Let $F\to E\to B$ be a fibration with $B$ simply-connected. Suppose all differentials in the cohomology Serre spectral sequence (corresponding to the above fibration) are zero maps. Then as a graded module,
$$
H^*(E)\cong H^*(F)\otimes H^*(B).
$$
Question 1: as cohomology rings with cup products, do we still have the isomorphism
$$
H^*(E)\cong H^*(F)\otimes H^*(B)?
$$
My idea of Question 1: there is a product in the $E_r$-page
$$
E^{p,q}_r\times E^{s,t}_r\to E^{p+q,s+t}_r
$$
which induces a product in the $E_{r+1}$-page
$$
E^{p,q}_{r+1}\times E^{s,t}_{r+1}\to E^{p+q,s+t}_{r+1}.
$$
Since all differentials are zero, the products on $E_r$-pages are same for each $r$. Hence there is a product on $E_\infty$-page which is the same as the product on $E_2$-page, given by the product of the tensor algebra $H^*(F)\otimes H^*(B)$. However, these products are in the pages of the Serre spectral sequence, but not in $H^*(E)$.
Question 2: suppose the cohomology coefficient is taken in $\mathbb{Z}_2$ and the Steenrod operations on $H^*(F;\mathbb{Z}_2)$, $H^*(B;\mathbb{Z}_2)$ are already known. Is it possible to determine the Steenrod operations on $H^*(E;\mathbb{Z}_2)$?
Best Answer
1) No in general. A counterexample is the projective space bundle associated to a vector bundle. For a rank $n$ vector bundle, the fiber, $\mathbb C \mathbb P^{n-1}$, has cohomology ring $\mathbb Z[x]/(x^n)$ with $x$ in degree $2$. But in the total space, the usual lift of $x$ satisfies the equation
$$\sum_{i=0}^n (-1)^i c_i (V) x^{n-i}=0$$
Since this equation is not the same as $x^n=0$, you don't get an isomorphism unless the Chern classes all vanish. (You can get an isomorphism by changing $x$ if the Chern classes satisfy a very specific equation, but for a general vector bundle it will fail.)
I think an explicit explicit example where it fails is the fibrartion of the complete flag variety in $\mathbb C^3$ over $\mathbb P^2$ by $\mathbb P^1$.
2) I don't know - maybe real vector bundles give a similar counterexample?