I have seen it claimed that a non-constant holomorphic map $f:X \rightarrow Y$ between compact connected Riemann surfaces is a branched covering i.e. surjective and there is a finite set $\Sigma \subset Y$ and $r \in \mathbb{Z}_+$ such that $|f^{-1}(q)|=r$ for all $q \in Y \setminus \Sigma$. I can see why such a map is surjective, but I don't understand why the rest of the statement is true.
Complex Analysis – Holomorphic Map Between Compact Connected Riemann Surfaces
complex-analysisriemann-surfaces
Related Solutions
First, in your definition you should say that $S$ is closed with empty interior (instead of $S$ being dense which is surely not what you meant).
Next, I also do not think there is a standard terminology in general. The very least people require is that $f: X\to Y$ is a local homeomorphism away from some nowhere dense closed subset $Z$ of $X$. Nobody (as far as I know) requires $Z$ to be finite (unless dealing with compact Riemann surfaces). The definition you gave (with my correction) probably will not be accepted by most algebraic geometers since it would allow a blow-down map to be regarded as a branched covering.
To remedy this, one can make the following definition (motivated by the notion of the orbifold covering):
Definition. An open continuous map $f: X\to Y$ of topological spaces is called a branched covering if the following conditions hold:
There exists an open and dense subset $A\subset Y$ such that $f| f^{-1}(A)\to A$ is a covering map.
For every point $x\in X$ there exists an (open) neighborhood $V_x\subset X$ and a finite group of homeomorphisms $G_x$ acting on $V_x$ and a homeomorphism $U_x=f(V_x) \to V_x/G_x$ which makes the following diagram commutative $$ \begin{array}{ccc} V_x & \stackrel{id}{\to} &V_x\\ f\downarrow & ~ & \downarrow\\ U_x & \to & V_x/G_x \end{array} $$
In other words, locally, $f$ is modelled on the quotient by a finite group action.
Edit. One last thing: depending on the category you are working in, all maps and group actions should be from this category. For instance, if you are working in the category of complex manifolds, all maps should be holomorphic (or biholomorphic when needed) and group actions should be holomorphic too.
Here you have to use two facts:
- The $\textit{Open mapping theorem}$ tells you that if $f$ is non-constant, then it’s an open map, so that $Im(f)$ is an open set on $Y$. (The open mapping theorem from complex analysis carries over to Riemann surfaces basically immediately)
- The image $Im(f)$ is also closed on $Y$ because if $f(x_n)\to y$ then $\{x_n\}$ is a sequence on the compact space $X$, so admits a subsequence (that we denote always with $\{x_n\}$) such that $x_n\to x$. By continuity of $f$ you get $f(x_n)\to f(x)$ and $f(x_n)\to y$. However $Y$ is Hausdorff, so the limit has to be unique, that means $y=f(x)$.
Now $Im(f)$ is open, closed and non-empty, so it has to be $Im(f)=Y$ (remember that $Y$ is a connected space)
Best Answer
First of all, your definition of a branched covering (between surfaces) is incomplete. The correct definition is: It is a map $f: X\to Y$ such that there exists a finite subset $W\subset Y$ such that for $Z:=f^{-1}(W)$, the restriction $$ f|_{X - Z}: X- Z\to Y- W $$ is a covering map.
To prove that every nonconstant holomorphic map between compact connected Riemann surfaces is a branched covering, let $W\subset Y$ denote the set of critical values of $f$ (i.e. points $w\in Y$ such that there exists $z\in f^{-1}(w)$ so that $f'(z)=0$; such $z$ is called a critical point of $f$). Then observe that since $X$ is compact and $f$ is nonconstant, it has only finitely many critical points (otherwise, the set of critical points has an accumulation point in $X$ which is impossible). Now, the restriction $f|_{X- Z}$ is a local diffeomorphism. It is also a proper map, i.e. preimages of compact subsets in $Y -W$ are compact. (This follows easily from compactness of $X$.) Furthermore, by the maximum principle, nonconstant holomorphic maps are open. Since $X$ is compact, $f(X)$ is closed in $Y$; since $Y$ is connected, it follows that $f(X)=Y$. Thus, by the construction, $f|_{X- Z}: X- Z\to Y- W$ is surjective. Lastly, applying Ehresmann's "Stack of Records" theorem, we conclude that
$f|_{X- Z}: X- Z\to Y- W$ is a covering map.