$\newcommand{\i}{\iota}$ The general notion that I am trying to disprove is that if we are given a fibration $X \to Y$ with fiber $F$ such that the delooping $BF$ exists, that there is a map $Y \to BF$ such that $F \to X \to Y \to BF$ is a fibration sequence.

This question came up when I was trying to see what went wrong when trying to define postnikov towers for nonsimply connected spaces.

Here is my proposed counterexample:

Let $X$ be a nonsimply connected space with the action of $\pi_1(X) \curvearrowright \pi_2(X)$ nontrivial. Since the meaning of a postnikov tower here is ambiguous let me explicitly define $X_1=K(\pi_1(X),1)$ and $X_2$ the space obtained by killing all homotopy groups above dimension $2$, and let $X_2 \to X_1$ be the map obtained by just adding cells to $X_2$ to kill homotopy at dimension 2.

I want to show that there is no map $X_1 \to K(\pi_2(X),3)=BK(\pi_2(X),2)$ such that $K( \pi_2(X),2) \to X_2 \to X_1 \to K(\pi_2(X),3)$ is a fibration sequence.

Intuitively, the only candidate for such a map would be an element of $[X_1, K(\pi_2(X),3)]=H^3(X_1, \pi_2(X))$ arising as the transgression of the fundamental class of $K(\pi_2(X),2)$ in the Serre Spectral Sequence of fibration $K( \pi_2(X),2) \hookrightarrow X_2 \to X_1$.

In order for this to make sense the fundamental class $\i$ of $K(\pi_2(X),2)$ needs to be in the domain of the transgression, and it is not:

We can view $\i =Id \in Hom(\pi_2(X), \pi_2(X))$, $E_2^{0,2}=Hom(\pi_2(X), \pi_2(X))^{\pi_1(X)}$ where $\pi_1(X)$ acts on the first $\pi_2(X)$, and the transgression is $d_3$, which in view of the vanishing of $E_2^{1,1}$, acts on all of $E_2^{0,2}$.

Then we have $\iota \in Hom(\pi_2(X), \pi_2(X))$ is not $\pi_1(X)$ invariant, and so it is not in the domain of the transgression.

This leaves the following question on how I can make this argument rigorous:

How can I show that any candidate map $X_1 \to K(\pi_2(X),3)$ would have to arise as the transgression of the fundamental class of $K(\pi_2(X),2)$?

  • 1
    $\begingroup$ You might like to check out T. Ganea's paper "Induced Fibrations and Cofibrations". $\endgroup$
    – Tyrone
    Dec 18, 2016 at 15:26
  • 1
    $\begingroup$ $\mathbb Z/3$ is a group. Consider a fibration over a wedge of circles with fiber $\mathbb Z/3$ such that the full monodromy group is $S_3$. Obviously it can't be represented by a map to $B\mathbb Z/3$. In general, any fibration $F\to Y \to X$ can be represented by $X \to B\mathrm{Aut}\ F$, where $\mathrm{Aut}\ F$ is the group of homotopy autoequivalences. If $F$ is a group then we have $BF \stackrel{f}{\to} B\mathrm{Aut}\ F$ (left self-action), and if the classifying map doesn't factor through $f$, then representation by $X \to BF$ is impossible. $\endgroup$ Dec 20, 2016 at 5:54

2 Answers 2


I haven't thought about the example that you describe, but here is a different kind of example.

Consider a topological group $F$, and let $M$ denote the monoid of weak equivalences $F\to F$, so there is an evident inclusion $F\to M$. Let $X$ be any based space, and let $i\colon X\to CX$ be the inclusion of the base of the cone. Now suppose we have a based map $f\colon X\to M$. Take two copies of $CX\times F$ and identify $(i(x),u)$ in the first copy with $(i(x),f(x)(u))$ in the second. This gives a space $E(f)$ with evident maps $F\xrightarrow{j}E(f)\xrightarrow{q}\Sigma X$. I think that this is always a fibration up to homotopy. If the map $f$ does not factor (up to homotopy) through $F$, then we have a fibration that is not classified by a map $\Sigma X\to BF$.

Note here that $f$ is supposed to be a based map when we take the identity as the basepoint in $f$. One way to define such an $f$ is to take a based map $m\colon X\wedge F\to F$, and put $f(x)(u)=m(x\wedge u).u$.

For a minimal example of this, take $F=S^3=SU(2)$ and $X=S^1$. Let $\eta\colon S^3\to S^2$ be the Hopf map, and take $m=\Sigma\eta\colon X\wedge F\to F$. The resulting map $f\colon X\to M$ is nontrivial in homotopy, but $[X,F]=\pi_1(S^3)=0$.

  • $\begingroup$ Dear Professor Strickland, Thank you for giving me a better idea of when a fibration is classified by a map from the base space. I think I have been failing to generalize this to my example because this criterion of lifting to something that has a multiplication, seems specific to spaces that are suspensions. I hope to come back later when I have something more intelligent to say about this wonderful answer. Best Regards, Hari (B.A. 2017 Courant Institute) $\endgroup$ Dec 22, 2016 at 4:59

By hypothesis in my question, $\pi_1(K(\pi_1(X),1))=\pi_1(X)$ acts nontrivially on $H^2(K(\pi_2(X),2),\pi_2(X))$.

The path fibration $K(\pi_2(X),2) \hookrightarrow P \to K(\pi_2(X),3)$ has trivial monodromy and so does any pullback of this fibration. Therefore my fibration is not a pullback of a path loop fibration. In other words there is no map that classifies my fibration.


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