Dimension of homology of boundary is twice the dimension of kernel induced by inclusion.

Question

Let $M$ be a compact connected $n$-manifold with boundary $\partial M$, where $n \ge 2$. Assume that $M$ is orientable. Let $n = 2m + 1$ and let $K$ be the kernel of the homomorphism $H_m(\partial M) \to H_m(M)$ induced by the inclusion, where homology is taken with coefficients in a field. How do I see that$$\text{dim}\,H_m(\partial M) = 2 \,\text{dim}\,K?$$

Answer 1

Consider the long exact sequence of the pair $(M,\partial M)$ and notice that Lefschetz duality gives an isomorphism $H_k(M)\cong H^{n-k}(M,\partial M)\cong H_{n-k}(M,\partial M)$, using that $M$ is orientable and the coefficients are over a field.

Consider the truncated exact sequence $H_m(\partial M)\to H_m(M)\to H_m(M,\partial M) \to H_{m-1}(\partial M)\to \cdots.$ Because this sequence is exact, except in the beginning where there is a kernel $K$, it's euler characteristic is $(-1)^m\dim (K)$. Write the full long exact sequence as $$L\to H_m(\partial M)\to R,$$ where $L$ and $R$ stand for the appropriate truncated sequences. Then $\chi(L)=\chi(R)=\dim (K)$ by Lefschetz duality. On the other hand, the whole sequence is exact, so the Euler characteristic is $0$. So $\chi(L)-\dim(H_m(\partial M))+\chi(L)=0$. Thus $\dim(H_m(\partial M))=2\dim(K)$.