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 $H^\*(E;\mathbb{Z}\_2)$?



cup product and Steenrod operations in Serre spectral sequence
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?


