Let $G$ be a topological group, let $A,B$ be closed subgroups of $G$ such that $A$ is an open subgroup in $B$. Then we have an open continuous map $f:G/A\to G/B$, with typical fiber $B/A$. I read the claim that $f$ is a covering map (which sounded very convincing at first) but I was unable to prove this.

This is easy if $B$ is a discrete subgroup, and I would be happy with a proof that works if $B$ is locally compact.

So what can be said about $f$ in general?

(Everything is Hausdorff.)

Let $G$ be a (Hausdorff) topological group, let $A,B$ be closed subgroups of $G$ such that $A$ is an open subgroup in $B$. Then we have an open continuous map $f:G/A\to G/B$, with typical fiber $B/A$. I read the claim that $f$ is a covering map (which sounded very convincing at first) but I was unable to prove this.

This is easy if $B$ is a discrete subgroup, and I would be happy with a proof that works if $B$ is locally compact.

So what can be said about $f$ in general?

Edit: If $G$ is a Lie group, then it is well-known that $f$ is a locally trivial fiber bundle. If $G$ is locally compact and almost connected, then $f$ is a fibration by [Skljarenko, The topological structure of locally bicompact groups and their quotients].

(An example of a quotient map where no local section exists is $G=SU(2)^I$ for some infinite set $I$, and $K=Z(G)=\{\pm1\}^I$. In this case $p:G\to G/K$ admits no local section. By Skljarenko's result mentioned above, $p$ is nevertheless a fibration.)

The claim above about $f$ is made in [Wigner, Algebraic cohomology of topological groups].