What I want to prove is exactly as stated it the title:

Let $f:\Omega\to \Omega$ be a holomorphic function and $\Omega\subsetneq \mathbb{C}$ a connected open subset of $\mathbb{C}$ such that its complement contains at least two point. Assume also that $\exists a,b\in \Omega:f(a)=a\neq b=f(b)$. Prove that $f$ is conformal and there exists a natural number $n:f^{[n]}=f\circ \dots\circ f=\text{Id}$

I proved it as follows: since $\Omega$ avoids at least two points, it is hyperbolic (i.e. its universal cover is the unit disc $\mathbb{D}$). Thus we have the following diagram ($\pi$ is the covering map):

$$\require{AMScd} \begin{CD} \mathbb{D} @>{\tilde f}>> \mathbb{D}\\ @VVV @VVV \\ \Omega @>{f}>> \Omega; \end{CD}$$ where $\tilde f$ is the lifting of $f\circ \pi$ such that $\tilde f(\tilde b)=\tilde b$ with $\pi(\tilde b)=b$. Let $\phi$ be an automorphism of $\mathbb{D}$ sending $\tilde b$ to $0$ and viceversa. We define $g:=\phi \tilde f\phi$. By Schwarz's lemma, $|g'(0)|\le 1$ and it is $1$ iff $g$ is a rotation. Suppose $|g'(0)|<1$. By Montel's theorem we can extract a convergent sequence of $g^{[n_k]}$, which is equal to $\phi \tilde{f}^{[n_k]}\phi$. Since $|g'(0)|<1$, $0$ is an attracting fixed point for a neighbourhood of itsel, which implies that $\lim g^{[n_k]}$ is constant in a neighbourhood of $0$ and so it is constant on $\mathbb{D}$. Thus, $\lim \tilde{f}^{[n_k]}=\lim \phi g^{[n_k]} \phi=\phi \left(\lim g^{[n_k]}\right)\phi\equiv \phi(0)=\tilde b$. However, by definition $\tilde{f}(\pi^{-1}(a))\subseteq \pi^{-1}(a)$, and since this is a discrete set not containing $\tilde b$, we get a contradiction. So $|g'(0)|=1$, and thus $g$ is a rotation. Suppose its angle is not a rational multiple of $\pi$. Then $\overline{\{g^{n}(z_0)\}}=\{z:|z|=|z_0|\}$. This is true in particular for every element of $\phi(\pi^{-1}(a))$. This implies that $\overline{\pi^{-1}(a)}$ contains at least one circle, contradicting the fact that $\pi^{-1}(a)$ is discrete. Now the result follows, since $\pi=\pi \tilde f^{n}= f^n\pi$, and inverting locally we get $f^{n}=Id$, which implies by the identity principle $f^{n}=Id$

Is my proof correct? Does this theorem have a name? I have seen something similar (a holomorphic map $f:\Omega\to \Omega$ on a bounded region, with two fixed points is conformal) being referred to as Cartan's theorem, but I am not sure whethere the attribution is correct.


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