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Let $X,Y$ be two compact Riemann surfaces. Let $a \in X$ and $f : X-\{a\} \to Y$ be a injective holomorphic map. Prove that $a$ is removable singularity of $f$.

I want to apply Riemann removable singularity theorem somehow but I am unable to get it. Any help will be appreciated.

Mayuresh L
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2 Answers2

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I have an idea, but (UPDATE) with a gap:

Note that $f:X-\{a\}\to f(X-\{a\})$ is biholomorphic.

Let $\phi: U \to D$ be a chart from $U\subseteq X$ to the unit disc $D$ such that $a\in U$ and $\phi(a)=0$. Then $\psi:f(U-\{a\})\to D^\times$ defined by $\psi=\phi\circ f^{-1}$ is a biholomorphism from $f(U-\{a\})$ to the punctured unit disc. Define the sequence $z_n:=\frac{1}{1+n}$ in $D^\times$. Since $Y$ is compact, the sequence $\psi^{-1}(z_n)$ has a convergent subsequence with limit point $b$.

Here's the gap: I want to argue that $V:=f(U-\{a\})\cup\{b\}$ is open. Intuitively, $b$ fills the puncture in the open set $f(U-\{a\})$, just as zero fills the puncture in $D^\times$. However I struggle to translate that thought to topology.

If we assume it is true, then $\psi$ is a bounded holomorphic function on $V-\{b\}$ and thus, by the removable singularity theorem, $\psi$ can be extended to a holomorphic function on the open set $V$. In order for the map to be continuous, $b$ must be sent to zero: $\overline{\psi}:V\to D$ with $\overline{\psi}(b)=0$.

From then on, it would be simple: $\overline{\psi}^{-1}\circ \phi$ is a biholomorphism between $U$ and $V$. Hence the mapping $$F:X\to f(X-\{a\})\cup\{b\},\quad x\mapsto\begin{cases}f(x), & x\neq a\\ (\overline{\psi}^{-1}\circ \phi)(a)=b,& x=a\end{cases}$$ is biholomorphic. Since $X$ is compact and open, $F(X)=f(X-\{a\})\cup\{b\}$ is compact and open in $Y$, which is only possible if $F(X)=Y$.

mgns
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  • Sorry for such a late reply. Could you please elaborate on "Since $Y$ is compact, there is a $b \in Y$ such that $:=(−{})\cup {}$ is isomorphic to the unit disc. Let $\phi :\toℂ$ be a chart of $Y$ with $b \in W$. Then $\phi ( \cap )$ is an open neighborhood of $\phi ()$"? – Mayuresh L Feb 01 '22 at 13:23
  • No worries, you posted the question years ago, so, if anything, it's my answer that's late ;) – mgns Feb 02 '22 at 14:32
  • Take a sequence $(z_n){n\in\mathbb{N}}\subseteq D^\times$ on the punctured unit disc $D^\times$ that converges to zero: $\lim{n\to\infty}z_n=0$. By means of the isomorphism $\psi:D^\times\to f(U-{a})$, you get a sequence $\psi(z_n)$ of points in $f(U-{a})\subseteq Y$. Since $Y$ is compact, this sequence must have a limit point $\lim_{n\to\infty}\psi(z_n)=:b\in Y$ and we can naturally expand the isomorphism $\psi$ to $\overline{\psi}:D\to f(U-{a})\cup{b}$ by sending zero to $b$. – mgns Feb 02 '22 at 14:40
  • Then, since $V$ is isomorphic to the unit disc, $b$ is an inner point of $V$ and $V$ is an open neighborhood of $b$. Since $W$ is also an open neighborhood of $b$, the intersection $V\cup W$ is again an open neighborhood of $b$. Finally, charts are homeomorphisms and hence open maps. Therefore, in the image, $\phi(V\cup W)$ is an open set that contains $\phi(b)$, hence an open neighborhood of $\phi(b)$. – mgns Feb 02 '22 at 14:49
  • Everything in the above comments is clear except the uniqueness of $b$; $\psi(_)$ may have multiple limit points, right? – Mayuresh L Feb 05 '22 at 10:31
  • Well, I don't use the uniqueness of $b$ anywhere, so is there any point in showing it? – mgns Feb 05 '22 at 14:46
  • However, you could show it like this: let $(\tilde{z}n){n\in\mathbb{N}}$ be another choice of a sequence that converges to zero and assume that $\lim_{n\to\infty}\psi(z_n)\neq\lim_{n\to\infty}\psi(\tilde{z}n)$. Then define $(\omega_n){n\in\mathbb{N}}$ as the sequence $z_1,\tilde{z}1,z_2,\tilde{z}_2,z_3,\tilde{z}_3,...$. This sequence does converge to zero, but $\psi(\omega_n)$ does not converge due to $\lim{n\to\infty}\psi(z_n)\neq\lim_{n\to\infty}\psi(\tilde{z}_n)$. This contradicts the compactness of $Y$. – mgns Feb 05 '22 at 14:54
  • By compactness, you can only show that every sequence has a convergent subsequence, so I don’t see any contradiction. Also, if $b$ is not unique then how to apply the Riemann removable singularity a nbhd of $a$? – Mayuresh L Feb 06 '22 at 11:38
  • Ah yes, I see the problem... Riemann surfaces are metrizable, so compactness does imply completeness. Therefore the Cauchy-sequences must converge in $Y$. This is what I was thinking of. However, it is not directly clear that the Cauchy sequence on the unit disc maps to a Cauchy sequence on the surface. This might get unnecessarily complicated.

    However, we really only need the existence of $b$.

     – mgns Feb 06 '22 at 17:29
  • Removable Singularity Theorem:

    Let $X$ be a Riemann surface, $U\subseteq X$ an open subset and $a\in U$. Let $f:{U-{a}}\to {\mathbb{C}}$ be a holomorphic function. If there is a neighborhood $T\subseteq U$ of $a$ such that $f$ is bounded on $T-{a}$, then there exists a unique holomorphic function ${g}:{U}\to {\mathbb{C}}$ with ${g}|_{U-{a}}=f$.

     – mgns Feb 06 '22 at 17:32
  • I updated my answer, thinking I found a way to simplify it, only to find that it does in fact contain a larger gap... :( Maybe you have an idea how to fill it? – mgns Feb 06 '22 at 20:28
  • Somehow I think the key of this is proving that the complement of $f(X-\lbrace a\rbrace)$ in $Y$ is a finite set (for example by showing it's discrete). That way, it would be reduced to https://math.stackexchange.com/questions/2554046/extending-isomorphism-of-punctured-riemann-surfaces?noredirect=1&lq=1 – Compacto Feb 14 '22 at 23:04
  • But seeing it's not already part of the hypothesis makes me suspicious... Are you sure this assertion is true? @MayureshL – Compacto Feb 14 '22 at 23:06
  • Nevermind, the post I linked actually gives the solution (first answer). Advanced theorems like Riemann-Roch or Uniformization theorem were needed. @mgns Nice try, though. – Compacto Feb 14 '22 at 23:24
  • @Compacto Yes, the assertion is true. – Mayuresh L Feb 15 '22 at 07:05
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As the chain of comments is becoming too long, I'm gonna post as an answer that the problem was addressed (and solved) in this other post:

https://math.stackexchange.com/a/2554589/971932

It requires use of deeper theorems such as Riemann-Roch (or Riemann's existence theorem, I think) or Uniformization.

Compacto
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