Proof
Suppose we have a closed surface M2 in 3-space and we divide it up into triangles Tj, where j = 1, 2, ..., #T. For any individual triangle, the total Gaussian curvature is given by the Gauss-Bonnet Theorem:
 TΚdA +  Tκgds +  ext.angles = 2π
To find the total Gaussian curvature over the entire surface, we sum over all of the triangles.
M2ΚdA + j[Tjκgds + ext.angles] = j2π
Along a shared edge between two triangles, the lines
circulate in opposite directions, which causes the
line integrals of geodesic curvature to cancel each
other out. Therefore,
jTjκgds = 0
To justify the above statement, we add that since the surface itself is closed, it has no boundary, and therefore, the integral of geodesic curvature over the boundary of the entire surface equals zero. In any case, this leaves us with the following equation:
M2ΚdA + jext.angles = 2π(#T)
For any triangle, we know that
ext.angles = 3π - int.angles,
where the last term denotes the sum of the interior angles. So,
M2ΚdA + 3π(#T) - jint.angles = 2π(#T)
The sum of the interior angles for all of the triangles can be written in terms of the total number of vertices in the triangulation by using the fact that the sum of the interior angles at any given vertex is equal to 2π. This means that
M2ΚdA + 3π(#T) - 2π(#V) = 2π(#T)
In our final step, we observe the relationship between the total number of triangles in the triangulation and the total number of edges. Each triangle has 3 edges and there are #T triangles. This means that together there are 3(#T) edges for all of the triangles. But, since each edge is shared between two triangles, we are actually double counting every edge. Therefore, the number of edges is given by
#E = (3/2)#T
Substituting this last expression into our equation for the total Gaussian curvature of the surface yields the desired result.
M2ΚdA = 2π(#V) - 2π(#E) + 2π(#T), or
M2ΚdA = 2πχ(M2)
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