X → Y Between Riemann Surfaces

1. Riemann surfaces All the Riemann surfaces are assumed to be connected.

A continuous map f : X → Y between Riemann surfaces is holomorphic if for every complex coordinate system (U, z) on X and every complex coordinate system (V, w) on Y with U ∩ f −1(V ) 6= ∅, the mapping w ◦ f ◦ z−1 : z(U ∩ f −1(V )) → w(V ) is holomorphic. A holomorphic mapping f : X → C is called a holomorphic function while a holomorphic 1 mapping f : X → P = C ∪ {∞} is called a meromorphic function. Let U be any open subset of a Riemann surface X. Then U has a structure of Riemann surface; the notion of holomorphic functions on U are deﬁned. The set of all holomorphic functions on U is denoted by OX (U) while the set of all meromorphic functions on U is denoted by MX (U). Proposition 1.1. (Open Mapping Theorem) Let f : X → Y be a nonconstant holomorphic mapping between Riemann surfaces. Then f is an open mapping.

Proof. It follows from the fact that any holomorphic function F : D ⊂ C → C on an open set D of C is an open mapping. Proposition 1.2. (Coincidence Principle) Let f, h : X → Y be nonconstant holomorphic mapping between Riemann surfaces. If f = h on a subset S of X with a limit point in X, then f = h. Proof. We consider the subset W of X consisting of points x such that f(x) = h(x). Let U be the set of all interior points of W. Then U is an open subset of X. We can show that U contains the limit point of S and hence U is nonempty. Then we show that U is an open subset of X. We ﬁnd U is a nonempty and open and closed in X. By connectedness of X, the only nonempty closed and open subset of X is X itself. We ﬁnd U = X. Hence f = h on X. The proof is the same as that in the case when X is a region in C. The detail of the proof will be omitted here. Please see any textbook of complex analysis. Theorem 1.1. Let X and Y be Riemann surfaces with X compact. Suppose f : X → Y is holomorphic. Then f is either constant or surjective. In the later case, Y is also compact. Proof. Suppose f is not a constant function. Since f is a non constant holomorphic function, f is an open mapping. Hence f(X) is an open subset of Y. Since X is compact and f is continuous, f(X) is also compact and hence is closed in Y. Since f is non constant, f(X) is not empty. By connectedness of Y, f(X) = Y. (The only nonempty closed and open subset of Y is Y itself.) This implies that Y is also compact. Corollary 1.1. Any holomorphic function on a compact Riemann surface is a constant function.

Proof. Let f : X → C be a holomorphic function. Since C is connected and not compact, by the previous theorem, f must be a constant.