It is of course completely standard that closed orientable surfaces have even Euler characteristic. What is the most elementary proof of this?


More specifically, suppose I have a finite simplicial complex $K$ with vertices $V$, edges $E$ and faces $F$. I suppose that each edge is contained in precisely two faces, and that the link of each vertex is a cycle of size at least three. This implies that $|K|$ is a closed surface, and it is not hard to see that $3|F|=2|E|$, so $|F|$ is even. We want to show that $|V|-|E|+|F|$ is even, or equivalently that $|V|=|E|\pmod{2}$. This is not true in general if $K$ is not orientable, so we need to say something about orientations.


Let $D$ be the set of directed edges, and let $\chi\colon D\to D$ reverse direction. Let $S$ be the set of permutations $\sigma\colon D\to D$ such that


* For any $(u,v)\in D$ we have $\sigma(u,v)=(v,w)$ and $\sigma(v,w)=(w,u)$ and $\sigma(w,u)=(u,v)$ for some $w$ such that $\{u,v,w\}\in F$
* If $u,v,w$ are as above, and $x$ is the other vertex such that $\{u,v,x\}\in F$, then $\sigma(v,u)=(u,x)$ and $\sigma(u,x)=(x,v)$ and $\sigma(x,v)=(v,u)$.


We find that $S$ bijects with the set of orientations of $|K|$. Also, we have $\sigma^3=1$ and we can describe the composite $\rho=\sigma\chi$ as follows: the edge $\rho(u,v)$ starts at $u$, and is the next edge round $u$ after $(u,v)$ in clockwise order.


Given $K$ and $\sigma$ as above, it seems that there should be some very direct combinatorial argument to show that $|V|=|E|\pmod{2}$, but I am not seeing one.



Closed orientable surfaces have even Euler characteristic
The set of vertices can be split as the union $V=V\_{\text{odd}}\cup V\_{\text{even}}$ of vertices with odd and even degrees, respectively. Since the sum of degrees over all vertices is the same as twice the number of edges, we know that $|V\_{\text{odd}}|= 0\pmod 2$. Therefore we want to establish that $|V\_{\text{even}}|=|E|\pmod 2$.


The permutation $\sigma$ is the product of $|F|$ cycles of length 3, so $\operatorname{sign}(\sigma)=1$. The permutation $\chi$ is the product of $|E|$ cycles of length 2 so $\operatorname{sign}(\chi)=(-1)^{|E|}$. And finally the permutation $\rho$ is the product of $|V\_{\text{odd}}|$ cycles of odd length and $|V\_{\text{even}}|$ cycles of even length, therefore $\operatorname{sign}(\rho)=(-1)^{|V\_{even}|}$. Since $\rho=\sigma \chi$ we must have
$$\operatorname{sign}(\rho)=\operatorname{sign}(\sigma)\operatorname{sign}(\chi)\implies |E|=|V\_{\text{even}}|\pmod 2$$


