Proper map of Riemann surfaces

Question

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?

Answer 1

Here you have to use two facts:

1. The $\textit{Open mapping theorem}$ tells you that if $f$ is non-constant, then it’s an open map, so that $Im(f)$ is an open set on $Y$. (The open mapping theorem from complex analysis carries over to Riemann surfaces basically immediately)
2. The image $Im(f)$ is also closed on $Y$ because if $f(x_n)\to y$ then $\{x_n\}$ is a sequence on the compact space $X$, so admits a subsequence (that we denote always with $\{x_n\}$) such that $x_n\to x$. By continuity of $f$ you get $f(x_n)\to f(x)$ and $f(x_n)\to y$. However $Y$ is Hausdorff, so the limit has to be unique, that means $y=f(x)$.

Now $Im(f)$ is open, closed and non-empty, so it has to be $Im(f)=Y$ (remember that $Y$ is a connected space)