Given a discrete group $G$ (**not** assumed to be abelian), is there a nice construction of a topological space which classifies bundles of $K(G,1)$'s? I guess I should take something like $B\operatorname{Aut}(K(\pi,1))$ where $\operatorname{Aut}(K(\pi,1))$ is the monoid of self homotopy equivalences of some fixed $K(\pi,1)$. I don't know if this makes any sense, though.


Morally speaking, things are defined as follows. A bundle of $K(\pi,1)$'s over a space $X$ is a fibration $Y\to X$ whose fibers are $K(\pi,1)$'s. Two bundles of $K(\pi,1)$'s $Y\_1,Y\_2\to X$ are said to be isomorphic iff there is a map $Y\_1\to Y\_2$ of spaces over $X$ which is a homotopy equivalence on each fiber.



What space classifies bundles of K(pi,1)'s?
For any space $F$ you can form the topological monoid $Aut(F)$ and take its classifying space. That will classify Hurewicz fibrations with fiber $F$. A little care is needed since $Aut(F)$ won't generally have the homotopy type of a CW complex unless $F$ is a finite CW complex, which $K(G,1)$ is usually not. For the proof it is sensible to take base spaces of the homotopy types of CW complexes. One source for a complete proof is "Classifying spaces and fibrations", #15 on my web page. (There are two obvious notions of the right equivalence relation on fibrations, but these give the same answer, as shown in "Fiberwise localization and completion'', #31 on my web page.)


