# Linear fractional transformation which maps disc to disc.

$\textbf{Schwarz Lemma}$ : If $f$ is a holomorphic function defied on a unit disc $U(0,1)$ satisfying $|f(z)| < 1$ for all $|z| < 1$ and $f(0) = 0$, then $|f(z)| \leq |z|$ for $|z| < 1$ and $|f'(0)| \leq 1$. Furthermore, if $|f(c)| = |c|$ for some $c$ with $0 < |c| < 1$ or if $|f'(0)| = 1$, then there exists a $\alpha \in \mathbb{C}$ such that $|\alpha| = 1$ and $f(z) = \alpha z$ for all $|z| < 1.$

I try to verify the following statement , but not far accomplish :

According to Schwarz Lemma, every injective holomorphic mapping of a disc onto another disc is a liner fractional transformation (A liner fractional transformation $T$ is of the form $$T(z) = \frac{az + b}{cz + d}$$ where $a, b, c, d \in \mathbb{C}$ with $ad - bc \neq 0.$) Notice that the disc is also mean half plain with $\infty$ adjoined to the boundary.

Formulate the corresponding statement for bijective holomorphic self mapping of $1 < |z| < +\infty.$

I know that if $f$ is an injective holomorphic mapping of a disc to another disc, say $U(a, R_1)$ to $U(b, R_2)$, I can try to define $g(\frac{z-a}{R_1}) = \frac{f(z) -b}{R_2}$ to match Schwarz condition. But I cannot see the link between this and linear fractional transformation. I do not try for the half plain yet since it contain $\infty$ which might be harder to formulate $g$ to fit Schwarz condition. Now, for a bijective self mapping of $1 < |z| < +\infty$, I do not sure if Schwarz can be apply since the domain is the region outside the unit disc.

The domain $\Omega=\{z: 1 < |z| < \infty\}$ is a bit different from a disk or half-plane, because it is not simply-connected. So "the corresponding statement" will look different.

If $f:\Omega\to\Omega$ is bijective, then $g(z) = 1/f(1/z)$ is a bijective map of punctured unit disk $D\setminus \{0\}$ onto itself. The singularity at $0$ is removable because $g$ is bounded. The value $g(0) = \lim_{z\to 0}g(z)$ has to be a boundary point of the domain $D\setminus \{0\}$. It can't be a point on the unit circle because then $|g|$ would attain a local maximum at $0$. So, $g(0)=0$. We now have a bijective map of unit disk to itself, and since $g(0)=0$, it is of the form $g(z) = \alpha z$ with $|\alpha|=1$. This translates into a statement about $f$.