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As I was reading some text in Riemann Mapping theorem. There was an argument in the text that

In general There doesnot exist holomorphic map which takes multiply connected domain onto a simply connected domain.

I was not able to think of counter example nor the proper reasoning .

Please help. Thnx and regards

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closed as off-topic by Maria Mazur, JonMark Perry, TheGeekGreek, Nosrati, Did Dec 28 '17 at 0:13

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    $\begingroup$ Maybe a more complete quote of the argument/statement from the text you got it from, as to the conditions etc., would help the post. $\endgroup$ – coffeemath Dec 27 '17 at 10:03
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    $\begingroup$ In the context of RMT it would also make sense to state that "there is not bijective holomorphic map $f\colon X\rightarrow Y$ between two domains $X,Y\subset \mathbb{C}$ of which only one is simply connected." This is true, because such $f$ would have to be open, thus it would be a homeomorphism and simply connectedness is preserved under homeomorphisms. However, I don't know whether the satement is still true if one replaces bijective by onto. $\endgroup$ – Jan Bohr Dec 27 '17 at 10:24
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    $\begingroup$ There is another way to weaken the bijective condition, but this needs a little more topology: If you replace bijective by the three conditions proper, onto and ($f'(z)\neq 0$ for all $z\in X$), then the statement is still true. (This might help by ruling out possible counterexamples.) Such $f$ would have to be a covering map and hence injective on $\pi_1$, i.e. $Y$ could not be simply connected. $\endgroup$ – Jan Bohr Dec 27 '17 at 10:37
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    $\begingroup$ Sure, you should read my argument as part of a proof by contradiction. If such a map existed, then $f_*\colon \pi_1 X \rightarrow \pi_1 Y = 0$ would be injective, thus $\pi_1 X = 0$ and hence $X$ could not be multiply connected. $\endgroup$ – Jan Bohr Dec 27 '17 at 10:52
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Let me first rephrase the question:

Let $X,Y\subset \mathbb{C}$ be open and connected and $Y$ additionally be simply connected, let $f\colon X\rightarrow Y$ be holomorphic. Then under which conditions on $f$ is $X$ simply connected as well?

The question proposes that $f$ being onto is sufficient, this I can neither prove nor disprove. But here are some weaker results:


(1) If $f$ is bijective, then it is true.

Proof: Nonconstant holomorphic maps are open (by the open mapping theorem) and thus $f$ would be a homeomorphism. Simply connectedness is preserved under homeomorphisms, hence $X$ is simply connected.


(2) If $f$is proper, onto and has nowhere vanishing derivative, then it is also true.

Proof: If the derivative of $f$ does not vanish at $z \in X$, then by the inverse function theorem, there is a neighbourhood $U$ around $z$ such that $f\colon U \rightarrow f(U)$ is a diffeomorphism (even biholomorphic). Thus if the derivative is nowhere vanishing, then $f$ is a local homeomorphism. Since it is also proper and surjective, it is a covering map. But then $f_*\colon \pi_1X \rightarrow \pi_1Y$ is injective and $\pi_1Y$ being trivial forces $\pi_1 X$ to be trivial, in other words $X$ is simply connected.

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Someone should note that it can happen that a holomorphic functon maps a multiply connected domain onto a simply connected domain. For example $X=\mathbb D\setminus\{1/2\}$, $Y=\mathbb D$, $f(z)=z^2$.

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  • $\begingroup$ @DevendraSinghRana You suspect a disk with a point removed might be simply connected? $\endgroup$ – David C. Ullrich Dec 27 '17 at 19:57
  • $\begingroup$ @DevendraSinghRana It surely will be what? I can't tell whether you're saying it surely will be simply connected or surely will be multiply connected. $\endgroup$ – David C. Ullrich Dec 27 '17 at 20:11
  • $\begingroup$ "Multiply connected" means "not simply connected". Do you know what "simply connected" means? $\endgroup$ – David C. Ullrich Dec 27 '17 at 20:18
  • $\begingroup$ @DavidCUllrich your argument do seems to be correct but the text in the book suggests that this thing is not true "in general " i.e it may be true somewhere but not always and that is what I have written in the question. $\endgroup$ – Devendra Singh Rana Dec 28 '17 at 7:37

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