infinite covering of a compact surface corresponding to a handlebody subgroup Let $\Sigma$ be a closed orientable surface of genus $g \geq 2$.  Suppose that $\Sigma = \partial H$ where $H$ is a handlebody.  We then have the subgroup $N$ which is the kernel of the inclusion $\pi_1(\Sigma) \to \pi_1(H)$.  I know that this map is surjective on $\pi_1$ and since $\pi_1(H)$ is a free group of rank $g$, $N$ must be a countably generated free group.
What is the topology of the surface $\Sigma'$ covering $\Sigma$ corresponding to $N$?   I imagine that since the group must be free, there can not be any genus. 
 A reference for the result would be wonderful.  
I know that orientable open surfaces are classified by their genus (either finite or infinite) and their ideal boundary.  More generally I would love to know which of these types of surfaces can occur as (regular and/or irregular?) coverings of $\Sigma$.  
 A: $\Sigma'$ is homeomorphic to the sphere minus a Cantor set. 
Edited to address comments: One way to prove this is to prove that the universal cover $\tilde H$ of the handlebody $H$ is homeomorphic to the 3-ball $B^3$ minus a Cantor subset $C \subset \partial B^3 = S^2$, and that the restriction of the universal covering map $\tilde H \to H$ to the boundaries $\partial\tilde H \to \partial H = \Sigma$ is a regular covering map with deck transformation group $N=\pi_1 H$, hence $\Sigma' \approx S^2 - C$. 
And if you want to know why $\tilde H$ has the description that I say it has, a good way is to use 3-dimensional hyperbolic space $\mathbb{H}^3$. Using the ping-pong argument, there is an action of $F_g$ on $\mathbb{H}^3$ and a Cantor subset $C$ in the sphere at infinity $S^2_\infty = \partial \mathbb{H}^3$  (the set $C$ is called the "limit set" of the action) such that the restriction of $F_g$ to $\overline{\mathbb{H}}^3 - C \approx B^3 - C$ is free and properly discontinuous, and the quotient of this restricted action is homeomorphic to the genus $g$ handlebody.
Also, just because one is removing a Cantor subset $C \subset S^2$ it does not follow that $S^2 - C$ is homotopy equivalent to an uncountable wedge of circles. In fact $S^2-C$ is homotopy equivalent to a countable wedge of circles.
