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Let $G$ be a nice topological group, say a compact connected Liegroup. Then one can construct a model of its classifying space as $EG/G$ where $EG$ is any contractible space with free $G$ action.

On the other hand, one could take the geometric realization of the simplicial construction of the classifying space of the singular simplices of $G$:

$|\bar{W} (sing G)|$

Is it written down somewhere that these two spaces are weakly homotopy equivalent (or some variation of this statement)?

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What exactly do you mean by "the simplicial construction of the classifying space"? By the way, for two connected CW complexes X,Y any weak equivalence X->Y is already a homotopy equivalence. This is due to Whitehead. Classifying spaces of a group G can be characterised to be the homotopy types which have pi_1 the group and no other homotopy. Given two spaces X and Y that satisfy this property, one can construct an explicit morphism X->Y that will in turn be a weak equivalence. – Konrad Voelkel Feb 6 '12 at 18:58
By "the simplicial construction" I mean the for example the construction in Goer Jardine simplicial homotopy theory chapter V.4, but I think I would be fine with an answer refering to any construction. Your criterion with pi_1 makes only sense for discrete groups, for other groups one has pi_i+1(BG)=pi_i(G), but I don't think this characterizes the homotopy type of BG uniquely. – Jan Weidner Feb 7 '12 at 9:38
I see, I somehow missed "topological group" entirely... You're right that the shifting of the homotopy groups doesn't characterise the homotopy type of BG uniquely. So probably one should use the Quillen adjunction of the singular construction and the geometric realization. – Konrad Voelkel Feb 7 '12 at 14:11
One needs stronger assumptions that just a free action if you want EG/G to be well-defined up to homotopy. You need the projection EG-->G to be a principal G-bundle, so the action needs to have "slices." An extreme example is that every topological group G acts freely (by translation) on $G^{\textrm{ind}}$, the set $G$ with the indiscrete topology. This is certainly a contractible space, but $G^{\textrm{ind}}/G$ is a point. – Dan Ramras Feb 20 '12 at 0:58
up vote 1 down vote accepted

I think you should be able to prove this roughly as follows: first consider the loop space of your construction. For nice simplicial spaces, the loop space can be calculated level-wise (see May's Geometry of Iterated Loop Spaces, for instance), and hence the loop space of your construction is homotopy equivalent to the (realization of the) simplicial space $[n] \mapsto \Omega |\overline{W} Sing_n (G)|$. But this space is level-wise weakly equivalent to $[n] \mapsto Sing_n (G)$, whose realization is homotopy equivalent to $G$. So the loop space of your construction yields $G$, and now you need to apply some form of the statement: $B\Omega X \simeq X$.

Also, note that it's a standard fact (due to Segal's "Classifying spaces and spectral sequences" and/or May's "Classifying spaces and fibrations") that the simplicial bar construction applied directly to your topological group $G$ gives a model for $BG$ (this requires the inclusion of the identity of $G$ to be a cofibration, which is of course true for Lie groups).

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