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?
 A: I am reading Milnor's Lectures on h-cobordism Theorem right now, and let me explain here:
I prefer to write the diagram as
\begin{matrix} 
H_0(p)&\stackrel{j_*}{\longrightarrow}&H_0(N)\\ 
\downarrow{\psi}&&\downarrow{\psi}\\ 
H_r(\mathbb R^r,\mathbb R^r-\{0\})&\stackrel{i_*}{\rightarrow}&H_r(N\times\mathbb R^r,N\times(\mathbb R^r-\{0\}))\\
\downarrow{\cong}&&\downarrow{\cong}\\
H_r(U,U-p)&\stackrel{i_*}{\rightarrow}&H_r(V,V-N) 
\end{matrix}
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. 
