Odd dimensional manifold has $0$ Euler characteristic

Question

I am reading Cor 3.37 of Hatcher's book. This first proves that for orientable odd dimensional manifold the Euler characteristic is $0$, which is easy. Then for non-orientable manifold, to apply Poincare duality again, he choose the coefficient to be $\mathbb{Z}_2$ so that the manifold is $\mathbb{Z}_2$-orientable. Then the task is to show the sum of $\dim H_i(M;\mathbb{Z}_2)$ is equal to the sum of $\operatorname{rank}H_i(M;\mathbb{Z})$, for this part I am completely missing. Does any have idea to prove this?

By the way, to prove the statement actually we can use the fact that the orientable cover of $M$ has 2 times Euler characteristic of $M$'s Euler characteristic. I am just try to figure out another way. Thanks!

Answer 1

This isn't true for non-compact manifolds (e.g. $\chi(\Bbb R^3)=1$). Any compact manifold $M^n$ is $\Bbb Z_2$-orientable, by Poincare duality $H_i(M,\Bbb Z_2)\cong H^{n-i}(M,\Bbb Z_2)$. By the universal coefficient theorem, $H^{n-i}(M,\Bbb Z_2)\cong H_{n-i}(M,\Bbb Z_2)$ (given both are finitely generated). Hence $\chi(M)$$$=\sum_{i=0}^n(-1)^i\dim (H_i(M,\Bbb Z_2))=\sum_{i=0}^{(n-1)/2}(-1)^i\dim (H_i(M,\Bbb Z_2)) +\sum_{i=0}^{(n-1)/2}(-1)^{n-i}\dim (H_i(M,\Bbb Z_2))=0$$