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.



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


