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? $\endgroup$ Jan 7, 2020 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, 2020 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, 2020 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? $\endgroup$ Jan 7, 2020 at 5:16

3 Answers 3


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 $Z(G)$. $\endgroup$ Jan 7, 2020 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, 2020 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? $\endgroup$ Jan 8, 2020 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, 2020 at 1:44
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    $\begingroup$ Ugh, I have to learn about $\infty$-groupoids to understand something this classical? I'm a little skeptical that this is really necessary (or that it really gives the same answer when you're dealing with ordinary spaces rather than simplicial objects), but David Robert's comment seems like a good place to start learning the old-fashioned way to think about this, so I'll accept this answer. $\endgroup$
    – Tina
    Jan 15, 2020 at 1:49

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? $\endgroup$ Jan 7, 2020 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, 2020 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, 2020 at 9:54

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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