Relative cup product ($H^p(Y,U)\otimes H^q(Y,V)\to H^{p+q}(Y,U\cup V)$)

For open subsets $U,V$ of a topological space $Y$, the relative cup product givesCompatibility with the maps to absolute cohomology identifies its image with the ordinary cup product. More generally the construction requires an excisive pair of subsets. The open-set case can be proved using subdivision to compute on chains subordinate to the open cover.

If a pointed CW complex is covered by $l$ contractible subcomplexes containing the basepoint, every reduced class lifts to a relative class for each appropriate subcomplex. Their relative cup product lies in $h^{\ast }(X,{\bigcup }_j{A}_j)=h^{\ast }(X,X)=0$, so every product of $l$ reduced classes vanishes. The conclusion also holds if $X$ is path connected. Without either condition a two-point space covered by its two points is a counterexample in ordinary degree-zero cohomology.

Let $U_1,\dots ,U_k$ be an open cover of $Y$, and suppose ${\alpha }_i\in H^{d_i}(Y;R)$ restricts to zero on $U_i$. The pair long exact sequence lifts each ${\alpha }_i$ to $H^{d_i}(Y,U_i;R)$. Their relative cup product lies in $H^{\sum d_i}(Y,{\bigcup }_i{U}_i;R)=H^{\sum d_i}(Y,Y;R)=0$, so ${\alpha }_1⌣\cdots ⌣{\alpha }_k=0$. This proves characteristic-class products vanish from local trivializations without cancelling possible zero divisors in the coefficient ring.