A question regarding the proof of Hopf's theorem. This is a question regarding the proof of the Hopf theorem given in "Topology from a Differential Viewpoint" by Milnor:

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 $S^{m-1}$. In particular, $\sum{\iota}$ does not depend on the choice of $v$.

Remember that if $X\subset R^m$ is a compact manifold with boundary, then the Gauss mapping is $$g:\partial X\to S^{m-1}$$ which assigns to each $x\in\partial X$ the outward unit normal vector at $x$. 
The proof of Hopf's theorem, as given in the text is:

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 $-\sum{\iota}$ (the $-$ sign occurs because each sphere gets the wrong orientation.) Therefore $$\text{deg}(g)-\sum{\iota}=0$$

I have the following questions:


*

*What is the new boundary of the manifold after all those $\epsilon$-balls are removed? Does the new boundary include the boundary of the $\epsilon$-balls?

*If the vectors of $v(x)/\|v(x)\|$ can be arranged in a sphere, why does that imply that sum of the degrees of $v(x)/\|v(x)\|$, restricted to various boundary conditions, is zero?

*How does "every sphere get the wrong orientation"?

*Shouldn't the degree of every Gaussian map be just one? At least on a sphere?
 A: *

*If the isolated zeroes are $p_1,\cdots,p_n$ and the you remove sufficiently small balls $B_1,\cdots,B_n$ around them you will obtain the manifold $X\setminus (B_1\cup\cdots\cup B_n)$ which has boundary $\partial X\cup\partial B_1\cup\cdots\cup\partial B_n$ with $\partial B_j=S^{m-1}$.

*It follows from this proposition (earlier proven in the book, if I remember well): if you have a smooth map $f:\partial X\rightarrow Y$ which admits a smooth extension $F:X\rightarrow Y$ then $deg(f)=0$.

*The orientation of each sphere $\partial B_j=S^{m-1}$ is induced by the orientation of $X$ so it is the one given by the normal vector pointing inward the sphere, instead, when you calculate the index of $v$ around $p_j$, you use the opposite orientation on $\partial B_j$ given by the outer normal (by definition of index, chosen an oriented chart $(U,\phi)$ in $p_j$, the index of $v$ around the isolated zero $p_j$ is the degree of the map $\overline{v}=v/||v||$ defined from $\partial B_\epsilon$ to $S^{m-1}$ where $B_\epsilon$ is a small ball centered in $p_j$ oriented as the boundary of $B_\epsilon$, by the rule of the outer normal). 

*To conclude you observe that the degree of $\overline{v}$ has two contributions: the degree of $\overline{v}|_{\partial X}$ and the sum of the degrees of $\overline{v}|_{\partial B_j}$, it is easy to see that $\overline{v}|_{\partial X}$ is homotopic to $g$ so they have by a previous result the same degree, on the other hand $deg(\overline{v}|_{\partial B_j})=-ind(v,p_j)$ because of the different orientations on $\partial B_j$, so we have
$$
0=deg(\overline{v})=deg(\overline{v}|_{\partial X})+\sum_{j=1}^n{deg(\overline{v}|_{\partial B_j})}=deg(g)-\sum_{j=1}^n{ind(v,p_j)}
$$
I hope that answers your questions
