Let $M$ be a compact manifold, then I am trying to prove that if $\chi(M)=0$ then $M$ admits a nowhere vanishing vector field. To do this I am told to use the following 2 facts
If $\phi : S^{n-1}\rightarrow S^{n-1}$ has degree zero then it's homotopic to a constant map.
Given a finite collection of points $C=\{p_1,...,p_N\}$ we can find a diffeomorphism which sends $C$ to an open ball in $M$.
Now my attempt is the following : So we know that $M$ admits vector field $X$ with a finite number of zeros , $p_1,..,p_N$, and by the second statement we can assume that they are contained in an coordinate ball $(U,\phi)$. Now by the Poincare-Hopf theorem we know that $\chi(M)=\sum_{i=1}^{N}Ind_{p_i}X=0$. By definition we have that $Ind_{p_i}X=Ind_{\phi(p_i)}\phi_*X$ and this will be the degree of the map $G: S^{n-1}\rightarrow S^{n-1}$ that sends $p\rightarrow \frac{\phi^{-1}_* X(p)}{||\phi^{-1}_* X(p)||}$. Now we have that for every $p_i$ this map $G$ is the same and so we get that $Ndeg(G)=0$ and so $deg(G)=0$. This tells us that $G$ is homotopic to a constant and so we have an extension $G':\phi(U)\rightarrow S^{d-1}$, with this we construct the vector field that is $X$ outside $U$ and is $G'\circ \phi$ in $U$. (I am assuming the homotopy is smooth, but I guess an argument of differential topology would fix this problem if it wasn't).
What do you guys think ? Thanks in advance.
