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. 



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> >  **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.
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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.



Computing an explicit homotopy inverse for $B(*,H,*) \hookrightarrow B(*,G,G/H)$
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.


