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The following are well-known facts on the Serre spectral sequence

For a fibration $F \rightarrow E \rightarrow B$ we have the Serre spectral sequence (in cohomology with a coefficients in a field $k$) with $E_2$-term

$$E_2^{p,q} = H^p(B; \mathcal{H}^q(F)) \Rightarrow H^{p+q}(E) $$

If $B$ is simply connected, (or $\pi_1(B)$ acts trivially on the cohomology of the fiber) we have that

$$E_2^{p,q} = H^p(B) \otimes H^q(F) $$

Moreover, if the spectral sequence degenerates at the $E_2$-term ($d_r = 0$ for $r \geq 2$),

then $H^*(B) \otimes H^*(F) \cong H^*(E)$.

My question is the following,

Suppose that for a fibration $F \rightarrow E \rightarrow B$ we know that $H^*(E) \cong H^*(B) \otimes H^*(F)$, does it follows that in the spectral sequence $E_2^{p,q} = H^p(B) \otimes H^q(F) $ ? and also that $d_r = 0$ for $r \geq 2$ ?

I assume that it is not true and a counterxample should involve a non-trivial local coefficient system, but I do not know many "computable" examples where the base space is non-simply connected.

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    $\begingroup$ Your second claimed isomorpism is incorrect. If $\pi_1B$ acts trivially on $H^*F$ then the local coefficent system is trivial and you can replace local cohomology group $\mathcal{H}^*F$ with ordinary cohomology group $H^*F$, leaving $E_2^{p,q}\cong H^p(B;H^qF)$. From here you still need to apply the universal coefficent theorem. You will only get $E_2^{p,q}\cong H^pB\otimes H^qF$ if at least one of $H^qB$ or $H^q$ satisfies some flatness conditions over the ring $R$. $\endgroup$ – Tyrone Oct 30 '18 at 17:22
  • $\begingroup$ @Tyrone thanks, I will edit my assumptions (maybe work with field coefficients to make things smoother) $\endgroup$ – C. Zhihao Oct 30 '18 at 17:29

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