Action of a group on a Riemann surface If $G$ is a finite group of order $n$, and acting on a compact Riemann surface $X_{g'}$ ($g'$ is genus), then $X_{g'}/G$ is another compact Riemann surface, $X_g$, of genus $g$.
Also then $X_{g'} \rightarrow X_g$ is a branched covering. Are the regular covering space of $X_g$ lie between $X_{g'}$ and $X_g$?  
The numbers $g,g',n$ are related by
$2g-2=|G|.{(2g'-2)+|G|.\sum (1-\frac{1}{n_x})}$,
where $n_x$ is the branching order at branch point $x\in X$. Can one suggest a topological proof of this relation? (I want to look at the problem of symmetries of Riemann surface, in topological point of view.)  
 A: Another derivation can be given using the Lefschetz fixed point formula. Let $X$ be a Riemann surface, $G$ be a finite group acting faithfully on $X$. All cohomology groups are with complex coefficients. Then $G$ acts on $H^i(X)$, and the Lefschetz number of $f$ is 
$$L_f:= \sum_{i=0}^{2} (-1)^i Tr (f|_{H^i (X)})$$.
You need two rather difficult results, which are very useful in the study of group actions.


*

*The first is the Lefschetz fixed point formula (it is nicely discussed in the books by Bredon (Topology and Geometry) and also Greenberg-Harper (Lectures on Algebraic topology)). As far as I remember, there is also a proof in Farkas-Kra (for surfaces, which is what you want).
The fixed point formula asserts that for $f \neq id$, $L_f$ is the number $|F_f|$ of fixed points of $f$. 

*Then we need the knowledge that $H^i(X/G)=H^i(X)^G$ (the space of invariants). This is not easy. You can use the Hodge presentation of cohomology ($X \to X/G$ is holomorphic and it has to be injective in cohomology by looking what it does to a holomorphic $1$-form; any $G$-invariant holomorphic $1$-form on $X$ descends to one on $X/G$).
If you are comfortable with homological algebra, and you can find an alternative proof in a more general setting here:
Euler characteristic of orbifolds
Now we start the real proof. From elementary representation theory of finite groups, you deduce that
$$|G|\chi(X/G)=|G|\sum_{i} (-1)^i dim (H^i (X)^G) = \sum_{f \in G} L_f.$$
Using the fixed point formula and the obvious identity $L_{id}= \chi(X)$), write
$$\sum_{f \in G} L_f = L_{id} + \sum_{f \neq id} L_f = \chi(X) + \sum_{f \neq id} |F_f|.$$
Let $S \subset X$ be the union of the fixed point sets of all $id \neq f \in G$. This is finite, and let $G_p$ be the stabilizer subgroup at $p \in S$. A simple counting gives
$$\sum_{f \neq id} |F_f| = \sum_{p \in S} (|G_p|-1).$$
Hence 
$$|G|\chi(X/G)=\chi(X) + \sum_{p \in S} (|G_p|-1).$$
Reinterprete this formula in terms of branching numbers and you are done.
A: Set $X:=X_{g'}$ and $Y=X_g$, and let $X^0$ and $Y^0$ be the open sets where the covering $X \to Y$ is unramified. If $e$ denotes the topological Euler number, we have
$e(X^0)=n \cdot e(Y^0)$,
$e(Y)= e(Y^0)+ \sum_x 1$,
$e(X)= e(X^0) + \sum_x \frac{n}{n_x}$,
where the sums extend over the points with non-trivial branching order.
Using these relations, we obtain
$e(X)=n \cdot e(Y)+ n \cdot \sum (\frac{1}{n_x}-1)$,
and since 
$e(X)=2-2g(X), \quad e(Y)=2-2g(Y)$ 
we are done.
