Prove the following.
Theorem. Let \(G\) be a topological group with multiplication operation \(m: G \times G \to G\) and identity element \(e\). Assume \(p: \overline{G} \to G\) is a covering map. Given \(\tilde{e}\) with \(p(\tilde{e}) = e\), there is a unique multiplication operation on \(\overline{G}\) that makes it into a topological group such that \(\tilde{e}\) is the identity element and \(p\) is a homomorphism.
Let \(I: G \to G\) be the map \(I(g) = g^{-1}\). Show there exist unique maps \(\overline{m}: \overline{G} \times \overline{G} \to \overline{G}\) and \(\overline{I}: \overline{G} \to \overline{G}\) with \(\overline{m}(\tilde{e} \times \tilde{e}) = \tilde{e}\) and \(\overline{I}(\tilde{e}) = \tilde{e}\) such that \(p \circ \overline{m} = m \circ (p \times p)\) and \(p \circ \overline{I} = I \circ p\).
Show the maps \(\overline{G} \to \overline{G}\) given by \(\tilde{g} \to \overline{m}(\tilde{e} \times \tilde{g})\) and \(\tilde{g} \to \overline{m}(\tilde{g} \times \tilde{e})\) equal the identity map of \(\overline{G}\). [Hint: Use the uniqueness part of Lemma 79.1.]
Show the maps \(\overline{G} \to \overline{G}\) given by \(\tilde{g} \to \overline{m}(\tilde{g} \times \overline{I}(\tilde{g}))\) and \(\tilde{g} \to \overline{m}(\overline{I}(\tilde{g}) \times \tilde{g})\) map \(\overline{G}\) to \(\tilde{e}\).
Show the maps \(\overline{G} \times \overline{G} \times \overline{G} \to \overline{G}\) given by
\[ \tilde{g} \times \tilde{g}' \times \tilde{g}'' \to \overline{m}(\tilde{g} \times \overline{m}(\tilde{g}' \times \tilde{g}'')) \]
\[ \tilde{g} \times \tilde{g}' \times \tilde{g}'' \to \overline{m}(\overline{m}(\tilde{g} \times \tilde{g}') \times \tilde{g}'') \]
are equal.
- Complete the proof.
Solution
Let \[ H=p_*\pi_1(\overline G,\tilde e)\le\pi_1(G,e). \] For a topological group, \(\pi_1(G,e)\) is abelian, and the multiplication map induces addition: \[ m_*(u,v)=u+v, \] while inversion induces \(I_*(u)=-u\).
On fundamental groups, \[ (m\circ(p\times p))_*\pi_1(\overline G\times\overline G) =H+H=H, \] and \[ (I\circ p)_*\pi_1(\overline G)=-H=H. \] Thus the lifting criterion gives lifts \[ \overline m:\overline G\times\overline G\to\overline G, \qquad \overline I:\overline G\to\overline G \] with \[ p\overline m=m(p\times p),\qquad p\overline I=Ip, \] and with \(\overline m(\tilde e,\tilde e)=\tilde e\), \(\overline I(\tilde e)=\tilde e\). Uniqueness of based lifts makes these maps unique.
Define \[ L(\tilde g)=\overline m(\tilde e,\tilde g),\qquad R(\tilde g)=\overline m(\tilde g,\tilde e). \] Both satisfy \[ pL=p=pR \] and both send \(\tilde e\) to \(\tilde e\). The identity map of \(\overline G\) is another based lift of \(p\). By uniqueness of lifts, \[ L=R=1_{\overline G}. \] So \(\tilde e\) is a two-sided identity.
The maps \[ A(\tilde g)=\overline m(\tilde g,\overline I(\tilde g)),\qquad B(\tilde g)=\overline m(\overline I(\tilde g),\tilde g) \] both project to the constant map \(e\), since \[ pA(\tilde g)=p(\tilde g)p(\tilde g)^{-1}=e, \qquad pB(\tilde g)=e. \] They both take \(\tilde e\) to \(\tilde e\). The constant map \(\tilde g\mapsto\tilde e\) is the unique based lift of the constant map to \(e\). Hence \[ A=B\equiv\tilde e. \] Thus \(\overline I\) supplies two-sided inverses.
Let \[ F(\tilde g,\tilde g',\tilde g'') =\overline m(\tilde g,\overline m(\tilde g',\tilde g'')), \] \[ F'(\tilde g,\tilde g',\tilde g'') =\overline m(\overline m(\tilde g,\tilde g'),\tilde g''). \] Their projections are equal by associativity in \(G\): \[ pF=p(\tilde g)\bigl(p(\tilde g')p(\tilde g'')\bigr) =\bigl(p(\tilde g)p(\tilde g')\bigr)p(\tilde g'')=pF'. \] Both send \((\tilde e,\tilde e,\tilde e)\) to \(\tilde e\). By uniqueness of based lifts, \[ F=F'. \] Hence \(\overline m\) is associative.
Parts (b)–(d) show that \((\overline G,\overline m,\overline I,\tilde e)\) is a group. The operations are continuous because they were obtained as continuous lifts. By construction, \[ p(\overline m(x,y))=p(x)p(y),\qquad p(\overline I(x))=p(x)^{-1}, \] so \(p\) is a continuous group homomorphism.
Finally, any other multiplication making \(\tilde e\) the identity and \(p\) a homomorphism would be a based lift of \(m(p\times p)\), hence would equal \(\overline m\) by uniqueness. The inverse map is then forced as well. Thus the topological-group structure is unique.