# Automorphism of a compact Riemann surface with finitely many points removed maps punctured disc to punctured disc

I wonder if the following is true: If we have a compact Riemann surface $X$ where finitely many points are removed, so $X'=X - \{x_1,...x_k\}$ and an automorphism $f$ of $X'$, is it true that $f$ maps a punctured disc around a point $p\in \{x_1,...,x_k\}$ to a punctured disc around some point $q\in \{x_1,...,x_k\}$? By a punctured disc around $p$ I mean an open set $U\subset X'$ such that $U\cup \{p\}$ is biholomorphic to a disc in $\mathbb{C}$ via a chart $(U\cup \{p\}, \phi)$ of $X$.

I know that this is not the case if $X$ is a non compact Riemann surface. For example consider $X=\mathbb{C}$ and $X'=\mathbb{C}-\{0\}$. Then the automorphism $f(z)=\frac{1}{z}$ maps the punctured disc $B_1(0)-\{0\}$ to $\mathbb{C}-\overline{B_1(0)}$ which is not a punctured disc.

• So essentially the question is: Can $f$ be extended to $X\to X$? – Hagen von Eitzen Jun 10 '17 at 16:09
• @HagenvonEitzen: You're right. The background to this question is exactly this statement. I have an idea of a proof for which I need to know if the above question is true (and if so how to prove it). – user454042 Jun 10 '17 at 16:21
• math.stackexchange.com/questions/2554046/… – Moishe Kohan Dec 6 '17 at 22:25