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I'm wondering if there's anything that can be said about the cup product when computing the cohomology ring of a product space? An example would be $S^2\times S^4$, where the generator of $H_2(S^2\times S^4)$ is essentially contributed by $H_2(S^2)$ from the Kunneth formula and then from the universal coefficient theorem.

My guess would therefore be that if $\alpha$ is the generator of $H^2(S^2\times S^4)$, then $\alpha^2=0$, since the generator of $H^2(S^2)$ squares to zero. Could anyone elaborate on if anything useful can follows from this line of thinking. if not, then how would one try to approach the computation of the cup product of a product space assuming one knows how the cup product behaves for each factor in the product?

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The algebra $H^*(X\times Y)$ is isomorphic as an algebra to $H^*(X)\otimes H^*(Y)$ over coefficient rings over which the isomorphism holds as groups---for example, over fields.

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But this doesn't quite answers the question, right? (Embarassingly, I'm not sure I know the whole answer for cohomology over $\mathbb Z$, actually.) – Grigory M Jan 15 '12 at 20:56
I don't know if the cohomology rings of the factors determine the cohomology of the product, really. – Mariano Suárez-Alvarez Jan 17 '12 at 6:13
What are the sufficient conditions such that $H^\ast(X\times Y;R)\cong H^\ast(X;R)\otimes_R H^\ast(Y;R)$ as graded $R$-algebras? Over any commutative ring $R$ such that $H^k(Y;R)$ is free of finite rank (Hatcher, 3.17)? – Leon May 12 '14 at 8:43

I would point out that you can look at maps of spaces and note that they induce ring maps. Also, there is a spectral sequence that degenerates to the above isomorphism after some generous application of the UCT.

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