# Why is Lemma 6.3 of Milnor's Lectures on the h-cobordism Theorem True?

Milnor's statement is:

"Let $M^r$ and $N^s$ be sub-manifolds of $V^{r+s}$ which are all smooth, compact, oriented and without boundary. If $p$ is a point of $M^r$ contained in an $r$-cell $U$, naturality of the Thom isomorphism implies that the inclusion induced map

$H_r(U,U-p) \longrightarrow H_r(V,V-N)$

is an isomorphism given by $\gamma \rightarrow \epsilon \psi(\alpha)$ where $\gamma$ is the orientation generator of $H_r(U,U-p)$, $\psi:H_0(N) \longrightarrow H_r(V,V-N)$ is the Thom isomorphism, $\alpha$ is the canonical generator of $H_0(N)$ and $\epsilon$ is the intersection number of $M$ and $N$ at $p$."

I understand that naturality of the Thom isomorphism is the fact that this diagram commutes, \begin{matrix} H_0(p)&\stackrel{j_*}{\longrightarrow}&H_0(N)\\ \downarrow{\psi}&&\downarrow{\psi}\\ H_r(U,U-p)&\stackrel{i_*}{\rightarrow}&H_r(V,V-N) \end{matrix}

where $i_*$ and $j_*$ are induced by inclusion. By the commutativity of the diagram, it is clear to me that $i_*$ is an isomorphism.

But why is $i_*(\gamma)=\epsilon \psi(\alpha)$ true?

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where I assume $N$ has trivial normal bundle in $V$ and the indicated isomorphisms in the two columns are by excisions. We can certainly assume that the orientation on $\mathbb R^r$ is chosen so that the normal bundle $N\times \mathbb R^r$ has the same orientation as $V$, then the isomorphism $$H_r(N\times\mathbb R^r,N\times(\mathbb R^r-\{0\}))\xrightarrow{\cong}H_r(V,V-N)$$ send generator to generator. But now when you look at the left column isomorphism$$H_r(\mathbb R^r,\mathbb R^r-\{0\})\xrightarrow{\cong}H_r(U,U-p)$$ How does it send generator on $H_r(\mathbb R^r)$ to $H_r(U)$? It certainly depends on how we choose the orientation on $\mathbb R^r$. So by definition of intersection number, it is exactly sending generator on $H_r(\mathbb R^r)$ to $\epsilon\cdot$(generator of $H_r(U)$), and that's basically why the intersection number comes out here.