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The stable homotopy groups of the spheres $\pi_{*}^{s}$ assemble into a graded ring

$\pi_{*}^{s} = \bigoplus_{n\geq 0} \pi_{n}^{s},$

with the graded product defined `in terms of composition'. For example, if $\alpha \in \pi_{i}^{s} $ has representative $[f: S^{n+i} \rightarrow S^{n}]$ and $\beta \in \pi_{j}^{s} $ has representative $[g: S^{n+j} \rightarrow S^{n}]$, then the product $\alpha \cdot \beta \in \pi_{i+j}^{s}$ may be defined as

$\alpha \cdot \beta := [g \circ \Sigma^{j}f : S^{n+i+j} \rightarrow S^{n}]$.

But we could have also defined the product as

$\alpha \cdot \beta := [f \circ \Sigma^{i}g : S^{n+i+j} \rightarrow S^{n}]$.

(The point is that we define the product in terms of a composition, possibly after suspending.)

My question is how do you show that these two definitions define the same product?

Reading the comment to this question, one may try to approach this via an Eckmann-Hilton Duality argument. But using this argument as stated, one would end up proving that these products are commutative. This is not true, as they are graded commutative.

Maybe there is a `graded' Eckmann-Hilton duality argument I could apply, but my attempts so far have lead me to conclude the above two products only coincide up to sign, which clearly should not me true. I am clearly going wrong somewhere, but I can't figure out where.

I tried looking in the literature, but I could not find anyone who comments on this, so any help would be much appreciated.

Sunny Sood
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  • Does your argument show that the ``up to a sign'' is independent of what you are multiplying? If that is the case, I would just check the sign on the identity product which must trivially agree. – Connor Malin Feb 04 '23 at 16:55
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    It's not clear what you're asking. The stable homotopy groups of spheres form a graded commutative ring: elements commute up to a sign, not on the nose, so the order does matter. – John Palmieri Feb 04 '23 at 17:00
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    Also, I suspect the Eckmann-Hilton argument is as follows: in a symmetric monoidal category $(C,\otimes,1)$, the Hom set $C(1,1)$ is a monoid in two ways: through the tensor product and through composition. Now one can define the graded Spanier-Whitehead category to have objects the finite pointed spaces and Hom sets the direct sum $\bigoplus_k \mathrm{colim}_{n \rightarrow \infty}[\Sigma^n X,\Sigma^{n+k}Y]$, and this forms a symmetric monoidal category under smash product. Thus you can apply the theorem to the unit $S^0$, which solves your problem – Connor Malin Feb 04 '23 at 17:01
  • @JohnPalmieri I am asking whether the two possible definitions of product I have written in the question are the same? I know that the stable homotopy groups of spheres form a graded commutative ring. – Sunny Sood Feb 06 '23 at 11:12
  • @ConnorMalin Oh, nice. I will have to think about it, but this sounds good. Thanks for your comment :). – Sunny Sood Feb 06 '23 at 11:14
  • The smash product is also graded commutative, and that should probably inform the Eckmann-Hilton argument. – John Palmieri Feb 06 '23 at 18:26

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