# A question from Otto Forster's book

I'm tackling exercise 8.2 on page 59 which goes as follows:

Let $X$ and $Y$ be compact Riemann surfaces, $a_1,\dots,a_n\in X$ and $b_1,\dots,b_m\in Y$ distinct points and $X^{\prime}=X-\{a_1,\dots,a_n\}$, $Y^{\prime}=Y-\{b_1,\dots,b_n\}$. Suppose that $f\colon X^{\prime}\to Y^{\prime}$ is biholomorphic. Then it can be extended to a biholomorphic mapping $\tilde{f}\colon X\to Y$.

Earlier in the section (thm 8.4) it was proved that if $X$ and $Y^{\prime}$ are Riemann surfaces and $X^{\prime}=X\setminus A$ where $A$ is discrete and closed in X. Then every holomorphic and unbranched proper covering $f\colon Y^{\prime}\to X^{\prime}$ can be extended to a branched proper holomorphic covering $\tilde{f}\colon Y\to X$ i.e. $\exists Y$ Riemann surface and $i\colon Y^{\prime}\to Y\setminus \tilde{f}^{-1}(A)$ biholomorphic s.t. $f=\tilde{f}\circ i$.

I figured it shouldn't be too hard to deduce the claim in the question from this theorem, but I keep falling on technical problems with topology. What am I missing here?

Thanks.

• You want to look in local charts and convince yourself that the map has a removable singularity at each of the $a_i$ and that $f(a_i)$ must be one of the $b_j$. Equivalently, you can just apply the theorem and observe that the degree at potential branch points for the extended map must be 1. – AnonymousCoward Aug 24 '14 at 6:46
• @AnonymousCoward I was trying to do the first thing that you suggested and maybe show that if $X^{\prime}\ni x_n\to a_i$ then $f(x_n)\to b_j$ for some $j$. Compactness assures that there are partial limits and $f$'s biholomofity assures that they are all in $Y\setminus Y^{\prime}$ but I couldn't figure why there was just one. – peter Aug 24 '14 at 7:53

Since $Y$ is a manifold, for each $i$ we can take a neighborhood $U_i$ of $b_i$ s.t. all of the $U_i$'s are disjoint and homeomorphic to the unit disc, while $U_i\setminus\{b_i\}$ is homeomorphic to the punctured unit disc. Denote by $V$ the open set $\bigcup f^{-1}(U_i\setminus\{b_i\})$. I claim that $\forall i$, $V\cup\{a_i\}$ is open.
Suppose otherwise, then $X\setminus (V\cup\{a_i\})$ is not closed. Thus there exists a converging sequence $x_n\to x$ s.t. $\forall n$, $a_i\neq x_n\notin V$ and $x$ is either $a_i$ or is in $V$. It cannot be in $V$ for $V$ is open and so it is $a_i$. We can assume that $\forall n, x_n\in X^{\prime}$ and so, $f(x_n)$ is a sequence in $Y$. As $Y$ is compact, it has a convergent subsequence $f(x_{n_k})$ with $b\in Y$ for a limit. If $b\in Y^{\prime}$ then $f^{-1}(f(x_{n_k}))=x_{n_k}\to f^{-1}(b)\neq a_i$. Thus $b=b_j$ for some $j$. This means that $\exists N\in\mathbb{N}$ s.t. $\forall k>N, f(x_{n_k})\in U_i\setminus\{b_j\}$ and this contradicts the assumption that $x_n\notin V$ for all $n$.
Now we can take $V_i$ to be a neighborhood of $a_i$ homeomorphic to the unit disc s.t. $V_i\setminus\{a_i\}$ is homeomorphic to the punctured unit disc. And we can also assume that $V_i\subset V\cup\{a_i\}$. Thus, $V_i\setminus\{a_i\}\subset V$ and since the former is connected, there exists a unique $j$ s.t. $V_i\setminus\{a_i\}\subset f^{-1}(U_j\setminus\{b_j\})$. This gives us a canonical way to define $\tilde{f}(a_i)$ as $b_j$. This function will be continuous. Since $f$ is biholomorphic, this process can be repeated for $f^{-1}$ to show that the $\tilde{f}$ generated is a homeomorphism.