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 [ f1g 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. 1.