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?