Universal Cover of a Surface (with Boundary) I'm trying to see if there is a "nice-enough" way of describing/constructing the universal cover for a compact surface with $n$ boundary components. Clearly, if $n=0 $, the classification theorem for surfaces gives us a nice recipe; using either $S^1 $, or for genus $g$, using $\mathbb R^g$. I guess we could just cap a disk on each boundary component, end up with a genus$-g$ surface, find its cover, and then consider the restriction of the cover to the capped disks, but this does not seem to help. Any ideas?
Thanks.
 A: The universal covering space for all compact surfaces with nonempty boundary (except for the disk, annulus and Moebius band) is the following. Take the closed 2-disk and then remove a Cantor set from its boundary.  Here Cantor set is any set homeomorphic to the standard Cantor set. Describing the covering transformation group is much more difficult, although possible. 
Edit. Here are some steps towards the proof assuming you do not know any hyperbolic geometry. First of all, the universal cover is contractible and has nonempty boundary. One then sees that it's interior is homeomorphic to an open disk; call the boundary circle of the disk C. Then by working a bit more you see that it (let's call the universal cover U) is an open disk plus some disjoint union (call it A) of (open) boundary  arcs. 
Using the fact that fundamental group is free nonabelian you see that A consists of countably infinitely many boundary arcs. Then you check that topology of U does not  change if you collapse each component of C minus A to a point. Now the set C minus A is compact, metrizable, totally disconnected and infinite, call this set K. The next and the most difficult step is to show that K is perfect. Lastly, you quote a theorem from general topology that such K is homeomorphic to the standard Cantor set.  
Edit: As an alternative to the topological argument, you can use hyperbolic geometry as follows: First, you realize that your compact surface with boundary admits a complete hyperbolic metric with geodesic boundary. Then, represent the surface equipped with this metric as the quotient of a closed convex domain $D$ in the hyperbolic plane ${\mathbb H}^2$ by a "Fuchsian group of the 2nd kind" $\Gamma$. I will think of the hyperbolic plane as the open unit disk equipped with the Poincare metric.Consider the closure $\bar{D}$ of $D$ in the complex plane. It is homeomorphic to the closed unit disk. The frontier of $D$ is the disjoint union of two sets: A countably infinite disjoint union of open circular arcs and a certain set $\Lambda(\Gamma)$, the intersection of $\bar{D}$ with the unit circle. The set  $\Lambda(\Gamma)$ is compact, perfect, totally disconnected and, of course, metrizable. Therefore, it is homeomorphic to the standard Cantor set. See for instance the book
S. Katok, "Fuchsian Groups", 
especially Theorem 3.4.5 and section 3.6, for (many) more details. 
For the annulus and Moebius band the universal covering space is the product of a closed interval and the real line. The proof is elementary, by describing the covering map. In the case of annulus you can use polar coordinates to define this map. 
