The simply-connectedness of quotient space

Question

If $U$ is a Lie group with a closed subgroup $K$ such that both $U$ and $U/K$ are simply-connected, then can we conclude that $K$ is connected?

Answer 1

Yes, this follows from the long exact sequence on homotopy groups of the fibration sequence $K\to U\to U/K$. Specifically, we have an exact sequence $$\pi_1(U/K)\to \pi_0(K)\to \pi_0(U).$$ Since $U$ is connected and $U/K$ is simply connected, $\pi_1(U/K)$ and $\pi_0(U)$ are both trivial. Thus exactness of the sequence gives that $\pi_0(K)$ is trivial, so $K$ is connected.

Here's the argument in more geometric terms. If $K$ is disconnected, choose a path in $U$ between two points in different components of $K$ (this is possible since $U$ is connected). The image of this path in $U/K$ will then be a loop. If this loop were nullhomotopic, then the homotopy could be lifted to a homotopy of paths in $U$. But restricting this homotopy to an endpoint of the paths would then give a path in $K$ between the two different components of $K$, which is impossible. So the loop is not nullhomotopic, contradicting the assumption that $U/K$ is simply connected.