Conformal automorphism of disk implies fractional linear transformation

Statement

Suppose $D$ is the open unit disc in $\mathbb {C}$. Then, any conformal automorphism (a bijective biholomorphic mapping to itself) of $D$ comes as the restriction to $ofafractionallineartransformation.==Proof=====Proofoutline===Let<math>G=Aut(D)$ and $H$ be the subgroup of $G$ comprising those automorphisms that arise by restricting fractional linear transformations to $D$. We need to show that $H=G$. We prove this by showing two things:

• $H$ acts transitively on $D$
• $H$ contains the isotropy subgroup of $0$ in $G$

Combining these two facts, we see that $H=G$

Proof of transitivity

We'll show that given any element $w\in D$, there exists an element $\sigma \in H$ such that $\sigma (0)=w$. This'll show transitivity.

Define:

$\sigma :=z\mapsto {\frac {z-w}{{\overline {w}}z-1}}$

This is a fractional linear transformation. To see that it takes the disc to itself,