Let $f:X\rightarrow Y$ be a non-constant holomorphic map between compcat riemann surfaces, we need to show $f^{-1}(y),\forall y\in Y$ is finite and discrete subset of $X$.

What if $X$ and $Y$ are non compact?

well, $f$ is onto clearly, and I understand some how I need to use the fat that $Zeros$ of $f$ is a discrete set, but I am not able to write rigoriously the answer.

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    $\begingroup$ Your description of $S$ doesn't make sense (or describes the empty set in a benevolent interpretation). $\endgroup$ Jul 24, 2012 at 21:22
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    $\begingroup$ @Patience In a compact space any infinite set has an accumulation point. So any closed and discrete set is finite. $\endgroup$
    – azarel
    Jul 24, 2012 at 21:44
  • $\begingroup$ @GeorgesElencwajg I am sorry!! I will do that from next time, If $X$ and $Y$ are non compact, how will I conclude? $\endgroup$
    – Balbichi
    Jul 24, 2012 at 21:50

2 Answers 2


a) If $X$ and $Y$ are non compact it certainly is not clear that $f$ is onto since it is false: think of the inclusion of a disk into $\mathbb C$.
b) It is not true either that the fibers $f^{-1}(y)$ are finite: think of the sinus function $\sin:\mathbb C\to \mathbb C$ with $\sin^{-1}(0)=\pi \mathbb Z$, an infinite set.
c) It is however true that the fibers $f^{-1}(y)\; (y\in Y)$ are discrete closed subsets of $X$: closedness follows from continuity of $f$ and discreteness can be checked locally at points of $x\in X$.
This means that in order to check it you may assume that $X$ and $Y$ are disks containing the origin and that $x=0, f(x)=0$.
You may then invoke the result that zeros of non-constant holomorphic functions are isolated.

  • $\begingroup$ Can the fibers be uncountable though ? i.e. can $f^-1(c)$ be uncountable for some $c$ and some holomorphic $f$ $\endgroup$
    – Amr
    Jun 26 at 2:38
  • $\begingroup$ oh, never mind. answer should be no because of the identity theorem $\endgroup$
    – Amr
    Jun 26 at 3:04

I will add a little bit to George's fine answer. I believe that the property you're looking for is properness. A map is called proper if inverse images of compact subsets are again compact. As georges pointed out, the fibers will be discrete and since a point is clearly compact, the fiber will be compact discrete, hence finite.

When i learned about this theorem in class, the teacher gave it for compact surfaces and mentioned its extension to the proper case. Therefore i believe that this is the most general statement of the theorem.

Hope that helps.


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