Use cup products to compute the map \(H^*(\mathbb{CP}^n; \mathbb{Z}) \to H^*(\mathbb{CP}^n; \mathbb{Z})\) induced by the map \(\mathbb{CP}^n \to \mathbb{CP}^n\) that is a quotient of the map \(\mathbb{C}^{n+1} \to \mathbb{C}^{n+1}\) raising each coordinate to the \(d\)th power, \((z_0, \ldots, z_n) \mapsto (z_0^d, \ldots, z_n^d)\), for a fixed integer \(d > 0\). [First do the case \(n = 1\).]
Solution
1. For \(n = 1\), the map \(\mathbb{CP}^1 \to \mathbb{CP}^1\) is \([z_0 : z_1] \mapsto [z_0^d : z_1^d]\), which has degree \(d\).
on \(\mathbb{CP}^1 \cong S^2\), the map \(z \mapsto z^d\) (in the affine chart) has degree \(d\).
2. Hence on \(H^2(\mathbb{CP}^1) = \ZZ\) (generated by \(x\)), the induced map sends \(x \mapsto d x\).
1 (the induced map on top cohomology is multiplication by the degree).
3. For general \(n\), \(H^*(\mathbb{CP}^n) = \ZZ[x]/(x^{n+1})\) with \(|x| = 2\).
cohomology of \(\mathbb{CP}^n\).
4. The map \(f : \mathbb{CP}^n \to \mathbb{CP}^n\) satisfies \(f^*(x) = d x\).
the restriction of \(f\) to any \(\mathbb{CP}^1 \subseteq \mathbb{CP}^n\) (a coordinate line) is the degree-\(d\) map, so \(f^*(x)\) restricts to \(d x\) on each \(\mathbb{CP}^1\); since \(H^2(\mathbb{CP}^n) \to H^2(\mathbb{CP}^1)\) is an isomorphism, \(f^*(x) = d x\).
5. Hence \(f^*(x^k) = f^*(x)^k = (dx)^k = d^k x^k\).
4 and the ring homomorphism property (cup product).
6. Therefore the induced map \(f^* : H^*(\mathbb{CP}^n) \to H^*(\mathbb{CP}^n)\) is multiplication by \(d^k\) on \(H^{2k}\), i.e. \(f^*(x^k) = d^k x^k\).
5.
7. Q.E.D.
6.