular boundary map ∂n as

∂n : Cn (X) → Cn−1 (X)

n
X
σ 7→ (−1)i σ|[v0 ,...,v̂i ,...,vn ] .
i=1

CHAPTER 5. HOMOLOGY 83
 Lecture 23: Singular Homology

Definition 5.3.5 (Singular homology group). The singular homology groups are the homology groups
of this singular chain complex given as
.
Hn (X) = ker ∂n Im ∂ .
n+1

Remark. We now see that from the definition of homology group, we can put different structure on
which. But the idea is the same, namely we are taking Hn (X) being
.
Hn (X) := ker ∂n Im ∂ ,
n+1

where the difference is what structure we put on X which induces different chain complex (C∗ (X), ∂∗ ).
In this case, we have singular homology group since we are considering singular chain complex, while
we can also have simplicial homology group.
Since the generating sets for Cn (X) when considering singular chain complex are almost always hugely
uncountable from its definition, it’s almost impossible to compute with these. However, it does give us
a definition that does not depend on any other structure than the topology of X, making it useful for
developing theory.

Note. The heuristic is that, we interpret a chain σ1 ± σ2 ± · · · ± σk as a map from a ∆-complex to

X.

Example. For example, with σ1 + σ2 as below,

σ1 σ2 X

where we’ve glued [v1 , v2 ] of σ1 to [v0 , v2 ] of σ2 if σ1 |[v1 ,v2 ] and σ[v0 ,v2 ] are the same singular
n-chain with opposite signs.

With what we have defined, we now have some goals.

• Singular homology is a homotopy invariant. (Theorem 5.4.2)
• Singular and simplicial homology groups are isomorphic. (Theorem 5.5.6)

Exercise. Check that if X has path components {Xα } then

Hn (X) ∼
M
= Hn (Xα ).
α

Exercise. If X = {∗}, then (

Z, if n = 0;
Hn (X) =
0, if n ≥ 1.

Exercise. If X is path-connected, then H0 (X) ∼

= Z.

5.4 Functoriality and Homotopy Invariance

CHAPTER 5. HOMOLOGY 84
 Lecture 24: Chain Homotopy

Definition 5.4.1 (Induced map on singular chains). For a given continuous map f : X → Y , we can
consider the map f# induced by singular chains as

f# : Cn (X) → Cn (Y )
[σ : ∆n → X] 7→ [f ◦ σ : ∆n → Y ].

Note. Note that we’re considering singular chain groups specifically in this case.

Remark. We see that the functoriality doesn’t depend on any kind of ∆-complex structure.

Definition 5.4.2 (Chain map). Given two chain complexes (C∗ , ∂∗ ) and (D∗ , δ∗ ), a chain map be-
tween them is a collection of group homomorphisms fn : Cn → Dn such that the following diagram
commutes.
∂n+2 ∂n+1 ∂n ∂n−1
... Cn+1 Cn Cn−1 ...
fn+1 fn fn−1
δn+2 δn+1 δn δn−1
... Dn+1 Dn Dn−1 ...
i.e. we have that δn ◦ fn = fn−1 ◦ ∂n .

Exercise. We have that f# ∂ = ∂f# . In other words, f# is a chain map. Thus, by the homework f#
induces a group homomorphism on the homology groups. We write this as f∗ : Hn (X) → Hn (Y )
for all n.

Exercise. We have functoriality, i.e. (f ◦ g)∗ = f∗ ◦ g∗ and (idX )∗ = idHn (X) .

Theorem 5.4.1 (Homology group defines a functor). The n-th homology group Hn : X 7→ Hn (X)
gives a functor from Top to Ab.

Proof. This follows from the two exercises above. ■

Theorem 5.4.2 (Functoriality is homotopy invariant). If f, g : X → Y are homotopic, then they will
induce the same map on homology

f∗ = g∗ : Hn (X) → Hn (Y ).
The proof of Theorem 5.4.2 can be found here.

Exercise. Theorem 5.4.1 and Theorem 5.4.2 imply that Hn is a homotopy invariant.
I’m not sure whether the above discussion holds only for singular homology group or can be ex-
tended to general homology group. I link them to general homology group anyway.

Lecture 24: Chain Homotopy

To prove Theorem 5.4.2, we introduce some homological algebra. 09 Mar. 10:00

Definition 5.4.3 (Chain homotopy). Given chain complexes (A∗ , ∂∗A ) and (B∗ , ∂∗B ) and chain maps
f# , g# : A∗ → B∗ . A chain homotopy from f# to g# is a sequence of group homomorphisms
ψn : An → Bn+1 such that
B
fn − gn = ∂n+1 ◦ ψn + ψn−1 ◦ ∂nA .

CHAPTER 5. HOMOLOGY 85
 Lecture 24: Chain Homotopy

In diagram, letting hn := fn − gn , we have the following.

A A A A
∂n+2 ∂n+1 ∂n ∂n−1
... An+1 An An−1 ...
ψn ψn−1
hn+1 hn hn−1

B B B B
∂n+2 ∂n+1 ∂n ∂n−1
... Bn+1 Bn Bn−1 ...

This diagram does not commute, however, the red map is the sum of the blue maps composed up,
so it shows everything that is going on.

Theorem 5.4.3. If there is a chain homotopy ψ from f# to g# , then the induced maps f∗ , g∗ on
homology are equal.
Proof. Let σ ∈ An be an n-cycle, i.e. ∂nA σ = 0. Then we compute that:
B
(fn − gn )(σ) = ∂n+1 (ψn (σ)) + ψn−1 (∂nA (σ)) = ∂n+1
B B
(ψn (σ)) ∈ Im(∂n+1 ).

This tells us that (fn − gn )(σ) is a boundary, and so (fn − gn )(σ) = 0 when considered as an
element of the homology group (with degree n). Thus, fn (σ) = gn (σ) in the homology group, and
so f, g induce the same map as desired. ■
We now sketch the proof of Theorem 5.4.2 given in Hatcher [HPM02]. From this point in the course
many of the theorems require much more algebraic work than we are interested in. We instead want to
learn how to use the computational tools.
Proof sketch of Theorem 5.4.2. Suppose we have some homotopy F : I × X → Y from f to g.
The most difficulty in this proof is the combinatorial difficulty involved in the fact that the product
of a simplex in X and I is not a simplex.
We now consider

(a) Subdivide ∆n × I into (n + 1)-dimensional subsimplices.a

∆′

(b) We define the prism operator:

Pn : Cn (X) → Cn+1 (Y )
alternating sums of restrictions
 
σ×id F
[σ : ∆n → X] 7→  ∆n × I −−−→ X × I − →Y 
to each simplex in our subdivision

(c) We now need to check that

Y
∂n+1 Pn = g# − f# − Pn−1 ∂nX .

We have the following diagram.

CHAPTER 5. HOMOLOGY 86
 Lecture 25: Relative Homology

g#
∆n × {1}

P∂

∆n × {0}
f#

Thus P is a chain homotopy, and we’re done.

■
a We want to do this since the product between two simplices is not a simplex, as we just note.

Lecture 25: Relative Homology

We are now interested in the relationship between Hn (X), Hn (A), Hn (X / A). 11 Mar. 10:00

5.5 Relative Homology

Definition 5.5.1 (Reduced homology group). The reduced homology groups H

e n (X) = Hn (X) when
n > 0. When n = 0 we have that:
He 0 (X) ⊕ Z = H0 (X).

Remark. The usefulness of this is that for path-connected space X we have H

e 0 (X) = 0, and for
contractible spaces X we have Hn (X) = 0.
e

Definition 5.5.2 (Good pair). Let X be a space, and A ⊆ X. Then (X, A) is a good pair if A is
closed and nonempty, and also it is a deformation retract of a neighborhood in X.

Example. Let’s see some examples.

(a) If X is a CW complex and A is a nonempty subcomplex, then (X, A) is a good pair. The
proof is given in the Appendix of Hatcher [HPM02] and requires some point-set topology.
(b) If M is a smooth manifold, and N ⊆ M is a smooth submanifold which is nonempty, then
(M, N ) is a good pair.
(c) (Hawaiian earring, bad point) is not a good pair.
(d) (Rn , proper open set) is not a good pair.

Theorem 5.5.1 (Long exact sequence of a good pair). If (X, A) is a good pair, then there exists a long
exact sequence (exact at every n) on reduced homology groups given by the following commutative

CHAPTER 5. HOMOLOGY 87
 Lecture 25: Relative Homology

diagram.
i∗ j∗
... H
e n (A) H
e n (X) H
e n (X / A)
δ

i∗ j∗
H
e n−1 (A) H
e n−1 (X) H
e n−1 (X / A)
δ

i∗ j∗
... H
e 0 (X) H
e 0 (X / A) 0

where i : A ,→ X is the inclusion and j : X → X / A is the quotient map.

We see that both i∗ and j∗ is naturally induced, but not for δ. In fact, we’ll construct δ in the
proof! Specifically, we’ll see that Theorem 5.5.1 is just a special case of Theorem 5.5.3, hence rather
than proving Theorem 5.5.1 directly, we will prove Theorem 5.5.3 instead later.

Remark. The fact that this sequence is exact often means that if we know the homology groups of
two of the spaces we can compute the homology of the remaining space.
Before we see the proof of Theorem 5.5.1, we see one application.

Proposition 5.5.1. We have that

(
Z, if i = n;
e i (S n ) =
H
0, if i ̸= n.

Proof. Some facts we need:

• (Dn , ∂Dn ) is a good pair (since it is a CW complex and a subcomplex)
• Dn /∂Dn ∼
= Sn.

• H
e n (Dn ) = 0 for all n since Dn is contractible.

• ∂Dn ∼
= S n−1 .
We then proceed by induction on n. To start with, we need to verify the following.

Exercise. Verify Proposition 5.5.1 in the case n = 0, so S 0 is just 2 points.

Now, using the long exact sequence, we have

... e n (∂Dn )
H i∗ e n (Dn )
H j∗ e n (S n )
H
δ

e n−1 (∂Dn )
H i∗ e n−1 (Dn )
H j∗ e n−1 (S n )
H

... i∗ e 0 (Dn )
H j∗ e 0 (S n )
H 0

By induction, we have H e n−1 (S n−1 ) = Z, hence we can fill in some of these groups
e n−1 (∂Dn ) = H
as follows.
... 0 i∗ 0 j∗ He n (S n )

Z i∗ 0 j∗ e n−1 (S n )
H

... i∗ 0 j∗ e 0 (S n )
H 0

CHAPTER 5. HOMOLOGY 88
 Lecture 25: Relative Homology

In all, we have an exact sequence

e n (S n ) δ
0 H Z 0

By exactness, δ is an isomorphism, thus H e n (S n ) ∼

= Z. Now we must verify H
e i (S n ) = 0 when i ̸= n.
In that case the exact sequence looks like:

e i (Dn )
H e i (S n )
H e i−1 (∂Dn )
H

0 e i (S n )
H 0

Exactness then tells us that H

e i (S n ) = 0. ■

Theorem 5.5.2 (Brouwer’s fixed point theorem). ∂Dn is not a retract of Dn . Hence, every continuous
map f : Dn → Dn has a fixed point.
Proof. If r : Dn → ∂Dn were a retraction, then by definition this would give us
i r
∂Dn Dn ∂Dn
id∂Dn

Functoriality of homology implies

e n−1 (∂Dn ) i∗ e n−1 (Dn ) r∗ e n−1 (∂Dn )

H H H
id

So then:
i∗ r∗
Z 0 Z
id

which is impossible since the map idZ can’t be factored through 0. ■

Exercise. As with D2 , if f : Dn → Dn had no fixed point, we could build a retraction.

In order to proof Theorem 5.5.1, we introduce the concept of diagram chase.

Lemma 5.5.1 (The short five lemma). Suppose we have a commutative diagram

ψ φ
0 A B C 0
α β γ

′ ψ′ φ′
0 A B C′ 0

so that the rows are exact. Then:

(a) If α, γ are injective then β is injective.

(b) If α, γ are surjective then β is surjective.
(c) If α, γ are isomorphisms then β is an isomorphism
Proof. (a) and (b) imply (c). We leave (b) as an exercise. Let fix b ∈ B such that β(b) = 0, and

CHAPTER 5. HOMOLOGY 89
 Lecture 26: Continue on Relative Homology

we want to show that β = 0, which can be done by doing a diagram chase as

ψ φ
0 • b φ(b) 0
α β γ
ψ′ φ′
0 • 0 0 0

And thus by injectivity of γ we know φ(b) = 0. By exactness, b ∈ Im ψ. We then may write for
some a ∈ A such that the following diagram commutes.
ψ φ
0 a b 0 0
α β γ
ψ′ φ′
0 α(a) 0 0 0

Therefore ψ ′ (α(a)) = β(ψ(a)) = β(b) = 0 by commutativity. By exactness of the bottom row we

know that ψ ′ is an injection.
Thus, α(a) = 0, so since α is injective, a = 0. With this b = ψ(a) = ψ(0) = 0. With this,
ker(β) = 0, and β injects. ■

Lecture 26: Continue on Relative Homology

We start from a definition. 14 Mar. 10:00

Definition 5.5.3 (Relative chain group). Let X be a space and let A ⊆ X be a subspace. Then we
define the relative chain group as
.
Cn (X, A) = Cn (X) C (A),
n

which is a quotient of Abelian groups of the singular chain groups between X and A.

Definition 5.5.4 (Relative chain complex). The relative chain complex (C∗ , ∂∗ ) consists of relative
chain group and the usual differential associated with the singular chain groups which induces our
relative chain group.

Remark. We can indeed adapt Definition 5.5.4 by either singular chain complex structure or sim-
plicial chain complex structure.
It’s not entirely clear that whether Definition 5.5.4 is well-defined, hence we have the following
exercise.

Exercise. Since ∂n∗ (Cn (A)) ⊆ Cn−1 (A), hence there exists a well-defined map
. .
∂n : Cn (X) C (A) → Cn−1 (X) C (A).
n n−1

We can verify that ∂ 2 = 0. Then, since ∂ 2 = 0 we can conclude that these groups will in fact form
a chain complex (C∗ (X, A), ∂).

Definition 5.5.5 (Relative homology group). The homology groups of the relative chain complex
(C∗ (X, A), ∂) are denoted by Hn (X, A), and they are called relative homology groups.
We see that there are something interesting going on in relative chain group. Indeed, we can further
classify the cycles in which as follows.

CHAPTER 5. HOMOLOGY 90
 Lecture 26: Continue on Relative Homology

Definition. Let C∗ (X) be the relative chain complex.

Definition 5.5.6 (Relative cycle). Elements in ker ∂n are called relative n-cycles. These are
elements α ∈ Cn (X) such that ∂n α ∈ Cn−1 (A).

relative cycle

relative 2-cycle
X X

Definition 5.5.7 (Relative boundary). Elements α in Im ∂n+1 are called relative n-boundaries.
This means that α = ∂β + γ where β ∈ Cn (X) and γ ∈ Cn−1 (A).

