TITLE: Surjectivity of Lie group covering map
QUESTION [1 upvotes]: 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!

REPLY [1 votes]: 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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