This is probably has an obvious proof or a straightforward counterexample, but I'm having trouble finding either.

Let $p:E \to B$ be a fibre bundle, with fibre $F$. Assume that there is a spectrum $X$ and a homotopy equivalence of spectra

$$f: X \wedge \Sigma^{\infty} B_+ \to \Sigma^{\infty} E_+$$

(in particular, there is a stable section to $p$).

Can we conclude that $X \simeq \Sigma^{\infty} F_+$? If so, we may conclude that the bundle is stably trivial; that is,

$$\Sigma^{\infty} E_+ \simeq \Sigma^{\infty} (B\times F)_+.$$

If this is not the case, what sort of conditions do we need to demand of the fibration to make it hold? I'm happy to localize everything in site at your favorite prime, or for that matter, cohomology theory...

Edit: There is an immediate type of counterexample, gotten by taking $E = B \times F$ to be trivial, but where $\Sigma^\infty F_+ \simeq X \vee Y$, where $Y$ is $\Sigma^\infty B_+$-acyclic; i.e.,

$$\pi_*(Y \wedge \Sigma^\infty B_+) = 0$$

One can define this problem away by assuming that $X$ and $\Sigma^\infty F_+$ are $\Sigma^\infty B_+$-local, which I'm happy to do for now.

  • 1
    $\begingroup$ I think you're also missing a condition on $X$: perhaps $X$ should have $S^0$ as a retract? Otherwise, I don't see how you're going to conclude, as you write above, that $p$ has a stable section. $\endgroup$
    – John Klein
    Feb 15, 2012 at 12:17
  • $\begingroup$ Yup! Good point, John. $\endgroup$ Feb 15, 2012 at 20:57
  • $\begingroup$ I think that if you have $F = *$ you can hit it with an Eilenberg swindle. $\endgroup$ Feb 16, 2012 at 2:37
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    $\begingroup$ Also, if $B$ is nonempty then there are no $\Sigma^\infty B_+$-acyclics, because it has $S^0$ as a retract; in fact, I think that the existence of the map $f$ implies that if $E$ and $B$ are nonempty (and $X$ is connective) you might already be able to find a retract from $X$ to $S^0$. $\endgroup$ Feb 16, 2012 at 2:46
  • $\begingroup$ Ah, cool! I'm having trouble seeing how you use the Eilenberg swindle here, though... $\endgroup$ Feb 16, 2012 at 21:17

1 Answer 1


This is often not true. Here's an example where all components are finite complexes.

Let $E = S^3 \times S^2$ and $B = S^2$, with the map given by $S^3 \times S^2 \to S^3 \to S^2$ where the latter is the Hopf fibration. The suspension spectrum is equivalent to $\Sigma^\infty_+ S^3 \wedge \Sigma^\infty_+ S^2$, but the fiber is $S^1 \times S^2$ and doesn't have $\Sigma^\infty_+ S^3$ as the suspension spectrum.

(This doesn't come with a stable section to $p$, unfortunately. If you take $(S^3 \times S^2) \coprod S^2$, you can get an example with a stable section.)

What went wrong here is that we didn't take the "over $B$" structure into account. The suspension spectrum of $B$ inherits a coalgebra structure from the diagonal, and the suspension spectrum of $E$ inherits a comodule structure using the structure map. If $B$ is 1-connected and the weak equivalence $X \wedge \Sigma^\infty_+ B \simeq \Sigma^\infty_+ E$ is a weak equivalence of comodules, then you can construct a map of Eilenberg-Moore complexes $$ C(\Sigma^\infty_+ E, \Sigma^\infty_+ B, \Sigma^\infty_+ *) \to C(X \wedge \Sigma^\infty_+ B, \Sigma^\infty_+ B, \Sigma^\infty_+ *) $$ which becomes a weak equivalence $\Sigma^\infty_+ F \to X$ on Tot. If the base is not 1-connected, convergence is obviously a lot more delicate.

Unfortunately this basically assumes stable triviality of the bundle, rather than going in the reverse direction.

(In order to avoid saying a lot about homotopy theory for coalgebras and comodules, I'm not going to characterize such weak equivalences as anything better than "something where you can ultimately compare the cobar complexes".)


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