When I try to find the covering space of a punctured unit circle, I met some obstacles. The only one remain is given as follows:
Suppose $H$ is the left half plane, $D^*$ is the punctured unit disk with $D^*=\{z\in\mathbb{C}:0<|z|<1\}$. $X$ is a Riemann surface, where $f:X\to D^*$ is an unbranched holomorphic covering mapping. Since $\exp:H\to D^*$ is the universal covering, it follows that there exists a holomorphic mapping $\psi$ such that the following diagram commutes. $$\require{AMScd} \begin{CD} X @<{\psi}<< H\\ @V{f}VV @VV{\exp}V \\ D^* @<{id}<< D^* \end{CD}$$ My question is, is $\psi$ a surjective mapping?
This question arouse when I try to classify the covering space with the deck transformation subgroup $G=Deck(H/X)$. When $G$ contains only the identity, $\psi:H\to X$ is injective, then it's biholomorphic and onto. But when $G$ contains other stuff, I can't see why it's onto. Or can I prove that there is a biholomorphic mapping without using the condition that $\psi$ is onto?
Can I show that $\psi$ is actually a covering map? Since covering map is sure to be surjective. Forster's defines the universal covering like this:
Suppose $X$ and $Y$ are connected topological spaces and $p:Y\to X$ is covering map, $p:Y\to X$ is the universal covering of $X$ if it satisfies the following universal property. For every covering map $q:Z\to X$, with $Z$ connected, and every choice of points $y_0\in Y,z_0\in Z$ with $p(y_0)=q(z_0)$, there exists exactly one continuous fiber-preserving mapping $\psi:Y\to Z$ s.t. $\psi(y_0)=z_0$.
The definition of universal covering in Wikipedia is different, and the only difference is that $\psi$ should be a covering map. So I wonder are these two definitions equivalent? Since continuous fiber-preserving mapping does not implies covering, I can't see why they are equivalent. Thanks for attention!
With $f$ an unbranched unlimited covering, we have $\psi(H) = X$, since
$\psi$ is a non-constant holomorphic mapping (and $H$ is connected), so $\psi(H)$ is open, and
$X\setminus \psi(H)$ is also open (see below), hence, since $X$ is connected too, we must have $\psi(H) = X$.
To see that $X \setminus \psi(H)$ is open, consider an $x \in X\setminus\psi(H)$.
$f(x)$ has an evenly covered (by $f$) connected neighbourhood $U$. We may assume that $U$ is also evenly covered by $\exp$ (otherwise choose a smaller disk around $f(x)$). Let $V$ be the connected component of $f^{-1}(U)$ that contains $x$. If there was a $z \in H$ with $\psi(z) \in V$, then let $W$ be the component of $\exp^{-1}(U)$ containing $z$. Then $(f\lvert_V)^{-1} \circ (\exp\lvert_W)$ and $\psi\lvert_W$ coincide in a neighbourhood of $z$, hence in all of $W$, and thus it follows that $x\in (f\lvert_V)^{-1}\circ (\exp\lvert_W)(W) = \psi(W) \subset \psi(H)$ contradicting the assumption $x \notin \psi(H)$. Hence $V \cap \psi(H) = \varnothing$, and $X\setminus \psi(H)$ is recognised as open.