Two models for the classifying space of a subgroup via the geometric bar construction

Question

Let $H$ be a topological group which is a subgroup of two other topological groups $G$ and $G'$. It follows (from Rmk 8.9 in May - Classifying spaces and fibrations (MSN, free)) that there exist weak equivalences $B(*,G,G/H)\to BH$ and $B(*,G',G'/H)\to BH$.

One of the reasons one would like to look at such a model for $BH$ would be if one understands $G$ and $G/H$ (and $G'$ and $G'/H$) better than $H$. Now of course I could realize the weak equivalence between $B(*,G,G/H)$ and $B(*,G',G'/H)$ by the following zig-zag of weak equivalences $$B(*,G,G/H)\xrightarrow\sim BH\xleftarrow\sim B(*,G',G'/H). $$ However, since I don't "understand" $H$ I would like to realize the weak equivalence between $B(*,G,G/H)$ and $B(*,G',G'/H)$ without using $H$, but rather using constructions that uses $G$, $G'$, $G/H$, $G'/H$ etc.… Is that even possible?

Answer 1

One way to think about this is as follows. One has a topological groupoid $\mathcal G$ with objects $G/H$ and morphisms $G \times G/H$. Similarly one has the groupoids $\mathcal G^{\prime}$, and $\mathcal H$ (with just one object $H/H$). Your bar constructions are just the classifying spaces of these three groupoids, and equivalences between these categories will induce homotopy equivalences.

There are evident inlclusions of groupoids $\mathcal G \leftarrow \mathcal H \rightarrow \mathcal G^{\prime}$, and thus maps $B\mathcal G \leftarrow B\mathcal H \rightarrow B\mathcal G^{\prime}$. May apparently gives you maps in the other direction, but you can also do this by finding morphisms (i.e. functors) of groupoids $\mathcal G \rightarrow \mathcal H \leftarrow \mathcal G^{\prime}$. I am going to be lazy at the this point, and assume a discrete topology. A morphism $\mathcal G \rightarrow \mathcal H$ will involve choosing coset representatives - I don't think you can avoid this - but it isn't hard. The induced maps $B\mathcal G \rightarrow B \mathcal H$ will be independent of these choices in the homotopy category.

Now can one bypass $B\mathcal H$? You should look for equivalences of groupoids $\mathcal G \rightarrow \mathcal G^{\prime}$. I've just outlined how one can always find such equivalences that factor through $\mathcal H$. Depending on the particular groups you are looking at there may (or may not - consider $H=e$, $G=C_2$, $G^{\prime} = C_3$) be others.