Automorphism of unit disk without zero 

Let $S$ be the unit disk without $0$. Find all $f \in Auto(S)$


I got the following idea. By Riemann 0 is a removable singularity. Since for $g\in Auto(D)$ where $D$ is the unit disk. $g(z)= e^{i{\phi}} \frac{z-\omega}{\bar\omega z -1}$
 Then $f(z)=e^{i{\phi}} \frac{z-\omega}{\bar\omega z -1}$ and $ w \neq 0$ are all the elements of $Aut(S)$
 A: There are two problems. First, you need to show that when you remove the singularity the resulting function is in fact an automorphism of $D$. Second, once you've done that, not every automorphism of $D$ arises this way.
Say $f$ is an automorphism of $S$. Then yes, $f$ has a removable singularity at $0$. Say $g$ is the extended function. The Maximum Modulus Theorem shows that $|g(0)|<1$, so $g:D\to D$.
Now the major missing step: It follows that $g(0)=0$. Suppose not; $g(0)=\alpha\ne0$. Since $f$ is an automorphism of $S$, there exists $\beta\in S$ with $g(\beta)=\alpha$. Now the Open Mapping Theorem shows that if $\alpha'$ is close enough to $\alpha$ then there exist $z$ near $0$ and $\beta'$ near $\beta$ with $g(z)=g(\beta')=\alpha'$. So $f$ is not one-to-one in $S$, contradiction.
So $g(0)=0$. Since $g$ maps $S$ bijectively to itself it follows that $g$ maps $D$ bijectively to itself. So $g$ is an automorphism of the disk, and also $g(0)=0$. So $g$ is a rotation: $g(z)=e^{i\phi}z$.
A: You're right that $f$ extends to an automorphism $g$ of $D$. However, one must have $g(0)=0$, i.e., $w=0$. Hence $f(z)=e^{i\phi}z$, for some $\phi\in\mathbf R$.
