Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 23H / Solution

In local coordinates around $p\in R$ and $f(p)\in S$, a nonconstant analytic map has the formAfter shrinking the chart, $h$ has an analytic $m$th root, so a change of coordinate makes the map $z\mapsto z^m$. It maps small discs onto neighborhoods of the origin; hence every nonconstant analytic map of Riemann surfaces is open. If $R$ is compact and $S$ connected, $f(R)$ is compact and therefore closed, and it is also nonempty and open. Thus $f(R)=S$.

For $f:\mathbb{D}\to \mathbb{D}$ with $f(0)=0$, the removable-singularity theorem makes $g=f/z$ analytic. On $|z|=r$, $|g(z)|=|f(z)|/r\le 1/r$, and the maximum principle gives the same bound throughout ${\mathbb{D}}_r$. Applying the Schwarz lemma to an automorphism $h$ and to $h^{-1}$ shows that every automorphism fixing zero is

If $F:\mathbb{C}\to \mathbb{C}$ is an analytic isomorphism fixing zero, continuity of $F^{-1}$ implies that the preimage of every closed disc is compact. Hence $|F(z)|\to \infty$ as $|z|\to \infty$, and defining $F(\infty )=\infty$ gives an analytic isomorphism of the Riemann sphere.

Consequently an automorphism of $\mathbb{D}$ fixing zero does extend to $\mathbb{C}$. A general disc automorphism need not: for $a\ne 0$,has a finite pole and cannot extend to an entire isomorphism.

Solved by gpt-5.6-sol high.