Winter term 2024/25 Karlsruhe Institute of Technology
Algebraic Topology JProf. Dr. Manuel Krannich
Solutions for Exercise Sheet 1 Dr. Florian Kranhold
Exercise 1.1 (Rotations). We can regard the circle S1 as the space {z ∈ C; |z| = 1} of unit
length complex numbers. For a fixed angle α ∈ S1 , we let rotα : S1 → S1 , z 7→ α · z be the
rotation by α. Show that rotα is homotopic to the identity on S1 .
Solution. We find 0 ⩽ s < 2π with α = eis . This gives rise to a homotopy H : I × S1 → S1
by H (t, z) = eits · z, which satisfies H (0, −) = idS1 and H (1, −) = rotα .
Exercise 1.2 (Filling spheres). Let D n := {z ∈ Rn ; |z| ⩽ 1} be the n-dimensional disc; its
boundary is Sn−1 := {z ∈ Rn ; |z| = 1}. Let X be a space and let f : Sn−1 → X be a map.
Show that the following statements are equivalent:
(1) f is nullhomotopic, i.e. homotopic to a constant map.
(2) f can be extended to a map f˜ : D n → X, that is: f˜|Sn−1 = f .
Solution. We prove the two directions separately:
• For “(2) ⇒ (1)”, we note that the map H : I × Sn−1 → X defined by H (t, z) = f˜(t · z)
is a homotopy from the constant map z 7→ f˜(0) to f .
• For “(1) ⇒ (2)”, we assume a homotopy H : I × Sn−1 → X from some constant map
z 7→ x0 and f . Then we consider the map f˜ : D n → X with
(
H |z|, |zz|
for z ̸= 0,
f˜(z) =
x0 for z = 0.
For any z ∈ Sn−1 , we have f˜(z) = H (1, z) = f (z), so f˜ is indeed an extension of f ;
however, it remains to show that f˜ is continuous. To do so, we consider the surjective
continuous map π : I × Sn−1 → D n taking (t, z) to t · z. Since I × Sn−1 is compact
and D n is Hausdorff, it is a quotient map1 (i.e. the topology on D n is the quotient
topology with respect to π), so by the universal property of the quotient topology, f˜
is continuous if and only if f˜ ◦ π is continuous. But f˜ ◦ π is exactly the homotopy H.
Exercise 1.3 (Bouquets of spheres). This exercise is meant to be solved in pictures. Let X and Y
be spaces, with basepoints x0 and y0 . We define the bouquet X ∨ Y := ( X ⊔ Y )/( x0 ∼ y0 ),
with basepoint [ x0 ] = [y0 ], and we define recursively 0 X := ∗ and n X := X ∨ n−1 X.
W W W
Moreover, we fix basepoints for all spheres.
W3 2
(a) Draw S1 ∨ S1 , S1 ∨ S2 , and S = S2 ∨ S2 ∨ S2 .
(b) Let A ⊊ Rn be finite. Find k, d such that Rn ∖ A is homotopy equivalent to
Wk d
S .
1 Indeed, if A ⊆ D n is a subset such that π −1 ( A) ⊆ I × Sn−1 is closed, then π −1 ( A) is compact as a closed
subspace of a compactum. Thus, A = π (π −1 ( A)) is compact, and hence closed in the Hausdorff space D n .
1
Wk d
(c) Let A ⊊ Sn be finite. Find k, d such that Sn ∖ A is homotopy equivalent to S .
(d) Let T 2 := S1 × S1 be the 2-torus and let ω ∈ T 2 . Find k and d such that T 2 ∖ {ω } is
homotopy equivalent to k Sd .
W
(e) Let z, z′ ∈ S2 be two distinct points of the 2-sphere and let X := S2 /(z ∼ z′ ), i.e. we
glue the two points together. Argue that X is homotopy equivalent to S1 ∨ S2 .
Solution. Handwritten sheet at the end of the document.
Exercise 1.4 (The homotopy category of spaces). Recall the homotopy category hTop of
spaces: its objects are spaces and its morphisms are homotopy classes of maps.
(a) Show that two spaces are isomorphic in the category hTop if and only if they are
homotopy equivalent.
(b) Let C be some category and let F : Top → C be a functor. Show that the following
statements are equivalent (in this case, we call F homotopy invariant):
(1) If f , f ′ : X → Y are homotopic, then F ( f ) = F ( f ′ ).
(2) There is a functor F̄ : hTop → C such that F̄ ◦ pr = F, where pr : Top → hTop
is the functor that sends a space to itself and map f to its homotopy class [ f ].
(3) If f : X → Y is a homotopy equivalence, then F ( f ) is an isomorphism.
(Hint. Consider the two inclusions η0 , η1 : X ,→ I × X at time 0 and 1.)
Solution. (a) Two spaces X and Y are homotopy equivalent if and only if there are maps
f : X → Y and g : Y → X with g ◦ f ≃ idX and f ◦ g ≃ idY . However, this is the
same as providing homotopy classes of maps [ f ] : X → Y and [ g] : Y → X with
[ g] ◦ [ f ] = [ g ◦ f ] = idX and [ f ] ◦ [ g] = idY in hTop.
(b) We show “(1) ⇒ (2) ⇒ (3) ⇒ (1)”:
• For “(1) ⇒ (2)”, we define F̄ ( X ) := F ( X ) and F̄ ([ f ]) = F ( f ), which is well-
defined by assumption. Then F̄ ([idX ]) = F (idX ) = idF̄(X ) and
F̄ ([ g] ◦ [ f ]) = F̄ ([ g ◦ f ]) = F ( g ◦ f ) = F ( g) ◦ F ( g) = F̄ ([ g]) ◦ F̄ ([ f ]),
so F̄ is indeed a functor, and it satisfies F̄ ◦ pr = F by construction.
• For “(2) ⇒ (1)”, we use that the proof of (a) in particular shows that if f : X → Y
is a homotopy equivalence, then [ f ] is an isomorphism in hTop. Now we use
Lemma ii.2.4 to see that F ( f ) = F̄ ([ f ]) is again an isomorphism.
• For “(3) ⇒ (1)”, we consider the two inclusions η0 , η1 : X ,→ I × X at time 0 and
1, respectively. If p : I × X → X is the projection, then p ◦ η0 = p ◦ η1 . Since p is
a homotopy equivalence, F ( p) is an isomorphism; hence
F ( p ) ◦ F ( η0 ) = F ( p ◦ η0 ) = F ( p ◦ η1 ) = F ( p ) ◦ F ( η1 )
2
implies F (η0 ) = F (η1 ). If H : I × X → Y is a homotopy from f to f ′ , then
F ( f ) = F ( H ◦ η0 ) = F ( H ) ◦ F ( η0 ) = F ( H ) ◦ F ( η1 ) = F ( H ◦ η1 ) = F ( f ′ ) .
