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I need to answer the following question, hopefully in the negative.

Question: Does there exist a conformal map $f$ of degree $1$ from the annulus $\{1<|z|<R\}$ to the punctured disk $\{0<|z|<r\}$, such that $f$ extends to a continuous map $\{1\leq|z|<R\}\rightarrow\{|z|<r\}$ sending the inner circle $\{|z|=1\}$ to 0 ?

Remark: It can be seen in a geometric way that there does exist surjective conformal maps $\{1<|z|<R\}\rightarrow\{0<|z|<r\}$, which however map $\{|z|=1\}$ to self-intersecting curves passing through $0$.

Of course, if one can show that any hypothetical such $f$ is injective on some neighborhood $\{1<|z|<1+\epsilon\}$ of the inner circle, then the classical Schottky Theorem on conformal equivalences between annuli immediately yields a negative answer. However, I don't have any idea to prove this. I also tried to deduce a contradiction by examining the extremal lengths, but wasn't able to achieve it. Can someone help?

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  • $\begingroup$ What is your definition of a conformal map? For me a conformal map is always injective, but you seem to allow non-injective conformal maps. If a conformal map means an analytic injection, then injectivity on $\{1<|z|<1+\epsilon\}$ is trivial. $\endgroup$ – Joonas Ilmavirta Sep 13 '14 at 21:23
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    $\begingroup$ By a conformal map, I mean a holomorphic map with nowhere vanishing derivative, as in Wikipedia. $\endgroup$ – Xin Nie Sep 13 '14 at 21:49
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    $\begingroup$ @JoonasIlmavirta, I do think it is often that "conformal" is merely a local property, although, as you note, we are often interested in global bijectiveness. $\endgroup$ – paul garrett Sep 13 '14 at 22:28
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    $\begingroup$ It seems that the definition of conformal mappings in the literature is inconsistent: Ahlfors and Rudin allow conformal maps which are not 1-1, while, say, Stein and Shakarchi assume 1-1. $\endgroup$ – Misha Sep 13 '14 at 22:36
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If a bounded holomorphic function $f$ on the unit disk has boundary value zero on a positive measure subset, then $f\equiv 0$ (this is a well known fact from the theory of Hardy spaces, and it holds more generally).

This rules out the existence of functions such as the ones you describe above (by conformal mapping of a simply connected piece of your annulus).

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  • $\begingroup$ Thanks! I didn't expect such a simple solution. Could you give me some references about the fact that you referred to? $\endgroup$ – Xin Nie Sep 13 '14 at 22:08
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    $\begingroup$ You can use Cartan-Rado theorem. R.Narasimhan, Complex Analysis in one variable, Chapter 11, section 8. $\endgroup$ – Misha Sep 13 '14 at 22:22
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    $\begingroup$ @XinNie: The general version is that if $f$ is Nevanlinna class (aka bounded type), then $\int_{\partial D} \ln |f(e^{it})|\, dt > -\infty$ (which in particular says that $|f|>0$ a.e.). This is Theorem 20.2.11 in vol. 2 of Conway's book. $\endgroup$ – Christian Remling Sep 13 '14 at 22:22
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Here is another way to get the answer: Any map as in your question is a covering map. Hence, if the degree is one, it is in fact a conformal isomorphism.

Note that this implies (as does Christian's answer) that the assumption of being degree $1$ is not necessary.

It also shows that, in general, the only locally conformal maps between annuli that map boundaries to boundaries are the obvious ones, with the corresponding relations between moduli.

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