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Consider the following Puppe sequence. $$\cdots\to K(\mathbb{Z}/2\mathbb{Z},2)\to \Omega X\to K(\mathbb{Z}/2\mathbb{Z},1)\stackrel{a}{\to} K(\mathbb{Z}/2\mathbb{Z},3)\to X\to K(\mathbb{Z}/2\mathbb{Z},2)\stackrel{b}{\to} K(\mathbb{Z}/2\mathbb{Z},4)\to \cdots$$ If $b=\mathrm{Sq}^2$, the stability of the Steenrod square implies $a=\mathrm{Sq}^2$. However, this seems inconsistent to me because $b$ is not trivial but $a$ is trivial. Namely, we have $\Omega X\simeq K(\mathbb{Z}/2\mathbb{Z},2)\times K(\mathbb{Z}/2\mathbb{Z},1)$ but $X\not\simeq K(\mathbb{Z}/2\mathbb{Z},3)\times K(\mathbb{Z}/2\mathbb{Z},2)$, which looks inconsistent.

Question: I must have made stupid mistakes somewhere. Could anyone help me to figure it out?

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  • $\begingroup$ Why do you think that looks inconsistent? $\endgroup$
    – Tyrone
    Commented Jun 26, 2023 at 10:03
  • $\begingroup$ @Tyrone I think $\Omega (X\times Y)\simeq \Omega X\times \Omega Y$. $\endgroup$
    – Leo
    Commented Jun 26, 2023 at 10:08
  • $\begingroup$ That certainly holds. It is still possible that $\Omega Z\simeq \Omega X\times \Omega Y$ without $Z$ being homotopy equivalent to $X\times Y$. $\endgroup$
    – Tyrone
    Commented Jun 26, 2023 at 13:50
  • $\begingroup$ @Tyrone I think the delooping of a deloopable space is unique (up to weak homotopy equivalence). So $Z$ and $X\times Y$ should be equivalent if both of them are connected. $\endgroup$
    – Leo
    Commented Jun 26, 2023 at 14:07
  • $\begingroup$ You have a perfectly good counterexample to that thought above. $\endgroup$
    – Tyrone
    Commented Jun 26, 2023 at 14:17

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