= Relative cup product
{title2=$H^p(Y,U)\otimes H^q(Y,V)\to H^{p+q}(Y,U\cup V)$}
For open subsets $U,V$ of a <topological space> $Y$, the relative <cup product> gives
$$
H^p(Y,U;R)\otimes H^q(Y,V;R)\longrightarrow H^{p+q}(Y,U\cup V;R).
$$
Compatibility 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.
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