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We know that $H^p(S^2 \vee S^4) = H^p(S^2)\oplus H^p(S^4)$ for $p\neq 0$. I want to show that this space has different ring structure than $CP^2$. So, given a generator in $H^2(S^2 \vee S^4)$ I want to cup it with itself and get 0. My idea is to use the generator from $H^2(S^2)$(which obviously is zero when squared). How should I go from here? Is this even the correct idea?

($\vee$ here is one-point intersection).

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Yes, that is the right idea. Use that the cup product is "natural" with respect to pull backs.

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    @M.B. It is not very obvious how to rigorously show that the generator is mapped to the generator. You can use cellular model for cohomology and explicit definition of the pull back...2010-12-13