Consider the cohomology spectral sequence of the trivial fibration $F \hookrightarrow F \times B \to B$. I showed in my answer here Why all differentials are $0$ for Serre Spectral Sequence of trivial fibration?, that differentials after $d_1$ are 0.
Question: Why is $Assgrd H^*(B \times F)=H^*(B \times F)$?
I expect the extension problem to be trivial because, if I assume it is then $H^n(E)=\oplus_{p+q=n} H^p(B,H^q(F))$ which takes into account the Tor terms which are $\oplus_{p+q=n} Tor( H^{p+1}(B), H^q(F))$.
I understand why one of the extensions are trivial here: The extension $0 \to F_{1,n-1} \to F_{0,n} \to E_{\infty}^{0,n} \subset E_2^{0,n}$ has a splitting because $H^n(E)=F_{0,n} \to E_{\infty}^{0,n} \subset E_2^{0,n}=H^n(F)$ is the edge map, which has the obvious splitting.