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Let $f\colon S^n \rightarrow S^n$ be a basepoint preserving map of degree $d \geq 1$. We then get an induced map $\Omega \Sigma(f)\colon \Omega \Sigma S^n \rightarrow \Omega \Sigma S^n$. There is another map $g\colon \Omega \Sigma S^n \rightarrow \Omega \Sigma S^n$ taking a based loop $\gamma$ in $\Sigma S^n$ to $\gamma^d$, where $\gamma^d$ uses the loop product (and thus goes around $\gamma$ a total of $d$ times).

These maps seem pretty similar. Question: is $\Omega \Sigma(f)$ homotopic to $g$? Or at least is the diagram

$$\require{AMScd}\begin{CD} S^n @>{f}>> S^n \\ @VVV @VVV \\ \Omega \Sigma S^n @>{g}>> \Omega \Sigma S^n \end{CD}$$

commutative up to homotopy?

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    $\begingroup$ Let $f,g\colon S^n\to S^m$ be maps. I think that it is more generally true that $\Omega \Sigma (f+g)$ and the loop product of $\Omega \Sigma(f)$ and $\Omega\Sigma(g)$ agree up to homotopy as maps $\Omega\Sigma S^n \to \Omega \Sigma S^m$. This is based on a variant of the Eckmann--Hilton argument (or the argument that the different produts on $\pi_nX$ agree for $n\geq 2$.) $\endgroup$
    – Lennart Meier
    Commented Jan 5, 2023 at 11:14
  • $\begingroup$ @LennartMeier You can check that $\Omega\Sigma(1-1)=0$ and $\Omega\Sigma(1)+\Omega\Sigma(-1)=1+\Omega\Sigma(-1)$ do not induce the same map on homology. $\endgroup$
    – Tyrone
    Commented Jan 5, 2023 at 11:59
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    $\begingroup$ Long story short: the diagram commutes by an Eckmann-Hilton argument. The maps $\Omega\Sigma(f)$ and $g$ are not homotopic (take $n=1$ and consider their action on homotopy groups. $g$ is multiplication by $d$, but $\Omega\Sigma f_*\tilde\eta=d^2\cdot\tilde\eta$, where $\tilde\eta:S^2\rightarrow\Omega S^2$ is adjoint to the Hopf map). $\endgroup$
    – Tyrone
    Commented Jan 5, 2023 at 12:11
  • $\begingroup$ @Tyrone You're right: $\Omega \Sigma f$ and $g$ are not homotopic (postcomposition vs. precomposition). But on homology: $H_*(\Omega\Sigma S^n)\cong\mathbb{Z}[x]$ with $|x|=n$. Given maps $d, e: S^n\to S^n$, we can write the loop product of $\Omega\Sigma d$ and $\Omega \Sigma e$ as $\Omega\Sigma S^n \xrightarrow{d,e}\Omega\Sigma S^n\times \Omega\Sigma S^n \to \Omega\Sigma S^n$. The first map sends $x^m$ to $\sum_k \binom{m}{k}(dx)^k(ex)^{m-k}$ (Hatcher 3.C.1). The second is multiplication and thus the composite is $\sum_k \binom{m}{k}d^ke^{m-k}x^m = (d+e)^mx^m$. Am I wrong again? $\endgroup$
    – Lennart Meier
    Commented Jan 6, 2023 at 7:40
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    $\begingroup$ Hi @LennartMeier. The class $x$ lives in degree $n$, so you have to keep track of signs. The formula you wrote is correct when $n$ is even, but breaks down when $n$ is odd (e.g. when $k=2$ you end up with $(d^2+e^2)x^2$). Returning to your original notation, the relation is essentially $\Omega\Sigma(f+g)=\Omega\Sigma(f)+\Omega\Sigma(g)+\overline{[f,g]}\circ h_2$, where $h_2$ is the second Hilton-Hopf invariant. It's a little more complicated, but refer to $\S4$ of Neisendorfer's book for the details. $\endgroup$
    – Tyrone
    Commented Jan 6, 2023 at 8:22

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