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Let $G$ be a Lie group, $\tilde{G}$ its simply connected universal cover and $\pi: \tilde{G} \longrightarrow G$ the associated covering map. Then $\tilde{G}$ is also a Lie group and $\pi$ is a Lie group homomorphism.

Because $\pi$ is surjective, it is clear that $\pi$ induces a bijective group homomorphism between $\tilde{G}/\ker$ and $G$. Why is this homomorphism also a homeomorphism?

Sebastiano
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Drops
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    you can even see it is a diffeomorphism by applying the smooth orbit-stabilizer theorem to a surjective Lie group homomorphism $\pi:G\to H$ with the group action of $G$ on $H$ given by $(g,h)\mapsto \pi(g)h$. Apply the orbit-stabilizer theorem to $e_H$, i.e $H=\text{orbit}(e_H)\cong G/\text{Stab}(e_H)=G/\ker \pi$. (Note that at the purely algebraic level, I think this is a nice way to look at the first isomorphism theorem). – peek-a-boo Jan 02 '24 at 10:55

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Let $\pi^\star$ be that map. Since it is a continuous bijection, in order to prove that it is a homeomorphism all that is needed is to prove that $\pi^\star$ is an open map. Let $A$ be an open subset of $\widetilde G/\ker\pi$. Then $\pi^{-1}(A)$ is an open subset of $\widetilde G$. But $\pi^\star(A)=\pi\left(\pi^{-1}(A)\right)$ and, which is an open set, since $\pi$ is an open map (since it is a covering map) and $\pi^{-1}(A)$ is open.

José Carlos Santos
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  • Thank you. I missed the fact, that covering maps are open, since i was mainly looking at the definition from Hall's Lie Groups, Lie Algebras, and Representations. It defines a universal cover as a simply connected Lie group $\tilde{G}$ together with Lie group homomorphism $\pi:\tilde{G} \longrightarrow G$ such that the associated Lie algebra homomorphism $\Pi: Lie(\tilde{G}) \longrightarrow Lie(G)$ is a Lie algebra isomorphism. From this definition, it it not that obvious (for me at least) that the covering map is open. Is it because the exponential map for a connected Lie group is open? – Drops Jan 02 '24 at 11:18
  • @Drops I tried to answer this with a comment, but didn't have enough space, so it has turned into a post below. – krm2233 Jan 02 '24 at 14:07
  • @krm2233 That was nice. I have a suggestion for both of you: Drops could post that question as a separate question and then krm2233 could post their answer as an answer to that question (and delete the answer posted here). – José Carlos Santos Jan 02 '24 at 14:10
  • @JoséCarlosSantos happy to do that if Drops would like to! – krm2233 Jan 02 '24 at 19:11
  • Thank you very much. I have opened another question here https://math.stackexchange.com/questions/4838739/the-universal-cover-phi-h-longrightarrow-g-of-a-connected-lie-group-g-ind

    In the answer below, the notation is a little mixed up in some places (i think mainly because i used a capital letter for the Lie algebra homomorphism). The new questions has new notation.

     – Drops Jan 04 '24 at 17:36
  • Nice! I have already upvoted it. And if @krm2233 posts there the answer which was posted below, I will upvote it too. – José Carlos Santos Jan 04 '24 at 17:40