(May 2013) 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
- Identifying two distinct points in a path-connected CW complex adds a circle up to homotopy. Applying this to the north and south poles of \(S^2\) gives \[ X\simeq S^2\vee S^1. \] One can see this directly by choosing a CW decomposition with the two poles as vertices; after the identification and collapse of a maximal tree, one \(1\)-cell remains from the identification while the \(2\)-sphere remains as a sphere summand.
Thus \[ \pi_1(X)\cong\mathbb Z. \]
- Connected covers correspond to subgroups of \(\mathbb Z\), namely \(n\mathbb Z\) for \(n\ge1\) and the trivial subgroup. The universal cover is a bi-infinite chain of spheres \[ \cdots\cup S^2_{-1}\cup S^2_0\cup S^2_1\cup\cdots \] with the north pole of \(S^2_i\) identified with the south pole of \(S^2_{i+1}\). The generator of the deck group shifts the chain by one sphere.
For \(n\ge1\), quotienting by the subgroup \(n\mathbb Z\) produces an \(n\)-sheeted cover: a cyclic necklace of \(n\) spheres, with successive north and south poles identified cyclically. The trivial subgroup gives the infinite chain. These are all connected covering spaces up to equivalence.