I'm pretty stuck on the following problem:
Let $U \subset \mathbb{C}$ be an open connected, hyperbolic set, i.e. its (holomorphic) universal cover is the unit disk $\mathbb{D}$. Suppose $f : U \rightarrow U$ is a holomorphic map fixing two distinct points $a,b \in U$. Show that $f$ is an automorphism.
I know that it is sufficient to show that $f$ is a holomorphic covering map , since a holomorphic covering map of a hyperbolic region onto itself that fixes a point is an automorphism (in fact, of finite order). Therefore it is sufficient for me to show the following:
Let $f : U \rightarrow U$ be a holomorphic map fixing two distinct points $a,b \in U$, let $h : \mathbb{D} \rightarrow U$ be a holomorphic covering map such that $h(0) = a$, and suppose $\tilde{f}(0) = 0$ is the (holomorphic) lift of $f \circ h$ with respect to $h$ fixing $0$, i.e. $\tilde{f} : \mathbb{D} \rightarrow \mathbb{D}$ fixes $0$, is holomorphic and satisfies $f \circ h = h \circ \tilde{f}$. Now it is sufficient to show that $\tilde{f} : \mathbb{D} \rightarrow \mathbb{D}$ is a covering map, which occurs if and only if $f \in \mathcal{A}(\mathbb{D},0)$ (i.e. $f$ is a conformal automorphism fixing $0$, so a rotation), since once I show this I can also show that $\tilde{f}$ has finite order and therefore $f$ has finite order, but I am not sure how to use the additional fixed point $b$ to show $\tilde{f}$ is a rotation.
Thanks in advance for any help.
Edit: Just for some more information, if $U$ was $\mathbb{D}$ (or simply connected) the result would follow (via identifying $\mathbb{D}$ with $U$) because it follows for $U = \mathbb{D}$ from the schwarz lemma on $\mathbb{D}$, i.e. let $g : \mathbb{D} \rightarrow \mathbb{D}$ be holomorphic and fix two points $u,v \in \mathbb{D}$, define an automorphism $h$ of $\mathbb{D}$ sending $0 \rightarrow u$, then $h^{-1} \circ g \circ h : \mathbb{D} \rightarrow \mathbb{D}$ fixes $0$ and one other point, so by the schwarz lemma it is a rotation, therefore $g$ is an automorphism of $\mathbb{D}$.
Best Answer
Consider the map $\tilde{f} : \mathbb{D} \rightarrow \mathbb{D}$ (it is the lift mentioned in the question above). It fixes the origin and is holomorphic. Consider the set $h^{-1}(b) \subset \mathbb{D}$, it is closed and discrete, since $0 \notin h^{-1}(b)$ there is some minimum positive (euclidean) distance $r>0$ of the points in $h^{-1}(b)$, consider $M = h^{-1}(b) \cap \{|z|=r \}$. We clearly have that $\tilde{f}(h^{-1}(b)) \subset h^{-1}(b)$ since $b$ is a fixed point of $f$ and $f \circ h = h \circ \tilde{f}$. Moreover by the schwarz-pick lemma, $\tilde{f}$ is either a contraction of the hyperbolic distance on $\mathbb{D}$ or an isometry. It follows that $\tilde{f}(M) \subset M$ , because $\tilde{f}$ fixes $0$ and the hyperbolic distance of a point from $0$ in $\mathbb{D}$ is monotonic increasing relative to the euclidean distance. But then $|\tilde{f}(z)| = |z|$ for some $z \neq 0$ and by the schwarz lemma, $\tilde{f}$ is an automorphism.
Since $M$ is finite and $\tilde{f}$ permutes $M$, $\tilde{f}$ is a rotation of finite order, therefore $f$ is of finite order.
This gives the result.