I'm trying to solve a problem found in a topology book. The problem is :
Let $G$ be a connected topological group and let $H$ be a closed, locally path-connected subgroup of $G$. Prove that if $G/H$ is simply connected, then $H$ is connected.
The book has hints that go like:
- Let $H_{0}$ be the connected component of the neuter element of $H$. Then $H_{0}$ is normal and $H/H_{0}$ is discrete.
- $G/H_{0}\rightarrow G/H$ is a covering
- conclude that $H=H_{0}$
I think I managed the first two by using only the fact that $H$ is closed. However I can't manage to prove the third one. The book has several variations on the theorem that if a covering space is locally path-connected and the base is simply connected, then it is a homeomorphism, but they always need that the space itself be locally path-connected and $G$ is not necessarily path-connected.
I tried considering only the path-connected component containing the neuter element of $G$ , let us say $G_{0}$ , but could not prove that $G_{0}/H$ is simply connected even if $G/H$ is. So I know that $G/H$ is connected and simply connected, but not path-connected.
I also seem to not use at all the fact that $H$ is locally path-connected.