Gauss Bonnet theorem, degree of surface with genus I'm reading Frankel's The Geometry of Physics and I have been trying to understand the proof of the Gauss Bonnet theorem but I'm stucked with one part of the proof that is left as an exercise.
I have arrived until
$$\frac{1}{4\pi}\int\int K dS = deg(n: M^{2} \rightarrow S^{2})$$
with $K$ the curvature and the right part the (Brouwer) degree of the Gauss normal map. What I don't know is how to prove that the degree of the map in case of a surface of genus $g$ ($g$ holes) would be $1 - g$, so:
$$\frac{1}{4\pi}\int\int K dS = 1 - g$$
As an extra, could you suggest a reference to read a proof of the Gauss Bonnet for physicists? Thanks.
 A: It is very simple. For clarity I’ll use  g = 3 and draw everything in rectangular shape.Picture for demonstration The  frame (in red) with 3 rectangular cavities can be reshaped differentiably ( perhaps except for the edges) from the original  surface of g = 3 and the genus remains unchanged. Place a cube (in black) which encloses the three cavities but surrounded by the outermost boarder of the frame. Apparently the cube can be reshaped differentiably from the unit sphere (perhaps except for the edges as well) and is also of genus 0.
Next construct the map $\phi$ from the frame to the cube surface:
1.Project the vertical surfaces of the frame horizontally to the corresponding sides of the cube, e.g. the left surfaces of the frame $L_{0}, L_{1}, L_{2}, L_{3}$ should be projected to the $S_{L}$ surface of the cube, and the right surfaces $R_{0}, R_{1}, R_{2}, R_{3}$ should be projected on  $S_{R}$, etc.
2.Map the horizontal surfaces of the frame to the upper and lower sides of the cube (it is a simple affine transform and will of course leave holes on the cube).
Now we have mapped the frame differentiably (except for the edges, which can be ignored here) to the surface of the cube. Take a point $y\in S_{L}\bigcap\phi(L_{0})\neq\emptyset$, apparently there is a unique $x_{i}$ on each $L_{i}$, $i=0,1,2,3$, such that $\phi(i)=y$. Apparently the orientation at $x_{0}$ on $L_{0}$ is positive, and the orientation at $x_{i}$ on $L_{i}$ ($i=1,2,3$) is negative because they point inward. 
 Therefore $\text{sign}\phi(x_{0})=1$, and $\text{sign}\phi(x_{i})=-1$, $i=1,2,3$. 
Therefore $\text{deg}(\phi)=1-3=-2$. For general g you can do the same thing. Q.E.D.
It in fact describes how the surfaces $L_{0}, L_{1}, \ldots L_{g}$ change their orientation when mapped on $S_{L}$. Of course you can construct a formal argument, but here is the geometric picture that should be born in mind.
A: Following up on Ted Shifrin's nice comment, trying to make it general:
Pick a point $\vec{N}$ in $S^2$ at which both the Gauss map $n$ AND it's negative $-n$ are regular (this exists by Sard's theorem*). The degree of $n$ is the signed number of preimages of $\vec{N}$ in $M^2$. Now define a smooth section $s$ of $TM$ (a vector field tangent to $M$) by projecting $\vec{N}$  to $TM_p$ at every $p\in M$. This vanishes precisely when $n(p)=\pm \vec{N}$. One checks that $\pm \vec{N}$ being regular value is equivalent to $s$ being transverse to the zero section, and that the sign of the $p \in im(s)\pitchfork 0$ is the same as that of $p \in n^{-1}(\pm\vec{N})$. We conclude that the signed number of zeroes (aka the Euler number) is twice the degree of $n$. So $2-2g=2deg(n)$, or $deg(n)=1-g$ as wanted.
*Either note that the set of critical points of $n$ is of zero measure and so it's union with it's central reflection also is of zero measure; or apply Sard directly to the composition of $n$ with the projection to $\mathbb{R}P^2$.
