Let \(G\) be a path-connected, locally path-connected topological group with identity \(e\) and multiplication \[ \mu:G\times G\to G. \] Let \[ p:(H,e')\to(G,e) \] be a universal covering map.
Show that there is a continuous map \[ \mu':H\times H\to H \] such that \[ p\circ\mu'=\mu\circ(p\times p). \]
Solution
1. The space \(H\times H\) is path connected and simply connected.
The universal covering space \(H\) is path connected and simply connected. Products of path-connected spaces are path connected, and \[ \pi_1(H\times H,(e',e')) \cong \pi_1(H,e')\times\pi_1(H,e') =0. \]
2. Define\[ F\definedas\mu\circ(p\times p):H\times H\to G. \]Then\[ F(e',e')=e. \]
Since \(p(e')=e\) and \(e\) is the identity of \(G\), \[ F(e',e') =\mu(p(e'),p(e')) =\mu(e,e) =e. \]
3. The based map\[ F:(H\times H,(e',e'))\to(G,e) \]has a unique lift\[ \mu':(H\times H,(e',e'))\to(H,e') \]through \(p\).
The covering-space lifting criterion says that a based map \(F:(X,x_0)\to(G,e)\) lifts to \((H,e')\) if and only if \[ F_*\pi_1(X,x_0) \subseteq p_*\pi_1(H,e'). \] Here both groups are trivial: the left one by 1 and the right one because \(H\) is simply connected. Thus a lift exists. Its value at the basepoint is prescribed to be \(e'\), so uniqueness of based lifts gives uniqueness.
4. The lift \(\mu'\) has the required compatibility with multiplication on \(G\).
By the definition of a lift in 3, \[ p\circ\mu'=F. \] Substitute the definition of \(F\) from 2 to obtain \[ p\circ\mu'=\mu\circ(p\times p). \] In particular, \[ p(\mu'(a,b))=p(a)p(b) \qquad(a,b\in H). \] Continuity of \(\mu'\) is part of the lifting theorem, so no path-dependent definition of the product is required.