I am reading Appendix B in Fulton's Young Tableaux about Borel-Moore homology. In particular, I'd like to understand why for compact manifolds the Borel-Moore homology groups are isomorphic to singular homology groups. Fulton's argument goes as follows:
We have a lemma: if $X$ is embedded as a closed subspace of an oriented differentiable manifold $M$, then there is a canonical isomorphism $H_i^{BM} \cong H^{m-i}(M,M\backslash X)$, where $m = dim(M)$.
So if $M$ is an oriented $n$-dim. manifold we can embed $M$ in itself and we get $H_i^{BM} = H^{n-i}(M,M\backslash M) = H^{n-i}(M)$. Hence $H_i^{BM} = 0$ for $i >n$ and $H_n^{BM} = H^0M$ is a free $\mathbb{Z}$-module with one generator for each connected component of $M$.
Then Fultons states that if $M$ is compact, the Poincare duality for singular homology shows that $H^{n-i} \cong H_iM$ and so $H_i^{BM} \cong H^{n-i}M \cong H_iM$.
My problem is that I don't understand the connection between compactness and Poincare duality. The hypothesis in the Poincare duality theorem is that the manifold is closed and orientable. So does compactness imply that $M$ is closed and orientable? Or compactness together with the fact that $H^0$ is a free $\mathbb{Z}$-module for each connected component? I would appreciate help with understanding the implicit assumptions here.