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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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The usual terminology is the wedge product and it's denoted $S^2 \vee S^4$. $S^2 \wedge S^4$ is the usual terminology for the smash product. –  Ryan Budney Dec 12 '10 at 23:41
    
Fixed. (also commented in next answer) –  M.B. Dec 13 '10 at 0:35

1 Answer 1

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

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Thank you. Could you please be a little more explicit? I'm not so used to the concept of "naturality" yet. –  M.B. Dec 12 '10 at 23:34
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Look at $\pi:S^2\vee S^4\rightarrow S^2$ which collapses the $S^4$ to a point. Use the fact that $\pi$ induces a ring map $\pi^*:H^2(S^2)\rightarrow H^2(S^2\vee S^4)$. (Isn't $\vee$ usually used instead of $\wedge$ to mean the one point intersection?) –  Jason DeVito Dec 12 '10 at 23:42
    
It is, I just wrote "\wedge" since I tend to call it "wedge" rather than one-point intersection. –  M.B. Dec 13 '10 at 0:12
    
Jason: So, I let $\pi$ be the projection map which gives me the induced ring homomorphism $\pi^*$. Since $H^2(S^2) \cong H^2(S^2 \vee S^4)\cong \mathbb{Z}$ then a generator is mapped to a generator? –  M.B. Dec 13 '10 at 0:18
    
Yes, that is the idea. Now look use the fact, as stated above, that $\pi^*$ is a ring map. –  Sean Tilson Dec 13 '10 at 2:47

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