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Question: Let $p(z)$ be a polynomial over $\mathbb{C}$. Is it true that $p:\mathbb{C} \to \mathbb{C}$ is a covering map ?

Partial answer: Let us look first at the points where $p'(z)\ne 0$. There the inverse function theorem tells us much more -- that $p$ is locally invertible, and the inverse is in fact holomorphic.

At the points where $p'(z) = 0$, however, $p$ will not be $1-1$ in any neighborhood, and certainly we cannot find an open neighborhood $U$ such that $p^{-1}(U)$ is a family of disjoint open sets, each homeomorphic to $U$.

Is the reasoning correct ? any comments are welcome.

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Yes. If the degree of $p$ is not $1$, then $p$ will never be a covering map. However, for degree greater than $1$, if you remove the bad points, then you do get a covering map. –  Zhen Lin Sep 2 '12 at 14:43

2 Answers 2

up vote 1 down vote accepted

Yes. $p$ defines instead a branched cover.

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The above answers are suffice for the purpose. I just want to remark that any polynomial $p(z)$ can be transformed by a suitable biholomorphic coordinate transformation $z=f(w)$ into $q(w)=w^{n}$(can you fill the detail yourself?). So you only need to deal with simple power maps.

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Thanks, that's helpful. –  Teddy Sep 3 '12 at 7:05
    
OK, do you have a ref. for deriving that biholomorphic transformation ? –  Teddy Sep 3 '12 at 13:29
    
Hint: search "fundamental theorem of algebra" and "winding number". –  Bombyx mori Sep 3 '12 at 13:53
    
@Teddy: I don't think this is true. See here –  Mike F Nov 25 '13 at 3:52

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