In the proof of a theorem we did in a class (namely: if $M$ is an odd-dimensional, closed manifold, then $\chi(M)=0$), there's the following step:
$$H_k(M;\mathbb{Z}_2)\cong H^k(M;\mathbb{Z}_2)$$
and then we proceed using Poincaré duality and all the usual homological algebra machinery. It might be something stupid that I'm missing, but I really can't see why the isomorphism above should hold true. Any help?
Use the universal coefficient theorem ($\mathbb{Z}_2$ is a PID):
$$0 \to \operatorname{Ext}^1_{\mathbb{Z}_2}(H_{i-1}(M; \mathbb{Z}_2), \mathbb{Z}_2) \to H^i(M; \mathbb{Z}_2) \to \hom_{\mathbb{Z}_2}(H_i(M; \mathbb{Z}_2), \mathbb{Z}_2) \to 0$$
Since $\mathbb{Z}_2$ is a field, the $\operatorname{Ext}^1$ term is zero, and since $M$ is a closed (thanks Jack Lee) finite dimensional manifold, the homology groups are finitely generated and so are isomorphic to their ($\mathbb{Z}_2$-)dual.