# 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$$\times S^4 and CP^3 has same cohomology groups but they have different ring structure.I understand that they have same cohomology group but I dont' understand why square of generator of H^2(S^2\times S^4 ) is zero.Book gives some argument using pullback but I am not getting that. I will appreciate if somebody clarify that argument. Thanks. Note:I am considering Cohomology with integral coefficient. ## 1 Answer The generator of H^2(S^2\times S^4) is \pi^*(\alpha) where \pi:S^2\times S^4\rightarrow S^2 is the projection on the first factor, and \alpha is a generator of H^2(S^2). Then \gamma:=\pi^*(\alpha)\smile\pi^*(\alpha)=\pi^*(\alpha\smile\alpha) by functoriality. But \alpha\smile\alpha=0 as H^4(S^2)=0, so \gamma=0 in H^4(S^2\times S^4). edit: A bit about the classes in the Künneth formula (see also Hatcher page 218). I'm surpressing the coefficients There is a map \times:H^*(X_1)\otimes H^*(X_2)\rightarrow H^*(X_1\times X_2), which is given by \alpha\times\beta:=\times(\alpha\otimes \beta)=\pi_1^*(\alpha)\smile \pi_2^*(\beta). The Künneth formula states that under favourable conditions (depending on torsion and finiteness, satisfied here) this map is an isomorphism. Here this means that H^2(S^2\times S^4)=\mathbb{Z} and is generated by \pi_1^*(\alpha)\smile \pi_2^*(1) where 1 is the generator of H^0(S^4). But \pi_2^*(1)=1 (where the second 1 is the generator of H^0(S^2\times S^4)). This class 1 is the identity of the cup product so we get the required result. • Could you please explain why generator of H^2(S^2$$\times$$S^4$) is $\pi^*(\alpha)$? Aug 9, 2015 at 9:54
• Do you know the Künneth formula? Aug 9, 2015 at 9:59
• yes I know that formula... Aug 9, 2015 at 10:02
• okkk...I think I understood it...Thanks..... Aug 9, 2015 at 10:15
• Ripan: was that cynical? I can explain more. Aug 9, 2015 at 10:28