# Proper map of Riemann surfaces

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?

• The Proposition seems to be missing a hypothesis, i.e. nonvanishing derivative. For example, the formula $f(z)=z^2$ defines a proper, nonconstant holomorphic map $f : \mathbb C \to \mathbb C$ that is surjective but is not a covering map. Nonetheless, your question about surjectivity is valid even without the assumption of nonvanishing derivative. Commented May 10, 2022 at 15:06
• @LeeMosher Good point! I added that $f$ is unramified.
– Zuy
Commented May 10, 2022 at 15:11

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)

• Why is $X$ compact?
– Zuy
Commented May 10, 2022 at 11:57
• @Zuy Usually the definition of a Riemann Surface X contains the requests that X is also compact 😀 Commented May 10, 2022 at 14:00
• I see. For us, compactness is not part of the definition unfortunately. Still, your answer is quite nice, thanks for that.
– Zuy
Commented May 10, 2022 at 14:43
• Your answer can easily be promoted to answer the actual question. If $f(x_n) \to y$ then, choosing $B \subset Y$ to be a closed and therefore compact ball centered on $y$, it follows from properness of $f$ that $f^{-1}(B)$ is a compact subset of $X$, and that subset contains all but finitely many terms of the sequence $(x_n)$. Commented May 10, 2022 at 16:02
• @LeeMosher beautiful Commented May 10, 2022 at 16:16