Let \(\Sigma X\) be the suspension of a space \(X\), and let \(U, V\) be the open sets obtained as the complements of the two suspension points. Show that \(H^*(\Sigma X, U) \cong \widetilde{H}^*(\Sigma X)\). By considering the cup product \[ H^*(\Sigma X, U) \times H^*(\Sigma X, V) \to H^*(\Sigma X, U \cup V), \] deduce that the cup product of any two cohomology classes of positive degree on \(\Sigma X\) is zero.
Solution
1. The open set \(U=\Sigma X\setminus\{\text{one suspension point}\}\) is contractible, and similarly for \(V\).
Removing one suspension point leaves an open cone on \(X\), which contracts toward the opposite suspension point.
2. The map in the long exact sequence of the pair gives a natural isomorphism\[ \boxed{H^*(\Sigma X,U)\cong\widetilde H^*(\Sigma X)}. \]
Since \(U\) is nonempty and contractible, its reduced cohomology vanishes. The reduced long exact sequence for the pair \((\Sigma X,U)\) therefore identifies relative cohomology with reduced cohomology of \(\Sigma X\) in every degree.
3. Every positive-degree class \(a\in H^*(\Sigma X)\) has a lift\[ \bar a\in H^*(\Sigma X,U), \]and every positive-degree class \(b\) has a lift\[ \bar b\in H^*(\Sigma X,V). \]
Apply 2 to \(U\) and to \(V\). In positive degree reduced and ordinary cohomology agree.
4. Their relative cup product lies in\[ H^*(\Sigma X,U\cup V)=H^*(\Sigma X,\Sigma X)=0. \]
The relative cup product has the form \[ H^p(\Sigma X,U)\times H^q(\Sigma X,V) \to H^{p+q}(\Sigma X,U\cup V). \] Since the two complements together cover the suspension, \(U\cup V=\Sigma X\).
5. Hence every cup product of two positive-degree classes on a suspension is zero.
The natural map from the relative product in 4 to absolute cohomology sends \(\bar a\smile\bar b\) to \(a\smile b\). The relative product is zero, so \[ \boxed{a\smile b=0.} \]