If $M$ is a non-orientable closed connected manifold with a single $n$-cell, then the torsion subgroup of $H_{n-1}(M;\mathbb{Z})$ is $\mathbb{Z}_2$ The following is a paraphrased passage from pages 238-239 of Hatcher's Algebraic Topology book.
Suppose $M$ is a non-orientable closed connected manifold that has a CW structure with a single $n$-cell. Consider the ceullar boundary map $d_n : C_n(M;\mathbb{Z}) \to C_{n-1}(M;\mathbb{Z})$. Hatcher then goes on to say the following:

$d$ must take a generator of $C_n(M;\mathbb{Z})$ to twice a generator $\alpha$ of a $\mathbb{Z}$ summand of $C_{n-1}(M;\mathbb{Z})$ in order for $H_n(M;\mathbb{Z}_p)$ to be zero for odd primes $p$ and $\mathbb{Z}_2$ for $p = 2$. The cellular chain $\alpha$ must be a cycle since $2\alpha$ is a boundary and hence a cycle. It follows that the torsion subgroup of $H_{n-1}(M;\mathbb{Z})$ must be a $\mathbb{Z}_2$ generated by $\alpha$.

There are a few parts on this that I am confused by.

*

*I can see that non-orientability of $M$ would imply that $H_n(M;\mathbb{Z}_p)$ is zero for odd primes $p$ and $\mathbb{Z}_2$ for $p = 2$, but I do not see why this implies that $d$ should take a generator of $C_n(M;\mathbb{Z})$ to twice a generator $\alpha$ of a $\mathbb{Z}$ summand of $C_{n-1}(M;\mathbb{Z})$

*I can see that $2\alpha$ is a boundary and hence a cycle, but I do not see why $\alpha$ should also be a cycle.

*I do not see how it follows that the torsion subgroup of $H_{n-1}(M;\mathbb{Z})$ must be a $\mathbb{Z}_2$ generated by $\alpha$.

Regarding the first bullet point, I should say that if I consider $d : C_n(M;\mathbb{Z}_2) \to C_n(M;\mathbb{Z}_2)$ then the fact that $H_n(M;\mathbb{Z}_2) \cong \mathbb{Z}_2$ implies that $\ker d = \mathbb{Z}_2$ which implies that $d$ is the zero map having domain $\mathbb{Z}_2$ which (I guess?) implies that $d$ is multiplication by $2^k$ for some $k$. But I don't really see how this influences $d : C_n(M;\mathbb{Z}) \to C_{n-1}(M;\mathbb{Z})$.
 A: There is a direct summand into which the image of the boundary $C_n(M, \mathbb Z) \to C_{n-1}(M, \mathbb Z)$ falls. This image forms a subgroup of some index $k$ there. The map $\mathbb Z \to \mathbb Z_p$ induces a map of chain complexes $C_*(M, \mathbb Z) \to C_*(M, \mathbb Z_p)$, so the the map $C_n(M, \mathbb Z_p) \to C_{n-1}(M, \mathbb Z_p)$ will also send the generator to $k$ times the generator of this summand, but the summand will look like $\mathbb Z_p$ now. For which $k$ does the map $\times k: \mathbb Z_p \to \mathbb Z_p$ have the kernel $0$ for all $p \neq 2$? As you've noticed $k = 2^m.$ Looking at $H_n(M, \mathbb Z_4)$ also excludes the possibility $m > 1$. Looking at $H_n(M, \mathbb Z_2)$ leads to $m=1$.
$d \alpha$ lives in a free group $C_{n-2}(M, \mathbb Z).$ So $2d \alpha =0$ implies $d \alpha =0$.
$H_{n-1}(M,\mathbb{Z})$ equals $Z_{n-1}(M,\mathbb{Z})/B_{n-1}(M,\mathbb{Z}).$ The numerator is free, and the denominator looks there like $2\mathbb Z$ as we've found out. This means that the torsion will only come from this $2\mathbb Z$ and whence will be $\mathbb Z_2.$