X
β
A

Figure 5.2: We see that we have α + γ = ∂β, where α is a relative boundary, and γ ∈ Cn−1 (A).

Theorem 5.5.3 (Long exact sequence of a pair). Let A ⊆ X be spaces, then there exists a long exact
sequence
... H
e n (A) i∗
He n (X) q H e n (X, A)
∂

i∗ q
H
e n−1 (A) ... H
e 0 (X, A) 0

where i∗ is induced by A ,→ X, and q is induced by Cn (X) ↠ Cn (X) / Cn (A).

We will prove that when (X, A) is a good pair, then Hn (X, A) ∼
=H e n (X / A). Then Theorem 5.5.1 is
a special case of Theorem 5.5.3. The key to the proof of Theorem 5.5.3 above is the following remark.

Remark. A short exact sequence of chain complexes gives rise to a long exact sequence of homology
groups. Namely, given a short exact sequence of chain complexes (A∗ , ∂ A ), (B∗ , ∂ B ), (C∗ , ∂ C ) such
that
ι q
0 A∗ B∗ C∗ 0
where ι, q are chain maps such that
ιn qn
0 An Bn Cn 0

CHAPTER 5. HOMOLOGY 91
 Lecture 27: Excision

is exact for all n. Then Theorem 5.5.1 will follow from a short exact sequence

0 C
e∗ (A) C
e∗ (X) C
e∗ (X, A) 0

where C
e∗ denotes the augmented chain complex (the one with Z after it, as in Definition 5.5.1).

Exercise. If A is a single point in X, then Hn (X, A) = H

e n (X / A) = H
e n (X).

Lecture 27: Excision

Let’s start with a theorem. 16 Mar. 10:00

Theorem 5.5.4 (Excision). Suppose we have subspace Z ⊆ A ⊆ X such that Z ⊆ Int(A). Then the
inclusion
(X − Z, A − Z) ,→ (X, A)
induces isomorphisms
∼
=
Hn (X − Z, A − Z) −
→ Hn (X, A).
Proof Sketch. We first see an equivalent formulation of Theorem 5.5.4.

Remark. Equivalently, for subspaces A, B ⊆ X whose interiors cover X, the inclusion

(B, A ∩ B) ,→ (X, A)

induces an isomorphism
∼
=
Hn (B, A ∩ B) −
→ Hn (X, A)

Proof. We see that this follows from

B := X \ Z, Z = X \ B,

then we see that A ∩ B = A − Z and the condition requires from Theorem 5.5.4, Z ⊆ Int(A)
is then equivalent to
X = Int(A) ∪ Int(B)
since X \ Int(B) = Z.

A A
X Z X Z

Hn (X, A) Hn (X / Z, A / Z)

⊛
We now sketch the proof of the above equivalent form of Theorem 5.5.4, which is notorious for

CHAPTER 5. HOMOLOGY 92
 Lecture 28: Singular Homology v.s. Simplicial Homology

being hairy.
• Given a relative cycle x in (X, A), subdivide the simplices to make x a linear combination of
chains on smaller simplices, each contained in Int(A) or X \ Z.

Z A X Z A X

Figure 5.3: ∆n → X subdivide into subsimplices with images in.

This means x is homologous to sum of subsimplices with images in Int(A) or X \ Z. One of

the things we use is that simplices are compact, so this process takes finite time.
The key is that the Subdivision operator is chain homotopic to the identity.
• Since we are working relative to A, the chains with image in A are zero, thus we have a relative
cycle homologous to x with all simplices contained in X \ Z.

Exercise. Show that H∗ (Y, y0 ) ∼

=He ∗ (Y ).

Theorem 5.5.5. For good pairs (X, A), the quotient map q : (X, A) → (X / A, A / A) induces iso-
morphisms
∼
q∗ : Hn (X, A) = Hn (X / A, A / A) = ∼H e n (X / A)

for all n.
Proof Sketch. Let A ⊆ V ⊆ X where V is a neighborhood of A that deformation retracts onto A.
Using excision, we obtain a commutative diagram
∼
= ∼
=
Hn (X, A) Hn (X, V ) Hn (X − A, V − A)
q∗ q∗ ∼
= q∗
∼
= ∼
=
Hn (X / A, A / A) Hn (X / A, V / A) Hn (X / A − A / A, V / A − A / A)

Done if we can prove all the colored isomorphisms.

• ∼
= is an isomorphism by excision.
• ∼
= is an isomorphism by direct calculation (since q is a homeomorphism on the complement
of A).
• ∼
= on Homework, since V deformation retracts to A.

Remark. The last equality is from the above exercise with A / A = {∗}.

CHAPTER 5. HOMOLOGY 93
 Lecture 29: Proof of Theorem 5.5.6

Lecture 28: Singular Homology v.s. Simplicial Homology

Remark. If M is a smooth manifold and N is an embedded smooth closed submanifold, then (M, N ) 18 Mar. 10:00
is a good pair. Why? Well this follows from the tubular neighborhood theorem, which should be
proven in a course like MATH 591. We will only use the result in obvious cases, and simply assert
that certain pairs are good pairs.

Example. With pairs like (Rn+1 , S n ), you can just assert that this is a good pair.

Proof. If we want to be rigorous, we need to show that S n is a smooth submanifold of Rn+1 , but
we simply assert this in our course. ⊛

Example. Manifolds and their boundary always form a good pair.

Theorem 5.5.6 (Singular homology agrees with simplicial homology). Let X be a ∆-complex. We
use ∆n (X) to represent the simplicial chain groups on X, and Cn (X) to denote the singular chain
groups. Likewise, we denote .
∆n (X, A) = ∆n (X) ∆ (A)
n

and .
Cn (X, A) = Cn (X) C (A).
n

The inclusion ∆∗ (X, A) ,→ C∗ (X, A) given by

[σ : ∆n → X] 7→ [σ : ∆n → X]

induces an isomorphism on homology such that

Hn∆ (X, A) ∼
= Hn (X, A).

If we consider the case that A = ∅, we recover the case of absolute homology

Hn∆ (X) ∼
= Hn (X).
The proof of Theorem 5.5.6 uses the following lemma.

Lemma 5.5.2 (The five lemma). If we have a commutative diagram with exact rows as following,

i j k ℓ
A B C D E
α β γ δ ϵ

A′ B′ C′ D′ E′
i′ j′ k′ ℓ′

If α, β, δ, ϵ are isomorphisms, then so is γ.

Proof. Diagram chase! ■

Lecture 29: Proof of Theorem 5.5.6

We now give a proof sketch for Theorem 5.5.6. 21 Mar. 10:00
Proof Sketch of Theorem 5.5.6. The idea is as follows.
• We can use the long exact sequence of a pair and the Lemma 5.5.2 to reduce to proving the
result for absolute homology groups (and we will recover the general result).
• Because the image ∆n → X is compact, it is contained in some finite skeleton X k . Use this

CHAPTER 5. HOMOLOGY 94
 Lecture 29: Proof of Theorem 5.5.6

to reduce the proof to the finite skeleton X k of X, namely we can use induction.
From the long exact sequence of a pair we get

∆
Hn+1 (X k , X k−1 ) Hn∆ (X k−1 ) Hn∆ (X k ) Hn∆ (X k , X k−1 ) ∆
Hn−1 (X k−1 )
α β γ δ ϵ

Hn+1 (X k , X k−1 ) Hn (X k−1 ) Hn (X k ) Hn (X k , X k−1 ) Hn−1 (X k−1 )

The Goal is to prove γ is an isomorphism using the Lemma 5.5.2.

We assume that β, ϵ are isomorphisms by induction, checking the case manually for X 0 (which
will be a discrete set of points).

Exercise. Check the base case, namely when X 0 .

It remains to show that α, δ are isomorphisms. We know then that
(
k k−1 Z[k-simplices], if k = n; ∼ ∆ k k−1
∆n (X , X )= = Hn (X , X ).
0, otherwise

We claim that Hn (X k , X k−1 ) are also free Abelian on the singular k-simplices defined by the
characteristic maps ∆k → X k when n = k, and 0 otherwise. Consider the map
a
Φ: (∆kα , ∂∆kα ) → (X k , X k−1 )
α

defined by the characteristic map. This induces an isomorphism on homology since

a
∆kα a = Xk
∼
. .
k −→
α ∂∆α X k−1 .

This reduces to check that

(
k k 0, if n ̸= k;
Hn (∆ , ∂∆ ) =
Z, if n = k

generated by the identity map ∆k → ∆k . ■

Corollary 5.5.1. If X has a ∆-complex structure (or is homotopy equivalent to one), then we have
the followings.
(a) If the dimension is ≤ d, then Hn (X) = 0 for all n > d.
(b) If X has no cells of dimension p, then Hp (X) = 0.
(c) If X has no cells of dimension p, then Hp−1 (X) is free Abelian.

Corollary 5.5.2. Given a singular homology class on X, without loss of generality we can choose
a ∆-complex structure on X, and we then we can assume the class is represented by a simplicial
n-cycle.

Remark. Recall the definition of homology class, as we noted before, this means we can view
singular chain complex as some kind of geometric subjects. The construction can be found in
Hatcher [HPM02].

5.6 Degree

CHAPTER 5. HOMOLOGY 95
 Lecture 29: Proof of Theorem 5.5.6

Definition 5.6.1 (Degree). Let f : S n → S n , then

f∗ : Z ∼
= Hn (S n ) → Hn (S n ) ∼
=Z

is a multiplication by some integera d ∈ Z, which we call it as the degree, denotes as deg(f ) of f .

a This just follows from group theory.

Remark (Properties of Degree). We first see some properties of degree.

(a) deg(idS n ) = 1 since (idSn )∗ = idZ .

(b) If f : S n → S n , n ≥ 0 is not surjective, then deg(f ) = 0. To see this, we know that f∗ factors
as
Hn (S n ) Hn (S n − {∗}) = 0 Hn (S n )
f∗

And since the middle group is zero because S n \ {∗} is contractible, f∗ = 0, so is its degree.
(c) If f ≃ g, then f∗ = g∗ , so deg(f ) = deg(g).

Note. The converse is true! We’ll see this later.

(d) (f ◦ g)∗ = f∗ ◦ g∗ , and so deg(f ◦ g) = deg(f ) deg(g). Consequently, if f is a homotopy

equivalence then f ◦ g ≃ id, hence

deg f deg g = deg id = 1,

so deg f = ±1.
(e) If f is a reflection fixing the equator, and swapping the 2-cells, then deg f = −1.

Exercise. It is possible to put a ∆-complex structure with 2 n-cells, ∆1 and ∆2 glued

together along their boundary (∼ = S n−1 ), and

Hn (S n ) = ⟨∆1 − ∆2 ⟩.

∆1

f , reflection
∆2

Answer. By doing so, we see that the reflection f interchanges ∆1 and ∆2 , hence the
generator is now its negative. ⊛

(f) We now have the following linear algebra exercise.

Exercise. The antipodal map − id : S n+1 → S n+1 given by x 7→ −x is the composite of

(n + 1) reflections.

So the antipodal map − id : S n → S n given by x 7→ −x has degree which is the product of

n + 1 copies of (−1), and so it has degree (−1)n+1 . (i.e., since the (n + 1) × (n + 1) scalar
matrix (−1) is composition of (n + 1) reflections.)

(g) We see the following.

CHAPTER 5. HOMOLOGY 96
 Lecture 30: Degree

Exercise. If f : S n → S n has no fixed points, then we can homotope f to the antipodal

map via
(1 − t)f (x) − tx
ft (x) = .
∥(1 − t)f (x) − tx∥

Therefore, deg f = (−1)n+1 .

Lecture 30: Degree

With the definition of degree and some of its properties, we have the following theorems. 23 Mar. 10:00

Theorem 5.6.1 (Hairy ball theorem). The sphere S n admits a nonvanishing continuous tangent
vector field if and only if n is odd.
Proof. Recall that a tangent vector field to the unit sphere S n ⊆ Rn+1 is a continuous map

v : S n → Rn+1

such that v(x) is tangent to S n at x, i.e., v(x) is perpendicular to the vector x for each x. Let v(x)
be a nonvanishing tangent vector field on the sphere S n , then we define
 
v(x)
ft (x) := cos(πt) + sin(πt) ,
∥v(x)∥

which is a homotopy from the identity map idS n : S n → S n to the antipodal map − idS n : S n → S n .
This simply follows from varying t from 0 to 1, where we have
 
v(x)
f0 (x) = cos(0)x + sin(0) = x ⇒ f0 = idS n ,
∥v(x)∥

while  
v(x)
f1 (x) = cos(π)x + sin(π) = −x ⇒ f1 = − idS n .
∥v(x)∥
The last thing needs to be verified is that ft (x) is continuous, but this is trivial.
From the property of degree, we know that it’s a homotopy invariant, hence

deg(− idS n ) = deg(idS n ),

which implies
(−1)n+1 = 1,
so n must be odd.
Conversely, if n is odd, say n = 2k − 1, we can define

v(x1 , x2 , . . . , x2k−1 , x2k ) = (−x2 , x1 , . . . , −x2k , x2k−1 ).

Then v(x) is orthogonal to x, so v is a tangent vector field on S n , and |v(x)| = 1 for all x ∈ S n . ■

Theorem 5.6.2 (Groups acting on S 2n ). If G acts on S 2n freely, then

.
G = Z 2Z or 1.

Proof. There exists a homomorphism given by

d : G → {±1}
g 7→ deg(τg )

Where τg is the action of g ∈ G on S 2n as a map S 2n → S 2n . We know this map is well-defined

CHAPTER 5. HOMOLOGY 97
 Lecture 30: Degree

since τg is invertible (simply take τg−1 ) for each g ∈ G. Also, we see that d(g1 g2 ) = d(g1 ) · d(g2 ),
hence it is a homomorphism.
We want to show that the kernel is trivial, since then by the first isomorphism theorem G ∼ = Im,
and the image is either trivial or Z / 2Z. Suppose that g is a nontrivial element of G, then since G
acts freely we know that τg has no fixed points. With this in mind we have

deg τg = (−1)2n+1 = −1.

Thus, g ̸∈ ker, hence the kernel is trivial as desired. ■

Corollary 5.6.1. S 2n has only the trivial cover S 2n → S 2n or degree 2 cover (for example, S 2n →
RP 2n ).
Proof. This follows since any covering space action acts freely, then we simply apply Theorem 5.6.2.
■

Definition 5.6.2 (Local degree). Let f : S n → S n (n > 0). Suppose there exists y ∈ S n such that
f −1 (y) is finite, say, {x1 , . . . , xm }. Then let U1 , . . . , Um be disjoint neighborhoods of x1 , . . . , xm
that are mapped by f to some neighborhood V of y.

x1 y
U1 f V

x2
U2

The local degree of f at xi , denote as deg f |xi , is the degree of the map

f∗ : Z ∼
= Hn (Ui , Ui − {xi }) → Hn (V, V − {y}) ∼
= Z.

