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Let $A$ be an abelian group and let $n \geq 2$. For any connected CW complex $X$, it is standard that a fibration $f\colon E \rightarrow X$ whose fibers are homotopy equivalent to a $K(A,n)$ is fiberwise homotopy equivalent to the pull-back of the loop-space fibration over a $K(A,n+1)$ if and only if $\pi_1(X)$ acts trivially on the fibers of $f$ (up to homotopy equivalence).

Question: Consider the more general set of fibrations $f\colon E \rightarrow X$ whose fibers are homotopy equivalent to a $K(A,n)$, but with a possibly nontrivial monodromy action, up to fiber homotopy equivalence. Is this functor representable?

From the first paragraph, if it is representatable by a space $Z$ then the universal cover of $Z$ is a $K(A,n+1)$.

Of course, there are many variants here (e.g. of the definition of a fibration, or the equivalence relation on the set), and I'd be interested in any of these.

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    $\begingroup$ Do the hypotheses of Brown representability not apply? $\endgroup$ – John Greenwood Jan 7 at 4:17
  • $\begingroup$ @JohnGreenwood: It's not clear to me. The issue is that you have to be able to glue fibrations that are isomorphic on a subcomplex, and since the gluing map is just a fiberwise homotopy equivalence (and thus not a homeomorphism) this seems subtle. $\endgroup$ – Tina Jan 7 at 4:19
  • $\begingroup$ (if we could assume that these fibrations were actually locally trivial fiber bundles and isomorphisms were fiber bundle isomorphisms, then this would not be a problem; however, I don't know if this is possible). $\endgroup$ – Tina Jan 7 at 4:21
  • $\begingroup$ My mistake! Gluing fibrations seems tricky...what if you glue with the homotopy equivalence anyway, and then force the result to be a fibration? $\endgroup$ – John Greenwood Jan 7 at 5:16
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Mark Grant's excellent answer already resolves the question. However, let me sketch how this arises as a special case of the more general problem of classifying fibrations with a given fiber.

For any space $X$, the homotopy automorphisms $\operatorname{hAut}(X)$ are defined as the self-homotopy equivalences of $X$ (topologized with the compact-open topology; note that they form a union of path components of the space of all self-maps of $X$). They form a group-like monoid, so there is a connected pointed space $B\operatorname{hAut}(X)$ such that there is an equivalence of $A_\infty$-spaces $\operatorname{hAut}(X)\simeq \Omega B\operatorname{hAut}(X)$. In fact, $B\operatorname{hAut}(X)$ can be built as the geometric realization of the simplicial space $B(*,\operatorname{hAut}(X),*) = * \leftarrow \operatorname{hAut}(X) \Leftarrow \operatorname{hAut}(X)\times\operatorname{hAut}(X) \Lleftarrow \dots$. Since $\operatorname{hAut}(X)$ acts on $X$ from the left, we can also build the simplicial space $B(*,\operatorname{hAut}(X),X) = X\leftarrow \operatorname{hAut}(X)\times X\Leftarrow\dots$, and the map $X\to *$ induces a fibration $X\to E_X\to B\operatorname{hAut}(X)$. This is the universal fibration with fiber $X$, that is, for every fibration $X\to F\to Y$ there is a unique homotopy class of maps $f: Y\to B\operatorname{hAut}(X)$ such that $F\simeq f^*E_X$.

If $X = K(A,n)$ is an Eilenberg-MacLane space, the grouplike monoid $\operatorname{hAut}(X)$ can be described quite explicitly: In this case, $X$ can also be given the structure of a grouplike monoid with identity $e$, so that the map $\operatorname{hAut}(X)\to X, f\mapsto f(e)$ has a homotopy right inverse given by sending $x$ to (left, say) translation by $x$. Thus there is a homotopy equivalence $\operatorname{hAut}(X)\simeq \operatorname{hAut}_*(X)\times X$, where $\operatorname{hAut}_*(X)$ is the submonoid of self homotopy equivalences that preserve the basepoint (note, however, that this is not compatible with the monoid structure). Now it is straightforward to show that $\operatorname{hAut}_*(K(A,n))\simeq K(\operatorname{Aut}(A),0)$ is homotopy equivalent to a discrete space, and there is a retraction $\operatorname{hAut}(X) \to \pi_0(\operatorname{hAut}(X))\cong \operatorname{Aut}(A)$. All in all, we obtain an equivalence of grouplike monoids $$ \operatorname{hAut}(K(A,n))\simeq \operatorname{Aut}(A)\ltimes K(A,n) $$ It follows that $B\operatorname{hAut}(K(A,n))$ has exactly two nonvanishing homotopy groups, namely $\pi_1(\operatorname{hAut}(K(A,n)))\cong \operatorname{Aut}(A)$ and $\pi_{n+1}(\operatorname{hAut}(K(A,n)))\cong A$, with the evident action of $\pi_1$ on $\pi_{n+1}$. In particular, a map $f:Y\to B\operatorname{hAut}(K(A,n))$ is determined by $\rho\in H^1(Y;\operatorname{Aut}(A))$, which determines a local system $A_\rho$ on $Y$, and a cohomology class $[f]\in H^{n+1}(Y;A_\rho)$.

