Closed curve divides into two or more regions compact and orientable surfaces with positive gaussian curvature

Question

Proposition. Every closed curve divides into two or more regions every compact and orientable surface with positive gaussian curvature.

I need to prove this sentences using Gauss-Bonnet theorem, but I'm really lost. Can someone give me a hint?

Answer 1

Assume the curve is connected, we will prove that it divides in at least two components.

If not the complement $S\setminus C$ is connected. Cut $S$ along $C$ and adding two copies of $C$ we find a closed oriented surface with boundary $C^{\pm}$. Consider two copies of this surface and glue them isometrically , gluing $C_1^+$ with $C_2^-$ , and $C_2^+$ with $C_1^-$(with obvious notations). The result is a closed connected riemanian surface, and the integral of the Gaussian curvature is twice the integral of the Gaussian curvature on $S$, ie. $4.2\pi$, with Euler characteristic 4, which do not exist.

In fact Gauss Bonnet is just used to prove the well known topological theorem of additivity of the Euler characteristic $\chi (A \cup B)=\chi (A)+\chi(B)-\chi (A\cap B)$, for open sets.