TITLE: Relationship between induced maps at homotopy groups level for maps $f:S^2\to S^2$
QUESTION [8 upvotes]: It is a known result that based point maps $f:S^2\to S^2$ are classified by their degree. That is, by the induced map at $\pi_2$-level $f_{*2}:\pi_2(S^2)\to\pi_2(S^2)$ (the subindex $*2$ just means it is the induced map by $f$ at $\pi_2$-level). However, I am interested in knowing which is the relationship (if there is any) between the induced maps $f_{*2}:\pi_2(S^2)\to\pi_2(S^2)$ and $f_{*3}:\pi_3(S^2)\to\pi_3(S^2)$. For instance, if the former is the zero map it means that $f$ is null-homotopic and therefore the latter should also be the zero map. But, it they are not zero, how are they related? Thanks in advance.

REPLY [2 votes]: Another way to see this is by using Pontryagin-Thom.  The generator of $\pi_3(S^2)$ is represented by an unlink in $\mathbb{R}^3$ which twists around once (like a figure-8).  So precomposing $\eta$ with an element of $\pi_3(S^3)$ is a disjoint union of n figure-8's, while postcomposing with an element of $\pi_2(S^2)$ is cabling the figure-8 n-times.  The n-cable of the figure-8 has $n^2$ crossings, so it's cobordant to $n^2$ figure-8s.
The way I think about this is that the figure-8 is explaining a recipe for turning a 2-loop into a 3-loop (i.e. appear the loop and its inverse, braid them past each other using Eckman-Hilton, and then cancel).  If you first do the Hopf construction and then take its nth power as a 3-loop you're first doing a figure-8 and then taking its disjoint union n-times.  While if you take the power of a 2-loop and then apply the Hopf construction you're first taking the disjoint union of n points and then doing the figure-8 construction to all of them together.  This is why it's composing with an element of $\pi_2(S^2)$ which corresponds to the cabling.