Universal cover of $(S^1\times S^3)\# (S^1\times S^3)$ How can we determine the universal cover of $(S^1\times S^3)\# (S^1\times S^3)$, or more generally of 
$$
X_k:=\overset{k}{\underset{i=1}{\#}}(S^1\times S^3)
$$ 
for $k\in \mathbb{N}$? 
The case $X_1$ I am able to compute: $\tilde{X}_1\cong \mathbb{R}\times S^3$.
For $n>1$, according to this post, $\tilde{X}_k$ is obtained by "appropriately" gluing the universal covers of $(S^1\times S^3)\setminus \{\mathrm{pt}\}$. What is the appropriate gluing in this case?
Thanks.
 A: The fundamental group is a free group of rank $k$, as one can see by applying Van Kampen's Theorem. 
Therefore, the "appropriate" gluing pattern is to take the universal cover of a rank $k$ rose, which is a tree with every vertex of valence $2k$, then to place a copy of $\overset{k}{\underset{i=1}{\#}}([0,1]\times S^3)$ at each vertex, which is a 4-manifold with boundary whose boundary has $2k$ components, each homeomorphic to $S^3$, and then to glue together a pair of these $S^3$ boundaries for each edge, one $S^3$ boundary from one endpoint of the edge and the other $S^3$ boundary from the opposite endpoint of the edge.
A: Start with a collection of pairwise disjoint closed round balls $B_i, B_i', i=1,...,k,$ in $S^n$. Let $F$ denote the closure of the complement
$$
S^n - \bigcup_{i=1}^k (B_i\cup B_i'). 
$$
The boundary of $F$ has standard orientation induced from $F$. The boundary of $F$ is a disjoint union of 
round sphere $S_i=\partial B_i, S'_i=\partial B'_i$. Let $h_i: S_i\to S'_i, i=1,...,k$ be orientation-reversing diffeomorphisms. 
Let $M$ denote the $n$-dimensional manifold obtained by gluing boundary spheres of $F$ via   the maps $h_i, i=1,..,k$. 
It is a pleasant exercise to verify that $M$ is diffeomorphic to the $k$-fold connected sum of $S^n\times S^1$. 
I now impose more conditions on the maps $h_i$:  I will assume that all these are Moebius transformations of $S^n$. 
If you consider $S^n$ as the boundary of the unit ball model of the hyperbolic space ${\mathbb H}^{n+1}$ 
then extensions of isometries of ${\mathbb H}^{n+1}$ are precisely the Moebius transformations of $S^n$. 
Let $G$ denote the group of Moebius transformations of $S^n$ generated by $h_1,...,h_k$. One verifies (by induction on word-length) 
that $G$ is a free group on the generating set $h_1,...,h_k$ and that $F$ is a fundamental domain of $G$: The intersection 
$gF\cap F$ is either $F$ (precisely when $g=1$) or is a boundary sphere of $F$ (precisely when $g=h_i^{\pm 1}$ for some $i$) 
or is empty.  
Assuming that $n\ge 3$, the domain $F$ is simply-connected and so is the union
$$
\Omega=\bigcup_{g\in G} gF. 
$$
Thus, $\Omega$ is equivariantly diffeomorphic to the universal covering space of the manifold $M$. The domain $\Omega$ 
is the discontinuity domain of the group $G$. The complement $\Lambda=S^n -\Omega$ is the limit set of $G$. 
One verifies that $\Lambda$ is totally disconnected and perfect (contains no isolated points) provided that $k\ge 2$ which I will assume from now on. 
One way to prove it is as follows: Let $G_L$ denote the subset of $G$ consisting of words of length $\le L$ in  the generators 
$h_i^{\pm 1}$, $i=1,...,k$. Then observe that the boundary of the union
$$
F_L= \bigcup_{g\in G_L} gF
$$ 
consists of round spheres of radii $\le r_L$ and $\lim_{L\to\infty} r_L=0$. 
Thus, $\Lambda$ is homeomorphic to the classical Cantor set.  
One can do a bit better. Fix a round circle $S^1\subset S^n$ and assume that the spheres $S_i, S_i'$ are chosen to be orthogonal to $S^1$. Then the Moebius transformations $h_i$ can be chosen to preserve the circle $S^1$. It follows that the limit set $\Lambda$ is contained in $S^1$. Identify $S^n$ with the 1-point compactification of ${\mathbb R}^n$  and choose $S^1$ to be the one-point compactification of a straight line ${\mathbb R}^1$ in ${\mathbb R}^n$. Then there exists a homeomorphism $f: S^n\to S^n$ 
sending $S^1$ to itself and sending $\Lambda$ to the standard ternary Cantor subset $K$ of ${\mathbb R}^1$. 
We conclude that the universal covering space of $M$ is homeomorphic to ${\mathbb R}^n \cup \{\infty\} - K$. 
You can find more details in Chapter "Schottky Dance" in  
"Indra's Pearls: The Vision of Felix Klein" 
by David Mumford, Caroline Series and David Wright. 
