Show that each biholomorphic function $f:U\setminus M \longrightarrow U\setminus M$ has a biholomorphic extension $g:U \longrightarrow U$. Let $U\subseteq \mathbb{C}$ be an open and bounded set without isolated points on the border. Let $M\subset U$ be a set without limit points in $U$.
Show that each biholomorphic function $f:U\setminus M \longrightarrow U\setminus M$ has a biholomorphic extension $g:U \longrightarrow U$.
Remark: Remember us:

A function $f:U \longrightarrow U$ is biholomorphic if $f:U \longrightarrow U$ is biyective,  holomorphic and $f^{-1}:U \longrightarrow U$ is holomorphic.

My attempt: As $M\subset U$ is a set without limit points in $U$, then $M$ is discrete in $U$, therefore, if each point in $M$ is removable singularity of  $f$ and $f^{-1}$, then by Riemann extension theorem we have that there are holomorphic extensions $g_{1}$ and $g_{2}$ of $f$ and $f^{-1}$ respectively. Here is my first problem: How do I prove that $g_{1}^{-1}=g_{2}$?.
The other problem is: What should I do in the case where there are  points in $M$ that are not removable singularity?
 A: Since $M$ has no limit points in $U$, every point of $M$ is an isolated singularity of $f$. Since $U$ is bounded, by Riemann's removable singularity theorem the points of $M$ are removable singularities.
If we denote the holomorphic extension of $f$ to $U$ by $g_1$, then we know that $g_1\colon U \to \overline{U}$. Since $g_1$ is not constant - except in the case $U = \varnothing$ when the assertion is trivial - we know that $g_1(U)$ is open, so in fact $g_1 \colon U \to \overset{\Large\circ}{\overline{U}}$.
The same argument applies to $f^{-1}$, for which we get the extension $g_2$. To know that $g_2 \circ g_1$ is defined on all of $U$ (and ditto $g_1 \circ g_2$), we must see that actually $g_1(U) \subseteq U$ [the same argument then shows $g_2(U)\subseteq U$].
This is where the hypothesis that $U$ has no isolated boundary points comes into play. Let $m \in M$. By the open mapping theorem, for every neighbourhood $V \subseteq U$ of $m$ the set $g_1(V)$ is a neighbourhood of $g_1(m)$. If we had $g_1(m) \in \partial U$, then $g_1(V)$ would also contain further boundary points of $U$, since $\partial U$ has by assumption no isolated points. But we can choose $V$ so small that $V\cap M = \{m\}$, and then $g_1(V) \cap \partial U \subset \{g_1(m)\}$. Therefore $g_1(m) \in U$.
Thus $g_2\circ g_1$ and $g_1\circ g_2$ are defined on all of $U$, and the identity theorem - applied to every connected component of $U$ - shows $g_2\circ g_1 = \operatorname{id}_U = g_1 \circ g_2$, since these identities hold on $U \setminus M$ which intersects every component of $U$ in a nonempty open set.
