Let $d>0$ and $f:\mathbb{C}^{n+1}\rightarrow \mathbb{C}^{n+1}$ be given by $f(z_0,...,z_n)=(z_0^d,...,z_n^d)$. Let $F:\mathbb{CP}^n \rightarrow \mathbb{CP}^n$ be the induced map by $f$. Compute $F^*:H^*(\mathbb{CP}^n;\mathbb{Z}) \rightarrow H^*(\mathbb{CP}^n,\mathbb{Z})$.
Attempt:
Considering first the case when $n=1$, we know $\mathbb{CP}^1$ can be obtained as the quotient of the disk $D^2$ under the identifications $v\sim \lambda v$ for $v\in \partial D^2$.
i.e all points in $\mathbb{CP}^1$ can be represented by a point $\gamma =(w,\sqrt(1-|w|^2))$. When $|w|<1$ this representation is unique, and when $|w|=1$ we have the identification $\gamma \sim\lambda \gamma$, $\lambda \in \mathbb{C}$.
$\mathbb{CP}^1\cong \mathbb{S}^2$ so to see what $F_*$ does to $H_2(\mathbb{CP}^1)$ is the same as figuring out the degree of $F$. Using local degrees, we can see how many elements are in the pre-image of $\gamma =(w,\sqrt(1-|w|^2))$, where $|w|<1$. For the second component, we only want the pre-images of $\sqrt(1-|w|^2)$ that are positive real numbers, so we have a unique d$^{th}$ root for this component. For the second component we have $d$ complex roots (we need to rescale to get elememnts of norm 1, but we still end up with $d$ distinct roots).
Thus, $F_*$ is multiplication by $d$ on $H_2$. Using the fact that $F^*=(F_*)^*$ for maps on $\mathbb{CP}^n$, we have that $F^*:H^2(\mathbb{CP}^2) \rightarrow H^2(\mathbb{CP}^2)$ is equal to multiplication by $d$.
By cellular homology we know that $i:\mathbb{CP}^2 \rightarrow \mathbb{CP}^n$ induces an isomorphism $i_*:H_2(\mathbb{CP}^2) \rightarrow H_2(\mathbb{CP}^n)$ . So we get a commutative diagram of the maps $i_*$ and $F_*$, giving us that $F_*$ and $F^*$ are multiplication by $d$ on $H_2(\mathbb{CP}^n)$ and $H^2(\mathbb{CP}^n)$.
Finally, using cup product structure of $H^*(\mathbb{CP}^n)$, we get $F^*:H^{2k}(\mathbb{CP}^n) \rightarrow H^{2k}(\mathbb{CP}^n)$ is multiplication by $d^k$.
Is this correct? or is there another way to do this?