Cover of a finitely punctured plane

Question

Let $X_n$ be the plane with a finite number $n$ of punctures, and let $p : Y \rightarrow X_n$ be a covering map (it may have infinite degree). Can we say anything about the topology of $Y$? (I know that $Y$ has a natural structure of Riemann surface).

I should really read properly about the general theory of covering maps, which I only vaguely know. I've heard about a 1 to 1 correspondance between covering maps and automorphisms of the fundamental group (which is here the free group with $n$ generators) but I don't know anymore than that.

Answer 1

In fact, the following is true:

Let $X$ be a compact connected oriented surface (without boundary). Then there exist finite subsets $P\subset X$, $Q\subset S^2$ and a finite cover $p: (X-P)\to (S^2 - Q)$.

So see this, recall that every surface $X$ as above, admits a complex structure and, with respect to this structure, it admits a nonconstant holomorphic map $f: X\to S^2$. This map is not a covering map, but, let $P_o\subset X$ denote the set of critical points of $f$ (this is exactly the subset where $f$ fails to be a local homeomorphism), set $Q=f(P_o)$, $P=f^{-1}(Q)$. Then the restriction of $f$ $$ p: (X -P) \to (S^2 -Q) $$ is a local homeomorphism (clear) and is proper. (The latter is a nice exercise.) Therefore, by the "stack of records" theorem, $p$ is a covering map. The surface $S^2-P$ is, of course, homeomorphic to a punctured plane (if $Q$ is nonempty, which will be the case unless $X\cong S^2$).

For a more concrete example, consider $X=T^2$. Then there exists an involution $\tau$ of $T^2$ which has 4 fixed points, such that the quotient $T^2/\tau$ is homeomorphic to $S^2$. If you think of $T^2$ as $S^1\times S^1$, then $$ \tau(z, w)= (z^{-1}, w^{-1}) $$ where I identify $S^1$ with the unit circle in the complex plane. Then, $P\subset X$ consists of 4 fixed points and $Q\subset S^2$ is also a 4-point subset. Removing $P$ and $Q$ from $X$ and $S^2$, we obtain a 2-fold regular cover (with the group $Z_2$ of covering transformations). The involution $\tau$ is a special case of the "hyperelliptic involution": Compact oriented surface of any genus $\ge 1$ has a complex structure which admits such an involution $\tau$. The quotient $X/\tau$ is again $\cong S^2$.

To conclude, not much can be said about (even finite) covers of punctured planes. However, every such cover extends to a branched cover over $S^2$ and if you prescribe the ramification data at the punctures and the degree, you can use the Riemann-Hurwitz formula in order to compute the genus of the covering surface (and more).

Answer 2

$Y$ can be homeomorphic to a punctured Riemann surface ( @studiosus: thanks!). I guess we should think about the Riemann surfaces as they were first conceived.

As to the coverings and fundamental groups, there is a $1-1$ correspondence between ( conjugacy classes of ) subgroups of the fundamental group of $X_n$ and coverings of $X_n$. The fundamental group of $X_n$ is free on $n$ generators. Any subgroup of $\pi_1(X_n)$ is again free. So you know right away that $\pi_1(Y)$ is again free.