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    $S^2\times\RP^4$ is not homotopy equivalent to $S^4\times\RP^2$

    P-S7WVQ

    Area
    Topology
    Topics
    Cohomology, Homotopy, Product Topology
    Status
    draft

    Show that \(S^2 \cross \RP^4 \not\homotopic S^4 \cross \RP^2\).

    Solution

    Take coefficients in \(\FF_2\), apply Kunneth to get \[ H^*(X_1; \FF_2) = \Extalgebra_{\FF_2}{x_2} \tensor_{\FF_2} {\FF_2\adjoin{y_1} \over \gens{ y_1^5 }} \\ H^*(X_2; \FF_2) = \Extalgebra_{\FF_2}{w_4} \tensor_{\FF_2} {\FF_2\adjoin{z_1} \over \gens{ z_1^3 }} .\]

    Now use that any homotopy equivalence induces a graded ring isomorphism \(f\), where \(f(y_1) = z_1\) in \(H^1\), but these have different orders.

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