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Let $X$ be a topological group with a locally path-connected, path-connected covering $(\tilde{X},p)$. If we fix an $u\in p^{-1}(e)$, we should be able to deduce a unique group structure for $H=\tilde{X}$ such that the identity element is $u$, and $p$ is a morphism of groups.

Up to now, I've been able to use the lifting criterion to deduce the product $$g\,:\,H\times H\longrightarrow H,$$ and I've also shown using the uniqueness of liftings that $g(h,\, u)=h=g(u,\, h)$.

Now, I'm stuck on proving that the operation $g$ is associative and there are inverses. Do you have any hint? I know I should apply uniqueness of liftings but how in those cases?

EDIT: I managed to prove that the $g$ is associative, but I still cannot construct the right maps for the inverses.

EDIT2: Let $i_G$ be the inversion map for $G$, then we consider the composition $$i_Gp:H\longrightarrow G.$$ The problem should now be if it is possible to lift that map using the lifting criterion. But I cannot see why $(i_G)_*(p_*(\pi_1(H,\, u)))$ should be contained in $p_*(\pi_1(H,\, u))$

W4cc0
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  • Is your topological group $X$ semi-locally simply connected? – Sumanta Oct 14 '18 at 13:20
  • No. I know that would give an "universal cover", and than you can probably construct directly the product in the cover, but no, I'm looking for an answer without that hypothesis. – W4cc0 Oct 14 '18 at 13:54
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    To prove what you want, you can apply the Eckmann-Hilton argument to see that the two multiplications on $\pi_1(X,e)$ coincide – Maxime Ramzi Oct 14 '18 at 17:37
  • @Max That was unexpectedly helpful! – W4cc0 Oct 15 '18 at 10:45
  • @W4cc0 After lifting the map $i_G p$, how did you show that the lifted map is the inversion map of $H$? I got stuck in this step. – user302934 Oct 29 '19 at 14:40

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