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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 $$(z_1,\ldots,z_{n+1})\to (z_1^d,\ldots,z_{n+1}^d)\cdot (\sum_i |z_i|^{2d})^{-\frac{1}{2}}.$$ Clearly, this map is of degree $d^{n+1}$ (since the corresponding map $\mathbb C^{n+1}\to \mathbb C^{n+1}$ has degree $d^{n+1}$). 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.

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.

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 $(z_1,\ldots,z_{n+1})\to (z_1^d,\ldots,z_{n+1}^d)$. Clearly, this map is of degree $d^{n+1}$ (since the corresponding map $\mathbb C^{n+1}\to \mathbb C^{n+1}$ has degree $d^{n+1}$). 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.

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 $$(z_1,\ldots,z_{n+1})\to (z_1^d,\ldots,z_{n+1}^d)\cdot (\sum_i |z_i|^{2d})^{-\frac{1}{2}}.$$ Clearly, this map is of degree $d^{n+1}$ (since the corresponding map $\mathbb C^{n+1}\to \mathbb C^{n+1}$ has degree $d^{n+1}$). 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.

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 $(z_1,\ldots,z_{n+1})\to (z_1^d,\ldots,z_{n+1}^d)$. Clearly, this map is of degree $d^{n+1}$ (since the corresponding map $\mathbb C^{n+1}\to \mathbb C^{n+1}$ has degree $d^{n+1}$). On the other hand we have the Hopf map $S^{n+1}\to \mathbb CP^{n+1}$ 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.

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 $(z_1,\ldots,z_{n+1})\to (z_1^d,\ldots,z_{n+1}^d)$. Clearly, this map is of degree $d^{n+1}$ (since the corresponding map $\mathbb C^{n+1}\to \mathbb C^{n+1}$ has degree $d^{n+1}$). 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.