Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 23H / Solution

A Riemann surface is a connected one-complex-dimensional manifold with holomorphic chart transitions. A map between such surfaces is holomorphic when all its local coordinate expressions are holomorphic; it is biholomorphic when it is a bijection whose inverse is also holomorphic.

The open mapping theorem says a nonconstant holomorphic function on a connected complex domain maps open sets to open sets. Applied in charts, with the identity theorem excluding a locally constant portion, it makes a nonconstant holomorphic map of connected surfaces open. If the domain is compact, its image is compact and hence closed in the Hausdorff target. A nonempty set both open and closed in a connected target is the whole target. Therefore the map is surjective, without any appeal to degree.

For a biholomorphism $\varphi :{\Delta }^{\ast }\to {\Delta }^{\ast }$, both $\varphi$ and its inverse are bounded near the missing origin. They extend holomorphically across zero by removability: in a Laurent series a negative coefficient satisfies $|a_{-m}|\le M{r}^m$ on radius $r$, hence vanishes as $r\to 0$. Let the extensions be $F,G$. Their images lie in $\Delta$: a modulus-one interior value would force a bounded holomorphic map to be constant by the maximum modulus principle. The compositions $GF$ and $FG$ equal the identity on the punctured disk and thus on the disk by continuity. If $F(0)\ne 0$, then $G(F(0))$ would be nonzero because the original inverse maps ${\Delta }^{\ast }$ into itself, contradicting $G(F(0))=0$. Consequently

For the final assertion only continuity on $R$ and holomorphicity on $R\setminus \{p\}$ are needed; the printed stronger description is redundant. Choose a target chart about $f(p)$ and, by continuity, a small source chart mapped into it. The resulting scalar function is holomorphic on a punctured disk and bounded near its center, so the same Laurent argument removes its singularity. continuity makes the extension's central value equal the prescribed value. Thus $f$ is holomorphic also at $p$.