Let \(X\) and \(Y\) be connected Riemann surfaces with \(X\) compact, and let \(f\colon X\to Y\) be a nonconstant holomorphic map. Then
\(f\) is surjective,
\(Y\) is compact,
\(\inverseof{f}(q)\) is finite for every \(q\in Y\), and
the ramification locus of \(f\) in \(X\) and the branch locus \(f(\text{ramification locus})\) in \(Y\) are finite.
By the open mapping theorem, \(f(X)\) is open in \(Y\); as the continuous image of a compact space in a Hausdorff space it is also compact, hence closed. Since \(Y\) is connected and \(X\neq\emptyset\), \(f(X)=Y\), so \(f\) is surjective and \(Y=f(X)\) is compact. For \(q\in Y\), the fibre \(\inverseof{f}(q)\) is closed in \(X\), and it is discrete because a nonconstant holomorphic map on a connected Riemann surface has isolated preimages of each point; a closed discrete subset of a compact space is finite. In a local coordinate the ramification points are the zeros of the derivative of a nonconstant holomorphic function, so the ramification locus is closed and discrete in \(X\), hence finite, and its image, the branch locus, is finite.