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Let $M$ be a compact connected non-orientable 3-manifold. The goal is to show that $H_1(M)$, the first integral homology group, is infinite.
There is a proof of this assuming $\partial M = \emptyset$. Unfortunately, the argument doesn't extend to $\partial M \ne \emptyset$ case.
Here's what I have done so far.
Suppose that $H_1(M)$ is finite. Let $\ell := \mathrm{rk} H_2(M)$. Then the characteristic of $M$, $\chi(M) = \mathrm{rk} H_0(M) - \mathrm{rk} H_1(M) + \mathrm{rk} H_2(M) - \mathrm{rk} H_3(M) = 1 + \ell$.
Consider the oriented double cover $\tilde{M}$ of $M$. It follows that $\tilde{M}$ is compact and connected.
Now, I want to apply Lefschetz duality to $\tilde{M}$ and use the fact that $\chi(\tilde{M}) = 2 \chi(M) = 2(1 + \ell)$ to get a contradiction (e.g. deduce that $\chi(\tilde{M}) = 0$), but I am not exactly sure how to do this.

I would appreciate any help/hint.

Luke
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    @BrevanEllefsen Not really. Note that I am not assuming $M$ to be closed, so $\chi(M)$ may not be 0 (it may be, but would require a proof if true) – Luke Aug 23 '23 at 01:06

1 Answers1

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This is simply false. Consider $M=RP^2 \times [0,1]$.

Moishe Kohan
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