Can anyone give a reference (or explain here), why the group $[\Sigma\Sigma X,Y]_*$ is commutative? How is it related to the fact that $\Sigma X$ is a co-H-space?
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1Fancy answer $[\Sigma \Sigma X, Y]* \cong [X, \Omega^2 Y]*$, and a double loop space like $\Omega^2 Y$ is an $E_2$ algebra which is homotopy commutative (you can prove this by itself using the $H$-space structure), and thus homotopy classes of maps in gives a commutative monoid. The fact that it's a group is just because its a single loop space. – Justin Young Mar 08 '16 at 12:49
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This is not my area of expertise, so this is a rough idea of why $[\Sigma\Sigma X, Y]_{\ast}$ is a commutative group.
First of all, $\Sigma$ is left adjoint to $\Omega$ so $[\Sigma\Sigma X, Y]_{\ast} \cong [\Sigma X, \Omega Y]_{\ast}$. Now $\Sigma X$ is a cogroup object in $\mathsf{hTop}_{\ast}$ (which is even stronger than being a co-H-space) and $\Omega Y$ is a group object in $\mathsf{hTop}_{\ast}$, so $[\Sigma X, \Omega Y]_{\ast}$ has two natural group structures. It should then follow from the Eckmann-Hilton argument that the two group structures coincide and that the group is abelian.
May gives a sketch of a more direct proof which doesn't use any such language in a lemma at the end of Chapter $8$, section $2$ of A Concise Course in Algebraic Topology.
This fact can be used to show that higher homotopy groups are abelian: for $n \geq 2$,
$$\pi_n(Y) = [S^n, Y]_{\ast} = [\Sigma\Sigma S^{n-2}, Y]_*$$
so $\pi_n(Y)$ is abelian.
Michael Albanese
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