Relative cap products with quotient domains displayed

Let $A,B\subseteq X$, $R$ be a commutative unital ring, and put $N=C_{\ast }(A;R)+C_{\ast }(B;R)$. The cohomology-first cap formula induces $C^p(X,A;R){\otimes }_R(C_n(X;R)\mathrm{/}{N}_n)⟶C_{n-p}(X,B;R),\phi \otimes [c]⟼[\phi \cap c]\mathrm{.}$ For a simplex in $A$ its front-face evaluation is zero; for a simplex in $B$ its retained back face is in $B$. Thus changing $c$ by a chain in either summand of $N$ changes the output by zero modulo $C_{\ast }(B;R)$. These are all relations in the quotient, and bilinearity respects the tensor relations.

Use Relative singular homology for relative cycles and boundaries. Suppose $\delta \phi =0$ and $\mathrm{\partial }c\in N$. The identity of Cap product boundary identity gives $\mathrm{\partial }(\phi \cap c)=(-1{)}^p\phi \cap \mathrm{\partial }c$, which is zero modulo $B$ by the preceding calculation. If $c$ changes by $\mathrm{\partial }b$ plus an element of $N$, the change in the output is the relative boundary $(-1{)}^p\mathrm{\partial }(\phi \cap b)$. If $\phi$ changes by $\delta u$ with $u$ vanishing on $A$, then $\delta u\cap c=u\cap \mathrm{\partial }c+(-1{)}^p\mathrm{\partial }(u\cap c)\mathrm{.}$ Its first term is zero modulo $B$ because $\mathrm{\partial }c\in N$, so this too is a relative boundary. For $p=0$ the possible $u$ is zero. This proves descent in both variables.

Under the excisive hypotheses of Relative cup product for an excisive triad, its explicitly constructed quotient-chain equivalence gives $H_n(C_{\ast }(X;R)\mathrm{/}N)\cong H_n(X,A\cup B;R)\mathrm{.}$ Transporting the chain construction through this canonical quotient map defines the relative cap product $H^p(X,A;R){\otimes }_R{H}_n(X,A\cup B;R)⟶H_{n-p}(X,B;R)\mathrm{.}$ As with relative cup, openness of $A,B$ in their union suffices; any other neighborhood/subcomplex replacement must supply the indicated compatible comparison. An arbitrary triad is not silently assumed excisive.

In particular, taking $B=\mathrm{\varnothing }$ gives $H^p(X,A;R){\otimes }_R{H}_n(X,A;R)⟶H_{n-p}(X;R)\mathrm{.}$ Taking $B=A$ and precomposing the homology input with the quotient map from absolute homology gives $H^p(X,A;R){\otimes }_R{H}_n(X;R)⟶H_{n-p}(X,A;R)\mathrm{.}$ For $A=\mathrm{\varnothing }$ the general construction includes the usual action on relative homology of $(X,B)$. If $A=X$ the cohomology input is zero; if $B=X$ the target is zero. Empty spaces, zero ring/inputs and point spaces cause no exception. When $n<p$ output chains are zero, and when $n=p$ they are vertex chains. Degenerate simplices obey the same containment tests. All operations and quotient comparisons are explicit, so no AC is required.