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



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> **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 ?
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**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?



Non-bijective conformal maps between annuli
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).


