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If M and N are compact Riemann surfaces, show that : $ f: M \to N $ biholomorphic mapping if and only if there is a finite set A, B, where $A \subset M, B\subset N$, such that $$ g: M-A \to N-B$$ is a biholomorphic mapping.

Here I have a similar problem: If M is a compact Riemann surface, and $ M-\{p\} \cong \mathbf{C}$, where p is a point in M, then $ M \cong \mathbf{C} \cup \{\infty\}$.

I have tried many times through definition. However, I still can't think clearly and give an isomorphic mapping.

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Hint: use local holomorphic charts and Riemann's removable singularities theorem.

Edit: here are a little more details.

Let $f: M \backslash \{p\} \to \mathbb C$ be an isomorphism. Let $\phi: U \to \mathbb C$ be a local holomorphic chart of $M$ near $p$. Let $g(z)=\frac{1}{f(\phi^{-1}(z))}$, and $V:=\phi(U)$, $z_0:=\phi(p)$. Then $g: V \backslash \{z_0\} \to \mathbb C$ is an injective holomorphic map. Can you then prove:

  • that up to restricting $U$ if necessary, $g(V \backslash \{z_0\})$ is bounded?

  • that $g$ extends holomorphically to $V$?

  • conclude.

This is for the second part of your question, which is a special case of the first. The more general method is the same, you just need to replace the chart $z \mapsto \frac{1}{z}$ by local charts for $N$.

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  • $\begingroup$ Is it similar method about charts with the $ \mathbf{S} \cong \mathbf{C}\cup \{{\infty}\}$? Why does it use the Riemann's removable singularities theorem? $\endgroup$
    – user469065
    Commented Mar 9, 2018 at 12:03
  • $\begingroup$ @Tinzoe-Yui see my edit $\endgroup$
    – Albert
    Commented Mar 9, 2018 at 12:49
  • $\begingroup$ The second step is to use the Riemann's removable singularities theorem to explain that $g$ can be extended to $V$. Here I have a question: how can I explain that $M$ can be isomorphic to $\mathbf{C}\cup\{{\infty}\}$? $\endgroup$
    – user469065
    Commented Mar 9, 2018 at 18:05
  • $\begingroup$ Please take a look at my thought. If $g$ is unbounded on the disc $U=\{ z: \left| z-z_{0} \right| <r \}. then $\frac{1}{g}$ is bounded, by the Liouville theorem, $f(\phi(z))$ is a constant, but this contradiction. Complete the first step you mentioned. $\endgroup$
    – user469065
    Commented Mar 9, 2018 at 18:14
  • $\begingroup$ not quite. if $g$ is unbounded, it doesn't necessarily mean that $1/g$ is bounded. however, recall that $f$ is an isomorphism, so you can just take $U \subset M$ such that $cl(f(U)) \cap \{0\} = \emptyset$ $\endgroup$
    – Albert
    Commented Mar 9, 2018 at 22:47

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