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Let $M$ be a compact oriented $m$-dimensional manifold with a boundary $N \neq \emptyset$. Then $N$ is a closed oriented $m-1$-dimensional manifold.

Is it always true that the homology class $[N]_M$ has to vanish in $H_{m-1}(M, \mathbb Z)$?

Terry Black
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    Let $M$ be a compact connected orienteted $n$-manifold with non-empty boundary $\partial M$. Let $N_1,..., N_k$ be connected components of $\partial M$, and using induced orientation on $\partial M$, let $[N_1],...,[N_k]$ be the generators of $H_{n-1}(N_1;\Bbb Z),...,H_{n-1}(N_k;\Bbb Z)$, respectively. Then homology long exact sequence for the pair gives – Sumanta Feb 23 '22 at 15:10
  • $$\cdots \to \underset{\cong 0}{\underbrace{H_n(M;\Bbb Z)}}\to \underset{\cong \Bbb Z}{\underbrace{H_n(M,\partial M;\Bbb Z)}}\xrightarrow{1\longmapsto (1,...,1)} \underset{\cong \bigoplus_{i=1}^k\Bbb Z}{\underbrace{H_{n-1}(\partial M;\Bbb Z)=\bigoplus_{i=1}^kH_{n-1}(N_i;\Bbb Z)}}\to H_{n-1}(M;\Bbb Z)\to \cdots$$ – Sumanta Feb 23 '22 at 15:10
  • That is the generator $[M]$ of $H_n(M,\partial M;\Bbb Z)$ goes to the element $[\partial M]:=\bigoplus_{i=1}^k [N_i]$ of $H_{n-1}(\partial M;\Bbb Z)$. – Sumanta Feb 23 '22 at 15:29
  • You should elaborate your comments in an answer? – Terry Black Feb 24 '22 at 10:44

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If $M$ is a compact oriented $m$-dimensional manifold with non-empty boundary $N$, then the image of the fundamental class of $[N]$ with respect to the natural inclusion $i(n)=n$ lives in $H_{m-1}(M)$, not in top homology of $M$ (you wrote $n$ but I assume you meant to write $m$). I think it is true that the image will always be zero, since it is a boundary. If you assume that everything is smooth and you work with real coefficients, then the image will be zero by Stokes theorem. More concretely, if you have a closed $m-1$ form $\omega$, then $\int_N \omega = \int_{\partial M} \omega = \int_M d\omega=0$, hence the image of $[N]$ is zero.

Laci
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