Ytukyg

I am readng Milnor's ''Topology From a Differentiable Viewpoint'' and in the chapter about vector fields, page 38, there is a Theorem that states that:

Given any vecotr field $v$ on $M subset mathbb{R^n}$, ($M$ an m-dimensional, compact boundaryless manifold) with only nondegenerate zeros, then the index sum of $v$ is equal to the degree of the Gauss mapping.

So, I am going to avoid writting the proof (if anyone needs it to give a better answer I will write it, just prefer to avoid this if not necessary).

You have:

1. $N_{epsilon}$ a closed $epsilon$-neighboorhood of $M$




2. $r$: $N_{epsilon} to M$, a differentiable map that maps $x$ to the point in $M$, closest to $x$. (Making $epsilon $ small enough this works well)




3. $w$ a vector field in $N_{epsilon}$ that extends $v$, given by: $w(x)= (x-r(x))+v(r(x))$

So.. What I think Milnor is trying to do, is use Hopf's lemma (Lemma 3 in the book),you can see that $w$ points outwards along the boundary and if it has a zero, it must be a zero of $v$ (since $x-r(x)$ and $v(r(x))$ are orthogonal), so all its zeros are isolated, then you are in condition to apply Hopf's lemma.

Now he states that:

$d_zw(h)= d_zv(h) $ in $T_zM$

and

$d_zw(h)=h$ in $T_zM$'s orthogonal complement.

So... maybe its easy to see, but I cannot see why this is. Trying to calculate $frac{partial w_i}{partial x_j}$ for arbitrary $i,j$ seems usless, since I do not know how to calculate the derivative of $r(x)$. This is my first doubt.

My second doubt.. Supposing true the previous statement, you have that $d_zw$ and $d_zv$ have the same determinant at any $z$, zero of $w$, hence the same index. Now Milnor then uses Hopf's Lemma, to prove that thindex sum of $w$ is equal the Gauss mapping. To finish this off I need to see that $v$ has the same zeros than $w$, why is this true? (You know that a zero of $w$ is a zero of $z$, why is the reciprocal true?)

I am really stuck with this, any help would be appreciated, thanks in advanced.