Remark. The homomorphism f∗ is a multiplication by an integer, which is the local degree as we

just defined, arises from the following natural diagram.
f∗
Hn (Ui , Ui − {xi }) Hn (V, V − {y})
∼
=
ki ∼
=
pi f∗
Hn (S n , S n − {xi }) Hn (S n , S n − f −1 (y)) Hn (S n , S n − {y})
j ∼
=
∼
=
f∗
Hn (S n ) Hn (S n )

The two isomorphisms in the upper half come from excision, and the lower two isomorphisms come
from exact sequences of pairs.

Theorem 5.6.3. Let f : S n → S n with f −1 (y) = {x1 , . . . , xm } as in Definition 5.6.2, then we have
m
X
deg f = deg f |xi .
i=1

Remark. Thus, we can compute the degree of f by computing these local degrees.
Let’s work with some examples for our edification.

CHAPTER 5. HOMOLOGY 98
 Lecture 31: Local Degree and Local Homology

Example. Consider S n and choose m disks in S n . Namely, we first collapse the complement of the
m disks to a point, and then we identify each of the wedged n-spheres with the n-sphere itself.

D1
D1 D3

D2

D3 D2

The result will be a map of degree m. We can see this by computing local degree.

x2 y
x1

x3

By choosing a good point in the codomain, we get one point for each disk in the preimage, and
the map is a local homeomorphism around these points which is orientation preserving. We could
likewise compose the maps to S n from the wedge with a reflection to construct a map of degree
−m.

Remark. We see that from the above construction, we can produce a map S n → S n in any degree.

Lecture 31: Local Degree and Local Homology

We first see another example of the application of Theorem 5.6.3. 25 Mar. 10:00

Example. Consider the composition of the quotient maps below

Sn RP n RP n / RP n−1 ∼
= Sn
f

We want to compute the degree of this map.

Sn RP n RP n / RP n−1 ∼
= Sn

x1 y y

x2
local homeomorphism
in neighborhood of
x1 , x2 ∈ f −1 (y)

CHAPTER 5. HOMOLOGY 99
 Lecture 31: Local Degree and Local Homology

Note that this restricts to a homeomorphism on each component of S n \ equator as a map to

RP n / RP n−1 . Suppose we’ve oriented our copies of S n in such a way that the homeomorphism on
the top hemisphere is orientation-preserving. The homeomorphism on the bottom hemisphere is
given by taking the antipodal map and composing with the homeomorphism of the top hemisphere
(
0, if n even;
deg = deg(id) = deg(antipodal) = 1 + (−1)n+1 =
2, if n odd.
We can now prove Theorem 5.6.3.
Proof of Theorem 5.6.3. If f : S n → S n and we have some y ∈ S n with f −1 ({y}) = {x1 , . . . , xm },
then we have a nice commutative diagram as follows.
f∗
Hn (S n ) Hn (S n )
LES of pair

m
Z = Hn (S n , S n − {x1 , . . . , xm })
L
∼
= LES of a pair
i=1

excision ∼
=
 m̀ m̀

Hn Ui , (Ui − {xi }) Hn (S n , S n − {y})
i=1 i=1
∼
= excision
homology of disjoint union ∼
=
L
Hn (Ui , Ui − {xi }) Hn (V, V − {y})
i

1 deg(f∗ )

P
(1, 1, . . . , 1) deg f = deg f |xi

where we trace around the outside of the diagram at the bottom, which just proves the result. ■
With degree, we have a very efficient way for computing the homology groups of CW complexes,
which is so-called cellular homology. But before we dive into this, we first grab some intuition about the
essential of which, namely,
what really is local homology?

Problem 5.6.1. What is local homology?

Answer. By excision, there is an isomorphism Hn (S n , S n \ {xi }) ∼

= Hn (U, U \ {xi }) for any open
neighborhood U of xi . The long exact sequence of a pair also gives us

... Hk (S n \ {xi }) Hk (S n ) Hk (S n , S n \ {xi }) Hk−1 (S n \ {xi }) ...

Since S n \{xi } is homeomorphic to an open n-ball, we see that Hk (S n \{xi }) = Hk−1 (S n \{xi }) = 0.
With this in mind, j∗ is an isomorphism.
We want to think about what j∗ does when k = n, i.e., when this is an isomorphism Z ∼ =
Hn (S n ) → Hn (S n , S n \ {xi }) ∼
= Z.
We see that ∆1 − ∆2 generate Hn (S n ), where ∆1 , ∆2 are the top and bottom hemisphere
indicated below.

CHAPTER 5. HOMOLOGY 100

 Lecture 31: Local Degree and Local Homology

Hn (S n , S n \ {x})
x1 ∆1

∆1 − ∆2 = ∆1
∆2

∆1 − ∆2

We then understand that j∗ (∆1 − ∆2 ) = ∆1 − ∆2 = ∆1 since ∆2 = 0 in Cn (S n )/Cn (S n \ {xi }).

The upshot is that Hn (S n , S n \ {x}) is generated by an n-simplex with x in its interior.
Suppose M is an n-manifold. Then Hn (M, M \ {x}) ∼ = Hn (U, U \ {x}), where U is a small ball
around x. Because U is a ball homeomrphic to Rn , we see that

Hn (M, M \ {x}) ∼
= Hn (U, U \ {x}) ∼
= Hn (Rn , Rn \ {x}).

By the long exact sequence of a pair

0 = Hn (Rn ) Hn (Rn , Rn \ {x}) Hn−1 (Rn \ {x}) Hn−1 (Rn ) = 0

And since Rn \{x} is homotopy equivalent to an n−1 sphere, this means that Hn (Rn , Rn \{x}) ∼
= Z.
By homework, this connecting homomorphism is given by taking the boundary of a relative cycle
as below.

u u

x δ x

Hn (Rn , Rn − x) Hn (Rn − x)

∆ ∂∆
an n − 1-sphere around x in u

We intuitively want to use this idea to compute degree using this idea. We use naturality of the
long exact sequence, namely the fact that where f : (Ui , Ui \ {xi }) → (V, y) is a map of pairs, then
the following diagram commutes.

... Hn (Ui , Ui \ {xi }) Hn−1 (Ui , Ui \ {xi }) ...

f∗ f∗

... Hn (V, V \ {y}) Hn−1 (V, V \ {y}) ...

By naturality of the long exact sequence and the isomorphism discussed above, we can compute the
local degree of a map S n → S n at a point x by computing the degree of the map

Hn−1 (U \ {x}) Hn−1 (V − {y})

In fact the local degree will be the degree restricted to a small S n−1 n the neighborhood U .

CHAPTER 5. HOMOLOGY 101

 Lecture 32: Cellular Homology

U V
n−1
S

x f y

f (S n−1 )

Example (Degree of z n ). Consider

e→C
f: C e
z 7→ z n .

We see that
deg f |0 = n.

Proof. We look at the illustration of f (z) = z n .

zn wound n times

0
C C

Lecture 32: Cellular Homology

28 Mar. 10:00
5.7 Cellular Homology
Suppose that X is a CW complex, then (X n , X n−1 ) is a good pair for all n > 1, and X n / X n−1 is a
wedge of n-spheres, one for each n-cell enα . Hence,
(
n n−1 ∼ 0, if k =
̸ n;
Hk (X , X )=
⟨eα | eα is an n-cell⟩, if k = n.
n n

Definition 5.7.1 (Cellular chain complex). The cellular chain complex on X is the chain complex
with cellular chain group and cellular boundary map defined as follows.

Definition 5.7.2 (Cellular chain group). The chain groups Cn (X) defined as

Cn (X) := Z⟨enα | enα an n-cells of X⟩(∼

= Hn (X n , X n−1 ))

with X −1 = ∅ is called cellular chain group.

CHAPTER 5. HOMOLOGY 102

 Lecture 32: Cellular Homology

Definition 5.7.3 (Cellular boundary map). For n = 0, we have

∂1 : C1 (X) → C0 (X)
⟨1-cell⟩ → ⟨0-cell⟩,

which is the usual simplicial boundary map.a For n > 1, the boundary map ∂n are defined as
X
∂n (enα ) = ∂αβ en−1
β
β

where ∂αβ is the degree of the map

attaching quotient by
∂enα = Sαn−1 map X n−1 Sβn−1
X n−1 \ en−1
β

a i.e., ∂1 : C1 (X) = H1 (X 1 , X 0 ) → C0 (X) = H0 (X 0 ) is just ∆1 (X) → ∆0 (X).

Example. In pictures, the degree of a function is given as the following.

eβn−1
en−1
β
enα

Sαn−1 X n−1

∂αβ = deg

Remark. We see that

Cn (X) ∼
= Hn (X n , X n−1 )
since (X n , X n−1 ) is a good pair, so Hn (X n , X n−1 ) ∼
= Hn (X n / X n−1 ), which is just the wedge of 1
n-sphere for each n-cell of X.
Furthermore, the orientations on spheres are defined by identifying the domains of characteristic
maps Dαn → X with an (oriented) disk in Rn . i.e., we need to choose a generator of

Hn−1 (∂Dαn ) ∼
= Hn−1 (S n−1 ) ∼
= Z.

Note. In Hatcher [HPM02], the approach of the definition of cellular chain complex is aP bit different,
especially for how we define the boundary maps. Here we simply define ∂n (enα ) := β ∂αβ eβn−1 ,
where this is so-called cellular boundary formula in Hatcher [HPM02]. Here, we just defined ∂n in
this way instead, but we should still check that this is well-defined of this definition. The proof is
given in Appendix A.3.

Definition 5.7.4 (Cellular homology group). We define the so-called cellular homology group by cel-
lular chain complex in our usual way of defining homology group.

Remark. We sometimes denote the cellular homology group as HnCW (X) if it causes confusion.

Theorem 5.7.1. Definition 5.7.1 indeed forms a chain complex.

CHAPTER 5. HOMOLOGY 103

 Lecture 32: Cellular Homology

Proof. We need to check two things, namely the chain group Hn (X n , X n−1 ) defined in Defini-
tion 5.7.1 is indeed free Abelian with basis in each n-cell. But this is trivial since we have an
one-to-one correspondence with the n-cells of X as we have shown, and we can think of elements
of Hn (X n , X n−1 ) as linear combinations of n-cells of X.
The fact that the boundary map defined in Definition 5.7.1 has the property ∂ 2 = 0 will be
proved in Theorem 5.7.2. ■

Theorem 5.7.2 (Cellular homology agrees with singular homology). The cellular homology groups
coincide with the singular homology groups, i.e.,

HnCW (X) ∼
= Hn (X).

Note. i.e., the isomorphism commutes ωf∗ for all continuous f : X → Y .

Theorem 5.7.2 implies the following.

Corollary 5.7.1. We have the followings.

• Hn (X) = 0 if X has a CW complex structure with no n-cells.
• If X has a CW complex with k n-cells, then Hn (X) is generated by at most k elements.
• If Hn (X) is a group with a minimum of k generators, then any CW complex structure on X
must have at least k n-cells.
• If X has a CW complex with no n-cells, then

Hn−1 (X) = ker(∂n−1 ),

which is free Abelian.

• If X has a CW complex with no cells in consecutive dimensions, then all ∂n = 0. Its homology
are free Abelian on its n-cells, namely the cellular chain groups.

Example. The last point in Corollary 5.7.1 is quite useful, as the following examples will show.
(a) S n , n ≥ 2. Since if we have S n with n ≥ 2, using the CW complex structure of en attached
to a single point x0 . The cellular chain complex is given as

0 0 ⟨en ⟩ 0 ... 0 ⟨x0 ⟩ 0

So then all the boundary maps are zero, and we see that
(
n Z, if k = 0, n;
Hk (S ) =
0, otherwise.

(b) CP n , ∀n. In this case, we can let CP n equipped with a CW complex structure with one cell
of each even dimension 2k ≤ 2n, thus
(
n ∼ Z, if k = 0, 2, . . . , 2n;
Hk (CP ) =
0, otherwise.

(c) S n × S n , n > 1. We let S n × S n has the product CW structure consisting of a 0-cell, two

CHAPTER 5. HOMOLOGY 104

 Lecture 33: Cellular Homology Examples

n-cells, and a 2n-cell. Thus, we have

if k = 0, 2n;

Z,

n ∼
Hk (CP ) = Z2 , if k = n;
otherwise.

0,


Exercise. Redo this calculation for S n with other CW complex structures on S n , e.g. glue 2 k-cells
onto S k−1 and proceed inductively.

Lecture 33: Cellular Homology Examples

Example (Cellular homology group of torus). Calculate the cellular homology group of a torus. 30 Mar. 10:00

Proof. Let the torus equips with the following CW complex structure.

a D a

x b

The cellular chain complex looks like

0 ⟨D⟩ ⟨a, b⟩ ⟨x⟩ 0

where we choose x as a base point (i.e. the 0-cell).

For ∂1 , since this is defined as the same as the usual simplicial boundary map, hence by a 7→
x − x = 0 and b 7→ x − x = 0, we have ∂1 = 0.
Now for ∂2 , since D is glued along aba−1 b−1 , so we look at the composed up maps

a
a

b deg 0
∂D

We wind forwards then backwards around a,a so the degree is zero. The same thing happens
for b, so
0 · a + |{z}
∂2 D = |{z} 0 · b = 0,
∂αβa a ∂αβb b

where we assume that α is the index of D, and βa is the index of a and same for b.
This gives a nice principle, namely if a 2-cells D is glued down via some words w (this only
makes sense for 2-cells), then the coefficientb to a letter a in ∂2 D is the sum of the exponents of a
in w. In this case, for both a and b, the coefficients for are both 1 + (−1) = 0.
Now we just have that the homology groups are equal to the chain groups because the boundary

CHAPTER 5. HOMOLOGY 105

 Lecture 34: Proof of Theorem 5.7.2

maps are all zero. Hence, we have

if k = 0, 2;

Z,

Hk (T ) = Z2 , if k = 1;
otherwise.

0,


⊛
a Intuitively, since we quotient out b, hence the gluing map is homotopy to constant maps.
b i.e. ∂αβ (a) where α is the index of a.

Example (Cellular homology group of Σg ). Calculate the cellular homology group of a genus g surface
Σg .

Proof. A genus g surface Σg has the CW complex structure as

• 1 0-cell x.
• 2g 1-cells a1 , b1 , a2 , b2 , . . ..

• 1 2-cell D glued along [a1 , b2 ][a2 , b2 ] · · · [ag , bg ] (a product of commutators)

For ∂1 , we have
∂1 (ai ) = ∂1 (bi ) = x − x = 0.
Furthermore, by the principle discussed above, we know that every 1-cell appears once in the
word, and its inverse appears once, so all the coefficients of 1-cells in ∂2 (D) are zero, so ∂2 (D) = 0.
This means we have a chain complex
0 0
0 Z Z2g Z 0

And so then we have that

if k = 0, 2;

Z,

Hk (Σg ) = Z2g , if k = 1;
otherwise.

