Math 215a Homework #4 solutions
1. Hatcher 2.1.14. Let’s do the last part first. I claim that if A is an
abelian group, then there exists a short exact sequence
0 −→ Z −→ A −→ Zn −→ 0
iff A ' Z ⊕ Zm where m is a positive integer dividing n. To see this,
suppose first that such an exact sequence exists. Let x ∈ A be the
image of a generator of Z. Let y ∈ A map to a generator of Zn . Then
ny maps to 0 ∈ Zn , so ny = kx for some integer k. It follows that the
map Z2 → A sending (1, 0) 7→ x and (0, 1) 7→ y induces an isomorphism
Z2
A' .
Z{(−k, n)}
We can find a change of basis of Z2 sending (−k, n) to (0, m) where m
is the greatest common divisor of k and n. So the right hand side of
the above equation is isomorphic to Z ⊕ Zm , and by definition m is a
divisor of n. Conversely, if n = md, then we can define a short exact
sequence
f g
0 −→ Z −→ Z ⊕ Zm −→ Zn −→ 0
by f (1) := (d, −1) and g(a, b) := a + bd.
By a similar but slightly more complicated argument, if A is an abelian
group, then there exists a short exact sequence
0 −→ Zpm −→ A −→ Zpn −→ 0
iff A ' Zpa ⊕ Zpb where a and b are nonnegative integers such that
a ≥ max(m, n) and a + b = m + n.
2. Hatcher 2.1.17(b). It follows easily from the long exact sequence that
Hi (X, A) = Hi (X, B) = 0 for i 6= 1, 2. From the long exact sequence
of the pair (X, A), we obtain an exact sequence
0 → H2 (X) → H2 (X, A) → H1 (A) → H1 (X) → H1 (X, A) → 0.
Since A separates X, it follows that H1 (A) maps to zero1 in H1 (X).
By the exact sequence, H2 (X, A) ' Z2 and H1 (X, A) ' Z 2g .
1
One way to see this using what we know so far is to consider the Mayer-Vietoris
sequence relating the homology of X to the homology of the left and right halves and the
homology of A.
1
Likewise, there is an exact sequence
0 → H2 (X) → H2 (X, B) → H1 (B) → H1 (X) → H1 (X, B) → 0.
In this case, one can choose identifications H1 (B) ' Z and H1 (X) '
Z2g so that the map H1 (B) → H1 (X) sends2 1 7→ (1, 0, . . . , 0). It
then follows from the exact sequence that H1 (X, B) ' Z2g−1 and
H2 (X, B) ' Z.
3. Hatcher 2.1.18. To compute H1 (R, Q), we use the long exact sequence
∗ i
0 = H1 (R) −→ H1 (R, Q) −→ H0 (Q) −→ H0 (R)
where i : Q → R is the inclusion map. It follows from this exact
sequence that H1 (R, Q) = Ker(i∗ ). Now H0 (Q) = ⊕Q Z. Moreover if
σq denotes the 0-simplex mapping to q ∈ Q, then
i∗ σq = 1 ∈ Z = H0 (R).
So the kernel of i∗ consists of finite integer linear combinations of σq ’s
with total coefficient zero. A basis for this is given by
{σ0 − σq | q ∈ Q \ {0}}.
4. Hatcher 2.1.27.
(a) This follows by applying the five-lemma to the diagram
Hn (A) −−−→ Hn (X) −−−→ Hn (X, A) −−−→ Hn−1 (A) −−−→ Hn−1 (X)
' ' ' '
y y y y y
Hn (B) −−−→ Hn (Y ) −−−→ Hn (Y, B) −−−→ Hn−1 (B) −−−→ Hn−1 (Y ).
(b) If g : (Dn , Dn \ {0}) → (Dn , S n−1 ), let g1 : Dn \ {0} → S n−1
denote the restriction. Since g is continuous at 0, it follows that
g maps all of Dn to S n−1 . Hence g1 induces 0 on H e n−1 , since the
n n−1
map g1 factors through D . Since Hn−1 (S ) 6= 0, it follows that
e
g1 does not induce an isomorphism on reduced homology, so g1
cannot be a homotopy equivalence.
2
One way to show this is to identify X with the quotient of a 4g-gon so that B corre-
sponds to one of the edges. We will see a more general way to prove this after introducing
the intersection pairing.
2
5. Hatcher 2.2.2. Let f : S 2n → S 2n . If f (x) 6= x for all x, then f is
homotopic to the antipodal map, so deg(f ) = (−1)2n+1 . If f (x) 6= −x
for all x, then f is homotopic to the identity, so deg(f ) = 1. Since
(−1)2n+1 6= 1, these possibilities are mutually exclusive, so there must
exist x ∈ S 2n with f (x) = x or f (x) = −x.
Next let f : RP2n → RP2n . Since S 2n is the universal covering space
of RP2n , it follows from the Lifting Criterion that f lifts to a map
RP2n → S 2n . We can then pull this back to a map fe : S 2n → S 2n , so
that the diagram
fe
S 2n −−−→ S 2n
y y
f
RP2n −−−→ RP2n
commutes, i.e. f ([x]) = [fe(x)] ∈ RP2n for all x ∈ S 2n . By the previous
paragraph, there exists x ∈ S 2n such that fe(x) = ±x, which means
that f ([x]) = [x].
0 −1
Taking the direct sum of n copies of the 2 × 2 matrix gives
1 0
a linear map R2n → R2n with no real eigenvalue, and hence a map
RP2n−1 → RP2n−1 with no fixed point.
6. Extra problem.
(a) The proof that ∂T = T ∂ is a straightforward matter of comparing
signs.
(b) We will construct a natural chain homotopy K : C∗ (X) → C∗+1 (X)
such that
∂K + K∂ = 1 − T. (1)
Let In ∈ Cn (∆n ) denote the identity map of the standard n-
simplex ∆n . It is enough to define KIn ∈ Cn+1 (∆n ) such that
∂(KIn ) = In − T In − K∂In . (2)
If we have solved equation (2), then for σ : ∆n → X we can define
Kσ := σ# KIn ,
3
and equation (1) will hold.
We solve equation (2) by induction on n. Since Hn+1 (∆n ) = 0,
a solution KIn to equation (2) exists iff the right hand side is a
cycle. To check this, we compute
∂(In − T In − K∂In ) = ∂In − ∂T In − ∂K∂In
= ∂In − ∂T In − (1 − T − K∂)∂In
= −∂T In + T ∂In
= 0.
(This argument, and similar arguments which we did in class, can
be formalized as the “acyclic models theorem”.)
