Show that an H-space or topological group structure on a path-connected, locally path-connected space can be lifted to such a structure on its universal cover. [For the group \(SO(n)\) considered in the next section, the universal cover for \(n > 2\) is a 2-sheeted cover, a group called \(\operatorname{Spin}(n)\).]
Solution
Let \[ p:\widetilde X\to X \] be the universal covering and choose \(\widetilde e\in p^{-1}(e)\).
For an H-space multiplication \(\mu:X\times X\to X\), consider \[ \mu\circ(p\times p):\widetilde X\times\widetilde X\to X. \] The domain is simply connected, so this map has a unique lift \[ \widetilde\mu:\widetilde X\times\widetilde X\to\widetilde X \] with \(\widetilde\mu(\widetilde e,\widetilde e)=\widetilde e\). The homotopies expressing the left and right identity laws for \(\mu\) lift uniquely, starting from the corresponding restrictions of \(\widetilde\mu\), and end at the identity map of \(\widetilde X\). Hence \(\widetilde X\) is an H-space.
If \(X\) is a topological group, apply the same construction to multiplication. The inversion map \[ \iota:X\to X, \qquad x\mapsto x^{-1}, \] also lifts uniquely to a map \(\widetilde\iota:\widetilde X\to\widetilde X\) fixing \(\widetilde e\). Associativity of \(\widetilde\mu\) follows from uniqueness of lifts: the two maps \[ \widetilde\mu(\widetilde\mu(-,-),-), \qquad \widetilde\mu(-,\widetilde\mu(-,-)) \] are lifts of the same map \(\widetilde X^3\to X\) and agree at \((\widetilde e,\widetilde e,\widetilde e)\), so they are equal. The strict identity and inverse laws follow in exactly the same way by comparing the relevant lifts. Thus \(\widetilde X\) is a topological group and \(p\) is a homomorphism.