I read the book of Miranda on Algebraic curves and Riemann surfaces, I read the proof of Riemann-Hurwitz formula page 52-53. I mean for holomorphic map $F:X\to Y$ between Riemann Surfaces, we have $$\chi(X)=d(F)\cdot\chi(Y)-R_F$$ with $\chi(X)$ is the Euler characteristic and $d(F)$ is the degree of $F$ and $R_F=\sum_{p \in X} (e_p(F)-1)$.
This result is proved in three steps:
First, we prove the set of ramification points is finite (easy), so $R_F$ is a finite sum. And we take a triangulation of $Y$, such that each branch point of $F$ is a vertex. I don't understand this, is this because any triangle is triangulated? I want to prove it formally.
Step 2: Assume there are $v_Y$ vertices, $e_Y$ edges, and $t_Y$ triangles. Lift this triangulation to $X$ via the map $F$ (how we can do it?), and there are $v_X$ (resp. $e_X , t_X$) vertices (resp. edges, triangles) on $X$. Since there are no ramification points over the general point of any triangle, each triangle of $Y$ lift to $\deg(F)$ triangle in $X$ (why, really two days to understand that and I can't). Thus $t_X = t_Y . d(F)$ Similarly $e_X =e_Y .d(F)$. Now fix a vertex $q\in Y$. The number of preimages of $q$ in $X$ is $F^{-1}(q)$, which we can rewrite $$|F^{-1}(q)|=d(F)+\sum_{p\in F^{-1}(q)}1-e_p (F).$$ Therefore the total number of preimages of vertices of $Y$, which is the number $v_X$ is $$v_X = \sum_{vertex \ q \ of \ Y}(d(F)+\sum_{p\in F^{-1}(q)}1-e_p (F))=d(F)\cdot v_Y - \sum_{vertex \ q\ of \ Y}\sum_{p \in F^{-1}(q)}e_p(F)-1)\\ = d(F)\cdot v_Y - \sum_{vertex \ p \ of \ X}e_p(F)-1$$ Also, I think the last inequality is true because the set of vertex $p$ of $X$ is the set of preimages of $q$ vertex in $Y$. It's true?
Last step is to calculate $\chi(X)$, we have $$\chi(X)=v_X -e_X + t_X=d(F)v_Y - d(F) e_Y + d(F)\cdot t_Y-\sum_{p \ vertex \ of \ X}e_p (F)-1,$$ I think last inequality holding because every ramification point of $F$ is a vertex of $X$.