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For each prime $p\geq 3$ let $\alpha_p:S^{2p}\to S^3$ denote a representative of $\pi_{2p}S^3$ of order $p$. Berstein and Hilton showed that for each $p$ the homotopy cofiber $C_{\alpha_p}$ of $\alpha_p$ is a co-H-space which does not have the homotopy type of a suspension space.

The maps $\alpha_p$ give rise to a map $\alpha:\bigvee_{p\geq 3} S^{2p}\to S^3$ whose cofiber $C_\alpha$ is a co-H-space. Is it known whether $C_\alpha$ has the homotopy type of a suspension space?

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  • $\begingroup$ Have you tried extending the proof of Lemma 3.6 in Berstein and Hilton to show that if $C_\alpha$ is a suspension then each of the $\alpha_p$ must be suspensions? $\endgroup$
    – Mark Grant
    Dec 22, 2015 at 8:23
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    $\begingroup$ I gave that a shot, but at some point Berstein and Hilton require a particular map to be injective on homotopy groups; I couldn't convince myself that the analogs of that map that arise in the construction of $C_\alpha$ enjoy this property. $\endgroup$
    – James Schwass
    Dec 22, 2015 at 15:42

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Isn't the localization of a 2-connected suspension also a suspension? Then this cannot be a suspension. (Brayton Gray pointed this out to me.)

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    $\begingroup$ This is true, since (i) localization commutes with homotopy pushouts, and (ii) a 2-connected suspension is a suspension of a 1-connected space, by the argument in Lemma 3.6 of Berstein and Hilton $\endgroup$
    – Mark Grant
    Dec 22, 2015 at 12:09
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    $\begingroup$ This seems like a perfectly good answer to me. Is the rhetorical question the problem? $\endgroup$
    – Gustavo Granja
    Dec 22, 2015 at 15:09

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