Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This should probably be easy, but I can't prove it.

Let $f : U^{\prime} \rightarrow U$ be a holomorphic, proper map of degree $d$. Here, $U^{\prime}, U \subseteq \mathbb{C}$ are open sets, both homeomorphic to discs, and $\overline{U^{\prime}}$ is compact.

Question: how do you prove that all critical points of $f$ must lie inside some compact set $L \subseteq U^{\prime}$?

I can see that they cannot accumulate inside $U^{\prime}$, but I can't see why can't we have a sequence $\{z_n\}_{n \in \mathbb{N}}$ of critical points converging to $\partial{}U^{\prime}$. For polynomial mappings, $f^{\prime}$ has degree $d-1$. If this was true in general (that is, for the setting above), then my question would easily follow. But then again, I don't know how to prove this either.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

$f$ is a proper map of degree $d$ is equivalent to say that $f$ is a branched covering map of degree $d$, so $f$ has only $d$ critical points (counting as multiplicity). If you know this, of course, all critical points of $f$ must lie inside some compact set $L⊆U′$

share|improve this answer
    
Well, when I first asked the question I didn't know about it, nor wanted to use more sophisticated approaches. But in the end, that's what I ended up doing; maybe there's no simple argument. –  student Sep 6 '12 at 14:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.