I read that the degree of the gauss map for a $M$ compact orientable $2-$manifold (connected to use the fact that those are only the $g-$torus) in $\mathbb{R}^{3}$ should be $(1-g)$, which is $\frac{1}{2} \chi(\Sigma_g)$, where $\Sigma_g$ is the genus $g$ surface; which implies, among other things, the surjectivity of the gauss map when $g \ne 1$.
It should be a corollary of this.
Is there a proof of the fact which involves degree theory ? I found this paper but it uses form-theory which I don't know yet.
Edit : I've been thinking about the problem since a while now and I realize that the easiest way to do it is actually produce a tangent outward to the boundary vector field with a proper number of isolated zeros,maybe $g-1$ saddle, which sums up to $\frac{1}{2}\chi(\Sigma_{g}) = 1-g$ and use the Hopf's Lemma in the following form :
Hopf's Lemma : Let $M$ be a compact $n-$manifold in $\mathbb{R}^{n}$ with boundary, and let $X$ be a smooth vector field with isolated zeros that points outward to the boundary, i.e $\forall \hspace{0.1cm} > p \in \partial M, X(p) \in T_p M - C_p M$, then $\sum\limits_{X(p) = > 0} \text{ind}_X(p) = \text{deg}(\nu)$ where $\nu$ is a gaupp map for the boundary of $M$.
The definition of degree I'm using is : Let $M$ be compact $N$ connected with dim$M=$dim$N$, both without boundary, orientable and oriented. Take $f : M \longmapsto N$ of class $C^{\infty}$ then deg($f$) := $\sum\limits_{x \in f^{-1}(y)} \text{sgn}df_x$ for $y \in \text{RegVal}(f)$.
The index of an isolated zero $p$ is defined as the pullback through a parametrization of the manifold of the index on an open set of $\mathbb{R}^{n}$ so I'm going to give this definition : $\text{ind}_X(p) := \text{deg}(f_{r})$, where $f_r : \partial B(p,r) \longmapsto \mathbb{S}^{n-1}$ with $f_r(q) := \frac{X(q)}{\lvert \lvert X(q) \rvert \rvert}$ for every $r \in (0,\epsilon)$ given $\epsilon$ the radius of the ball where there are no other zeros.
Any help on how to costruct the vector field would be appreciated
