# Proof verification: if $f:\Omega\to \Omega$ has two fixed points, is conformal and $f^n=\text{Id}$

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.