(THIS QUESTION HAS BEEN EDITED)
I am reading 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 vector 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 writing 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:
$N_{\epsilon}$ a closed $\epsilon$-neighborhood of $M$
$r$: $N_{\epsilon} \to M$, a differentiable map that maps $x$ to the point in $M$, closest to $x$. (Making $\epsilon $ small enough this works well)
$w$ 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 to 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: For any $z \in M$
$d_zw(h)= d_zv(h) $ in $T_zM$
and
$d_zw(h)=h$ in $T_zM$'s orthogonal complement.
So... maybe it is 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 that the previous statement is true, do you have that $d_zw$ and $d_zv$ have the same determinant at any $z$, zero of $w$, hence the same index. Now Milnor uses Hopf's Lemma, to prove that the index 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 $v$, why is the reciprocal true?)
I am really stuck with this, any help would be appreciated, thanks in advance.
(FROM HERE ON I EDITED IT)
So... I am going to answer my second doubt: If $z$ is a zero of $v$, then $z \in M$ so $r(z)=z$ and thus $w(z)=v(z)=0$
And for my first doubt, I was thinking that I never used that $v$ has non-degenerate zeros (Or so I think), so maybe that is a hint to anwering my first doubt, cannot seem to see it though. If I had used that $v$ has only non-degenerate zeros, please tell me where. Thanks in advance.
This question is old but has been upvoted a lot, here is an answer to the first doubt.
Suppose first that $h$ is in the orthogonal complement of $T_zM$. Note that for a scalar $t$ small enough, $r(z+th)=z$, since $z+th$ is perpendicular to $T_zM$. Then using the definition of derivative, we have : \begin{align*} d_zw(h) = \lim_{t\to 0}\frac{w(z+ht)-w(z)}{t}&=\lim_{t\to0}\frac{(z+th)-r(z+th)+v(r(z+th))-z+r(z)-v(z)}{t}\\ &=\lim_{t\to0}\frac{(z+th)-z+0-z+z-0}{t}\\ &=h \end{align*}
If $h$ is in $T_zM$, it suffices to notice that $d_zw(h)=d_zw|_{T_zM}(h)=d_z(w|_M)(h)$. The second equality is a consequence of the fact that commutative diagram of smooth map give rise to commutative diagram of their derivative (See chapter 1) .Then $w|_M=v$ as for any $x\in M$ we have $w(x)=x-r(x)+v(r(x))=x-x+v(x)=v(x)$.
LS