Compute the map $H^*(CP^n; \mathbb{Z}) \rightarrow H^*(CP^n, \mathbb{Z})$ I'm trying to solve problem 3.2.6 in Hatcher. The problem is stated:
Use cup products to compute the map  $H^*(CP^n; \mathbb{Z}) \rightarrow H^*(CP^n, \mathbb{Z})$ induced by the map $CP^n \rightarrow CP^n$ that is a quotient of the map $C^{n+1} \rightarrow C^{n+1}$ raising each coordinate to the $d^{th}$ powder, $(z_0, ..., z_n) \rightarrow (z_0^d, ... , z_n^d)$, for a fixed integer $d > 0$. First do the case $n = 1$. 
I'm guessing this is going to be some sort of induction proof after I calculate the $n = 1$ case? But I'm not sure how to do that. So if that's true could someone possibly help me with the $n = 1$ case and I could probably figure it out from there?
Many thanks. 

Le $f:P^n\to P^n$ be the map you describe and let $f^*:H^*(P^n)\to H^*(P^n)$ be the induced map on cohomology. Since $f^*$ is a map of rings and $H^*(P^n)$ is generated as a ring by its degree two component, it is enough to describe $f^2:H^2(P^n)\to H^2(P^n)$.
Moreover, $H^2(P^n)$ is a free $\mathbb Z$-module or rank $1$, generated by one class $\alpha\in H^2(P^n)$ which can be described quite explicitly (for example, using the usual CW structure on $P^n$ with one cell in each even dimension (and this CW structure happens to play very nicely with our map)), so it is enough to compute $f^*(\alpha)$.
Can you do that?
