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    $\mathbb{RP}^3$ is not homotopy equivalent to $\mathbb{RP}^2\vee S^3$

    E-HAT-3.2-7

    Area
    Topology
    Topics
    Cohomology, Cup Product
    Status
    draft

    Use cup products to show that \(\mathbb{RP}^3\) is not homotopy equivalent to \(\mathbb{RP}^2 \vee S^3\).

    Solution

    With \(\mathbb Z_2\) coefficients, \[ H^*(\mathbb{RP}^3;\mathbb Z_2) \cong \mathbb Z_2[\alpha]/(\alpha^4), \qquad |\alpha|=1, \] so \(\alpha^3\ne0\).

    For the wedge \(Y=\mathbb{RP}^2\vee S^3\), reduced cohomology splits additively as the direct sum of the reduced cohomologies of the two wedge summands, and products between positive-dimensional classes coming from different summands vanish. The unique nonzero class \(x\in H^1(Y;\mathbb Z_2)\) comes from \(\mathbb{RP}^2\), so \[ x^3=0 \] because \(H^3(\mathbb{RP}^2;\mathbb Z_2)=0\).

    Thus the mod-\(2\) cohomology rings are not isomorphic: in one ring the nonzero degree-one class has nonzero cube, while in the other its cube is zero. Hence the spaces cannot be homotopy equivalent.

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