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    $S^2\vee S^2$ does not cover $S^2$

    P-TOP-WORKSHOP-D8-CW2

    Area
    Topology
    Topics
    Counterexamples, Covering Spaces
    Status
    draft

    Prove or disprove: \(S^2\vee S^2\) is a covering space of \(S^2\).

    Solution

    The statement is false. The space \(S^2\vee S^2\) is connected. If it covered \(S^2\), then because \(S^2\) is simply connected, every connected covering of \(S^2\) would be one-sheeted and therefore a homeomorphism onto \(S^2\).

    But \(S^2\vee S^2\) is not homeomorphic to \(S^2\); for example, \[ H_2(S^2\vee S^2)\cong\mathbb Z\oplus\mathbb Z, \qquad H_2(S^2)\cong\mathbb Z. \] Therefore \(S^2\vee S^2\) cannot be a covering space of \(S^2\).

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