# Classifying space BG and contractable space EG

Choose a arbitrary discrete group $$G$$. The classifying space $$BG$$ of $$G$$ is 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. In other words is it possible to construct a homotopy $$h_t: EG \times I \to EG$$ which contracts $$EG$$ to a point explicitly? That is $$h_0(EG) =EG, h_1(EG) = \{*\}$$. How looks it concretely geometrically?

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 $$EG$$?

2. To prove contractibility of Milnor's $$EG$$, fix the vertex $$p=[1]$$ of $$EG$$. By the construction, $$[1]$$ together with any simplex $$s=[g_0,...,g_n]$$ in $$EG$$ span the simplex $$\Delta_s=[1,g_0,...,g_n]$$ in $$EG$$ (unless $$g_0=1$$ in which case we take $$\Delta_s=[1,g_1,...,g_n]$$). In particular, for every point $$q$$ in $$EG$$ we obtain a canonical affine line segment $$pq\subset \Delta_s$$, where $$q$$ is in $$s$$.
Thus, one defines the "straight-line homotopy" of the identity map to the constant map $$EG\to \{p\}$$ using the above line segments. Since $$G$$ is discrete, this homotopy is continuous. This is your explicit contraction.
Addendum. In fact, Milnor in his paper also constructs a contractible $$G$$-complex for an arbitrary topological group $$G$$ (not necessarily discrete) by taking a countably infinite join of copies of $$G$$. Milnor proves weak contractibility of his $$EG$$. For a proof of contractibility, see for instance Proposition 14.4.6 in