0,


Exercise. Calculate the cellular homology group of RP n .

Example (Torus example: ∂2 in more detail). We’re going to work through this example a bit more
carefully.

y
x2 x1
want degree of this map

S1

Let’s zoom in on these two preimage points and use local homology to compute this: Fill this
up!

Lecture 34: Proof of Theorem 5.7.2

1 Apr. 10:00

CHAPTER 5. HOMOLOGY 106

 Lecture 34: Proof of Theorem 5.7.2

We’re now going to work towards proving that cellular homology agrees with singular homology. First
we need some nontrivial preliminaries.

Lemma 5.7.1. We have that

(
0, if k ̸= n;
(a) Hk (X , X
n n−1
)=
⟨n-cells⟩, if k = n.

(b) Hk (X n ) = 0 for all k > n. If X is finite dimensional, then Hk (X n ) = 0 for all k > dim X.

(c) The inclusion X n ,→ X induces Hk (X n ) → Hk (X). Then this map is

• an isomorphism for k < n
• surjective for k = n
• zero for k > n.
Proof. For (a), we see that

Xn
.
∼
X n−1 = wedge of one n-sphere for each n-cell.

The result then follows from Theorem 5.5.5 and its immediately corollary, namely
!
e n (Xα ) ∼
M _
H =Hen Xα
α α

provided that the wedge sum is formed

` at ` basepoints xα ∈ Xα such that (Xα , xα ) are good, and
then we simply consider (X, A) = ( α Xα , α {xα }).
Now we prove (b) and (c), We consider the long exact sequence of a pair for fixed n,

... Hk+1 (X n , X n−1 ) Hk (X n−1 )

∼
=

Hk (X n ) Hk (X n , X n−1 ) ...

When k + 1 < n or k > n then Hk+1 (X n , X n−1 ) = 0 and Hk (X n , X n−1 ) = 0, so the above
map Hk (X n−1 ) → Hk (X n ) is an isomorphism. We also get sequences telling us the injective and
surjective maps when k = n or k = n − 1,

... 0 = Hn+1 (X n , X n−1 ) Hn (X n−1 ) Hn (X n )

Hn (X n , X n−1 ) Hn−1 (X n−1 ) Hn−1 (X n )

Hn−1 (X n , X n−1 ) = 0 ...

So the maps Hn (X n−1 ) → Hn (X n ) is injective, and the map Hn−1 (X n−1 ) → Hn−1 (X n ) is surjec-
tive.
Fix k, then we get a pile of maps induced by the inclusions X n ,→ X n+1

Hk (X 0 ) ∼
= Hk (X 1 ) ∼
= Hk (X 2 ) ∼
= ...
∼
=

Hk (X k−1 ) inj. Hk (X k ) surj. Hk (X k+1 )

∼
=

Hk (X k+2 ) ∼
= Hk (X k+3 ) ∼
= ...

CHAPTER 5. HOMOLOGY 107

 Lecture 34: Proof of Theorem 5.7.2

Note. This sequence is not exact. Descriptions of maps (in red) follow from our analysis of
the long exact sequence of a pair above.
To prove (b),
• k = 0, we do this by hand.
• k ≥ 1, then Hk (X 0 ) = 0, so we have that Hk (X 0 ), . . . , Hk (X k−1 ) are all zero from the
isomorphisms above. That is the k-th homology Hk (X n ) = Hk (X n ) is zero for every n-
skeleton where n < k, just as desired.
We also have the following collection of maps for fixed k
surj. ∼
= ∼
=
Hk (X k ) Hk (X k+1 ) Hk (X k+2 ) ...

This implies (c) when X is finite dimensional. For general X, we use the fact that every simplex
has image contained in some finite skeleton (since image is compact). ■

Exercise. Check (b) and (c) in Lemma 5.7.1 directly in the case that the CW complex structure is
a ∆-complex structure using simplicial chains.
We now prove Theorem 5.7.2.
Proof of Theorem 5.7.2. We get some exact sequences from our preliminaries,

0 = Hn+1 (X n ) Hn (X n ) Hn (X n , X n+1 ) Hn−1 (X n,−1 )

Hn+1 (X n+1 , X n ) Hn (X n ) Hn (X n+1 ) Hn (X n+1 , X n ) = 0

These come from the long exact sequences of a pair combined with the things we’ve deduced in the
preliminaries. We can paste these together into a diagram, we have
0

0 Hn (X n+1 ) ∼
= Hn (X)

Hn (X n )
∂n+1 jn

dn+1
... Hn+1 (X n+1 , X n ) Hn (X n , X n−1 ) dn
Hn−1 (X n−1 , X n−2 ) ...

∂n jn−1
Hn−1 (X n−1 )

0
Hatcher [HPM02] tells us this diagram commutes, and what we’ve done here tells us that the two
red diagonal pieces crossing at Hn (X n ) are exact. We also have exactness of the bottom right
diagonal by just going down a degree.
Then the horizontal row has to at least be a chain complex since the diagram commutes, and
we have
dn ◦ dn+1 = (jn−1 ◦ ∂n ) ◦ (jn ◦∂n+1 ) = 0,
| {z }
0

hence we see that d = 0.

2 a

By exactness, we know that if ι∗ : Hn (X n ) → Hn (X n+1 ), then using the first isomorphism

theorem,
n . n .
Hn (X) ∼ = Hn (X ) ker ι = Hn (X ) Im ∂
= Hn (X n+1 ) = Im ι∗ ∼ .
∗ n+1

CHAPTER 5. HOMOLOGY 108

 Lecture 35: Eilenberg-Steenrod Axioms

Since jn injects by exactness,

∼
=
jn : Hn (X n ) −
→ jn (Hn (X n ))
∼
=
Im ∂n+1 −
→ Im(jn ◦ ∂n+1 ) = Im dn+1 ,

so jn−1 must also inject by exactness, and by applying exactness, we have

ker dn = ker ∂n = Im jn .

Then we just do some group theory, the n-th cellular homology group is

Hn (X n )
. . .
ker dn ∼ Im jn ∼ ∼
Im dn+1 = Im(jn ◦ ∂n+1 ) = Im ∂n+1 = Hn (X).

There is one thing left to show, namely commutativity of this map. We claim that the differentials
dn = jn ◦ ∂n+1 satisfy the formula (in terms of degree) that we stated. This is done by direct
analysis of definitions of maps; details in Hatcher [HPM02]. ■
a This is the missing part of the proof of Theorem 5.7.1.

Lecture 35: Eilenberg-Steenrod Axioms

4 Apr. 10:00
5.8 Eilenberg-Steenrod Axioms
We can approach the homology theory in an axiomatic way. Specifically, we’re interested in the
Eilenberg-Steenrod axioms. To start with, we first see some definitions.

Definition 5.8.1 (Natural transformation). Given two functors

F, G : C → D,

a natural transformation η : F → G is a collection of maps ηX : F (X) → G(X) lying in D for every

X ∈ C so that for any map f : X → Y , we have a commutative diagram
ηX
F (X) G(X)
F (f ) G(f )

F (Y ) ηY
G(Y )

Definition 5.8.2 (Homology theory). A homology theory is a sequence of functors

Hn : pairs (X, A) of spaces → Abelian groups

equipped with natural transformations ∂ : Hn (X, A) → Hn−1 (A), where Hn−1 (A) := Hn−1 (A, ∅),
is called the boundary map.
Naturality here means that for any map f : (X, A) → (Y, B) we have a commutative diagram

∂
Hn (X, A) Hn−1 (A)
f∗ f∗

Hn (Y, B) ∂
Hn−1 (B)

These must satisfy the following 5 axioms.

(a) (Homotopy) If f, g : (X, A) → (Y, B) and f ≃ g, then f∗ = g∗ .

CHAPTER 5. HOMOLOGY 109

 Lecture 35: Eilenberg-Steenrod Axioms

(b) (Excision) If U ⊆ A ⊆ X such that U ⊆ Int(A), then

ι : (X \ U, A \ U ) ,→ (X, A)

induces isomorphisms on Hn .

(c) (Dimension) Hn (∗) = 0 for all n ̸= 0.

(d) (Additivity) Hn ( α Xα ) = α Hn (Xα ).
` L

(e) (Exactness) If we have an inclusion ι : A ,→ X a and j : X → (X, A) induces a long exact

sequence
ι∗ j∗ ∂
... Hn (A) Hn (X) Hn (X, A) Hn−1 (A) ...
a Note that we use X := (X, ∅) for every space X.

Definition 5.8.3 (Extraordinary homology theory). If H∗ satisfies all axioms but dimension, it is called
an extraordinary homology theory.

Example. Topological K-theory, bordism, and cobordism.a

a https://en.wikipedia.org/wiki/Cobordism

Theorem 5.8.1. If Hn : CW pairs → Ab is a homology theory and H0 (∗) = Z, then Hn are exactly
the singular homology functors up to a natural isomorphism of functors.
More generally, if H0 (∗) = G, then Hn are exactly the singular homology functors with coeffi-
cients in the Abelian group G.
Proof. Given H∗ , reconstruct the cellular chain groups Hn (X n , X n−1 ) using the axioms.

• Show the homology of this chain complex are the cellular homology groups of X.
• Show these agree with Hn (X n , X n−1 ). The exact same argument in Theorem 5.7.2 applies.
We then check that the cellular homology groups we just constructed satisfies the degree formula
as in our last step. This is a bit more difficult, but we won’t get into it. ■

CHAPTER 5. HOMOLOGY 110

Chapter 6

Lefschetz Fixed Point Theorem

After developing homology theory rigorously, we now see an incredibly useful and interesting application
and some of its implication as well.

6.1 Trace
Definition 6.1.1 (Trace). Let φ : Zn → Zn be a group homomorphism, we may represent this with
a matrix A = [aij ]i,j with trace being

tr(A) := a11 + · · · + ann .

For a group homomorphism φ : M → M where M is a finitely generated Abelian group, we define

the trace of φ to be the trace of the induced map φ : M / MT → M / MT , where MT is the torsion
subgroup of M .

Exercise. We have tr(AB) = tr(BA) and tr(A) = tr(BAB −1 ). Thus, trace is independent of change
of basis of Zn .

Lecture 36: Lefschetz Fixed Point Theorem

6 Apr. 10:00
6.2 Lefschetz Fixed Point Theorem
Definition 6.2.1 (Lefschetz number). Let X be a space with the assumption that Hk (X) is
L
k
finitely generated.a Then the Lefschetz number τ (f ) of a map f : X → X is
X
τ (f ) := (−1)k tr(f∗ : Hk (X) → Hk (X)).
k
a That
is, each homology group is finitely generated, and there are finitely many nonzero homology groups. For
example X could be a finite CW complex.

Remark. In particular, we can also write tr(f∗ ⟳ Hk (X)).

Example (Euler characteristic via homology). When f ≃ idX , we have f∗ = idHk (X) for all k. Then

tr(f∗ : Hk (X) → Hk (X)) = rank(Hk (X)).

Hence, we further have X

τ (f ) = (−1)k rank(Hk (X)) =: χ(X),
k

111
 Lecture 37: Simplicial Approximation

where χ(X) is the Euler characteristic.

This is not what we would use typically. Traditionally, we have the following.

Definition 6.2.2 (Euler characteristic). Let X be a finite CW complex, then the Euler characteristic
χ(X) of x is defined by the alternating sum
X
χ(X) = (−1)n cn
n

where cn is the number of n-cells of X.

Proposition 6.2.1. Definition 6.2.2 and the definition given in this example coincides
Proof. This is immediately by the fact that the homology group can be calculated by cellular
homology, hence cn is just rank(Hn (X)) for all n, so the result follows. ■

Theorem 6.2.1 (Lefschetz Fixed Point Theorem). Suppose X admits a finite triangulation,a or more
generally, X is a retract of a finite simplicial complex. If f : X → X is a map with τ (f ) ̸= 0, then
f has a fixed point.
a i.e. a finite simplicial complex structure.

Note. Note that the converse does not hold. And in particular, we have
X
τ (f ) = tr(f# ⟳ CkCW (X)).
k

Theorem 6.2.2. If X is a compact, locally contractible space that can be embedded in Rn for some
n, then X is a retract of a finite simplicial complex.

Remark. This includes compact manifolds and finite CW complexes. Hence, we see that Theo-
rem 6.2.1 applies on compact manifolds and finite CW complexes in particular.

Definition 6.2.3. Let F be a field, and let Hk (X; F) be the k-th homology of X with coefficients in
F. Then Hk (X; F) is always a vector space over F. Define τ F (X) be
X
(−1)k tr(f∗ : Hk (X; F) → Hk (X; F)).
k

Remark. The Lefschetz fixed point theorem still holds if we replace τ (X) ̸= 0 with τ F (X) ̸= 0.

Lecture 37: Simplicial Approximation

Example. Let f : S n → S n be a degree d map. Then τ (f ) is 8 Apr. 10:00

(−1)0 tr(f∗ : H0 (S n ) → H0 (S n )) + (−1)n tr(f∗ : Hn (S n ) → Hn (S n )).

Since f∗ : H0 (S n ) → H0 (S n ) is the identity, and f∗ : Hn (S n ) → Hn (S n ) is given by the 1 × 1 matrix

with entry d. And then we have
τ (f ) = 1 + (−1)n d.

CHAPTER 6. LEFSCHETZ FIXED POINT THEOREM 112

 Lecture 37: Simplicial Approximation

Corollary 6.2.1. f : S n → S n has a fixed point whenever 1 + (−1)n ̸= 0. Namely, whenever d ̸=

(−1)n+1 . That is f has a fixed point if its degree is not equal to the degree of the antipodal map.

Exercise. If f : X → X, then tr(f∗ : H0 (X) → H0 (X)) is equal to the number of path-components

of X mapped to themselves.

Exercise. If X is contractible, then for f : X → X, τ (f ) = 1.

Answer. Since the homology of f is concentrated in degree zero, the result follows. ⊛

Exercise. If X is a contractible compact manifold or finite CW complex, every f : X → X has a

fixed point.

Answer. From the last exercise, we have τ (f ) = 1. And since we have compactness, by Theo-
rem 6.2.1, the result follows. In particular, this recovers Brouwer’s Fixed Point Theorem. ⊛

Example. If we consider the map f : R → R given by translation by x ̸= 0, then τ (f ) = 1, but f

does not have a fixed point.

Proof. The key here is that R is not compact. ⊛

Example (QR May 2016). Let X be a finite, connected CW complex. X e is its universal cover, and
e is compact. Show that X
X e cannot be contractible unless X is contractible.

Proof. We actually have two different approaches.

