Relationship between induced maps at homotopy groups level for maps $f:S^2\to S^2$

Question

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.

Answer 1

I'll write $n_d$ for the degree $n$ map on $S^d$, and $\eta$ for the Hopf map $S^3\to S^2$. It is well-known that $\eta_*\colon\pi_3(S^3)\to\pi_3(S^2)$ is an isomorphism, so that $\pi_3(S^2)=\{\eta\circ n_3:n\in\mathbb{Z}\}$, and $\eta\circ n_3$ is just $n$ times $\eta$ with respect to the standard abelian group structure on $\pi_3(S^2)$. However, this is different from $n_2\circ\eta$: in fact we have $n_2\circ\eta=\eta\circ n_3^2$. To see this, we use the following models for the relevant homotopy types. We put $X=\{(z,w)\in\mathbb{C}^2:\max(|z|,|w|)=1\}$, which is a model for $S^3$. We take $Y=\mathbb{C}\cup\{\infty\}$, which is a model for $S^2$. We define $\eta\colon X\to Y$ by $\eta(z,w)=z/w$, with the convention $z/0=\infty$ when $|z|=1$; this is well-known to be a model for the Hopf map. For $n>0$ we define $f_n\colon X\to X$ by $f_n(z,w)=(z^n,w^n)$ and $g_n\colon Y\to Y$ by $g_n(u)=u^n$ (with the convention $\infty^n=\infty$). The degree of a map can be characterised as the number of preimages of a generic point, counted with appropriate multiplicity. For $f_n$ and $g_n$ one can check that all multiplicities are equal to one and so $\deg(f_n)=n^2$ and $\deg(g_n)=n$. It is clear that $g_n\circ\eta=\eta\circ f_n$, and our claim follows for $n>0$. For negative $n$, it will now suffice to treat the case $n=-1$. The map $z\mapsto z^{-1}$ actually has degree one on $Y$, but we can instead define $g_{-1}(u)=\overline{u}$ and $f_{-1}(z,w)=(\overline{z},\overline{w})$. We again gave $g_{-1}\circ\eta=\eta\circ f_{-1}$. It is not hard to see that $g_{-1}$ has degree $-1$. The map $f_{-1}$ comes from an $\mathbb{R}$-linear automorphism of $\mathbb{C}^2=\mathbb{R}^4$, and in that context the degree is the sign of the determinant, which is $+1$ in this case. So the claim holds for $n=-1$ as well.

Answer 2

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.