If $M^{2n}$ is oriented, then $H_n(M)\to H_n(M,\partial M)$ can be identified with the matrix of the intersection form

Question

If $M$ is an oriented $2n$-dimensional manifold with boundary $\partial M$, then the natural map $H_n(M)\to H_n(M,\partial M)$ can be identified (via Lefschetz duality) with a map $H_n(M)\to H^n(M)$. If, moreover, $H_{n-1}(M)$ is torsion-free or we take coefficients in a field, we can further identify this with a map $H_n(M)\to \operatorname{Hom}(H_n(M),\Bbb Z)$. Now, according to Madsen & Milgram (page 165), this last map can be taken as the intersection form on $H_n(M)$ (i.e., $\sigma\mapsto \sigma\cap -$). This seems plausible to me, but I cannot see why it is actually true. To summarize:

Why can we take the map $H_n(M)\to H_n(M,\partial M)$, appearing in the long exact sequence of $(M,\partial M)$, to be the map given by the intersection form $H_n(M)\otimes H_n(M)\to \Bbb Z$ (under the identification $H_n(M,\partial M)\cong H^n(M)\cong\operatorname{Hom}(H_n(M),\Bbb Z)$)?

This sort of geometric reasoning has never come easily to me, so I'm hoping people could bring their own perspectives here.

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Background:

This occurs during construction of an exotic $7$-sphere ($\partial M$). The intersection pairing on $H_n(M)$ is easily determined, and the idea is to use this to determine the homology of $\partial M$. In particular, $M^{2n}$ is homotopy equivalent to a wedge of $n$-spheres and the claim is that, if the Gram matrix of the intersection form on $H_n(M)$ is unimodular, then the exact sequence $$0\to H_n(\partial M)\to H_n(M)\xrightarrow{*} H_n(M,\partial M)\to H_{n-1}(\partial M)\to 0$$ forces $H_n(\partial M)=H_{n-1}(\partial M)=0$ (where $*$ is given by the matrix of the intersection form). After this, it follows easily that $\partial M$ is a homotopy sphere and thus (via the h-cobordism theorem) a topological sphere.

The book by Madsen & Milgram I referenced is "The Classifying Spaces for Surgery and Cobordism of Manifolds" (1979).

Answer 1

Naturally after posting a bounty the question answered itself. It turns out that Bredon lays this out quite carefully in his book "Topology and Geometry." As @Lee Mosher and @Tyrone commented, the proof just involves using duality and naturality correctly. A key part in making this explicit is the following commuting diagram used to prove Lefschetz duality:

$$\begin{matrix} \\ &&&& \Bbb Z^r && \Bbb Z^r& &&&\\ &&&& || && || & &&&\\ 0 & \to & H^{n-1}(\partial M) & \xrightarrow{\delta} & H^n(M,\partial M) & \xrightarrow{j^*} & H^n(M) & \xrightarrow{i^*} & H^n(\partial M) & \to & 0\\ & & \downarrow\tiny{\frown[\partial M]} & & \downarrow\tiny{\frown[M]} & & \downarrow\tiny{\frown[M]} & & \downarrow\tiny{\frown[\partial M]} & & \\ 0 & \to & H_n(\partial M) & \xrightarrow{i_*} & H_n(M) & \xrightarrow{j_*} & H_n(M,\partial M) & \xrightarrow{\partial}& H_{n-1}(\partial M) & \to & 0\\ &&&& || && || & &&&\\ &&&& \Bbb Z^r && \Bbb Z^r& &&& \\ \\ \end{matrix}$$

As I said in my post, we can replace $j_*$ with the composition $$H_n(M)\xrightarrow{j_*} H_n(M,\partial M)\xrightarrow{D} H^n(M)\xrightarrow{h} \operatorname{Hom}(H_n(M),\Bbb Z)$$

where $D$ denotes the inverse of the duality map $-\frown[M]$ and $h$ is the obvious map from the universal coefficient theorem. It remains to identify the image of $a\in H_n(M)$ as a function $H_n(M)\to \Bbb Z$. For $b\in H_n(M)$, we have \begin{align} hDj_*a(b)&=\langle Dj_*a,b\rangle & (1)\\ &=\langle j^*Da,b\rangle & (2)\\ &=\langle Da,j_*b\rangle & (3)\\ &=\langle Da,j^*Db\frown[M]\rangle & (4)\\ &=\langle Da\smile j^*Db,[M]\rangle &(5)\\ &=\langle Da\smile Db,[M]\rangle. & (6) \end{align}

where steps (2) and (4) follow from commutativity of our diagram, (3) and (5) are standard properties of the Kronecker product $\langle -,-\rangle$, and (6) is easily verified at the cochain level (since $j$ is just the identity map).