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I'm having trouble understanding the problem 8.2. in Förster's book on Riemann surfaces.

Let $X$ and $Y$ be compact Riemann surfaces, $a_1, \dots, a_n \in X$ and $b_1, \dots, b_m \in Y$ and $X' := X \setminus \{a_1, \dots, a_n\}$, $Y' := X \setminus \{ b_1, \dots, b_m \}$. Show that every isomorphism $f: X' \to Y'$ extends to an isomorphism $\tilde{f} : X \to Y$.

This question has been asked here before, but none of the answers explained why the number $m$ can be different from $n$.

If $m = n$, I think I've got a general idea of how to prove this: we can continuously extend the function $f$ on $X$, since it is obviously continuous on a dense subset of $X$. Then the resulting extension is holomorphic using some argument with Riemann removable singularity theorem + charts. But then every point in the image of $f$ has the same multiplicity, which must be $1$.

But if $m \neq n$, then to me the statement seems obviously false. If $f: X' \to Y'$ is a bijection, how can we possibly extend it to a bijection $\tilde{f}: X' \cup \{a_1, \dots, a_n \} \to Y' \cup \{b_1, \dots, b_m\}$ if we add a different number of points to $X'$ and $Y'$? Since all the asked questions on this site allow the possibility of $n \neq m$, I realize I must be missing something glaringly obvious, but I can't figure out what.

Any suggestions would be very welcome. Thanks!

the_dude
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  • The rather thorough discussion here makes no presumptions about (indeed, does not mention) $m$ and $n$. – Ted Shifrin Apr 08 '25 at 22:51
  • @TedShifrin Yes indeed, that's exactly why I'm asking this question. As I already stated in the post, it seems to me that some presumptions need to be made about $m, n$ (namely, that they are equal). – the_dude Apr 08 '25 at 23:11
  • Well, can you prove (just by topological considerations) that if $X’$ and $Y’$ are homeomorphic, then $X$ and $Y$ must be as well? – Ted Shifrin Apr 08 '25 at 23:45
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    It's a general topological fact that if two surfaces of finite type (i.e. obtained from compact surfaces by removing finite subsets) are homeomeomorphic then they have the same number of punctures. – Moishe Kohan Apr 09 '25 at 00:00
  • @MoisheKohan I see, I think I know how to prove this. Do we need the compactness assumption? – the_dude Apr 09 '25 at 00:25
  • No, but then instead of punctures you have to state the result in terms of ends. This is completely general and works for manifolds of all dimensions and even more complicated spaces. – Moishe Kohan Apr 09 '25 at 00:52
  • https://en.wikipedia.org/wiki/End_(topology) – Moishe Kohan Apr 09 '25 at 01:12
  • @MoisheKohan thanks for your answers! Do you know if this problem requires compactness of $X$, $Y$? Because the way I'm thinking right now, we can holomorphically extend $f: X' \to Y'$ to $\widetilde{f}: X \to Y$ and $f^{-1}: Y' \to X'$ to $\widetilde{g}: Y \to X$, then $\widetilde{g} \circ \widetilde{f} = \mathrm{id}_X$ and $\widetilde{f} \circ \widetilde{g} = \mathrm{id}_Y$ by the identity principle. So $\widetilde{f}: X \to Y$ is a biholomorphism, but this argument doesn't use compactness of $X$ or $Y$, which worries me. – the_dude Apr 09 '25 at 13:36
  • No, compactness is not needed, but it simplifies things a bit. – Moishe Kohan Apr 09 '25 at 14:03
  • @MoisheKohan Is there anything wrong with my argument in the comments? – the_dude Apr 09 '25 at 15:05
  • No, there is nothing wrong with the argument, but it should be a part of your post. Comments are not meant for this. – Moishe Kohan Apr 09 '25 at 15:13

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