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.
On
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.
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 $F(X)=Y$.