Is this proof that $M$ is orientable correct?

Question

The exercise is the following:

Let $M$ be a closed and connected topological manifold of dimension $n \geq 2$. If $H^1(M;\mathbb{Z}_2) = 0$, then $M$ is orientable.

These are my thoughts: by Poincaré duality, we know that $H_{n-1}(M;\mathbb{Z}_2) \cong H^1(M;\mathbb{Z}_2) = 0$. The Universal Coefficient Theorem for homology gives that

$$ 0 = H_{n-1}(M;\mathbb{Z}_2) \cong (H_{n-1}(M) \otimes \mathbb{Z}_2) \oplus \operatorname{Tor}(H_{n-2}(M), \mathbb{Z}_2) $$

In particular, $H_{n-1}(M) \otimes \mathbb{Z}_2 = 0$. Since $M$ is compact, its homology groups are finitely generated. Let us write

$$H_{n-1}(M) = \mathbb{Z}^d \oplus \mathbb{Z}_{q_1} \oplus \cdots \oplus \mathbb{Z}_{q_r}$$

for some $d \geq 0$ and powers of primes $q_i$. Now,

$$0 = H_{n-1}(M) \otimes \mathbb{Z}_2 = \mathbb{Z}_2^d \oplus \mathbb{Z}_{d_1} \oplus \cdots \oplus \mathbb{Z}_{d_r}$$

where $d_i = \gcd(2,q_i)$. We thus have $d = 0$ and $d_i = 1$ for each $i$.

Suppose that $M$ is not orientable. Then $H_n(M;\mathbb{Z}_2) = \mathbb{Z}_2$ and $H_n(M) = 0$. Again, the Universal Coefficient Theorem gives that

$$\mathbb{Z}_2 = (H_n(M) \otimes \mathbb{Z}_2) \oplus \operatorname{Tor}(H_{n-1}(M),\mathbb{Z}_2) = \operatorname{Tor}(H_{n-1}(M),\mathbb{Z}_2).$$

However,

$$\operatorname{Tor}(H_{n-1}(M),\mathbb{Z}_2) = \bigoplus_{i=1}^r \operatorname{Tor}(\mathbb{Z}_{q_i}, \mathbb{Z}_2) = \bigoplus_{i=1}^r Z_{d_i} = 0$$

which contradicts the previous equation.

Do you agree with this proof?

Answer 1

Seems good to me, however there is a proof that doesn't invoke Poincare Duality and also is quite a bit less tedious, in my opinion.

Recall that if a compact manifold has no subgroup of $\pi_1(M)$ with index 2, then the manifold is orientable. Clearly, $H_1(M,\mathbb{Z}_2)$ has no subgroup of index 2.

By the universal coefficient theorem, $H_1(M,\mathbb{Z}_2)$ is given by tensoring with $\mathbb{Z}_2$. Since tensoring is right exact we can deduce that if $0\rightarrow A \rightarrow B \rightarrow \mathbb{Z}/2 \rightarrow 0$ is exact, then $A \otimes \mathbb{Z}/2 \rightarrow B\otimes \mathbb{Z}/2 \rightarrow \mathbb{Z}/2 \rightarrow 0$ is exact. Hence the kernel of the map onto $\mathbb{Z}/2$ is index 2.

This means that there are no index two subgroups of $H_1(M, \mathbb{Z})$. Similarly, abelianization is a right exact functor which means that $\pi_1(M)$ also has no subgroup of index 2 which yields the result.