Covering Space of the Pearl Necklace Let $S^{2}_{1},\dots, S^{2}_{n}$ be disjoint copies of the unit  sphere, and, for each $i\in\{1,\dots, n\}$, let $p_i,q_i\in S^{2}_{i}$ be distinct points. Define an equivalence relation $E\subseteq (\bigcup_{i=1}^{n} S_{i}^{2})^2$ by 
$$pEq \Leftrightarrow \big ((p = q) \text{ or }  (p = p_i \text{ and } q = q_{i+1 \operatorname{mod} n})\big )$$
And let $X =\bigcup_{i=1}^{n} S_{i}^{2}/E $ be the n-pearl necklace. How can I determine a simply connected covering space for X.
 A: Consider the space $\widetilde X$ consisting of an infinite sequence of spheres, extending in both directions, with each sphere tangent to its two neighbors. This space is simply-connected (via Van Kampen, for example).
Consider the following action of $\mathbb Z$ on $\widetilde X$. For $k \in \mathbb Z$, each sphere is translated $kn$ times left or right depending on the sign of $k$. This is a covering space action. The resulting orbit space is a necklace of $n$ spheres.
A: The universal cover for this "pearl necklace" is the infinite string of beads, which is infinitely many copies of $S^2$, where the north pole of the $i^{th}$ sphere is glued to the south pole of the $i+1_{st}$ sphere. Proving that this is a covering space isn't too hard, and to prove that this is simply connected use the fact that the image of a compact set is compact, that compact subsets of euclidean space are closed and bounded (so take an embedding of your space into $\mathbb{R}^3$) and prove that a finite string of beads is simply connected. 
A: One way to get at this is to expand the pairs of identified points into edges. Then you have a cycle of edges and spheres joined end-to-end. Then ends of one of the edges can be dragged along the space and joined together to form a circle based at a point $*$. One is left with two spheres that only connect to one edge each. Drag the other ends of these edges so they emanate from the point $*$. Repeat until all of the spheres are connected to $*$ by a single edge. Now contract these edges to get
$$(\bigvee_{i=1}^n S^2)\vee S^1.$$