1. By homework, we know that since X

e is contractible and X
e has finitely many sheets d over X,

e = d · χ(X).
1 = χ(X)

Therefore, χ(X) = d = 1, and so p : X

e → X is a 1-sheeted cover, so it is a homeomorphism.
Therefore, X is contractible.
2. Since Xe is contractible, τ (f ) = 1 for all f : X e Furthermore, because X
e → X. e is compact and
covers a finite CW complex, it is a finite CW complex. Therefore, the Lefschetz fixed point
theorem applies, so any such map has a fixed point. If f is a deck map, then that means that
f = idXe from Corollary 4.2.1.
We have proved then that X ∼ =X e because p : X
e / G(X) e → X is normal, but then the deck
e ∼
group G(X) is trivial, so X = X, and we are done.
e

Exercise. A 1-sheeted cover is always injective and surjective. Furthermore, it’s a local homeomor-
phism. This suffices to show that a 1-sheeted cover is a homeomorphism.

Theorem 6.2.3. If X is a finite CW complex, with cellular chain groups Hn (X n , X n−1 ). If we have
a cellular map f : X → X, so f induces maps f∗ : Hn (X n , X n−1 ) → Hn (X n , X n−1 ). Then
X
τ (f ) = (−1)n tr(f∗ : Hn (X n , X n−1 ) → Hn (X n , X n−1 )).
n

Proof. Do some algebra! This is a purely algebraic fact

CHAPTER 6. LEFSCHETZ FIXED POINT THEOREM 113

 Lecture 38: Simplicial Approximation

Exercise. Given a commutative diagram with exact rows

0 A B C 0
α β γ

′
0 A B′ C′ 0

then tr(β) = tr(α) + tr(γ).

Using the above result, the theorem follows by an argument analogous to the argument for Euler
Characteristic in Homework. ■

6.3 Simplicial Approximation Theorem

As previously seen. Recall the definition of simplicial complex.

Definition 6.3.1 (Simplicial map). A simplicial map f : K → L is a continuous map that sends each
simplex of K to a (possibly smaller dimensional) simplex of L by a linear map in the form of
X X
ti vi 7→ ti f (vi ).

Remark. A simplicial map is completely determined by its restriction to the vertex set.

Lecture 38: Simplicial Approximation

Theorem 6.3.1 (Simplicial approximation theorem). Given any continuous map f : K → L where K 11 Apr. 10:00
is a finite simplicial complex and L is any simplicial complex. Then f is homotopic to a map that
is simplicial with respect to some iterated Barycentric subdivision of K.

Example (Barycentric subdivision). Here is barycentric subdivision in pictures.

new vertex of the new vertex of the new vertex of the

center of a 1-simplex center of a 2-simplex center of a 3-simplex

That is, we add a new vertex to the center of every subsimplex, filling things in like the above.

Note. This means that when we go one dimensional higher, the new vertex added in each
subsimplex carrying through to the next dimension as well.
For an n-simplex we end up with (n + 1)!-simplices with replace it.

CHAPTER 6. LEFSCHETZ FIXED POINT THEOREM 114

 Lecture 38: Simplicial Approximation

6.4 Proof of Lefschetz Fixed Point Theorem

We now have enough tools to prove Theorem 6.2.1
Proof Outline of Theorem 6.2.1. Fix a space X which is a finite simplicial complex (or a retract
of a finite simplicial complex) and a map f : X → X. Then we proceed step by step.

(a) We first reduce to the case of a finite simplicial complex X. Suppose K is a finite simplicial
complex, with r : K → X a retraction. First notice that the following composite of maps

r f ι
K X X K

has the same fixed points as f .

Exercise. r∗ : Hn (K) → Hn (X) is split surjective,a and so it has to be a projection onto

a direct summand.
a See ι∗ .

Exercise. It follows that tr(ι∗ ◦ f∗ ◦ r∗ ⟳ Hk (K)) = tr(f∗ ) on k th homology.

This implies that τ (f ) = τ (ι ◦ f ◦ r), therefore if we can prove the result for a finite simplicial
complex then we are done.
(b) Let X be a finite simplicial complex. We show that if f : X → X has no fixed points, then
τ (f ) = 0. The goal now is to find subdivisions K, L of X and g : K → L so that

• g is simplicial.
• g ≃ f , τ (f ) = τ (g).
• g(σ) ∩ σ = ∅ for all simplices σ.

Note. I.e., it moves every σ.

So this becomes a few steps, none of which we’ll justify too formally. Firstly, since trace is
given by diagonal entries of g in a matrix which respects to basis of simplices of K and L, if
g fixes no basis entries, then it has trace 0. With this, we have the following steps.

• Choose a metric d on X.
• Since X is compact, and f has no fixed point, then d(x, f (x)) has some minimum value
ϵ > 0.
• Subdivide all simplices of X until simplices have diameter smaller than ϵ/100.b Call this
subdivision L.
• Use the simplicial approximation theorem to obtain a map g : K → L, where K is a
subdivision of L and g ≃ f .
• By the proofc of simplicial approximation theorem, we can construct g so that for all
simplices σ, g(σ) is not too far from f (σ). We can then conclude that g(σ) ∩ σ = ∅.
• Consider g# ⟳ C∗CW (K), so then g is a cellular map K → K that moves every cell. We
can then check that
X
τ (f ) = τ (g) = (−1)n tr(g# : cellular n-chains → cellular n-chains) = 0
| {z }
g# ⟳C∗CW (K)

Because each g∗ has vanishing diagonal entries.

This proves the theorem.

CHAPTER 6. LEFSCHETZ FIXED POINT THEOREM 115

 Lecture 38: Simplicial Approximation

■
b Just a random constant!
c We omit the proof, see Hatcher [HPM02].

CHAPTER 6. LEFSCHETZ FIXED POINT THEOREM 116

Chapter 7

Epilogue

There are some final remarks we would like to make.

Lecture 39: Proof of Seifert-Van-Kampen Theorem

13 Apr. 10:00
7.1 Proof of Seifert-Van-Kampen Theorem
Let’s start to prove Theorem 3.6.1 which we omit before.
Proof of Theorem 3.6.1. The outline of the proof is the following. Let X = α Aα where Aα
S
are open, path-connected and contain the glue-point x0 . We also must guarantee that Aα ∩ Aβ is
path-connected.

1. Since we have a map induced by the inclusions

Φ : ∗ π1 (Aα , x0 ) → π1 (X, x0 ).
α

We want to show that Φ is surjective. Take some γ : I → X, then by the compactness of the
interval I, we can show that there is a partition I with s1 < · · · < sn so that

γ|si ,si+1 =: γi

has image in Aαi for some αi .

Exercise. Showing the above fact is a good exercise for point-set topology.

Specifically, since
• Aα is open for all α
• I is compact,
then for all i, we choose a path hi from x0 to γ(si ) in Aσi−1 ∩ Aαi , using path-connectedness
of the pairwise intersections. Now, take γ and write it as

γ = (γ1 · h1 ) · (h1 · γ2 ) · · · · · (γn−1 · hn−1 ) · (hn−1 · γn ).

Observe that each of these paths is fully contained in Aαi , so this implies that γ ∈ Im(Φ),
therefore Φ is surjective.
2. For the next step, we’ll show that the second part of Theorem 3.6.1. Assume that our triple
intersections are path-connected. We want to show that ker(Φ) is generated by

(iαβ )∗ (ω)(iβα )∗ (ω)−1 ,

117
 Lecture 39: Proof of Seifert-Van-Kampen Theorem

where iαβ : Aα ∩ Aβ ,→ Aα for all loops ω ∈ π1 (Aα ∩ Aβ , x0 ).

Before we go further, we’ll need a specific definition.

Definition 7.1.1 (Factorization). A factorization of a homotopy class [f ] ∈ π1 (X, x0 ) is a

formal product
[f1 ][f2 ] . . . [fℓ ]
with [fi ] ∈ π1 (Aα , x0 ) such that

f ≃ f1 · f2 · · · · · fℓ .

We showed that every [f ] has a factorization in step 1 already. Now we want to show that
two factorizations
[f1 ] · · · · · [fℓ ] and [f1′ ] · · · · · [fℓ′′ ]
of [f ] must be related by two moves:
(a) [fi ] · [fi+1 ] = [fi · fi+1 ] if [fi ], [fi+1 ] ∈ π1 (Aα , x0 ). Namely, the reaction defining the free
product of groups.
(b) [fi ] can be viewed as an element of π1 (Aα , x0 ) or π1 (Aβ , x0 ) whenever

[fi ] ∈ π1 (Aα ∩ Aβ , x0 ).

This is the relation defining the amalgamated free product.

Now, let Ft : I × I → X be a homotopy from f1 . . . fℓ to f1′ . . . fℓ′′ , since they both represent
[f ]. We subdivide I × I into rectangles Rij so that

F (Rij ) ⊆ Aαij =: Aij

for some αij using compactness. We also argue that we can perturb the corners of the squares
so that a corner lies only in three of the Aα ’s indexed by adjacent rectangles.

A31 A32 A33

A21 A22 A23

A11 A12 A13

We also argue that we can set up our subdivision so that the partition of the top and bottom
intervals must correspond with the two factorizations of [f ]. We then perform our homotopy
one rectangle at a time.

CHAPTER 7. EPILOGUE 118

 Lecture 40: Prepare for Final

ending loop f1′ · · · · · fℓ′′

starting loop f1 · · · · · fℓ step 1 step 2

Idea: Argue that homotoping over a single rectangle has the effect of using allowable moves
to modify the factorization.
At each triple intersection, choose a path from f (corner) to x0 which lies in the triple inter-
section, so we use the assumption that the triple intersections are path-connected.

In each intersection, choose

a path from f (corner) to x0

Choose path h from image of thise corner to x0

Along the top and bottom, we make choices compatible with the two factorizations. It’s now
an exercise to check that these choices result in homotoping across a rectangle gives a new
factorization related by an allowable move.

Lecture 40: Prepare for Final

15 Apr. 10:00
7.2 Review Problems
We now go through some problems to get prepared for the final!

Exercise (QR May 2019). T = S 1 × S 1 , T ′ = T / S × {1}. Find H∗ , π1 of T #T ′ (connected sum).

Answer. We see that

≃ ≃

collapse

We then see that T #T ′ is a wedge of T and S 1 , hence the fundamental group is just Z2 ∗ Z,
where Z2 is from T and Z is from S 1 . ⊛

CHAPTER 7. EPILOGUE 119

 Lecture 41: Prepare for Final

Exercise (QR Aug. 2019). Let X be a CW complex obtain from a k-sphere, k ≥ 1, by attaching
two (k + 1)-cells along attaching maps of deg m, n. Calculate H∗ (X).

Answer. We use cellular homology. Specifically, we have the following.

∂k+1
Z2 = Ck+1 Z = Ck

(1, 0) m

(0, 1) n

Hence, we see that ker(∂k+1 ) ∼

= Z, and Im(∂k+1 ) = gcd(m, n)Z. ⊛

Exercise (QR May 2017). Let S 1 be the complex numbers of absolute value 1 with induced topology.
Let K be the quotient of S 1 × [0, 1] by identifying (z, 0) with (z −2 , 1). Compute H∗ (K).

Answer. Again, we use cellular homology. We use the following gluing instruction.

b b

a D a

Exercise (QR Sep. 2016). Let Z = {(x, y) ∈ C2 | x = 0 or y = 0}. Find H∗ (C2 \ Z).

Answer. We see that

(C \ {0}) × (C \ {0}) ≃ S 1 × S 1 ,
which is a torus. ⊛

Exercise (QR May 2017). Let X be the connected CW complex, Hi (X) = 0 for all i > 0. Prove
that (
Z, if n = 0;
Hn (X × S k ) =
0, otherwise.

Answer. We can induct on k and use Mayer-Vietoris long sequence. Let U = X × upper hemisphere,
V = X × lower hemisphere, then we have

U ≃ X, V ≃ X,

and hence U ∩ V ≃ X × S k−1 . ⊛

Lecture 41: Prepare for Final

We now go through more problems to get prepared for the final! 18 Apr. 10:00

CHAPTER 7. EPILOGUE 120

 Lecture 41: Prepare for Final

Exercise (QR Jan. 2021). Let G be a connected topological space and w a topological group struc-
ture, i.e., continuous multiplication M : G × G → G and inverse i : G → G that define a group
structure. Assume G has finite CW complex structure. Show χ(G) = 0 unless G = {1}.

Answer. Let g ∈ G such that g ̸= id. We can then choose a path γ from g to id. Then we have

τg : G → G
h 7→ gh.

Then γ gives homotopy from τg to id such that h 7→ γ(t)h, so

τ (τg ) = τ (id) = χ(G) = 0,

hence by Theorem 6.2.1, τg has no fixed points. ⊛

Exercise (QR May 2019). For what g ≥ 0 is it true that for all h ≥ g, a compact oriented genus-g
surface X (no boundary) has covering f : Y → X where Y is a compact, oriented surface of genus-h?

Answer. From Euler characteristic, we have

(2 − 2g)d = 2 − 2h

for a degree d covering. This implies h ≡ 1 (mod g − 1). We see that the only solutions for all h if
g = 2. Then for a particular d, we have the following cover.

2π
d

Exercise (QR Jan. 2019). Let S1 , S2 be two disjoint copies of n-sphere, n > 1. Choose distinct
points Ai , Bi ∈ Si , and let Z is obtained by gluing A1 ∼ A2 , B1 ∼ B2 . What’s the lowest number
of cells in CW complex structure on Z.

Answer. In n = 2, we have the following figure.

CHAPTER 7. EPILOGUE 121

 Lecture 41: Prepare for Final

≃ ≃

Exercise (QR Jan. 2016). Let U, V ⊆ S n , n ≥ 2 be non-empty connected open sets such that
S n = U ∪ V . Show U ∩ V is connected.

Answer. We simply calculate H

e 0 via Mayer-Vietoris. ⊛

Exercise (QR Jan. 2016). Let p be a prime, and X be a finite CW complex, Z / pZ ⟳ X.

(a) If χ(X) is not divisible by p, show that the action has a (global) fixed point.

(b) Give an example of such an action that is fixed point free when χ(X) = 0.

Answer. We have the following.

(a) We use the Euler characteristic on the covering space action.a

(b) Rotation 2π/p on S 1 .

⊛
a Since we assume has no fixed point, with the fact that X admits a CW complex structure hence Hausdorff, so
Z / pZ is indeed a covering space action.

CHAPTER 7. EPILOGUE 122

Appendix

123
Appendix A

Additional Proofs

A.1 Seifert-Van Kampen Theorem on Groupoid

Theorem A.1.1 (Seifert-Van Kampen Theorem on groupoid). Given X0 , X1 , X as topological spaces
with X0 ∪ X1 = X. Then the functor Π : Top → Gpd maps the cocartesian diagram in Top∗ to a
cocartesian diagram in Gp as follows.

j0 Π(j0 )
(X0 ∩ X1 , x0 ) (X0 , x0 ) Π(X0 ∩ X1 ) Π(X0 )
Π
j1 i0 7−→ Π(j1 ) Π(i0 )

(X1 , x0 ) i1
(X, x0 ) Π(X1 ) Π(X)
Π(i1 )

Note. Notice that X0 , X1 , X don’t need to be path-connected in particular.

