holomorphic map between compact Riemann surfaces

Question

Why nonconstant holomorphic map between compact riemann surfaces is surjection? I don't understand: if open-closed connected subset X of connected space Y, then X is all Y??

Answer 1

Remember that nonconstant holomorphic maps are open (any open set of a compact Riemann surface is a union of open sets that are each holomorphically equivalent to an open disc in the complex plane; then use the open mapping theorem and the fact that taking images under a function preserves unions), and that continuous maps from a compact space to a Hausdorff space are closed. Hence the image $f(X)$ is both open and closed in $Y$. If your definition of compact Riemann surface includes connectedness, then the image must be all of $Y$.