Gauss–Bonnet theorem

Triangulate $M$ using geodesic triangles (triangles whose sides are shortest paths on the surface), obtaining $V$ vertices, $E$ edges and $F$ faces, related by $\chi(M) = V - E + F$.

The local Gauss–Bonnet formula for a single geodesic triangle $T$ with interior angles $\alpha,\beta,\gamma$ states $\int_T K\,dA = \alpha+\beta+\gamma-\pi$: the amount the angle sum exceeds the Euclidean value $\pi$ equals exactly the integral of curvature over that triangle, a fact obtained by applying Stokes' theorem to the rotation of a tangent vector parallel-transported around the triangle's boundary.

Summing $\int_T K\,dA = \alpha+\beta+\gamma-\pi$ over all $F$ triangles gives $\int_M K\,dA = 2\pi V - \pi F$: the triangles meeting at each of the $V$ vertices sweep out exactly one full turn $2\pi$ there, so all their angles together sum to $2\pi V$, while each of the $F$ triangles contributes a $-\pi$.

Each triangle has $3$ edges and each edge is shared by exactly $2$ triangles, so $3F = 2E$. Substituting $E = \frac{3F}{2}$ into $\chi(M) = V - E + F$ gives $2\pi\chi(M) = 2\pi V - \pi F$, which matches the previous step exactly. Hence $\int_M K\,dA = 2\pi\chi(M)$, proving the theorem.