Surprisingly, the proof of Appendix A.1 is much more elegant with the elementary proof of Theo-
rem 3.6.1, hence we give the proof here.
Proof. Let G ∈ Ob(Gpd) a groupoid, and given functors

F : Π(X0 ) → G , G : Π(X1 ) → G

such that
F ◦ Π(j0 ) = G ◦ Π(j1 ).
Π(j0 )
Π(X0 ∩ X1 ) Π1 (X0 )
Π(j1 ) Π(i0 )
F
Π1 (X1 ) Π1 (X)
Π(i1 )
∃!K
G G
We now only need to prove that there exists a unique functor K : Π(X) → G such that the above
diagram commutes.
We can define K as

• on objects: For all x ∈ Ob(Π(X)) = X,

(
F (x), if x ∈ X0 ;
K(x) =
G(x), if x ∈ X1 .

This is well-defined since the diagram (without K) commutes.

124
 Lecture 41: Prepare for Final

• on morphisms: For every p, q ∈ X, ⟨γ⟩ : p → q in HomΠ(X) (p, q), we need to define K(⟨γ⟩) ∈
HomG (K(p), K(q)). Our strategy is for every path γ from p to q, we define K(γ)
e ∈ HomG (K(p), K(q)).
Then if we also have K(γ) = K(γ ) for γ ≃ γ rel{0, 1}, then we can just let
e e ′ ′

K(⟨γ⟩) := K(γ).
e

Now we start to construct K.

e
Given a path γ : [0, 1] → X, γ(0) = p, γ(1) = q. Since Int(X0 ) ∪ Int(X1 ) = X, we see that

γ −1 (Int(X0 )) ∪ γ −1 (Int(X1 )) = [0, 1].

From Lebesgue Lemmaa , there exists a finite partition

0 = t0 < t1 < · · · < tm−1 < tm = 1

such that for every i,

γ([ti−1 , ti ]) ⊂ Int(X0 ) or Int(X1 ).

X
γ

0 t0 1 p q
γ(t0 )

X0 X1

Now, let γi : [0, 1] → X, t 7→ γ((1 − t)ti−1 + t · ti ), we see that γi is either a path in X0 or X1 .

We then define K(γ)e e m ) ◦ K(γ
:= K(γ e 1 ) ∈ HomG (K(P ), K(q)) such that
e m−1 ) ◦ · · · ◦ K(γ
(
F (⟨γi ⟩), if γi ⊂ X0 ;
K(γi ) =
e
G(⟨γi ⟩), if γi ⊂ X1 .

We need to prove that K(γ)

e does not depend on the partition. It’s sufficient to prove that for
any partition
0 = t0 < t1 < · · · < tm−1 < tm = 1,
we consider any finer partition

0 = t0 = t10 < t11 < · · · < t1K1 = t1 = t20 < t21 < · · · < tmKm = tm = 1.

As before, we denote γij : [0, 1] → X, t 7→ γ((1 − t)tij−1 + t · tij ). It’s clear that as long as

K(γ e iK ) ◦ K(γ
e i ) = K(γ e iK −1 ) ◦ · · · ◦ K(γ
e i0 ),
i i

then our claim is proved. But this is immediate since F and G are functor and for any i, we
only use either F or G all the time.
Now we prove γ ≃ γ ′ rel{0, 1}, then K(γ)
e e ′ ). This is best shown by some diagram.
= K(γ
H

APPENDIX A. ADDITIONAL PROOFS 125

 Lecture 41: Prepare for Final

H1 = γ ′ (1, 1)
X

H γ

γ′

(0, 0) H0 = γ

The left-hand side represents a partition P of [0, 1]×[0, 1] such that every small square’s image
in X under H is either entirely in X0 or in x1 . Consider all paths from (0, 0) to (1, 1) such
that it only goes right or up. We see that for any such path L, consider

γL : [0, 1] → L, t 7→ γL (t).

We let ΓL : H|L ◦ γL : [0, 1] → X, we see that ΓL is a path from p to q. Now, if for two paths
L1 and L2 such that they only differ from a square.

H1 = γ ′ (1, 1)

L1
L2

(0, 0) H0 = γ

We claim that γL1 , ΓL2 are two paths from p to q, and K(Γ e L ). Now, we denote
e L ) = K(Γ
1 2
Γ0 and Γ1 as follows.

H1 = γ ′ (1, 1) H1 = γ ′ (1, 1)

(0, 0) H0 = γ (0, 0) H0 = γ

Γ0 Γ1

Figure A.1: The definition of Γ0 and Γ1 .

It’s clearly that by only finitely many steps, we can transform Γ0 to Γ1 , hence

K(Γ
e 0 ) = K(Γ
e 1 ).

APPENDIX A. ADDITIONAL PROOFS 126

 Lecture 41: Prepare for Final

Finally, we observe that

K(γ
e 0 ) = K(Γ
e 0 ) = K(Γ
e 1 ) = K(γ
e 1 ).

If we now define K(⟨γ⟩) = K(γ),

e then K : Mor(Π(X)) → Mor(G ), then it’s well-defined.
We now prove K : Π(X) → G is indeed a functor. But this is immediate from the definition of
K, namely it’ll send identity to identity and the composition associates.
Also, we need to prove that the following diagram commutes.

Π(j0 )
Π(X0 ∩ X1 ) Π1 (X0 )
Π(j1 ) Π(i0 )
F
Π1 (X1 ) Π1 (X)
Π(i1 )
K
G G

But this is again trivial.

Finally, we need to show that such K is unique. This is the same as the proof of Lemma 2.2.1,
hence the proof is done. ■
a https://en.wikipedia.org/wiki/Lebesgue%27s_number_lemma

A.2 An alternative proof of Seifert Van-Kampen Theorem

Theorem A.2.1. We claim that the diagram

(j0 )∗
π1 (X0 ∩ X1 , x0 ) π1 (X0 , x0 )
(j1 )∗ (i0 )∗

π1 (X1 , x0 ) π1 (X, x0 )
(i1 )∗

is cocartesian.
Proof. The basic idea is that, for this diagram,

Π(X0 ∩ X1 ) Π(X0 )

Π(X1 ) Π(X)

we want to construct a morphism r : Π(Z) → π1 (Z, p) in Gpd such that Z = X0 ∩ X1 , X0 , X1 , X.

For every x ∈ Z, we fix a path γx such that it connects p and x and satisfies
(a) If x ∈ X0 ∩ X1 , then Im(γx ) ⊂ X0 ∩ X1
(b) If x ∈ X0 , then Im(γx ) ⊂ X0
(c) If x ∈ X1 , then Im(γx ) ⊂ X1

(d) γp = cp
The proof is given in https://www.bilibili.com/video/BV1P7411N7fW?p=38&spm_id_from=
pageDriver. ■ If have
time.

APPENDIX A. ADDITIONAL PROOFS 127

 Lecture 41: Prepare for Final

A.3 Cellular Boundary Formula in Definition 5.7.1

Theorem A.3.1. For n > 1, the boundary maps ∂n of cellular chain complex given by
X
∂n (enα ) = ∂αβ en−1
β
β

is well-defined.
Proof. Here we are identifying the cells enα and en−1
β with generators of the corresponding summands
of the cellular chain groups, namely Cn (X). The summation in the formula contains only finitely
many terms since the attaching map of enα has compact image, so this image meets only finitely
many cells en−1
β . To derive the cellular boundary formula, consider the following commutative
diagram.

∂ ∆αβ ∗
Hn (Dαn , ∂Dαn ) e n−1 (∂Dαn )
H e n−1 (S n−1 )
H
∼
= β

Φα ∗ φα ∗ qβ ∗

∂n q∗
Hn (X n , X n−1 ) e n−1 (X n−1 )
H e n−1 (X n−1 / X n−2 )
H
jn−1 ∼
=
dn
∼
=
Hn−1 (X n−1 , X n−2 ) Hn−1 (X n−1 / X n−2 , X n−2 / X n−2 )

where
• Φα is the characteristic map of the cell enα and φα is its attaching map.

• q : X n−1 → X n−1 / X n−2 is the quotient map.

• qβ : X n−1 / X n−2 → Sβn−1 collapses the complement of the cell en−1
β to a point, the resulting
quotient sphere being identified with Sβn−1 a via the characteristic map Φβ .

• ∆αβ : ∂Dαn → Sβn−1 is the composition qβ qφα , i.e., the attaching map of enα followed by the
quotient map X n−1 → Sβn−1 collapsing the complement of en−1 β in X n−1 to a point.
The map Φα∗ takes a chosen generator [Dαn ] ∈ Hn (Dαn , ∂Dαn ) to a generator of the Z summand
of Hn (X n , X n−1 ) corresponding to enα . Letting enα denote this generator, commutativity of the left
half of the diagram then gives
∂n (enα ) = jn−1 φα ∗ ∂[Dαn ].
In terms of the basis for Hn−1 (X n−1 , X n−2 ) corresponding to the cells en−1
β , the map qβ ∗ is the
projection of Hn−1 (X
e n−1
/X n−2
) onto its Z summand corresponding to eβ . Commutativity of
n−1

the diagram then yields the formula for ∂n given above. ■

a Which is just Dβn−1 / ∂Dβn−1 .

APPENDIX A. ADDITIONAL PROOFS 128

Appendix B

Algebra

This section aims to give some reference about the required algebra content, including Abelian groups,
and free Abelian group, which is used heavily when discuss homology, and also some homological algebra,
where we will focus on exact sequence specifically.

B.1 Abelian Group

Definition B.1.1 (Abelian group). A group (G, ·) is an Abelian group if for every a, b ∈ G, we have

a · b = b · a.

We often denote · as + if (G, ·) is a Abelian group.

Definition B.1.2 (Product of groups). Given two groups (G, ·), (H, ·), the product of G and H,
denoted by G × H is defined as

G × H = {(g, h) | g ∈ G, h ∈ H}

and
(g1 , h1 ) · (g2 , h2 ) := (g1 · g2 , h1 · h2 ).

Notation. For simplicity, given an index set I, we’ll denote the order pair (gα1 , gα2 , . . . ) as (gα )α∈I .
Note that the latter notation can handle the case that I is either countable or uncountable, while
the former can only handle the countable case.

Definition B.1.3 (Direct product). Given (Gα , +), α ∈ I as a collection of Abelian group, we define
their direct product as !
Y
Gα , + ,
α∈I

where Y
Gα = {(gα )α∈I | gα ∈ Gα }
α∈I

and ∀(gα ), (hα ) ∈

Q
Gα
α∈I
(gα ) + (hα ) := gα + hα
for all α ∈ I.
Specifically, if I is finite, namely there are only finely many Abelian groups (G1 , +), . . . , (Gn , +),

129
 Lecture 41: Prepare for Final

 n

and can be denoted as
Q
Gi , +
i=1

(G1 × · · · × Gn , +) .

sum). Given
a collection of Abelian groups {Gα }α∈I , the external

Definition B.1.4 (External direct L
direct sum of them, denoted as α∈I Gα , + as

M  
Gα := (gα )α∈I | ∀ gα ∈ Gα , # non-zero elements in gα < ∞ .
α∈I
α∈I

And for every (gα ), (hα ) ∈ Gα ,

L
α∈I

(gα ) + (hα ) := gα + hα

for all α ∈ I.a

a This may not be the best notation: What we’re really trying to say is (gα )α∈I + (hα )α∈I := gi + hi for all i ∈ I.

Note. We see that M Y

Gα ⊂ Gα .
α∈I α∈I

Additionally, we also have ! !

M Y
Gα , + < Gα , + .
α∈I α∈I

Remark. We see that the operation + is indeed closed since the sum of g, g ′ ∈ α∈I Gα will have
L
only finitely non-zero elements if g, g ′ both have only finitely many non-zero elements.
We see that if I is a finite index set, given a collection of Abelian group {Gα }α∈I , then

G1 × · · · × Gn = G1 ⊕ · · · ⊕ Gn .

Definition B.1.5 (Internal direct sum). Given an Abelian group G, and a collection of the subgroups
{Gα }α∈I of G, we say G is an internal direct sum of {Gα }α∈I if for any g ∈ G, we can write
X
g= gα
α∈I

uniquely, where gα ∈ Gα has only finitely many non-zero elements. In this case, we denote
M
G= Gα .
α∈I

Intuitively, the external direct sum is to build a new group based on the given collection of groups
{Gα }α∈I , while the internal direct sum is to express an already known group G with an already
known collection of groups {Gα }α∈I .

Remark (Relation between InternalLand External direct sum). Given an Abelian group G and its
internal direct sum decomposition α∈I Gα , G is isomorphic to the external direct sum of {Gα }α∈I .
We see this from the following group homomorphism:
X
∀ g= gα 7→ (gα )α∈I .
g∈G
α∈I

Conversely, given a collection of Abelian group {Gα }α∈I , and let G = Gα as the external
L
α∈I

APPENDIX B. ALGEBRA 130

 Lecture 41: Prepare for Final

direct sum of {Gα }, denote iα0 : Gα0 → Gα as a canonical embedding

L
α∈I

gα0 7→ iα0 (gα0 ) = (hα )α∈I ,

where (
gα0 , if α0 = α;
hα =
0, if α0 ̸= α
given α0 . Then M
G′α0 := iα0 (Gα0 ) < Gα
α∈I

and G is the internal direct sum of G′α0 , α0 ∈ I. This is because ∀g = (gα )α∈I ∈ G(=
L
α∈I Gα ),
we have X
g= iα (gα ).
α∈I

Note that the above sum is well-defined since there are only finitely many non-zero elements for
each gα . And additionally, we can see the uniqueness of this decomposition by defining πα0 such
that M
πα0 : Gα → Gα0 , (gα )α∈I 7→ gα0 ,
α∈I

then πα ◦ iα = idGα , πα ◦ iβ = 0 for all β ≠ α and

!
X X
πβ (g) = πβ iα (gα ) = πβ ◦ iα (gα ) = πβ ◦ iβ (gβ ) = gβ
α∈I α∈I

for all β ∈ I, where the second equality is because this summation is finite. Hence, we have
X
g= iα (πα (g)).
α∈I

Definition B.1.6. Given two Abelian groups G, H, we define Hom(G, H) as

Hom(G, H) := {f : G → H | f is a group homomorphism} ,

then we can define

+ : Hom(G, H) × Hom(G, H) → Hom(G, H)
(φ, ψ) 7→ φ + ψ,

where
(φ + ψ)(g) := φ(g) + ψ(g).

Remark (Relation between direct sum and direct product). Given a collection of Abelian groups
{Gα }α∈I , and another Abelian group H, there exists a φ such that
!
M Y
φ : Hom Gα , H → Hom(Gα , H)
α∈I α∈I
f 7→ φ(f ) := (fα )α∈I

where fα = f ◦ iα , where iα is the canonical embedding from Gα to α∈I Gα . We claim that φ is

L
an isomorphism.

