Show that two simple and closed geodesics $\gamma_1,\gamma_2$ intersects each oter in a compact, connected surface S with gaussian curvature $K>0$ Sketch of proof: We know that for $K>0$ we have $S$ homeomorphic to sphere. Suppose by absurd than $\gamma_1,\gamma_2$ (the two geodesics) doesn't intersect. Now, by hipotesis, $\gamma_1,\gamma_2$ form  the boundary of a region cause they closed and does not have auto-intersections(simple).
With Gauss-Bonnet, we have $$
\underset{R}{\int\int}Kd\sigma=2\pi\chi(R)=0
$$
Because $\gamma_1,\gamma_2$ are the boundary of region and the Euler characteristic is $\chi(R)=0$.
But this is an absurd, cause $K>0$.
My question: I can't understand why $\chi(R)=0$.
The idea that i have is the region $R$ is compact, connected, then the Euler characteristic is $\{2,0,-2,..\}$ with $2$ if they homeomorphic to sphere. But for this idea (that i'm not know even it's right), i have to garantee that $R$ isn't $\chi(R)<0$ and i don't even know how to do.
 A: Same idea, just a bit different perspective:
Consider a closed simple geodesic on a closed surface homeomorphic to the sphere.  It may not cut the surface into two parts of equal area, but if we consider the weighted area ( multiply the area form by $K$) then the two parts have equal "area". Indeed, apply Gauss-Bonet to both pieces, that have a common boundary ( a geodesic).
Now, assume moreover that $K>0$. Then if the two geodesic would not intersect, while each cut the surface into equal parts, we would get the "area" between the curves to be $0$, contradiction.
Obs: this would be the same with Gauss-Bonet for the region, but it's not intuitive. The intuition: a closed simple geodesic cuts the surface into two "equal" parts.
$\bf{Added:}$ Let's explain why the Euler characteristic is $0$. We need to divide the region which is topologically the lateral surface of a cylinder into several polygonal regions. Convenient choices can be made. Then count
$$\chi(R) = v - e + f$$
where $v$ is the number of vertices, $e$ the number of edges, and $f$ the number of faces. So cut the region $R$ into two rectangles by drawing two curves joining the top and bottom geodesics ( these curves do not need to be geodesic, just to be simple, and without intersections). We have $v = 4$ ( two on each geodesic), $e= 2 + 2 + 2$ ( $2$ on each geodesic, and the $2$ shunts), and $f=2$. Now we see
$$\chi(R)= 4 - 6 + 2 = 0$$
