Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 23F / Solution

Take an open neighbourhood $\tilde{U}$ on which $p$ is one-to-one and a homeomorphism onto its image. By invariance of domain for topological surfaces, that image is open. Shrink $\tilde{U}$ so the image lies in a complex chart $(V,z)$ of $R$, and use $z\circ p$ as a chart of $\tilde{R}$. These charts cover $\tilde{R}$. On any overlap their transition is the transition between the original charts of $R$, restricted to an open set, so is holomorphic with holomorphic inverse. They define the desired complex structure, and in these charts $p$ is locally the identity. Thus $p$ is holomorphic. The openness supplied by invariance of domain is important when the hypothesis is phrased only as a homeomorphism onto the image.

Let $f:\mathbb{C}\to Y$ be holomorphic and let $\pi :\mathbb{D}\to Y$ be the stated covering. Choose a lift of $f(0)$. The covering-space lifting theorem says that a map from a simply connected, locally path-connected domain has a unique lift after fixing this initial point: lift paths from zero, and homotopy lifting makes their endpoint independent of path. Since $\mathbb{C}$ is simply connected, this constructs $\tilde{f}:\mathbb{C}\to \mathbb{D}$ with $\pi \circ \tilde{f}=f$. On a small neighbourhood, the lift is a biholomorphic local inverse of $\pi$ composed with $f$, so it is holomorphic. Its modulus is less than one, and Liouville theorem therefore makes it constant. Hence every holomorphic map $\mathbb{C}\to Y$ is constant.