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I am interested in a variant of the Leray-Hirsch theorem. Let $\mathbb{E}^*$ be a complex-oriented cohomology theory and let $\mathbb{E}_*$ the coefficient ring. Suppose that $F \xrightarrow{i} P \to B$ is a fiber bundle with $F,P,B$ finite CW complexes. I assume that $B$ has only even dimensional cells (though I expect the claim to be true more generally). Suppose I have a splitting $\mathbb{E}^*(F) \to \mathbb{E}^*(P)$ of the restriction $i^*: \mathbb{E}^*(P) \to \mathbb{E}^*(F).$

Is the resulting map $$\mathbb{E}^*(F)\otimes_{\mathbb{E}_{*}}\mathbb{E}^*(B) \to \mathbb{E}^*(P) $$ an isomorphism? I suspect that if this is true, it can be proven with standard spectral sequence arguments. Is there a reference for the result in this form?

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    $\begingroup$ It's proved on pg. 430 of tom Dieck's book that this is an isomorphism of left $E^*(B)$-modules (it is generally not an algebra isomorphism). $\endgroup$
    – Tyrone
    Commented Apr 23 at 18:07
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    $\begingroup$ @Tyrone I do not wish to assume the cohomology of the fiber is free over $\mathbb{E}_*$. $\endgroup$
    – onefishtwofish
    Commented Apr 23 at 18:18
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    $\begingroup$ You definitely need some flatness assumption on the cohomology of $F$ or $B$. If $E$ is complex orientable, evenness of the cells of $B$ probably saves you there, but without assumptions on $E$ there should be counterexamples even with trivial fiber bundles $F\times B$ (for example with $E=\mathbb{S}$) $\endgroup$
    – Achim Krause
    Commented Apr 23 at 18:40
  • $\begingroup$ @AchimKrause I understand your point. Even in my restrictive case, to get $E^*(B)$ to be free over $E_*$, I need an assumption. I am happy to assume complex orientable as well and have adjusted the question. How should the argument go then? $\endgroup$
    – onefishtwofish
    Commented Apr 23 at 23:45
  • $\begingroup$ I'm not sure about a reference, but the Atiyah-Hirzebruch spectral sequence for a fibration proves this theorem. You need to know that $E^q(P)$ surjects onto $H^0(B,E^q(F))$ which your assumption buys you. If you know that $E_2^{p,q}=H^P(B,E^q(F)) \cong H^p(B,E^*) \otimes_{E^*} H^q(F,E^*)=E_2^{p,0} \otimes E_2^{0,q}$ [ For example if the the cohomology of the fiber is free as an $E^*$ module, but maybe Achim's rmk is ok] then the spectral sequence collapsing on the second page ($d_2=0$ on $E^{p,0}$) proves the claim. I saw your last comment but this is the SS argument I know $\endgroup$
    – Andres Mejia
    Commented May 6 at 21:50

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