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I came across this question: link. There it is established that if $G$ be a Lie group and let $F:G \to H$ a surjective Lie group homomorphim such that $\Gamma=\ker F$ is a discrete subgroup, then the orbit space $G/\Gamma$ is diffeomorphic to $H$. The proof in the question starts like this:

Let $\pi:G \to G/\Gamma$ be the quotient map. Define $\tilde{F}: G/\Gamma \to H$ by $\tilde{F}(\Gamma x) = F(x)$. This is a well defined bijection which is also a homeomorphism.

I see that $\tilde{F}$ is a continuous bijection, but why is it a homeomorphism? I've never studied Lie groups and their properties so could someone quickly explain why this follows? (and would it also hold if $G,H$ were only topological groups?)

Henry T.
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  • You should read not just the question but also the answer. – Moishe Kohan May 05 '24 at 00:59
  • I'm sorry if I'm overlooking something obvious but the answer just starts with "you proved that this is a homeomorphism..." but doesn't address how this was proved. – Henry T. May 05 '24 at 01:03
  • You have to read more carefully: it uses the inverse mapping theorem to prove a diffeomorphism. – Moishe Kohan May 05 '24 at 01:04
  • But there is also a general result about completely metrizable sigma-compact topological groups: every bijective continuous homomorphism is a homeomorphism. – Moishe Kohan May 05 '24 at 01:06
  • If you have a lie group homomorphism $f$ you can find charts s.t. $f$ looks like $(x_1,...,x_m)\mapsto(x_1,x_2,...)$ where the $...$ either goes on like this up to $m$ and is $0$ afterwards or stops before that because the dimension of the codomain is less than or equal to the dimension of the domain. –  May 05 '24 at 02:50
  • See my answer here. – Moishe Kohan May 05 '24 at 03:04

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