A holomorphic map $p: S \rightarrow T$ between Riemann surfaces is called a covering map if every $t \in T$ has a neighbourhood $V$ for which $p^{-1}(V)$ breaks up as a disjoint union of open subsets $U_{\alpha}$ on which $p: U_{\alpha} \rightarrow V$ is biholomorphic.

(a) Suppose that $f: R \rightarrow T$ is any holomorphic map of connected Riemann surfaces, $R$ is simply connected and $p: S \rightarrow T$ is a covering map. By considering the lifts of paths from $T$ to $S$, or otherwise, prove that $f$ lifts to a holomorphic map $\widetilde{f}: R \rightarrow S$, i.e. that there exists an $\widetilde{f}$ with $f=p \circ \widetilde{f}$.

(b) Write down a biholomorphic map from the unit disk $\Delta=\{z \in \mathbb{C}:|z|<1\}$ onto a half-plane. Show that the unit disk $\Delta$ uniformizes the punctured unit disk $\Delta^{\times}=\Delta-\{0\}$ by constructing an explicit covering map $p: \Delta \rightarrow \Delta^{\times}$.

(c) Using the uniformization theorem, or otherwise, prove that any holomorphic map from $\mathbb{C}$ to a compact Riemann surface of genus greater than one is constant.