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