(May ’13) Let \(X\) be the space obtained from the \(2\)-sphere \(S^2\) by identifying the north and south poles (i.e. by identifying two diametrically opposite points).
Show that \(X\) is homotopy equivalent to \(S^1\vee S^2\).
Describe all connected covering spaces of \(X\).
Solution
Give \(S^2\) a CW decomposition in which the north and south poles are \(0\)-cells. After identifying these two vertices, choose a maximal tree in the resulting \(1\)-skeleton and collapse it. The quotient CW complex reduces to one \(1\)-cell accounting for the identified pair of vertices and one \(2\)-sphere summand. Equivalently, the standard cofibration fact for a path-connected CW complex says that identifying two distinct \(0\)-cells adds a circle up to homotopy. Hence \[ X\simeq S^2\vee S^1. \] In particular, \[ \pi_1(X)\cong\mathbb Z. \]
Connected coverings are classified by subgroups of \(\pi_1(X)\cong\mathbb Z\). These are \[ n\mathbb Z\quad(n=1,2,3,\dots) \qquad\text{and}\qquad \{0\}. \] They admit a concrete geometric description. The universal cover is a bi-infinite chain \[ \cdots\cup S^2_{-1}\cup S^2_0\cup S^2_1\cup\cdots, \] where the north pole of \(S^2_i\) is identified with the south pole of \(S^2_{i+1}\). The deck generator shifts \(S^2_i\) to \(S^2_{i+1}\).
For \(n\ge1\), quotienting this universal cover by the subgroup \(n\mathbb Z\) gives an \(n\)-sheeted connected cover: a cyclic necklace of \(n\) spheres, with the north pole of each sphere identified with the south pole of the next, cyclically. For \(n=1\) this is \(X\) itself. The subgroup \(\{0\}\) gives the universal infinite chain. These exhaust all connected covering spaces up to equivalence.