In a topological group, is $G/A\to G/B$ a covering map if $A$ is open in $B$?

Question

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].

Answer 1

Let $C$ be the intersection of the conjugates $bAb^{-1}$. Assume the additional hypothesis that $C$ is open in $B$. This holds, for example, when $B$ is locally connected.

The set $B\backslash C$, complement of $C$ in $B$, is closed in $G$, and therefore every point in $G$ belongs to some open set $U$ such that $U^{-1}U\cap B\subset C$. This means that whenever $u\in U$ and $b\in B$ and $ub\in U$, then $b\in C$. That is, if $u_1B=u_2B$ then $u_1C=u_1C$ (and therefore $u_1A=u_2A$).

Claim: The open set $UB/B\subset G/B$ is evenly covered by the map $G/A\to G/B$.

Proof: For each coset $bA\in B/A$ we have the open set $UbA/A\subset G/A$. The union of these sets is $UB/A$, and this maps onto $UB/B$. In fact, for each $b\in B$ the map $UbA/A\to UB/B$ is surjective.

Let's show that each of the surjections $UbA/A\to UB/B$ is injective. Suppose that $u_1bA/A$ and $u_2bA/A$ go to the same point. That means $u_1bB/B=u_2bB/B$, i.e. $u_1B=u_2B$, which by the hypothesis on $U$ implies that $u_1=u_2c$ for some $c\in C$. Now we have $$ u_1bA=u_2cbA=u_2b(b^{-1}cb)A=u_2bA. $$

To see that the continuous bijection $UbA/A\to UB/B$ is a homeomorphism, observe that the inverse corresponds to a map $s:UB\to UbA/A$ which is continuous because for each $b_0\in B$ the restriction of $s$ to the open set $Ub_0\subset UB$ is given by $$ ub_0\mapsto ubA/A. $$

Now let's show that the sets $UbA/A$ (one for each coset $bA$) are pairwise disjoint. Assume that some point $u_1b_1A\in Ub_1A/A$ is the same as some point $u_2b_2A\in Ub_2A/A$. I want to show that $b_1A=b_2A$. By assumption $u_1b_1A=u_2b_2A$. In particular $u_1B=u_2B$, so as before $u_1=u_2c$ for some $c\in C$. Now $$ u_2b_2A=u_1b_1A=u_2cb_1A, $$ so that $$ b_2A=cb_1A=b_1(b_1^{-1}cb_1)A=b_1A. $$