In his book Milnor proof the following Lemma:
If $v:X\to \Bbb{R}^m$ is a smooth vector field with isolated zeroes, and if $v$ points out of $X$ along the boundary, then the index sum $\sum{\iota}$ is equal to the degree of the Gauss mapping from $\partial X$ to $S^{m-1}$.
The proof is given as: Removing an $\epsilon$-ball around each zero, we obtain a new manifold with boundary. The function $\overline{v}(x)=v(x)/\|v(x)\|$ maps this manifold into $S^{m-1}$. Hence the sum of the degrees of $\overline{v}$ restricted to the various boundary conditions is zero. But $\overline{v}|\partial X$ is homotopic to $g$, and the degrees on the other boundary conditions add up to $-\sum{\iota}$. Therefore $$\text{deg}(g)-\sum{\iota}=0$$
I don't understand 2 points in this proof:
Hence the sum of the degrees of $\overline{v}$ restricted to the various boundary conditions is zero. Why is this? It should be in relation with the following Theorem: If $f : M \to N$ extends to a smooth map $F : X \to N$, then $deg(f; y) = 0$ for every regular value $y$. But I don't see the connection between them.
Why is this true: $\overline{v}|\partial X$ is homotopic to $g$?
Many thanks for your help.
$M$ here is the union of the boundaries of the balls and $\partial X$, properly oriented. The $X$ is not $X$; it's what's left when you remove the balls.
The fact that $\bar\nu$ is outward-pointing on the boundary makes it easy to write down the homotopy to $g$: Take the straight-line homotopy (which can never pass through the origin) and project to the unit sphere.