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An H-space $(X,\mu,e)$, is a pointed topological space $(X,e)$ together with a continuous map $\mu:X\times X\to X$, $s.t.$, The maps $\mu(\cdot,e):X\to X$ and $\mu(e,\cdot):X\to X$ are both homotopy to the identity $id:X\to X$ relative to $\{e\}$.

In Hatcher's Exercise 3.C.1(p.291), he claim that For a CW complex, the definition can be weaken:

let $e \in X$ be a 0 -cell, then $X$ is an H-space if there is a map $\mu: X \times X \rightarrow X$ such that $\mu(\cdot, e)$ and $\mu(e, \cdot)$, are homotopic to the identity(NOT necessarily rel$\{e\}$).

Or strengthen

With the same hypothesis, $\mu$ can be homotoped so that $e$ is a strict identity.

I was looking for a solution and I found this. However, there is some mistake in his proof. The existence of his assembled homotopy is equivalent to that $X$ is an H-space. He used the result that he was going to prove.

Although he didn't make it clear, the idea is worth taken, that is, deduce some homotopy $(X\vee X)\times I\to X$ and extend it to $X\times X$ by HEP. I didn't see how to do this, can anyone give some help?

Ted
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    What's your question? If you're asking for a solution, please show us what you've tried. If you're asking whether the linked is in fact incorrect and how to fix it, please at least repeat the argument here and point out exactly which step you think is erroneous. Please clarify! – Ben Steffan Mar 08 '25 at 13:59
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    Your title didn't mean what you thought it meant, so I rewrote it. If you disagree, feel free to backtrack. But yes, also, please ask a question. – Lee Mosher Mar 08 '25 at 14:54
  • After some clarification, this question now seems to be a duplicate. – Tyrone Mar 09 '25 at 04:33

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