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?

  • $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Commented May 10, 2022 at 15:06
  • $\begingroup$ @LeeMosher Good point! I added that $f$ is unramified. $\endgroup$
    – Zuy
    Commented May 10, 2022 at 15:11

1 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)

  • $\begingroup$ Why is $X$ compact? $\endgroup$
    – Zuy
    Commented May 10, 2022 at 11:57
  • $\begingroup$ @Zuy Usually the definition of a Riemann Surface X contains the requests that X is also compact 😀 $\endgroup$ Commented May 10, 2022 at 14:00
  • 1
    $\begingroup$ I see. For us, compactness is not part of the definition unfortunately. Still, your answer is quite nice, thanks for that. $\endgroup$
    – Zuy
    Commented May 10, 2022 at 14:43
  • 2
    $\begingroup$ 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 $(x_n)$. $\endgroup$
    – Lee Mosher
    Commented May 10, 2022 at 16:02
  • $\begingroup$ @LeeMosher beautiful $\endgroup$ Commented May 10, 2022 at 16:16

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