The top singular cohomology group with coefficient $\mathbb{Z}_2$ for non-compact, non-orientable manifolds without boundary

Question

I want to know the top singular cohomology group with coefficient $\mathbb{Z}_2$ for non-compact, non-orientable manifolds without boundary. I think Poincare and Lefschetz duality may help. However, to my knowledge, the first one works for compact manifolds and the second one works for non-orientable manifolds. Is there a duality theory can contribute to my question?

Edit.

The Universal coefficient theorem also relates to my question. We have the following exact sequence.

$$0\rightarrow Ext_{\mathbb{Z}_2}^1(H_{k−1}(M;\mathbb{Z}_2),\mathbb{Z}_2)\rightarrow H^k(M;\mathbb{Z}_2)\rightarrow \hom_{\mathbb{Z}_2}(H_k(M;\mathbb{Z}_2),\mathbb{Z}_2)\rightarrow0$$

But I don't know how to deal with the Ext part, since I don't know what $H_{k−1}(M;\mathbb{Z}_2)$ and $H_k(M;\mathbb{Z}_2),\mathbb{Z}_2$ are, since $M$ is not compact.

Answer 1

A version of Poincare duality holds for noncompact manifolds: cap product with the fundamental class over a ring of coefficients $A$ gives an isomorphism $H^*_c(M, A) \to H_*(M, A)$, where $H^*_c$ denotes compactly-supported cohomology; see Section 3.3 of Hatcher, for example. Any manifold $M^n$ is orientable over $\mathbb{Z}_2$, so $H_n(M, \mathbb{Z}_2) = H^0_c(M, \mathbb{Z}_2)$. But $M$ is not compact, so it's easy to verify from the definition (see, for example, Example 3.32 in Hatcher) that $H^0_c$ vanishes (assuming $M$ is connected, at least).

Answer 2

Since $\mathbb Z_2$ is a field, the functor $Ext_{\mathbb Z_2}^1(-,-)$ is identically zero.