# Cartan Theorem

Under $\mathbb{C}^1$ case:

If $\Omega$ is a bounded domain which contains origin,

If $F:\Omega \rightarrow \Omega$ is analytic with $F(0)=0, F'(0)=1$,

then $F(z)=z$ for any $z\in \Omega$.

Try to solve it by Riemann Mapping Theorem.

Scheme of Proof:

Think about that if $\Omega$ is a disk. Then use Schwarz Lemma.

Riemann Mapping Theorem:

If $G$ is simple connected domain of $\mathbb{C}$, and $G\neq \mathbb{C}$. Then for any point in $G$, there exists unique function $f:G\rightarrow \mathbb{C}$, such that:

1. $f$ is holomorphic and univalent in $G$ .
2. $f(a)=0, f'(a)\textgreater 0$.
3. $f(G)=B(0,1)$.

Thus, if $\Omega=B(0,1)$, then $F(0)=0$, by Schwarz Lemma, $|F(z)|\le |z|$ with $|F'(0)|\le 1$. The equality holds for $F(z)=e^{i\theta}z$.

By Riemann Mapping Theorem:

There exists a unique holomorphic and univalent function $\phi:\Omega\rightarrow B(0,1)$, such that $\phi(0)=0,\phi'(0)\textgreater 0$.

Set $h=\phi\circ F\circ \phi^{-1}$, then $h$ is holomorphic in $B(0,1)$ and satisfies that $h(0)=0, h'(0)=1$, by Schwarz Lemma:

$h:B(0,1)\rightarrow B(0,1)$, and $h(z)=z$.

Thus consider that $\phi$ is univalent, $F(\phi^{-1}(z))=\phi^{-1}(z)$, for any $z\in B(0,1)$, therefore $F(z)=z$, for any $z\in \Omega$.