Biholomorphic continuation to bounded domain Let $M \subset \mathbb{C}$ be a bounded domain without isolated boundary points and let $K \subset M$ a subset which has no accumulation point in $M$. Prove that each biholomorphic (holomorphic and bijective) map $f: M\backslash K \to M\backslash K$ can be extended to a biholomorphic map from $M \to M$.
Since $M \subset \mathbb{C}$ is bounded,
$M\backslash K$ also bounded and $f(M\backslash K)$ would thus also be bounded
But I do not know which theorems I could use here and how to use the information that K has no accumulation point, and M does not have isolated boundary points.
 A: This is a comment that got too long and that indicates how to proceed:
The extension part is easy since if $z_n$ are the points in $K$ (which must be at most countable as otherwise, the hypothesis about no accumulation points doesn't hold) then they are isolated singularities of $f$ which is bounded so by the usual results $f$ extends to them holomorphically.
The injective part is also easy since if $f(z_1)=f(z_2)$ there are small neighborhoods of $z_1, z_2$ on which $f$ takes the same values, but the surjectivity is not immediate and there you need to use the hypothesis about no isolated boundary points as that implies that $V$ the preimage under $f$ of $U-z_0$ for small $U$ neighborhood in $M$ of $z_0 \in K$ that doesn't contain other points in $K$, doesn't intersect the boundary since that would precisely give you that the boundary of $M$ has an isolated point. Now picking any $w_n \in U, w_n \to z_0$ it follows that $f^{-1}(w_n)=y_n$ is in $V$ and any accumulation point $y_0$ of it is in $K$ because it cannot be on the boundary by the no isolated points hypothesis and of course is unique by injectivity, so $f(y_0)=z_0$ under the extension of $f$
