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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.

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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 $L⊆U′$

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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

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