Cup product of cohomology in a Serre spectral sequence How to use Serre spectral sequence to compute cup product structures? 
Let $F\to E\to B$ be a fibration. Suppose all the differentials of the corresponding Serre spectral sequence of cohomology are zero. Can we obtain that for any field $k$,
$$
H^*(E;k)\cong H^*(F;k)\otimes H^*(B,k)
$$
as rings?
For example, how to solve the following problem?

 A: Maybe the simplest example is the following. There are two fiber bundles with base and fiber both $S^2$. One is the product $S^2\times S^2$ and the other consists of two copies of the mapping cylinder of the Hopf map $S^3 \to S^2$ glued together via the identity map between their "source" ends $S^3$. The resulting space is the connected sum of two copies of ${\mathbb C}P^2$ (with opposite orientations). In terms of clutching functions this is the $S^2$-bundle associated to the nontrivial element of $\pi_1SO(3)$. The cup product structures in the two total spaces are different because in the nontrivial bundle obtained from Hopf maps there are classes in $H^2$ that square to a generator of $H^4$, as one can see from the quotient map collapsing one of the mapping cylinders to a point, with quotient space ${\mathbb C}P^2$, while for $S^2\times S^2$ the square of any class in $H^2$ is always an even multiple of a generator of $H^4$.
A: A good example showing the answer to question 1 is no is the case of the projectivization of a complex vector bundle $V$ over $B$. This has $F=\mathbb C P^{n-1}$, and it is a theorem (see e.g. Hatcher, Vector bundles and K-Theory) that all differentials vanish or equivalently $H^*E$ is a free $H^*B$-module on generators $1,x,\dots x^{n-1}$, where $x$ restricts to a generator of $H^2E$. However, one has the equation
$$\sum_{i=0}^n c_i(V)x^{n-i}=0,$$
where $c_i(V)$ are the Chern classes, not $x^n=0$.
So in general: the thing one knows is the Leray-Hirsch theorem: $H^*E$ is a free $H^*B$-module on generators $x_i$ which map to generators $y_i$ of $H^*F$.
If $y_iy_j=y_k$, then one only knows that $x_ix_j=x_k$ modulo terms of higher filtration.
