Consider a proper holomorphic map $f:X\to Y$ between two (connected, but not necessarily compact) Riemann surfaces. Is it true that $f$ is surjective whenever it is non-constant?
In a lecture about Riemann surfaces, we proved the following Proposition:
A proper, non-constant, holomorphic, unramified map $f:X\to Y$ between two connected Riemann surfaces induces a covering map between topological spaces.
However, when proving this result, we did not explicitly check whether $f$ is surjective, which is a necessity in order for $f$ to be a covering map.
Does surjectivity of $f$ need to be added as an assumption to the Proposition, or is it a consequence of the already present assumptions?
Best Answer
Here you have to use two facts:
Now $Im(f)$ is open, closed and non-empty, so it has to be $Im(f)=Y$ (remember that $Y$ is a connected space)