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    $\begingroup$ As a side note, for bundles whose fibers are $K(G,1)$ with $G$ is a nonabelian groupthere there is a similar analysis. But then $\pi_0(hAut)$ turns out to be the outer automorphism group $Out(G)$ and $\pi_1(hAut)$ is the center $Z(G)$. $\endgroup$ – Tyler Lawson Jan 7 at 17:00
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    $\begingroup$ There's also work of Wirth, summarised in this paper joint with Stasheff: arxiv.org/abs/math/0609220 on homotopy locally trivial fibrations. $\endgroup$ – David Roberts Jan 8 at 6:03
  • $\begingroup$ So is it right to think that, morally, the difference between fiber bundles and fibrations is the structure "group" being an actual group versus a group-up-to-homotopy? $\endgroup$ – John Greenwood Jan 8 at 19:08
  • $\begingroup$ This is wonderful! Do you know a good place to read about this? (ps: sorry for only replying now -- I am not on the internet very often). $\endgroup$ – Tina Jan 11 at 1:44
  • $\begingroup$ (pps: I'll accept your answer once I get a reference.) $\endgroup$ – Tina Jan 12 at 14:32
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Denote $\pi=\pi_1(X)$ and fix the monodromy action $\rho:\pi\to \operatorname{Aut}(A)$. There is a generalized Eilenberg-Mac Lane space $L_\rho(A,n+1)$, whose only non-trivial homotopy groups are $\pi_1(L_\rho(A,n+1))=\pi$ and $\pi_{n+1}(L_\rho(A,n+1))=A$ and such that the action of $\pi_1$ on $\pi_{n+1}$ is given by $\rho$. The construction appears in Section 7 of

Gitler, Samuel, Cohomology operations with local coefficients, Am. J. Math. 85, 156-188 (1963). ZBL0131.38006.

Essentially, $L_\rho(A,n+1)$ is the Borel construction $E\pi\times_\rho K(A,n+1)$ done with simplicial sets, so that everything is sufficiently natural (and the action of $\pi$ on $A$ gives an action of $\pi$ on $K(A,n)$ for all $n$). The path loop fibration is equivariant, giving a fibration sequence $$ K(A,n)\to E\pi\times_\rho PK(A,n+1) \to E\pi\times_\rho K(A,n+1)=L_\rho(A,n+1). $$ I would bet that fibrations over $X$ with fibre $K(A,n)$ and monodromy $\rho$ are classified by maps (Edit: from $X$) into $L_\rho(A,n+1)$ (Edit: inducing the identity on $\pi_1$) using this construction.

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    $\begingroup$ Arbitrary maps? Maybe rather only those inducing isomorphism on $\pi_1$, or something like that? $\endgroup$ – მამუკა ჯიბლაძე Jan 7 at 21:18
  • $\begingroup$ You're right, of course. For example, pulling back by the trivial map would give a fibration with trivial monodromy. In fact, the monodromy of the pullback should (I think) be the pullback of the monodromy, so we'd want the map to induce the identity on $\pi_1$; see my edit. $\endgroup$ – Mark Grant Jan 8 at 9:53
  • $\begingroup$ Bertram's answer is more general, and probably does a better job of answering the quesiton asked. $\endgroup$ – Mark Grant Jan 8 at 9:54
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One place this was done but in greater generality was by Blakers, Annals of Matyh.2nd series, 49 (2) 428-461. He defines what he calls "group systems". These are now called "crossed complexes". An account is in the book Nonabelian Algebraic Topology (EMS, 2011). A particular case $C$ is where the crossed complex has $C_0$ a singleton, $C_1$ is a group, say $G$, for some $n>1$, $C_n$ is a $G$-module, and otherwise $C_n$ is trivial.

Any crossed complex $C$ has a classifying space $BC$ whose simplicial definition in the case $C_0$ is a point is due to Blakers. Any filtered space, $X_*$, and in particular any, CW-complex $X$ with skeletal filtration $X_*$, has a fundamental crossed complex $\Pi X_*$, (also due to Blakers). In particular the cubical nerve $NC$ of $C$ can be defined in dimension $n$ by $$(NC)_n = Crs(\Pi I^n_*, C), $$ where $I^n$ is the standard $n$-cube. (The simplicial version of this was published with P.J. Higgins as Math. Proc. Camb. Phil. Soc. 110 (1991) 95-120.)

(I confess I rather suspect the homotopy classification theorems of our book have to be restricted to $X$ finite dimensional or $BC$ having only a finite number of non trivial homotopy groups,)

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