I wanted to understand the proof of this theorem:
If $M$ is a compact manifold and $\chi(M)=0$, then there is a non-vanishing vector field in M.
I was looking at a post in MathOverflow about this topic, and there is a comment in the accepted answer that says
Two zeroes of opposite index can be taken to lie in a coordinate chart (because the manifold is connected) and then they can be eliminated using that fact that a map $S^{n-1}\to S^{n-1}$ of degree zero extends to $D^n$.
I understand why the first part of the comment is true, but I'm not sure as to why a map having degree zero allows it to be extended from the boundary to its interior. Furthermore, how does this extension help us in eliminating the zeroes inside the ball?
A proof is rather lengthy, but you can regard this as a special case of Corollary 4.25 in
Hatcher, Allen, Algebraic topology, Cambridge: Cambridge University Press (ISBN 0-521-79540-0/pbk). xii, 544 p. (2002). ZBL1044.55001.
This corollary states that the degree map $\pi_n(S^n)\to {\mathbb Z}$ is an isomorphism. In particular, if a map $f: S^n\to S^n$ has zero degree, then it is null-homotopic, equivalently, extends to a map $B^{n+1}\to S^n$.