TITLE: Top homology of an oriented, compact, connected smooth manifold with boundary
QUESTION [16 upvotes]: Let $M$ be an oriented, compact, connected $n$-dimensional smooth manifold with boundary. Then, is it true that $n$-th singular homology of M, that is $H_n(M)$, is vanish? I can't make counterexamples for this statement, but I don't have the slightest idea for proof.

REPLY [10 votes]: You can prove that $H_n(M, \mathfrak d M)\cong \mathbb Z$ in essentially the same way as the special case $\mathfrak d M=\emptyset$, and also that the connecting homomorphism $\mathfrak d\colon H_n(M, \mathfrak d M)\to H_{n-1}(\mathfrak d M)$ takes a relative orientation cycle to an orientation cycle of the boundary. In particular, it is injective if $\mathfrak dM\ne\emptyset$ (and an isomorphism if the boundary is connected). The result now follows from looking at the long exact sequence
$$0=H_n(\mathfrak d M)\to H_n(M)\to H_n(M, \mathfrak d M)\xrightarrow{\text{inj.}} H_{n-1}(\mathfrak d M).$$