• φ is injective. This is obvious since ker(φ) = 0 from the fact that if φ(f ) = 0, then fα = 0 for
all α, hence f is 0.

APPENDIX B. ALGEBRA 131

 Lecture 41: Prepare for Final

• φ is surjective. For every (fα )α∈I ∈ Hom(Gα , H), we define

Q
α∈I
M
f: Gα → H
α∈I
X X
gα 7→ fα (gα ).
α∈I α∈I

We see that f ∈ Hom Gα , H and φ(f ) = (fα )α∈I .

L

α∈I

This shows that !

∼
M Y
Hom Gα , H = Hom(Gα , H).
α∈I α∈I

Exercise. We can show that

!
∼
Y Y
Hom H, Gα = Hom(H, Gα ).
α∈I α∈I

Note the order in the Hom matters.

B.2 Free Abelian Group

Definition B.2.1 (Free Abelian group). Given an Abelian group (G, +), we say G is a free Abelian
group if there exists a collection of elements {gα }α∈J in G such that {gα }α∈J forms a basis of G,
i.e., for all g ∈ G, ∃!nα ∈ Z for all α ∈ J such that
X
g= nα gα
α∈J

with finitely many non-zero nα .

Remark. If G is a free Abelian group, and {gα }α∈J is a basis, then for every α ∈ J, ⟨gα ⟩ is an
infinite cyclic group since
n · gα = 0 = 0 · gα ⇒ n = 0.
And from Definition B.2.1, we have M
G= ⟨gα ⟩.
α∈J

Conversely, assume there are a collection of infinite cyclic group ⟨gα ⟩ for α ∈ I in G such that
M
G= ⟨gα ⟩,
α∈I

then {gα }α∈I is a basis of G, hence G is a free Abelian group.

Proposition B.2.1. If G is an Abelian group, then the following are equivalent.

(a) G is a free Abelian group.

(b) G is an internal direct sum of some infinite cyclic groups.

(c) G is isomorphic to the external direct sum of some additive groups of integers Z.
Proof. We see that 1. ⇔ 2. is already proved. And for 2. ⇔ 3., this follows directly from the
relation between internal and external direct sum. ■

APPENDIX B. ALGEBRA 132

 Lecture 41: Prepare for Final

Now, consider G as a free Abelian group, then

∼
= L
u: G Z
α∈I

for some I. Denote Z, where iα : Z → Z is the canonical embedding, i.e.,

L L
L eα := iα (1) ∈ α∈I α∈I
eα = (gα )α∈I ∈ α∈I Z, where (
1, if β = α;
gβ =
0, if β ̸= α.

Moreover, denote ϵα as the image of eα under the isomorphism u, namely ϵα = u−1 (eα ), then {ϵα }α∈I
is a basis of G.
Now, for every Abelian group H, we have
 
◦u
f ◦ u−1
L
Hom(G, H) ∼
=
Hom Z, H f
α∈I

∼
= φ

(f ◦ u−1 ◦ iα )α∈I
Q
∼
=
Hom (Z, H)
α∈I

∼
=

f ◦ u−1 ◦ iα (1) α∈I

Q

H
α∈I

where φ is the homeomorphism defined in here, and the homeomorphism

Y ∼
=
Y
Hom(Z, H) −→ H
α∈I α∈I

is trivial since every f ∈ Hom(Z, H) corresponds to f (1) ∈ H uniquely. We see that

Q
α∈I

f ◦ u−1 ◦ iα (1) = f ◦ u−1 (eα ) = f (ϵα ).

In other words, for all Abelian group H, a morphism from the set {ϵα }α∈I to H can be uniquely extended
to the group a homomorphism from G to H.

Remark. This means, to determine Hom(G, H), we only need to determine where each base element
in G will map to in H, and this is why it’s free.
We now want to generate free Abelian group by a set. Roughly speaking, given a set S, we can
generate a free Abelian group Z by defining
( )
X
Z := nx x | nx ∈ Z, # non-zero elements in nx < ∞
x∈S

with the naturally defined +. Formally, we have the following.

Definition B.2.2 (Free Abelian group generated by a set). Given a set S, the free Abelian group
generated by S (Z, +) is defined as

Z := {f : S → Z | only finitely many x ∈ S such that f (x) ̸= 0} ,

with
+: Z × Z → Z
(f, g) 7→ f + g.

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 Lecture 41: Prepare for Final

Remark. {ϕx | x ∈ S} forms a basis of Z, where ϕx : S → Z such that

(
1, if y = x;
y 7→ ϕx (y) =
0, if y ̸= x

is the characteristic function at x. We see this by for all f ∈ S, f = f (x)ϕx , which is uniquely
P
x∈S
defined. Hence, (Z, +) is a free Abelian group.

Note. Note that

1:1
S ←→ {ϕx | x ∈ S}
x 7→ ϕx .

Hence, we often denote the element in Z as

P
n x ϕx
x∈S |{z}
f (x)

X
nx · x.
x∈S

Theorem B.2.1 (The universal property of free Abelian group generated by a set). Denote a canonical
embedding i : S → Z, x 7→ ϕx . Then for all Abelian group H and f : S → H, there exists a unique
group homomorphism
fe: Z → H
such that fe ◦ i = f .
Proof. We define !
X X
fe nx · x := nx f (x),
x∈S x∈S

and the uniqueness is obvious. ■

Note that we can use the above universal property to describe a free Abelian group since we have the
following.

Proposition B.2.2. Given Z ′ as another Abelian group and i′ : S → Z ′ as another canonical embed-
ding such that for all Abelian group H and f : S → H, there exists a unique group homomorphism
fe: Z ′ → H such that fe ◦ i′ = f , then
Z′ ∼
= Z.
Namely, we can describe a free Abelian group by its universal property uniquely up to isomorphism.

Theorem B.2.2. Assume G is a free Abelian group. Assume there exists a finite basis {g1 , . . . , gn }
of G, and also assume that there exists another basis {hα }α∈I . Then we have

card(I) < ∞,

specifically, we have
card(I) = n.
Proof. Firstly, we observe that if we can show

card(I) ≤ n,

then by swapping {hα }α∈I and {gα }α∈I , we will have card(I) = n.
Suppose I is an infinite set, then we can find hα1 , . . . , hαm such that m > n and hαi ̸= hαj for

APPENDIX B. ALGEBRA 134

 Lecture 41: Prepare for Final

i ̸= j. Then since {gα }α∈I is a basis, we have

n
X
hαi = kij gj , ∀i = 1, . . . , m.
j=1

Specifically, we have   1
k12 . . . k1n
  
hα1 k1 g1
 ..   .. .. ..   ..  ,
 . = . . .  . 
1 2 n
hαm km km . . . km gn
| {z }
K∈Mm×n (Z)⊂Mm×n (Q)

where kij ∈ Z. From linear algebra, we know that there exists (r1 , . . . , rm ) ∈ Qm \ {⃗0} such that

(r1 , . . . , rm )K = (0, . . . , 0).

Multiplying both sides with the common multiple of the denominator of ri , we see that there exists
(ℓ1 , . . . ℓm ) ∈ Zm \ {⃗0} such that

(ℓ1 , . . . ℓm )K = (0, . . . , 0)
   
hα1 g1
 ..   .. 
⇒(ℓ1 , . . . , ℓm )  .  = (ℓ1 , . . . , ℓm )K  .  = (0, . . . , 0)
hαm gn
m
X
⇒ ℓi hαi = ⃗0 for (ℓ1 , . . . , ℓm ) ∈ Zm \ {⃗0}
i=1
⇒ card(I) < ∞.

From the same argument, we see that card(I) ≤ n ⇒ card(I) = n. ■

Remark. Furthermore, one can prove that if G is a free Abelian group, then we can prove that any
two bases of G are equinumerous, which handle the case that the basis is an infinite set.
This induces the following definition.

Definition B.2.3 (Rank). Let G ba a free Abelian group, the rank of G is the cardinality of any
basis of G.

B.3 Finitely Generated Abelian Group

Since we’re going to encounter some group as
.
Z ⊕ Z 2Z,

so it’s useful to look into those finitely generated Abelian group.

Let’s start with a definition.

Definition B.3.1 (Torsion subgroup). Given an Abelian group G, we say that g ∈ G has finite order
if ∃n ∈ Z such that n · g = 0. Specifically, we say that

T := {g ∈ G | g has finite order}

is a torsion subgroup.
If T = 0 given G, we say that G is torsion free.

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 Lecture 41: Prepare for Final

Note. Note that T is indeed a subgroup, since for any g1 , g2 ∈ T , g1 + g2 ∈ T from the fact that it
still has finite order.

Remark. If G is a free Abelian group, then G is torsion free. Conversely, if G is torsion free, we can’t
deduce G is a free Abelian group. We see this from (Q, +). Firstly, we see that Q is torsion free,
Now, suppose Q is a free Abelian group, then there exists a basis {rα }α∈I of Q such that |I| > 1.
Now, consider α1 , α2 ∈ I such that α1 , α2 ∈ I, for rα1 , rα2 , there exists n, m ∈ Z and n, m ̸= 0 such
that
nrα1 + mrα2 = 0 ⇒ n = m = 0

B.3.1 Classification of Finitely generated Abelian Group

Given a finitely generated Abelian group G, we may assume its generators are g1 , . . . , gn . Let F be

F := Z ⊕ · · · ⊕ Z,
| {z }
n times

then there are a natural surjective homomorphism

φ : F → G, ei 7→ gi

where ei = (0, . . . , 0, 1 , 0, . . . , 0). Now, let K := ker φ, we have

ith
.
= F K.
G∼

Then we have the following lemma.

Lemma B.3.1. K is a finitely generated Abelian group.

Proof.
Z is Noetherian, F is a finitely generated Z-module
⇒F is Noetherian module
⇒K as a submodule of F is a finitely generated Z-module
⇒K is a finitely generated Abelian group.

Please refer all the concepts above from [AM94]. ■

Hence, we may assume the generators of K as b1 , . . . , bm . From the definition of K, we can further
express bi as  
e1
bi = (bi1 , bi2 , . . . , bin )  ... 
 

en n×n

for all i = 1, . . . , m. Denote all such row vectors bi in a matrix B, namely

 
b11 . . . b1n
B :=  ... ..
.
..  ∈ M
.  m×n (Z),


bm1 . . . bmn

then we have    
b1 e1
 ..   .. 
 .  = B  . .
bm en

APPENDIX B. ALGEBRA 136

 Lecture 41: Prepare for Final

Multiply a matrix on the right-hand side. Now, consider a p ∈ GL(n; Z), then
     ′
b1 e1 e1
 ..  −1  ..   .. 
 .  = B · P P  .  = (BP ) ·  .  ,
bm en e′n
| {z }
new basis

where    ′
e1 e1
−1  ..   .. 
P  .  =:  .  .
en e′n
We see that B · P is the coefficient matrix of generators b1 , . . . , bm under the new basis e′1 , . . . , e′n .

Multiply a matrix on the left-hand side. For a A ∈ GL(m; Z), then

 ′    
b1 b1 e1
 ..   ..   .. 
 .  = Q  .  = QB  .  ,
b′m bm en

since Q is invertible, hence b′1 , . . . , b′m are also generators of K. We see that QB is the coefficient matrix
of new generators b′1 , . . . , b′m under basis e1 , . . . , en .

Generally Q · B · P is the matrix representation of a particular set of F ’s generators under a particular

basis.

Proposition B.3.1. There exists P ∈ GL(n; Z) and Q ∈ GL(m; Z) such that

 
d1
..
.
 
 
 
Q·B·P =  d k
,

 0 
..
 
.

where di ∈ Z+ and d1 | d2 | · · · | dk .
Proof. In fact, P, Q can be taken as the multiplication of the following three types of square
matrices:
• Pij :
 
1
..
.
 
 
 
 0 1(ij) 
..
 
Pij =  . ,
 
 
 1(ji) 0 
..
 
.
 
 
1
where the effect of multiplying Pij from the right is swapping column i, j.

APPENDIX B. ALGEBRA 137

 Lecture 41: Prepare for Final

• Pi (c), where c is the identity in Z, i.e., c = ±1:

 
1
..
.
 
 
 
Pi (c) =  c(ii) ,
..
 
.
 
 
1

where the effect of multiplying Pi (c) from the right is multiplying c to column i.
• Pij (a), a ∈ Z:  
1
..
.
 
 a(ij) 
Pij =  ,
..
.
 
 
1
where the effect of multiplying Pij (a) from the right is adding a times column i to column j.

We see that these are elementary column transformations in linear algebra. In particular, if we
multiply these matrices from the left, then it’s called elementary row transformations.
That is to say, we’re going to show
 
b11 . . . b1n
B =  ... ..
.
.. 
. 


bm1 ... bmn

can become  
d1
..
.
 
 
 

 dk ,

 0 
..
 
.
di ∈ Z+ , d1 | d2 | · · · | dk from elementary column/row transformations.
We now show the steps to make this happens.
• Step 1. Using elementary transformations, we make b11 > 0.
• Step 2. Using elementary transformations, we make b11 become a divisor of all elements in
the first column and row.
We see that if b11 ∤ b1i for i ̸= 1, we have b1i = r · b11 + s where 0 < s < b11 . Then we add
(−r) times the 1th column to the ith column and swapping the 1th and the ith column, which
makes B becomes !
s ...
.. . . ,
. .
for 0 < s < b11 . Since card({n ∈ Z | 0 < n < b11 }) < ∞, hence in finitely many steps we can
make B becomes !
d1 . . .
.. . . ,
. .
where d1 is a divisor of all other elements in the first column and row.

APPENDIX B. ALGEBRA 138

 Lecture 41: Prepare for Final

• Step 3. Using elementary transformations, we can multiply the first row by a proper integer
and add it to the other rows, do the same but for columns also, then we can make B becomes
 
d1 0 . . . 0
0 
 .. .
 
. B1 
0

• Step 4. We iteratively apply Step 1. to step 3., we make B into

 
d1
..
.
 
 
 

 d k
,

 0 
..
 
.

where di ∈ Z+ .
• Step 5. Using elementary transformations, by swapping columns and rows, we may assume
d1 ≤ d2 ≤ · · · ≤ dk .
• Step 6. Using elementary transformations, we can make B into
 ′ 
d1
..
.
 
 
′
 

 dℓ


 0 
..
 
.

such that 0 < d′1 ≤ · · · ≤ d′ℓ , d′1 | d′2 | · · · | d′ℓ since if d1 ∤ di for some i ∈ {2, . . . , k}, then
   
d1 d1 di
.. ..
. .
   
   
   
dk → dk
   ,
 
 0   0 
.. ..
   
. .

then from Step 2., we have !

s ...
.. ..
. .
where 0 < s < d1 and s is a divisor of all other elements in the first row and column. Now,
we repeat Step 3. to Step 5., we obtain
e 
d1
..
.
 
 
 

 dej 


 0 

..
.

where de1 ≤ · · · ≤ dej such that de1 < d1 . Since there are only finitely many integers which is

APPENDIX B. ALGEBRA 139

 Lecture 41: Prepare for Final

smaller than d1 , we see that by repeating these steps, we can always make
 
d1
..
.
 
