TITLE: Universal covers of non-prime 3-manifolds
QUESTION [8 upvotes]: Let $M$ be a closed, connected, oriented 3-manifold. If $M$ is prime, then we know what the universal cover of $M$ looks like: it is either $S^3, \mathbb{R}^3$ or $S^2 \times  \mathbb{R}$ depending on whether the 3-manifold is spherical, aspherical or $S^2 \times  S^1$.
If $M$ is non-prime, what does its universal cover look like? It must be a simply connected, non-compact 3-manifold (without boundary), but I do not know whether they are well-understood.
Perhaps something more concrete: what is the universal cover of the connected sum of lens spaces?

REPLY [3 votes]: Here is a version of Ryan's answer (in the comments).
Suppose that $M$ and $N$ are non-trivial lens spaces. Note that $M \# N$ is covered by $D_g$, the connect sum of $g$ copies of $T = S^2 \times S^1$, for some $g > 0$.  (Also, $g = 1$ if and only if $M = N = \mathbb{RP}^3$).
Suppose that we are not in that special case.  Then $g > 1$ and $D_g$ covers $D_2 = T \# T$.  So it suffices to understand the universal cover of $D_2$.  The manifold $D_2$ is obtained by doubling a genus two handlebody $U_2$.  Thus the universal cover of $D_2$ is obtained by taking the universal cover of $U_2$, and doubling.  However, the universal cover of $U_2$ is homeomorphic to a closed ball, minus a Cantor set from its boundary.  Thus the universal cover of $D_2$ is, as claimed, a copy of $S^3$ minus a Cantor set from its equatorial two-sphere.

This is somewhat similar to my answer here.