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.
 A: $\mathbb{C}-\overline{B_1(0)}$ is isomorphic to a punctured disc though, by the same mapping $\frac{1}{z}$. If you view $X$ as $\mathbb{C}^\infty-\{\infty\}$ and $X^\prime=\mathbb{C}^\infty-\{0,\infty\}$, it is the punctured disc around $\infty$. That is probably clear.
The general situation is as follows, i think. Let $D$ be a punctured disc around $a\in \{x_1,...,x_k\}$. Since $D$ is a subset of $X^\prime$ and $D$ is isomorphic to $\overline{B_1(0)}-\{0\}$ via a biholomorphism $\varphi$, for any $f\in\operatorname{Aut}(X^\prime)$, the image $f(D)\subseteq X^\prime$ is still isomorphic to $\overline{B_1(0)}-\{0\}$ via $\varphi\circ f^{-1}$. Now if there was no point $b$ in $X$ to fill $f(D)\subseteq X$  such that $f(D)\cup\{b\}\cong\overline{B_1(0)}$, then $X$ has a puncture itself and is not compact, a contradiction. Also, $b\in X-X^\prime=\{x_1,...,x_k\}$ since otherwise $f^{-1}(b)\in X^\prime-D$ which would violate the continuity of $f$.
Please correct me if I made any errors :) Cheers
