$M$ not orientable implying results about $H_{n-1}(M, \mathbb{Z})$, $H_n(M, \mathbb{Z}_q)$?

Question

Let $M$ be a compact connected $n$-manifold without boundary, where $n \ge 2$. How do I see that if $M$ is not orientable, then the torsion subgroup of $H_{n-1}(M, \mathbb{Z})$ is cyclic of order $2$ and $H_n(M, \mathbb{Z}_q)$ is zero if $q$ is odd and is cyclic of order $2$ if $q$ is even?

Answer 1

Observe that a non-orientable manifold $M$ is $R$-orientable iff $R$ contains a unit of order $2$, which is basically same as $2=0$ in $R$. So if $M$ is not orientable , then $M$ is not $R$ orientable for $Z_m$ where $m\geq 3$.

Now if $M$ (closed connected n-manifold) is not $R$ orientable then as theorem $3.26$ in Hatcher (pg 236) suggested that there exists an injective map $H_n(M;R) \to R$ with image {$r\in R | 2r=0$}. SO if $m$ is odd then the image is trivially zero. And if $m$ is even then it contains only two elements, so $H_n(M,Z_{2m})= Z_2$.

Now $H_n(M,\mathbb{Z})=0$. So universal co-efficient theorem for homology implies that $Tor(H_{n-1}(M), Z_m)= H_n(M,Z_m)$. Now if $H_{n-1}$ has some torsion other than $Z_2$, then there exists some integer $k>2$ s.t $Tor(H_{n-1}(M),Z_k)$ is (grater than) $Z_k$ ( which contradicts our last observation). So it is forced that torsion of $H_{n-1}(M,\mathbb{Z})=Z_2$.

Now for your last line use the property that $Tor(Z_m,Z_n)= Ker\{Z_n\to_m Z_n\}$. (for details see Hacher's pg 265).

Answer 2

The transfer homomorphism of the orientation cover, composed with the pushforward, is multiplication by $2$. The image is free abelian since the $(n-1)$-integral homology of the orientation cover is by my answer here, so the only torsion in $H_{n-1}(X, \mathbb{Z})$ is $2$-torsion. Applying the universal coefficient theorem for $H_n$ with $\mathbb{Z}/2$ coefficients by $\mathbb{Z}/2$-orientability of every manifold, we see that the $\text{Tor}$ term is precisely $\mathbb{Z}/2$, hence the torsion in $1$ degree lower is precisely $\mathbb{Z}/2$. Applying the same theorem for $\mathbb{Z}/q$, we find the desired conclusion.