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Determine with justification, whether $S^2 \times S^4$ is homeomorphic to $\mathbb{C}P^{3}$.

I know that they are not, but I do not know how to justify it , I got a hint that squaring a generator in $S^2 \times S^4$ is zero but in $\mathbb{C}P^{3}$ is not . could anyone help me in understanding this hint please?

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  • $\begingroup$ Do you know the cohomology rings for both these spaces? Not just as graded groups, but also the cup product structure? $\endgroup$
    – William
    Mar 20, 2020 at 14:30
  • $\begingroup$ I am using AT book and "Modern Classical Homotopy Theory" and they are difficult for me. so can you point out on which pages in any one of those books should I look please? $\endgroup$
    – Secretly
    Mar 20, 2020 at 14:33
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    $\begingroup$ For $H^*(S^2 \times S^4)$ look at Hatcher's section on the Künneth formula starting on 214, especially look at the end of 215/start of 216 for the ring structure on a tensor product of graded rings. Look at Theorem 3.19 on page 220 for $H^*(\mathbb{C}P^3)$. $\endgroup$
    – William
    Mar 20, 2020 at 14:54

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If you know something about homotopy theory, you should know that $\pi _4(CP^3)=0$. To prove this, consider the fibration $S^7\to CP^3$, and look at the long exact sequence on homotopy. On the other hand $\pi _4(S^4)=\mathbb Z\neq 0$. So $S^2\times S^4$ contains a 4 sphere which is not homotopic to $0$, unlike $CP^3$

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