# Steenrod squares in a Puppe sequence

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?

• Why do you think that looks inconsistent? Commented Jun 26, 2023 at 10:03
• @Tyrone I think $\Omega (X\times Y)\simeq \Omega X\times \Omega Y$.
– Leo
Commented Jun 26, 2023 at 10:08
• That certainly holds. It is still possible that $\Omega Z\simeq \Omega X\times \Omega Y$ without $Z$ being homotopy equivalent to $X\times Y$. Commented Jun 26, 2023 at 13:50
• @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.
– Leo
Commented Jun 26, 2023 at 14:07
• You have a perfectly good counterexample to that thought above. Commented Jun 26, 2023 at 14:17