Show that \(S^n\) is an H-space if and only if the attaching map of the \(2n\)-cell of \(J_2(S^n)\) is homotopically trivial.
Solution
Let \[ S^n\vee S^n\hookrightarrow S^n\times S^n \] be the standard inclusion. The product has a CW structure obtained from the wedge by attaching one \(2n\)-cell. Let \[ \omega:S^{2n-1}\to S^n\vee S^n \] be its attaching map, and let \[ \nabla:S^n\vee S^n\to S^n \] be the fold map, equal to the identity on each summand.
A multiplication \[ \mu:S^n\times S^n\to S^n \] with \(e\) as a strict two-sided identity is exactly an extension of \(\nabla\) across the \(2n\)-cell. Such an extension exists if and only if \[ \nabla\circ\omega:S^{2n-1}\to S^n \] is nullhomotopic.
By the definition of the second James stage \[ J_2(S^n)=S^n\times S^n/\bigl((x,e)\sim x\sim(e,x)\bigr), \] its \(n\)-skeleton is the single copy of \(S^n\) obtained from the wedge by the fold map, and its unique \(2n\)-cell is attached precisely by \[ \nabla\circ\omega. \] Hence the attaching map of the \(2n\)-cell of \(J_2(S^n)\) is nullhomotopic if and only if the fold map extends over \(S^n\times S^n\).
If it is nullhomotopic, choose such an extension \(\mu\); then \(\mu(x,e)=x=\mu(e,x)\), so \(S^n\) is an H-space. Conversely, if \(S^n\) is an H-space, Exercise 1 allows its multiplication to be homotoped to one with strict identity, hence to an extension of the fold map, forcing the attaching map to be nullhomotopic. Therefore \[ \boxed{S^n\text{ is an H-space}\iff \text{the }2n\text{-cell of }J_2(S^n)\text{ is attached trivially}.} \]