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.



Stable triviality of fiber bundles
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".)


