TITLE: Is it possible to construct K(G, 1) out of a subgroup and its quotient?
QUESTION [7 upvotes]: Suppose we have a short exact sequence $1 \to K \to G \to Q \to 1$ and let $X_K$ and $X_Q$ be $K(K, 1)$ and $K(Q, 1)$, respectively. Is it possible to construct (in a "natural" way) a model of $K(G, 1)$ out of $X_K$ and $X_Q$?
For example, if $G$ is a direct product of $K$ and $Q$, then it acts on $\tilde{X}_K \times \tilde{X}_Q$ in a decent way. Is it possible to generalize it somehow?
I know that something can be done for cohomology (there is this Lyndon–Hochschild–Serre spectral sequence). But I need a topological space that is a model of $K(G, 1)$ (or rather, I am interested in the universal cover of that, and the action by deck transformations).

REPLY [2 votes]: If the total spaces $EQ$ and $EK$ are given as $Q$ and $K$ CW-complexes respectively, you can also use a construction of C.T.C. Wall (see C. T. C. Wall, Resolutions for extensions of groups, Proc. Cambridge Phil. Soc. 57 (1961), 251-255, I also have a short summary in this paper here, in Section 3: http://arxiv.org/pdf/1006.0129v2.pdf) in the following way:
The spaces $EQ$ and $EK$ will give you free resolutions over $\mathbb{Z}Q$ and $\mathbb{Z}K$ (which are contractible over $\mathbb{Z}$).
Wall then explains how to construct a free resolution for $\mathbb{Z}G$ out of the two resolutions. This is a twisting of the idea that if $G=K\times Q$ you would just pick the product of the two resolutions. In fact, the number of $n$-cells in the new resolution will still be $\sum_{i+j=n} a_ib_{j}$ where $a_i$ is the number of $i$-cells in $EK$ and $b_j$ the number of $j$-cells in $EQ$. In the general case, where the sequence does not split, one needs to be more careful when choosing the differentials. From this free resolution you can construct (I believe) an $EG$ space and then also $BG$.