cohomology ring of $S^2$ $\times$ $S^4$ and $CP^3$. I was studying Hatcher's algebraic topology book.In page number 251,book says $S^2$$\times$ $S^4$ and $CP^3$ has same cohomology groups but they have different ring structure.I understand that they have same cohomology group but I dont' understand why square of generator of $H^2$($S^2\times$ $S^4$ ) is zero.Book gives some argument using pullback but I am not getting that.
I will appreciate if somebody clarify that argument.
Thanks.
Note:I am considering Cohomology with integral coefficient.
 A: The generator of $H^2(S^2\times S^4)$ is $\pi^*(\alpha)$ where $\pi:S^2\times S^4\rightarrow S^2$ is the projection on the first factor, and $\alpha$ is a generator of $H^2(S^2)$. 
Then $\gamma:=\pi^*(\alpha)\smile\pi^*(\alpha)=\pi^*(\alpha\smile\alpha)$ by functoriality. But $\alpha\smile\alpha=0$ as $H^4(S^2)=0$, so $\gamma=0$ in $H^4(S^2\times S^4)$. 
edit: A bit about the classes in the Künneth formula (see also Hatcher page 218). I'm surpressing the coefficients
There is a map $\times:H^*(X_1)\otimes H^*(X_2)\rightarrow H^*(X_1\times X_2)$, which is given by
$\alpha\times\beta:=\times(\alpha\otimes \beta)=\pi_1^*(\alpha)\smile \pi_2^*(\beta)$. The Künneth formula states that under favourable conditions (depending on torsion and finiteness, satisfied here) this map is an isomorphism. 
Here this means that $H^2(S^2\times S^4)=\mathbb{Z}$ and is generated by $\pi_1^*(\alpha)\smile \pi_2^*(1)$ where $1$ is the generator of $H^0(S^4)$. But $\pi_2^*(1)=1$ (where the second $1$ is the generator of $H^0(S^2\times S^4)$). This class $1$ is the identity of the cup product so we get the required result. 
