Euler characteristic of odd dimensional manifold - Hatcher

Question

I ran into some trouble while reading through Hatcher's proof of the following:

Corollary 3.37. A closed manifold of odd dimension has Euler characteristic zero.

There is only one part of the proof that I don't understand. He writes

"Each $\mathbb{Z}_m$ summand of $H_i(M;\mathbb{Z})$ with $m$ even gives $\mathbb{Z}_2$ summands of $H^i(M;\mathbb{Z}_2)$ and $H^{i-1}(M;\mathbb{Z}_2)$, [...]."

I have been able to show that we indeed get $\mathbb{Z}_2$ summands in $H^i(M;\mathbb{Z}_2)$ (if helpful, I can edit this post to include this proof), but I fail to see that this is also the case in $H^{i-1}(M;\mathbb{Z}_2)$. My best guess is that it follows from the UCT, but I haven't been able to show it.

Is there something obvious that I'm missing?

Answer 1

There is a mistake in your version of the book (or you read it wrong). It gives a copy of $\mathbb{Z}_2$ in $H^{i+1}$, not $H^{i-1}$. In my version of the book it is right.

This is a consequence of the shift in the index of $\operatorname{Ext}$ in the UCT: $$ H^n(C;\mathbb{Z}_2)\cong \operatorname{Hom}(H_n(C),\mathbb{Z}_2)\oplus \operatorname{Ext}(H_{n-1}(C),\mathbb{Z}_2),$$ notice the indices.

Hence, $\operatorname{Hom}(\mathbb{Z}_m,\mathbb{Z}_2)\cong \mathbb{Z}_2$ and $\operatorname{Ext}(\mathbb{Z}_m,\mathbb{Z}_2)\cong \mathbb{Z}_2/m\mathbb{Z}_2\cong\mathbb{Z}_2$ if $m$ is even.