Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 23F (Riemann Surfaces)

Here “covering map” uses the course convention of a local homeomorphism; “regular” adds the evenly-covered product condition. The derivative of $p(z)=z^4$ is $4z^3$, which never vanishes on the upper half-plane $\mathbb{H}$. The holomorphic inverse function theorem therefore makes $p$ a local biholomorphism and hence a covering map in this sense.

It is not regular. The point $1$ has only the preimage $i$ in $\mathbb{H}$, whereas $e^{i\epsilon }$ for small positive $\epsilon$ has the two preimages with arguments $\epsilon /4$ and $\pi /2+\epsilon /4$. Thus the fibre cardinality is not locally constant near $1$, so no neighbourhood of $1$ has a preimage of the form $U\times \Delta$ for a fixed discrete set $\Delta$.

For $|z|<1$, sum the geometric series and its derivative:ThereforeThis rational continuation is holomorphic across every point of the unit circle except $z=-1$, where it has a double pole. The set of singular points is consequently

The map $f$ is bounded by one near the puncture, so the Riemann removable singularity theorem extends it holomorphically to a map $\tilde{f}:\Delta \to \mathbb{C}$. The maximum modulus principle shows that its image still lies in $\Delta$: if $|\tilde{f}(0)|=1$, the extension would be constant, contradicting its values in $\Delta \setminus \{0\}$ unless that constant also had modulus less than one.

If $f$ is biholomorphic, its inverse $g:\Delta \setminus \{0\}\to \Delta \setminus \{0\}$ also extends holomorphically. Were $a=\tilde{f}(0)\ne 0$, continuity of $\tilde{g}\circ \tilde{f}$ and the identity $g(f(z))=z$ on the punctured disk would give $g(a)=0$. This is impossible because $a\in \Delta \setminus \{0\}$ and $g$ takes values in the punctured disk. Hence

View infinity as the puncture of the Riemann sphere and consider $g(1/z)$ near $z=0$. If this map had an essential singularity, the Great Picard theorem in a coordinate chart would make almost every nearby value occur infinitely often, contradicting the injectivity of $g$. The singularity is therefore removable as a map into the compact Riemann surface $R$, and $g$ extends to a nonconstant holomorphic mapThe extension is injective. A collision between $\tilde{g}(\infty )$ and $g(z_0)$, or local degree greater than one at infinity, would give nearby values two distinct preimages by the local normal form of a nonconstant holomorphic map, again contradicting injectivity on $\mathbb{C}$. Thus $\tilde{g}$ is a holomorphic bijection between compact Riemann surfaces and hence a biholomorphism. Therefore $R\cong {\mathbb{P}}^1$, whose genus is zero.