Computing an explicit homotopy inverse for $B(*,H,*) \hookrightarrow B(*,G,G/H)$ Suppose that $G$ is a finite group, $M$ is a right $G$-set and $N$ is a left $G$-set. Then we have a simplicial set $B(M,G,N)$ whose $n$-simplicies are $M \times G^n \times N$. 
Now suppose that $H \subseteq G$ is a subgroup. There is a natural inclusion $B(*,H,*) \hookrightarrow B(*,G,G/H)$. Both of these spaces have fundamental group $H$ and no higher homotopy groups, and the inclusion induces an isomorphism on fundamental groups, so it is a weak equivalence. 


 Question:  Is there a way to write down an explicit inverse weak equivalence $B(*,G,G/H) \to B(*,H,*)$? In practice, I would be happy with just the $B_{\leq 2}(*,G,G/H) \to B_{\leq 2}(*,H,*)$ part of such a map.


My instincts tell me that the answer is no: simplicial sets are not sufficiently squishy enough to do this, but I want to be sure before I start trying harder stuff.
 A: Yes, there is an explicit algorithm for doing this. Pick a set of representatives $a_i \in G$ for the left cosets of $G/H$. Then the inverse map is as follows.
Given any element $(g_n,\dots,g_1, g_0H)$ in $G^n \times G/H$, let $a_i$ be the chosen representative of the left coset $g_i g_{i-1} \dots g_0 H$. Then $g_{i+1} a_i H = a_{i+1} H$, and so we get an element in $H^n$ given by
$$
(a_n^{-1} g_n a_{n-1}, \dots, a_2^{-1} g_2 a_1, a_1^{-1} g_1 a_0).
$$
One can check (and should!) that this is compatible with the face and degeneracy maps.
But really, what is going on here?
The simplicial set $B(*,H,*)$ is the nerve of a groupoid with one object and $H$ as its set of self-maps. The simplicial set $B(*,G,G/H)$ is the nerve of a translation groupoid with $G/H$ as its set of objects and maps $g: aH \to gaH$ for any $g \in G$. The inclusion of simplicial sets comes from an equivalence of groupoids.
To construct the inverse to this equivalence of categories, we choose for each object (a left coset $aH$) an isomorphism from $eH$ to $aH$: these are our chosen coset representatives. Any map $g: aH \to gaH = bH$ is then replaced, under these isomorphisms, with the self-map $b^{-1} g a: eH \to eH$ (provided that $a$ and $b$ were our chosen coset representatives).
The fact that these are groupoids is at the core of this argument, because it means that the associated simplicial sets are "fibrant": they have enough flexibility to construct most desired maps into them.