 
 

 dk 

 0 
..
 
.

into e 
d1
..
.
 
 
 

 dep 


 0 

..
.
such that d′1 | d′i for all i ̸= 1 and d′1 ≤ d′2 ≤ · · · ≤ d′p . By the same idea of Step 3., we have
the desired matrix.
Since all operations are elementary and there are only finitely many of them, hence the result
follows. ■
From the definition of Q · B · P and Proposition B.3.1, there exists a basis e′1 , . . . , e′n of F such that
K has finitely many generators d1 e′1 , . . . , dk e′k , hence
. . .
G∼= Z d Z ⊕ Z d Z ⊕ ··· ⊕ Z d Z ⊕ Z ⊕ ··· ⊕ Z.
1 2 | {z } k
n−k times

This leads to the following important theorem.

Theorem B.3.1 (Fundamental theorem of finitely generated Abelian group). Given a finitely generated
Abelian group, either G is a free Abelian group, or there exists a unique set of {mi ∈ Z | mi >
1, i = 1, . . . , t} such that m1 | m2 | · · · | mt and a unique non-negative integer s such that
. . .
G∼= Z m Z ⊕ Z m Z ⊕ ··· ⊕ Z m Z ⊕ Z ⊕ ··· ⊕ Z.
1 2 t | {z }
s times

Proof. We need to show both uniqueness and existence.

Existence. From Proposition B.3.1, we obtain a basis e′1 , . . . , e′n of F and a basis d1 e′1 , . . . , dk e′k
in K such that d1 | · · · | dk . Let

(d1 , . . . , dk ) = (1, . . . , 1, m1 , . . . , mt ),

which implies
.
=F K
G∼
. . .
∼
= Z d Z ⊕ Z d Z ⊕ ··· ⊕ Z d Z ⊕ Z ⊕ ··· ⊕ Z
1 2 k
. . . .
Z Z Z Z
= 1Z ⊕ · · · ⊕ 1Z ⊕ m1 Z ⊕ · · · ⊕ mt Z ⊕ Z ⊕ · · · ⊕ Z
. .
= Z m Z ⊕ ··· ⊕ Z m Z ⊕ Z ⊕ ··· ⊕ Z.
1 t | {z }
∃!s times

Uniqueness. Under the isomorphism Z / m1 Z ⊕ · · · ⊕ Z / mt Z ⊕ Z ⊕ · · · ⊕ Z, we see that

| {z }
s times
. .
Z ⊕ · · · ⊕ Z
m1 Z mt Z

APPENDIX B. ALGEBRA 140

 Lecture 41: Prepare for Final

corresponds to G’s torsion subgroup T , which implies

.
G ∼ ⊕ · · · ⊕ Z,
T =Z| {z }
s times

which further implies G / T is a free Abelian group with

 . 
rank G T = s,

which proves the uniqueness of s.

The proof of the uniqueness of mi are long and tedious, we refer to [Arm13]. ■

Definition B.3.2 (Invariant factor). We call m1 , . . . , mt obtained from Theorem B.3.1 the invariant
factor.

Lemma B.3.2. Given a positive integer m such that

m = pn1 1 · · · · · pns s

where p ∈ P are all prime and pi =

̸ pj for i ̸= j, with ni ∈ Z+ for all i. Then
. . .
Z ∼
= Z n ⊕ ··· ⊕ Z n .
mZ p1 Z
1
ps s Z
Proof. We define ϕ as
. . .
ϕ : Z mZ → Z pn1 Z ⊕ · · · ⊕ Z pns Z
1 s
n 7→ (n + ⟨pn1 1 ⟩, . . . , n + ⟨pns s ⟩).

Then n ∈ ker ϕ ⇔ ∀ pni i | n ⇔ m | n ⇔ n = 0. This means ker ϕ = 0, hence ϕ is an injection.

i
We now prove ϕ is a surjection. It’s sufficient to prove that for all i,
. .
(0, . . . , 0, 1 + ⟨pni i ⟩, 0, . . . , 0) ∈ Z pn1 Z ⊕ · · · ⊕ Z pns Z,
1 s

there exists an n such that

ϕ(n) = (0, . . . , 0, 1 + ⟨pni i ⟩, 0, . . . , 0).
n n
Notice that for all i ̸= j, ⟨pni i ⟩ + ⟨pj j ⟩ ∈ Z, hence there exists uj ∈ ⟨pni i ⟩ and vj ∈ ⟨pj j ⟩ such that
uj + vj = 1. Let n as Y
n= (1 − uj ),
i̸=j

then
n n
n + ⟨pni i ⟩ = 1 + ⟨pni i ⟩, n + ⟨pj j ⟩ = 0 + ⟨pj j ⟩.
Above implies
ϕ(n) = (0, . . . , 0, 1 + ⟨pni i ⟩, 0, . . . , 0),
hence ϕ surjects, so . . .
Z ∼
= Z n ⊕ ··· ⊕ Z n .
mZ p1 Z
1
ps s Z
■
Combine Theorem B.3.1 and Lemma B.3.2, we see that we now only have
. . .
G∼= Z m Z ⊕ Z m Z ⊕ ··· ⊕ Z m Z ⊕ Z ⊕ · · · ⊕ Z,
1 2 t | {z }
s times

APPENDIX B. ALGEBRA 141

 Lecture 41: Prepare for Final

we can further decompose G into

. .
G∼
= Z ps1 Z ⊕ · · · ⊕ Z psk Z ⊕ Z ⊕ · · · ⊕ Z,
1 k | {z }
s times

where p1 , . . . , pk are primes (which may includes repeated terms), si ∈ Z+ for all i.

Definition B.3.3 (Elementary divisors). The set

{ps11 , . . . , pskk }

are called elementary divisors of G.

Theorem B.3.2 (Uniqueness of elementary divisors). Elementary divisors of a group G is unique.

Proof. Please refer to [Arm13]. ■

B.4 Homological Algebra

B.4.1 Exact Sequence
As previously seen. Given two Abelian groups A, B and the group homomorphism φ : A → B, then
we have
• ker φ = {x ∈ A | φ(x) = 0}
• Im φ = {φ(x) | x ∈ A}
• coker φ := B / Im φ

• coIm φ := A / ker φ
Consider a sequence of Abelian group homomorphism
ϕi−1 ϕi
... Ai−1 Ai Ai+1 ...

We denote this sequence as S.

Definition B.4.1 (Exact). We say S is exact at Ai if

Im ϕi−1 = ker ϕi .

Remark. Definition B.4.1 is same as Definition 5.2.15.

Definition B.4.2 (Exact sequence). We call S is an exact sequence if it’s exact at Ai for all i.

Remark. Specifically, consider the following two situations.

• We say
A0 A1 A2 ...
is an exact sequence if it’s exact at Ai for all i ≥ 1.
• We say
... A−2 A−1 A0
is an exact sequence if it’s exact at Ai for all i ≤ −1.

APPENDIX B. ALGEBRA 142

 Lecture 41: Prepare for Final

Remark. Denote ◦ as a trivial Abelian group, then

ϕ
A B ◦ is an exact sequence ⇔ ϕ is a surjective homomorphism;

conversely,
ϕ
◦ B A is an exact sequence ⇔ ϕ is an injective homomorphism.

Definition B.4.3 (Short exact sequence). A short exact sequence is an exact sequence such that it
has the following form
ϕ ψ
◦ A B C ◦ .

ψ i
Remark. Let B −→ C as a surjective homomorphism and K = ker ψ, and we denote K −→ B as
an injection. Then
i ψ
◦ K B C ◦
is a short exact sequence. Conversely, if
ϕ ψ
◦ A B C ◦

is a short exact sequence, then ϕ is an injective homomorphism since it is exact at A, and ψ is

a surjective homomorphism since it is exact at C, and ϕ(A) = ker ψ since it is exact at B. This
implies ϕ : A → ϕ(A) = ker ψ is a group homeomorphism.

Example. Given A, B as Abelian groups, then

i Proj2
◦ A A⊕B B ◦
i
a (a, 0)
Proj2
(a, b) b

is a short exact sequence.

Example. We see that

i Proj2
◦ Z Z Z / nZ ◦

k k·n

for n ∈ Z≥1 is a short exact sequence.

Definition B.4.4 (Isomorphism between sequences). Given A· and B· defined as two sequences of
Abelian group homomorphisms
ϕi
A• : . . . Ai Ai+1 ...

and
ψi
B• : . . . Bi Bi+1 ...
And we say a morphism α from A• to B• is a series of group homomorphisms αi : Ai → Bi for

APPENDIX B. ALGEBRA 143

 Lecture 41: Prepare for Final

all i ∈ Z such that the following diagram commutes.

ϕi
... Ai Ai+1 ...
αi αi+1
ψi
... Bi Bi+1 ...

Additionally, if for all i, αi is a group homeomorphism, then we say α : A• → B• is a homeo-

morphism.

Definition B.4.5 (Split short exact sequence). Given a short exact sequence

ϕ ψ
◦ A B C ◦

we say it is split if there exists a group homeomorphism θ : B → A ⊕ C such that

ϕ ψ
◦ A B C ◦
id θ id

◦ A A⊕C C ◦

is the isomorphism between these two short exact sequences.

Remark. Given split short exact sequence

ϕ ψ
◦ A B C ◦

and θ defined in Definition B.4.5, let i : A → A ⊕ C, a 7→ (a, 0) and j : C → A ⊕ C, c 7→ (0, c) are

two canonical embeddings, then we have

A ⊕ C = i(A) ⊕ j(C).
∼
=
Consider θ−1 : A ⊕ C −→ B, then

B = θ−1 (i(A)) ⊕ θ−1 (j(C)).

Since the diagram in Definition B.4.5 commutes, hence

θ−1 (i(A)) = θ−1 ◦ i(A) = ϕ(A),

hence
B = ϕ(A) ⊕ θ−1 (j(C)),
| {z }
D

which implies ψ|D : D → C is a group homeomorphism. We see that

ϕ ψ
◦ A B C ◦
∼
=
split implies B = ϕ(A) ⊕ D and ψ|D : D −→ C.
∼
=
Conversely, if B = ϕ(A) ⊕ D and ψ|D : D −→ C, then there exists a θ

θ: B → A ⊕ C
ϕ(a) + d 7→ (a, ψ(d))

APPENDIX B. ALGEBRA 144

 Lecture 41: Prepare for Final

for a ∈ A, d ∈ D such that

ϕ ψ
◦ A B C ◦
id θ id

◦ A A⊕C C ◦

ϕ(a) + d ψ(d)

(a, ψ(d)) ψ(d)

commutes.
ϕ ψ
Hence, for a short exact sequence ◦ A B C ◦ is split if and only if
∼
=
B = ϕ(A) ⊕ D and ψ|D : D −→ C.
ϕ ψ
Remarkably, let ◦ A B C ◦ is a split short exact sequence, then D
constructed above is not unique. To see this, consider

i Proj2
◦ Z Z⊕Z Z ◦

n (n, 0)

(n, m) m

We have Z ⊕ Z = i(Z) ⊕ j(Z) where j : Z → Z ⊕ Z, n 7→ (0, n). We see that we can let D := j(Z).
Meanwhile, we can also let
D := {(n, n) | n ∈ Z} < Z ⊕ Z
such that Z ⊕ Z = i(Z) ⊕ D.

Example (Non-split short exact sequence). We see that

i Proj2
◦ Z Z Z / nZ ◦

k k·n

is not a split short exact sequence, since if it is, then

.
Z ⊕ Z nZ ∼ =Z
(0, 1) 7→ k,

which is a contradiction since Z is torsion-free while Z ⊕ Z / nZ is not.

ϕ ψ
Lemma B.4.1 (Splitting lemma). If ◦ A B C ◦ is a short exact se-
quence, then the following are equivalent.
(a) This short exact sequence split.
(b) ∃p : B → A such that p ◦ ϕ = idA .

(c) ∃q : C → B such that ψ ◦ q = idC .

∼
=
Proof. • 1. ⇒ 2. Let θ : B −→ A ⊕ C such that it’s the isomorphism which makes the following

APPENDIX B. ALGEBRA 145

 Lecture 41: Prepare for Final

diagram commutes.
ϕ ψ
◦ A B C ◦
id θ id
i
◦ A A⊕C C ◦
Proj1

Then we let p := Proj1 ◦θ, then

p ◦ ϕ = Proj1 ◦θ ◦ ϕ = Proj1 ◦i = idA .

∼
=
• 1. ⇒ 3. Let θ : B −→ A ⊕ C such that it’s the isomorphism which makes the following diagram
commutes.
ϕ ψ
◦ A B C ◦
id θ id
Proj2
◦ A A⊕C C ◦
j

Then we let q := θ ◦ j, then for all c ∈ C, we have

−1

ψ ◦ q(c) = ψ θ−1 (j(c)) = Proj2 ◦θ θ−1 (j(c)) = Proj2 (j(c)) = c,

hence ψ ◦ q = idC .
• 2. ⇒ 1. We have
ϕ ψ
◦ A B C ◦
p

where p ◦ ϕ = idA . We claim that B = ϕ(A) ⊕ ker(p) since for every b ∈ B, ϕ(p(b)) ∈ ϕ(A),
and
b = ϕ(p(b)) + (b − ϕ(p(b)))
| {z } | {z }
∈ϕ(A) ∈ker(p)

from the fact that

p(b − ϕ(p(b))) = p(b) − p ◦ ϕ(p(b)) = p(b) − p(b) = 0.

We need to show the uniqueness also. Suppose b = ϕ(a1 ) + d1 = ϕ(a2 ) + d2 , a1 , a2 ∈ A,

d1 , d2 ∈ ker(p). We see that

ϕ(a1 − a2 ) = d2 − d1 ⇒ p(ϕ(a1 − a2 )) = 0 ⇒ a1 = a2 ⇒ d1 = d2 .

Finally, we claim that

ψ|ker(p) : ker(p) → C
is a group homeomorphism. But it’s obvious that ψ|ker(p) are both surjective and injective.
• 3. ⇒ 1. We have
ϕ ψ
◦ A B C ◦
q

where ψ ◦ q = idC . We claim that B = ϕ(A) ⊕ q(C) since for every b ∈ B,

b = (b − q(ψ(b))) + q(ψ(b)),
| {z } | {z }
∈ker(ψ)=Im(ϕ) ∈q(C)

APPENDIX B. ALGEBRA 146

 Lecture 41: Prepare for Final

which implies B = ϕ(A) + q(C). We can also prove that

B = ϕ(A) ⊕ q(C)

similarly.

APPENDIX B. ALGEBRA 147

Bibliography

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[Arm13] M.A. Armstrong. Basic Topology. Undergraduate Texts in Mathematics. Springer New York,
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[HPM02] A. Hatcher, Cambridge University Press, and Cornell University. Department of Mathe-
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