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This question is probably not research level that's why I asked it previously on MSE a week ago. Unfortunately it doesn't get much attention there and I thought I would try it here.

Choose a arbitrary discrete group $G$. The classifying space $BG$ of $G$ is classically constructed by forming a certain contractable $\Delta$-complex $EG$ (on concrete construction of $EG$: see below) endowed with an action by $G$. $BG$ is obtained by taking quotient $BG:= EG/G$.

The concrete construction of the $\Delta$-complex $EG$ works as follows: The $n$-simplices of $EG$ are the ordered tuples

$$[g_0,g_1,...,g_n] \cong \Delta_n =\left\{x\in \mathbb {R} ^{n}:x=\sum _{i=0}^{n}t_{i}v_{i}\ {\text{with}}\ 0\leq t_{i}\leq 1\ {\text{and}}\ \sum _{i=0}^{n}t_{i}=1 \right\}$$

with $g_i \in G$. The $v_i$ are spanning $\Delta_n$. We obtain a $\Delta$-complex by attaching $n$-simplices to the $(n − 1)$ simplices $[g_0,g_1,..., \hat{g}_i,...,g_n]$ in standard way as a standard simplex attaches to its faces. Here $\hat{g}_i$ means that this vertex is deleted.

Question. Does there exist a constructive way to show that $EG$ is contractable. By constructive I mean how to construct an explicite homotopy $h_t: EG \times I \to EG$ which contracts $EG$ to a point. That is $h_0(EG) =EG, h_1(EG) = \{*\}$. How looks it concretely geometrically?

I know some abstract arguments like Whitehead's theorem that also provide a reason for contrability of $EG$ but this is not what I'm looking for.

I tried to define such homotopy as follows: let $p \in [g_0,...,g_n]$. Then $h_t$ "slides" step by step $p$ along $(n+1)$-simplex $[e,g_0,...,g_n]$ to $[e]$ ($e \in G$ the identity element). What I mean by "step by step along $[e,g_0,...,g_n]$"?

If we use again the identification $[g_0,g_1,...,g_n] \cong \Delta_n$ then as long as we sitting "inside" $\Delta_n$ we can interpret the $g_i$ as spanning vectors $v_i$. Let $p = \sum _{i=0}^{n}t_{i}g_{i}$. As $[g_0,g_1,...,g_n] \subset [e, g_0,g_1,...,g_n]$ we can interpret $[g_0,g_1,...,g_n]$ as $x = t_{-1}e + \sum _{i=0}^{n}t_{i}g_{i} \in \Delta_{n+1}$ with $t_{-1}=0$. That is the "point" $p_e:= 1 \cdot e \in [e, g_0,g_1,...,g_n]$ is not contained in $[g_0,g_1,...,g_n]$ and we can define a unique line $l_pe$ which contains $p_e$ and is perpendicular to $[g_0,g_1,...,g_n]$ in our geometric picture $[g_0,g_1,...,g_n] \subset [e, g_0,g_1,...,g_n]$ corresponding to $\Delta_n \subset \Delta_{n+1}$.

This line uniquely intersects $[g_0,g_1,...,g_n]$ in a unique point $p_l$. Then we say that our homotopy slide $p$ along the unique line through $p $ and $p_l$ up to the first contact with a boundary of $[g_0,g_1,...,g_n]$.

Let this boundary be $[g_0,..., \hat{g}_i,...,g_n]$. Then we play the same game with $[g_0,..., \hat{g}_i,...,g_n]$ and $[e, g_0,..., \hat{g}_i,...,g_n]$.

Does this approach work? And is there known a more conventional "textbook" (that I still haven't found) way for the construction of $EG$?

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    $\begingroup$ The ``abstract nonsense'' explanation can be found here: ncatlab.org/nlab/show/decalage $\endgroup$ – Bertram Arnold Feb 3 at 13:05
  • $\begingroup$ @BertramArnold: so intuitively decalage "cleans" the obstuctions (like degenerations) of a simplicial set (more precisely it's geom realization) to be contractable? $\endgroup$ – MortyPB Feb 3 at 16:35
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The easiest way to construct an explicit contracting homotopy is to observe that EG is the geometric realization of the nerve of the groupoid G//G, which has G as its set of objects and exactly one morphism between any pair of objects.

The nerve functor sends equivalences of groupoids to homotopy equivalences of simplicial sets, and the geometric realization functor sends homotopy equivalences of simplicial sets to homotopy equivalences of topological spaces.

Thus, it remains to construct an equivalence between G//G and the trivial groupoid * with a single object and morphism. This is easy: there is just one functor G//G→*, and for the inverse functor take *→G//G that picks the identity element of G. The composition *→G//G→* is identity, whereas the composition G//G→*→G//G is isomorphic to the identity via the unique choices of morphisms in G//G.

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  • $\begingroup$ Could you lose few words on your last argument that the composition G//G→*→G//G is homotopic to identity "via the unique choices of morphisms in G//G". I not fully understand what you mean. By constuction between every $g,h \in G//G$ there exist exactly one map. Why does this imply the desired claim? $\endgroup$ – MortyPB Feb 3 at 16:39
  • $\begingroup$ @MortyPB: A natural transformation F→G (in our case, F=G//G→*→G//G and G=id) is a collection of morphisms F(X)→G(X) that satisfies a certain commutativity condition. The data of morphisms F(X)→G(X) is uniquely prescribed because we already know F(X) and G(X) and there is exactly one morphism of the form F(X)→G(X). Likewise, any square diagram (in fact, any diagram) in G//G automatically commutes because any pair of morphisms between the same objects coincides. $\endgroup$ – Dmitri Pavlov Feb 3 at 18:55

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