TITLE: $f: \Omega \rightarrow \Omega$ holomorphic, $f(0) = 0$, $f'(0) = 1$ implies $f(z) = z$
QUESTION [11 upvotes]: Let $\Omega$ be a bounded connected open subset of $\mathbb{C}$ containing $0$. Let $f: \Omega \rightarrow \Omega$ be holomorphic and $f(0) = 0$, $f'(0) = 1$. The problem I am working on is to show that $f(z) = z$.
If $\Omega = \mathbb{D}$, then this follows from the Schwarz Lemma. I also know of a solution (posted here) which involves looking at the power series coefficients of $f^n := f\circ f \circ f \circ \cdots \circ f$ ($n$ times) and using the Cauchy estimates, but is there a different way of doing this problem that doesn't involve taking the $k$th derivative of $f^n$?

REPLY [5 votes]: A  solution  which I find more natural than looking at iterations of $f$ involves a holomorphic covering map $\phi:\mathbb D\to\Omega$. Such a map exists for every domain $\Omega$ provided that $\mathbb C\setminus \Omega$ has at least two points. This is Theorem 16.5.1 in Conway's book Functions of one complex variable, vol. II. 
Normalize the covering map so that $\phi(0)=0$. Then $\phi^{-1}\circ f\circ \phi$ is a holomorphic map of disk to itself which fixes the origin and has derivative $1$ there. (Note: although $\phi$ is not globally invertible, it is locally invertible; therefore $\phi^{-1}\circ f\circ \phi$ is an analytic function, and since $\mathbb D$ is simply-connected, it has a single-valued holomorphic branch in $\mathbb D$.)
By the Schwarz lemma the  map $\phi^{-1}\circ f\circ \phi$ is the identity, and therefore $f$ is the identity. 

Key words for additional reading: uniformization theorem, hyperbolic metric.