By Exercise 35 in §4.2 there is a bundle \(S^2 \to \mathbb{CP}^3 \to S^4\). Let \(S^2 \to E_k \to S^4\) be the pullback of this bundle via a degree \(k\) map \(S^4 \to S^4\), \(k > 1\). Use the Leray-Hirsch theorem to show that \(H^*(E_k; \mathbb{Z})\) is additively isomorphic to \(H^*(\mathbb{CP}^3; \mathbb{Z})\) but has a different cup product structure in which the square of a generator of \(H^2(E_k; \mathbb{Z})\) is \(k\) times a generator of \(H^4(E_k; \mathbb{Z})\).
Solution
Let \(p:\mathbb{CP}^3\to S^4\) denote the bundle and let \(f:S^4\to S^4\) have degree \(k\). Form the pullback square \[ \begin{array}{ccc} E_k&\xrightarrow{F}&\mathbb{CP}^3\\ \downarrow p_k&&\downarrow p\\ S^4&\xrightarrow{f}&S^4. \end{array} \] Choose \(x\in H^2(\mathbb{CP}^3;\mathbb Z)\) restricting to a generator on each fiber \(S^2\). Then \(F^*x\), which we again call \(x\), restricts to a generator on each fiber of \(E_k\). Leray–Hirsch gives \[ H^*(E_k;\mathbb Z)\cong H^*(S^4;\mathbb Z)\{1,x\} \] as an \(H^*(S^4)\)-module. Thus additively there is one copy of \(\mathbb Z\) in degrees \(0,2,4,6\), exactly as for \(\mathbb{CP}^3\).
Let \(u\in H^4(S^4;\mathbb Z)\) be a generator and normalize signs so that in \(\mathbb{CP}^3\) \[ x^2=p^*u. \] Put \(v=p_k^*u\in H^4(E_k)\). Naturality in the pullback square gives \[ x^2=F^*(x^2)=F^*p^*u=p_k^*f^*u=k\,p_k^*u=kv. \] Hence \[ \boxed{x^2=kv,} \] where \(x\) and \(v\) generate \(H^2(E_k)\) and \(H^4(E_k)\). Since \(k>1\), this ring cannot be isomorphic to \(H^*(\mathbb{CP}^3;\mathbb Z)\), although the additive groups are the same.