If $M$ is a nonorientable $3$-manifold, why is $H_1(M, \mathbb{Z})$ infinite?

Question

Let $M$ be a compact connected $3$-manifold with boundary $\partial M$. If $M$ is nonorientable and $\partial M$ is empty, then how do I see that $H_1(M, \mathbb{Z})$ is infinite?

Answer 1

Every odd-dimensional manifold has vanishing Euler characteristic, so $$0 = \chi(M) = b_0 - b_1 + b_2 - b_3.$$We have $b_0 = 1$, and $b_3 = 0$ since $M$ is nonorientable. Hence, $b_1 > 0$ and thus $H_1(M, \mathbb{Z})$ is infinite.