Using the cup product \(H^k(X, A; R) \times H^\ell(X, B; R) \to H^{k+\ell}(X, A \cup B; R)\), show that if \(X\) is the union of contractible open subsets \(A\) and \(B\), then all cup products of positive-dimensional classes in \(H^*(X; R)\) are zero. This applies in particular if \(X\) is a suspension. Generalize to the situation that \(X\) is the union of \(n\) contractible open subsets, to show that all \(n\)-fold cup products of positive-dimensional classes are zero.
Solution
Suppose first that \(X=A\cup B\) with \(A,B\) contractible and let \(x\in H^p(X;R)\), \(y\in H^q(X;R)\) with \(p,q>0\).
Since \(H^p(A;R)=0\) for \(p>0\), exactness of the cohomology sequence of \((X,A)\) gives a lift \[ \bar x\in H^p(X,A;R) \] of \(x\). Likewise \(y\) lifts to some \(\bar y\in H^q(X,B;R)\). The relative cup product gives \[ \bar x\smile\bar y\in H^{p+q}(X,A\cup B;R)=H^{p+q}(X,X;R)=0. \] After forgetting relative structure this class maps to \(x\smile y\), so \(x\smile y=0\). In particular every product of positive-dimensional classes on a suspension vanishes, since a suspension is the union of its two contractible cones.
More generally suppose \[ X=A_1\cup\cdots\cup A_n \] with all \(A_i\) contractible, and let \(x_i\in H^{d_i}(X;R)\) with \(d_i>0\). Each \(x_i\) lifts to \[ \bar x_i\in H^{d_i}(X,A_i;R). \] Iterating the relative cup product gives \[ \bar x_1\smile\cdots\smile\bar x_n \in H^{d_1+\cdots+d_n}(X,A_1\cup\cdots\cup A_n;R) =H^*(X,X;R)=0. \] Its image in absolute cohomology is \(x_1\smile\cdots\smile x_n\), hence every \(n\)-fold product of positive-dimensional classes is zero.