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I am trying to solve Chapter's 3 exercise 11 in the "Lie Groups, Lie Algebras and Representations" Brian C. Hall's book. It reads as:

If $\tilde{G}$ is a universal cover of a connected Lie group $G$ with projection map $\Phi$, show that $\Phi$ maps $\tilde{G}$ onto $G$.

It looks like it has an easy and elegant proof but I am not arriving to it. Any help will be appreciated. Thanks!

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    $\begingroup$ Isn’t surjectivity in the definition of covering map? $\endgroup$
    – Aphelli
    Jul 10, 2019 at 7:36
  • $\begingroup$ I think not. The definition in this book says that $\Phi$ from $\tilde{G}$ to $G$ is only a Lie group homomorphism (such that the associated Lie Algebra homomorphism is a Lie Algebra isomorphism). $\endgroup$
    – Txordi
    Jul 10, 2019 at 18:56

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The map is open in a neighborhood of the identity (by inverse theorem, because is the identity at the level of its Lie algebra), and it is also a group homomorphism, so, its image is an open connected subgroup containing the identity. This implies that the image is the connected component in the identity

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  • $\begingroup$ I hope you'll consider this proof "easy and elegant" as required :) $\endgroup$ Jul 12, 2019 at 17:10
  • $\begingroup$ Sorry for the delay, I have not seen the notification. It is indeed great! However, I am not sure how do you use the inverse theorem. Shall I use that since $\Phi$ is also a smooth map, then it maps any open set in $\tilde{G}$ to an open set in $G$? $\endgroup$
    – Txordi
    Jul 14, 2019 at 16:30
  • $\begingroup$ The differential of the map, at the identity, is an isomorphism. $\endgroup$ Jul 14, 2019 at 23:51

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