Analytic map of compact Riemann surfaces is constant or surjective

Statement

Suppose $M$ is a compact Riemann surface and $N$ is a Riemann surface (note: Riemann surfaces are assumed to be connected), and $f:M\to N$ is an analytic map. Then, two possibilities arise:

• $f$ is a constant map
• $f$ is surjective. In this case, $N$ is also forced to be compact.

Proof

Proof outline

• By the open mapping theorem, $f$ is either constant, or an open map
• Thus, the image of $M$ in $N$ is open
• Since $f$ is an analytic map, $f$ is continuous
• Since $M$ is compact and $N$ is Hausdorff, the image of $M$ is $N$ is closed.

Further information: tps:Compact to Hausdorff implies closed

• $f(M)$ is thus both open and closed in $N$. Also, it is nonempty, and $N$ is connected. Thus, $f(M)=N$