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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).

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{EDIT: My initial answer to this question was wrong, or at best seriously misleading. I apologize to anyone who was confused by it. Here is my attempt to straighten things out.}

Let $EG$ be the universal cover of $K(G,1)$. So $EG$ is a contractible space with a free action of $G$, and the orbit space $EG/G$ is a model for $K(G,1)$.

It also is true that $EG/K$ is a model for $K(K,1)$, $Q$ acts on this model, and $K(G,1)$ is both the quotient and the homotopy quotient space of this action. By homotopy quotient I mean the space $K(K,1)\times_Q EQ$. The homotopy quotient has better homotopy invariance properties than the strict quotient, so it gives you more freedom in choosing the model for $K(K,1)$, as a space with an action of $Q$, that you want to work with.

So this is one answer to the question: the short exact sequence $1\to K \to G \to Q \to 1$ determines an action of $Q$ on $K(K,1)$ and $K(G,1)$ is the homotopy quotient of this action.

In particular, this gives rise to the homological spectral sequence.

For example, if the extension is trivial, then the action of $Q$ on $K(K,1)$ is trivial (or, if you prefer, is homotopy equivalent to the trivial action), and in this case the homotopy quotient is equivalent to $K(Q,1)\times K(K,1)$.

In general, it may be desirable to describe the action in terms of the extension. This is what I wrote in my initial answer (but it is wrong):

If $K$ is abelian then $Q$ acts on $K$ by lift-conjugation. So $Q$ acts on $K(K, 1)$, provided you use a functorial construction. The space $EQ\times_Q K(K,1)$ is a model for $K(G, 1)$. Here $EQ$ is a contractible space with a free action of $Q$.

This is only correct if the $G$ is a semi-direct product of $Q$ and $K$ (and in this case you need not assume that $K$ is abelian). In this case the extension is determined by the action of $Q$ on $K$, and therefore the action of $Q$ on $K(K,1)$ is determined by this as well.

To see that in general the action of $Q$ on $K$ does not determine the action on $K(K,1)$, consider the case when $G$ is a central extension of $Q$ by $K$. This is a case that is as far from semi-direct product as possible. In this case the action of $Q$ on $K$ is trivial, but the action of $Q$ on $K(K,1)$ is not. Indeed if it was, then my formula would say that $K(G,1)\simeq K(Q,1)\times K(K,1)$, which is false for non-trivial central extensions. A central extension is determined by an element of $H^2(Q; K)$, which is the same as a map $K(Q,1)\to K(K,2)$. $K(G,1)$ is the homotopy fiber of this map. This map can also be interpreted as a (roughly speaking) group homomorphism $Q\simeq \Omega K(Q,1)\to \Omega K(K,2)\simeq K(K,1)$. To be more precise, this data is enough to determine an action of $Q$ on $K(K,1)$, and $K(G,1)$ is equivalent to the homotopy quotient of this action.

Even more generally, a group extension of $Q$ by $K$ is the same thing as an action of $Q$ on the category of $K$-sets, as discussed, for example, in this MO question Group extensions and actions on categories. Since $K(K,1)$ can be identified with the geometric realizations of the category of $K$-sets isomorphic to $K$, an action of $Q$ on the category of $K$ sets induces an action on $K(K,1)$. In general $BG$ is equivalent to the homotopy quotient of this action.

Added later: Let me spell out in detail how you can associate an action of $Q$ on $K(K,1)$ with the short exact sequence $1\to K \to G \to Q \to 1$. Note that for any element $q\in Q$, its preimage in $G$ is a set with both a left and a right action of $K$. Define the following category: it has one object for each element of $Q$, and the set of morphisms from $q_1$ to $q_2$ is the set of functions from the preimage of $q_1$ in $G$ to the preimage of $q_2$, that respect the action of $K$ on the left. Equivalently, morphisms are given by right miltiplication by an element of $K$. This category is a groupoid, and the group of automorphisms of a single object is $K$. So the geometric realization of this category is $K(K,1)$. Furthermore, $Q$ acts on this category by left multiplication. This determines an action of $Q$ on $K(K,1)$. $K(G,1)$ is the homotopy quotient of this action.

The following MO question is related Classifying Space of a Group Extension

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  • $\begingroup$ Thank you for your answer. I do not understand what do you mean by "lift-conjugation". How do you define action of $Q$ on $K(K, 1)$? $\endgroup$ – Janusz Przewocki Aug 9 '16 at 7:00
  • $\begingroup$ Given an element $q\in Q$, choose a $g\in G$ that maps to $q$ (i.e., a lift of $q$), and let $g$ act on $K$ by conjugation. If $K$ is abelian, then this action is independent of the choice $g$, so it is a well-defined action of $Q$ on $K$ by homomorphisms, which in turn determines an action of $Q$ on $K(K,1)$. If $G$ is a semi-direct product of $Q$ and $K$ then $K(G,1)$ is the homotopy quotient of this action. For general extensions, it is true that $K(G,1)$ is the homotopy quotient of an action of $Q$ on $K(K,1)$, but this action is not determined by an action of $Q$ on $K$ by homomorphisms. $\endgroup$ – Gregory Arone Aug 9 '16 at 10:37
  • $\begingroup$ I have edited the answer once again, adding a bit of explanation. Hopefully it is clearer now. Apologies if it is too wordy... $\endgroup$ – Gregory Arone Aug 9 '16 at 10:38
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$K(G,1)$ is the classifying space of $G_d$, where $G_d$ is the group $G$ endowed with the discrete topology. So you have a fibration

$BK_d=K(K,1)\rightarrow BG_d=K(G,1)\rightarrow BQ_d=K(Q,1)$.

A fibration of classifying spaces

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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 $BG$.

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