When reading the "Lectures on Riemann Surfaces" by Otto Forster on page 37, he claimed that
Suppose $X$ is a Riemann surface and $f:X\to D^{*}$( $D^*$ is the punctured unit disk $\{z\in\mathbb{C}:0<|z|<1\}$) is an unbranched holomorphic covering map. Then one of the following holds:
(i) If the covering has an infinite number of sheets, then there exists a biholomorphic mapping $\varphi:X\to H$ of $X$ onto the left half plane such that the following diagram commutes $$ \require{AMScd} \begin{CD} X @>{\varphi}>> H\\ @V{f}VV @VV{\exp}V \\ D^* @>{id}>> D^* \end{CD} $$ (ii) If the covering is $k$-sheeted( $k<\infty$), then there exists a biholomorphic mapping $\varphi:X\to D^*$ such that following diagram commutes, where $p_k:D^*\to D^*$ is the mapping $z\to z^k$ . $$ \require{AMScd} \begin{CD} X @>{\varphi}>> D^*\\ @V{f}VV @VV{p_k}V \\ D^* @>{id}>> D^* \end{CD} $$
My question is, are there any deep explanations for this theorem? Why $D^*$ has such a good property that infinite and finite sheet can both be turned into some function we are familiar with? Can other Riemann surface other than $D^*$ have the similar property? Thank you for your help!
The universal covering space of the punctured disk $D^*$ in the category of Riemann surfaces is the upper-half plane, with the exponential map $H\to D^*$. The coverings $X \to D^*$ are classified by the subgroups of $\pi_1(D^*) = \mathbf Z$. The covering corresponding to $(0)$ is the full covering space $H \to D^*$ with the exponential map, whereas the the other coverings are $D^* \xrightarrow{z^k} D^*$. These, together, account for all of the subgroups of $\mathbf Z$, so all coverings of $D^*$ must fall in either category.