Suppose that $R_{1}$ and $R_{2}$ are Riemann surfaces, and $A$ is a discrete subset of $R_{1}$. For any continuous map $\alpha: R_{1} \rightarrow R_{2}$ which restricts to an analytic map of Riemann surfaces $R_{1} \backslash A \rightarrow R_{2}$, show that $\alpha$ is an analytic map.

Suppose that $f$ is a non-constant analytic function on a Riemann surface $R$. Show that there is a discrete subset $A \subset R$ such that, for $P \in R \backslash A, f$ defines a local chart on some neighbourhood of $P$.

Deduce that, if $\alpha: R_{1} \rightarrow R_{2}$ is a homeomorphism of Riemann surfaces and $f$ is a non-constant analytic function on $R_{2}$ for which the composite $f \circ \alpha$ is analytic on $R_{1}$, then $\alpha$ is a conformal equivalence. Give an example of a pair of Riemann surfaces which are homeomorphic but not conformally equivalent.

[You may assume standard results for analytic functions on domains in the complex plane.]