Let $S$ a compact and connected surface in $\mathbb{R}^3$, such as is not diffeomorphic to the sphere. I want to prove that it has point with positive, negative and $0$ Gauss curvature.
My work: I know that i have to use the global Gauss-Bonett theorem and the fact that there must be some invariant that is different from the sphere, since it's not diffeomorphic to it. But from here i don't know how to procede. Thanks in advance!
Remember that the Euler's characteristic of a compact and orientable surface is given by $2-2g$, where g is the genus of the surface (intuitively the number of holes of the surface). Since the surface is not diffeomorphic to the sphere, then $g\not=0$. Since $0\leq g$, it implies that $g>0$, so $\chi(S)=2-2g\leq0$. The Gauss theorem says that $\int_{S} K=2\pi\chi(S)$ for a compact orientable surface, hence $\int_{S} K\leq0$.
For being $S$ compact, it has at least one point $p$ with positive curvature, so it implies that there must be another $q$ point of negative curvature to "compensate" the value of the integral. From the continuity of the curvature function and the fact that the surface is connected there must be a point with curvature equal to $0$(applying the Bolzano's theorme to the restriction of $K$ to a curve that connects the $p$ and $q$).