Describing automorphisms of a punctured unit disc

Question

Here is my attempt :

Let $f \in \text {Aut}(D(0,1) \setminus \{0\})$.

Since $f$ is bounded on a deleted neighborhood of $0$, function $f$ has a removable singularity at $0$. Since $f$ is bijective, we must have $f(0)=0.$

By Schwarz's lemma we get $|f(z)| \le |z|.$

But $f^{-1} \in \text {Aut}(D(0,1) \setminus \{0\}) \; \text {and} f^{-1}(0)=0 \Rightarrow |z|=|f^{-1}(f(z))| \le |f(z)|.$

Thus $|f(z)|=|z| \Rightarrow f(z)=\lambda z \;$ for some $\lambda \in \Bbb C$ and $|\lambda|=1$.

We infer that all the elements of $\text {Aut}(D(0,1) \setminus \{0\})$ are rotations.

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Are my arguments correct?

Answer 1

The important missing detail is where you say "since $f$ is bijective, $f(0)=0$".

To explain what the problem is we need to be a little more careful with the notation (in fact you made the error because you're actually using the notation "$f$" for two different things. Not to say your notation was bad, it's fairly standard, but that's where the error came from).

Say $D$ is the disk and $D'$ is the punctured disk. You're given $f:D'\to D'$. Now $f$ has a removable singularity at the origin, great. So $f$ extends to a function $F:D\to\Bbb C$.

Now when you say "since $f$ is a bijection..." you're actually referring to the new function $F$. You're given that $f:D'\to D'$ is a bijection - how do you know that $F$ is a bijection?

For that matter, how do you even know that $F(0)\in D$?

Hints: A certain very basic result that you use all the time shows that $|F(0)|<1$. So you have $F:D\to D$. Now you can use the Open Mapping Theorem to show that since $f$ is a bijection we must have $F(0)=0$.

Answer 2

In my opinion one argument is missing: we have $|f(z)|=|z|$ for all $z \in D(0,1)$. Why we get $f(z)=\lambda z \;$ for some $ \lambda$ with $|\lambda|=1$ ?

Define $g : D(0,1) \to \mathbb C$ by $g(z)=f(z)/z$ if $z \ne 0$ and $g(0)=f'(0)$. Then $g$ is holomorphic and $|g(z)|=1$ for all $z \in D(0,1)$.

Hence $g(D(0,1))$ is not open, thus $g$ is constant.

Can you proceed ?