Let $f:\mathbb{S}^2\to \mathbb{S}^2$ with degree $d$.


It is well known that the induced map $$f\_\ast:\pi\_3(\mathbb{S}^2)=\mathbb{Z}\to \pi\_3(\mathbb{S}^2)=\mathbb{Z}$$ is given by multiplication by $d^2$.


But, $\mathbb{S}^2=\mathbb{C}P^1$ and $[\mathbb{C}P^n,\mathbb{C}P^n]=\mathbb{Z}$ for $n\geq 1$ and for any $f:\mathbb{C}P^n\to \mathbb{C}P^n$ determined by $d\in \mathbb{Z}$ the induced map $$f\_\sharp: H\_{2k}(\mathbb{C}P^n)=\mathbb{Z}\to H\_{2k}(\mathbb{C}P^n)=\mathbb{Z}$$ is given by multiplication by $d^k$ for $1\leq k\leq n$.


**Question:** is it true that the map $$f\_\ast : \pi\_{2n+1}(\mathbb{C}P^n)=\mathbb{Z}\to \pi\_{2n+1}(\mathbb{C}P^n)=\mathbb{Z}$$ is determined by multiplication by $d^{n+1}$?



Induced maps on homotopy groups by self maps of $\mathbb{CP}^n$
I guess it should be enough to prove this statement for one particular example (for each $d,n$), since homotopy classes of self-maps of $\mathbb CP^n$ are classified by their degree (aren't they?).


Let us identify $S^{2n+1}$ with $|z\_1|^2+\ldots+|z\_{n+1}|^2=1$.
Consider the map $\phi(n,d):S^{2n+1}\to S^{2n+1}$ which is given by 
$$(z\_1,\ldots,z\_{n+1})\to (z\_1^d,\ldots,z\_{n+1}^d)\cdot (\sum\_i |z\_i|^{2d})^{-\frac{1}{2}}.$$
It is not hard to see that this map is of degree $d^{n+1}$, for example, by counting the number of premiages of any point with all non-zero coordinates on unit $S^{2n+1}$. On the other hand we have the Hopf map $S^{2n+1}\to \mathbb CP^{n}$ and the map $\phi(n,d)$ clearly descends to a degree $d$ map $\mathbb CP^n\to \mathbb CP^n$. Thus, for this particular case the map $\phi(n,d)$ is a "geometric realisation" of the map $\pi\_{2n+1}(\mathbb CP^n)\to \pi\_{2n+1}(\mathbb CP^n)$. This basically completes the calculation in our example.


