Describe the cohomology rings \(H^*(X \sqcup Y)\) and \(H^*(X \vee Y)\) in terms of \(H^*(X)\) and \(H^*(Y)\).
Solution
1. For a disjoint union,\[ \boxed{H^*(X\sqcup Y;R)\cong H^*(X;R)\times H^*(Y;R)} \]as graded rings.
A singular simplex in \(X\sqcup Y\) lies entirely in exactly one component, so the cochain complex splits as the product of the cochain complexes of \(X\) and \(Y\). Cup products are computed componentwise, giving the product-ring structure.
2. For a wedge of based path-connected spaces,\[ \widetilde H^*(X\vee Y;R) \cong \widetilde H^*(X;R)\oplus\widetilde H^*(Y;R) \]as graded groups.
The wedge is the union of \(X\) and \(Y\) with contractible intersection consisting of the basepoint. The reduced Mayer–Vietoris sequence gives the direct-sum decomposition.
3. Under the decomposition in 2, products between positive-degree classes from different summands vanish, while products within a summand are its original cup products.
Let \(i_X:X\hookrightarrow X\vee Y\) and \(i_Y:Y\hookrightarrow X\vee Y\). A class coming from \(X\) restricts trivially to \(Y\) in positive degree, and vice versa. Hence the product of one class from each summand restricts to zero on both \(X\) and \(Y\), and therefore is zero under the Mayer–Vietoris identification. Naturality of cup product shows that products of two \(X\)-classes, or two \(Y\)-classes, agree with the original ring structures.
4. Equivalently, for path-connected \(X,Y\),\[ \boxed{H^*(X\vee Y;R)\cong H^*(X;R)\times_R H^*(Y;R),} \]where the fiber product identifies the two degree-zero copies of \(R\).
This is exactly the combination of the common unit in degree zero with the positive-degree direct sum and zero cross-products described in <1>2–<1>3.