$\operatorname{coker}(\phi)$ is discrete for a morphism of Lie groups

Question

suppose $\phi: G \to H$ a morphism of Lie groups such that $d\phi$ is surjective. Prove that $\operatorname{coker}\phi$ is discrete.

My attempts:

Prove that $\phi(G)$ is open which will lead to $H/\phi(G)$ is discrete: we have no topological information to use here.
prove that $Lie(H/\phi(G))=0$.

I don't know how to proceed and I don't see how to use the surjectivity of $d\phi$.

Thank you for your help.

Why do you say you have no topological information to use here? You know that $\phi$ is a submersion. Submersions are locally projections, and are thus open maps. — jgon, Apr 19 '19 at 14:44

score 2 · Accepted Answer · answered Apr 19 '19 at 14:44

2

Let $N$ be the kernel of $\phi$, $G/N$ is a Lie group and there exists a Lie hmomorphism $f:G/N\rightarrow H$, the differential of $f$ at any point of $G/N$ is an isomorphism, the local inverse mapping theorem implies that $f$ is open.

answered Apr 19 '19 at 14:44

Tsemo Aristide

87,475

Thank you @Aristide. Why is $d\phi$ a local isomorphism? it is surjective but is it injective? – Conjecture Apr 19 '19 at 14:58
$f$ the quotient map $G/N\rightarrow H$ is a local isomorphism. – Tsemo Aristide Apr 19 '19 at 15:00

$\operatorname{coker}(\phi)$ is discrete for a morphism of Lie groups

1 Answers1