When are two biholomorphisms from a domain to open unit disc equal?

Question

Apparently is is true that if $\Omega$ is a domain in $\mathbb{C}$ and $f,g: \Omega \to \mathbb{D}$ are biholomorphic maps which satisfy: there exists $q \in \Omega$ such that $f(q) = g(q)$ and $f'(q), g'(q) > 0$, then $f=g$ everywhere on $\Omega$! It seems magical (like many things in complex analysis) but how can I see that it's true? I suspect that it has something to do with conformal mappings/associated theory, but perhaps not.

Answer 1

This is just a souped-up version of the Schwarz lemma. First, we reduce to the case $\Omega=\mathbb{D}$ and $q=0$. Let $h:\mathbb{D}\to\mathbb{D}$ be a biholomorphism such that $h(f(q))=0$, and consider $F=h\circ g\circ f^{-1}\circ h^{-1}$. This is a biholomorphism $\mathbb{D}\to\mathbb{D}$, and $F(0)=0$. Moreover, $F'(0)>0$ by the chain rule, since the derivatives of $h$ and $h^{-1}$ will cancel out and the derivatives of $g$ and $f^{-1}$ will both be positive.

Now by the Schwarz lemma, $|F(z)|\leq|z|$ for all $z$, with $F(z)=az$ for some constant $a$ if equality ever holds for $z\neq 0$. But the Schwarz lemma also applies to $F^{-1}$, giving $|F^{-1}(w)|\leq |w|$ for all $w$. Letting $w=F(z)$, we get $|z|\leq|F(z)|$, so actually $|z|=|F(z)|$ for all $z$. Thus $F(z)=az$ for some constant $a$. Since $F'(0)>0$, we can only have $a=1$.

Thus $F$ is the identity map. Hence $g\circ f^{-1}=h^{-1}\circ F\circ h$ is also the identity map, and $f=g$.