Euler characteristic of closed surface

Question

Assume that you have a closed surface that can be covered by finitely many triangles. Then $K(p)= 6-val(P)$ where P is a vertex and $val(P)$ the number of edges that lead to this vertex. Now, I am supposed to show that $ \sum_{vertices \ p}K(p) = 6 \chi$ where $\chi$ is the Euler-characteristic of the closed surface. Now I know that this is $6\cdot(2-2g)$, but I just don't see the combinatorial argument, why these two things must be the same. I.e. I don't see how the genus is related to the valence.

Answer 1

$$\sum_{v\in V}K(v)=\sum_{v\in V}(6-val(v))=6|V|-\sum_{v\in V}val(v)$$ Since for every edge there are $2$ vertices $$\sum_{v\in V}val(v)=2|E|,$$ hence $$\sum_{v\in V}K(v)=6|V|-2|E|$$ Moreover, since every edge is the face of $2$ triangles and every triangle is bounded by $3$ edges you know that $$3|T|=2|E|\qquad\Rightarrow\qquad 3|T|-2|E|=0$$ Hence $$\sum_{v\in V}K(v)=6|V|-2|E|=6|V|-2|E|+2\left(3|T|-2|E|\right)=6\left(|V|-|E|+|T|\right)=6\chi$$