I have the following question
Let $G$ be a locally connected topological group and $H < G$ a closed locally connected subgroup, show that if $H$ is not connected then $G/H$ is not simply connected.
However the book does not provide the definition of simply connectedness in the context of topological groups, i know the the definition that is being path connected and with paths that can shrink to a point (something like that).
The hint is to use the identity connected component, so let $H_0$ be this component.
I know that $H_0$ is open since $H$ is loc connected, also $H_0$ is closed and proper.
My intuition tells me there is something to do with the action that takes $g \in G$ and goes to $gH_0 \in G/H_0$. But as i don't know what i should be looking to prove here i don't know how to procede. If anyone could enlighten me i would apreciate.
Edit: Apparently the way is showing that $G/H_0$ is homeomorphic to $G/H$ when $G/H$ is simply connected, once i can do this i will post the result.
Edit 2: $G$ needs to be connected