= Solution
Take an open neighbourhood $\widetilde U$ 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 $\widetilde U$ so the image lies in a complex chart $(V,z)$ of $R$, and use $z\circ p$ as a chart of $\widetilde R$. These charts cover $\widetilde R$. 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 \b[$p$ 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 $\widetilde f:\mathbb C\to\mathbb D$ with $\pi\circ\widetilde f=f$. 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 \b[every holomorphic map $\mathbb C\to Y$ is constant].
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