Relative cup product for an excisive triad

Let $A,B$ be subspaces of $X$ and $R$ a commutative unital ring. Set $U=A\cup B$ and $N=C_{\ast }(A;R)+C_{\ast }(B;R)\subset C_{\ast }(X;R)$. Use the relative cochains of Relative singular cochain complex, so $C^{\ast }(X,A;R)$ means cochains vanishing on $C_{\ast }(A;R)$. For $\phi \in C^p(X,A;R)$ and $\psi \in C^q(X,B;R)$, use the front/back formula of Singular cup product on cochains. A simplex wholly in $A$ has its front face in $A$, and one wholly in $B$ has its back face in $B$. Thus $\phi ⌣\psi$ vanishes on $N$ and defines a functional on $C_{\ast }(X;R)\mathrm{/}N$.

The Leibniz identity of Cup product Leibniz identity holds in these quotient-functional complexes. Its representative-change primitive also vanishes on $N$: each of its summands has a first factor vanishing on $A$ and a second vanishing on $B$. Consequently this construction descends to $H^p(X,A;R){\otimes }_R{H}^q(X,B;R)⟶H^{p+q}({\mathrm{Hom}}_R(C_{\ast }(X;R)\mathrm{/}N,R))\mathrm{.}$

Assume now that $A,B$ are open in $U$ (in particular they may be open in $X$). The relative cup product has target $H^{p+q}(X,U;R)$, using the following canonical comparison. By Cover-small chains for a two-open cover, $N$ is exactly the small-chain complex of the cover $U=A\cup B$. We need the following stronger data, not merely the abstract chain-homotopy-equivalence assertion of The cover-small inclusion is a chain homotopy equivalence. Let $S$ be barycentric subdivision and let $T$ satisfy $1-S=dT+Td$, as supplied by Subdivision is chain homotopic to the identity, Barycentric subdivision operator, and Subdivision prism homotopy. These operators preserve simplex images. For a simplex $\sigma$, let $a(\sigma )$ be the least subdivision count making it small, whose existence follows from Finite chains eventually become cover-small, and recursively set $m(\sigma )=\max (\{a(\sigma )\}\cup \{m(\sigma {\delta }_j):0\le j\le \dim \sigma \}),$ with $m=0$ on vertices. If $\sigma$ is already small then $m(\sigma )=0$. Define $D\sigma ={\sum }_{h=0}^{m(\sigma )-1}T{S}^h\sigma$ and $r=1-dD-Dd$. The telescoping identity shows that $r\sigma$ is small: besides $S^{m(\sigma )}\sigma$, each face correction is a sum of $T{S}^h(\sigma {\delta }_j)$ with $h\ge m(\sigma {\delta }_j)$ and is therefore small. Thus $r:C_{\ast }(U;R)\to N$ is a chain map, $ri=1$, $1-ir=dD+Dd$, and $D{\mathrm{\mid }}_N=0$. The integral construction extends to $R$ by tensoring, so all these identities remain valid and all operators preserve chains in $U$.

Here is why this gives the needed relative cohomology comparison, not just an absolute homology comparison. Extend $D$ to a degree-one map $E$ on $C_{\ast }(X;R)$ by the same formula on simplices with image in $U$, and by zero on all other simplex generators. This is an unambiguous linear extension; it need not itself commute with $d$. Put $P=1-dE-Ed$. Then $P$ is a chain map, $P{\mathrm{\mid }}_N=1$, and $P(C_{\ast }(U;R))\subset N$. It therefore induces $\bar{P}:C_{\ast }(X;R)\mathrm{/}{C}_{\ast }(U;R)⟶C_{\ast }(X;R)\mathrm{/}N\mathrm{.}$ The quotient map $q:C_{\ast }(X;R)\mathrm{/}N\to C_{\ast }(X;R)\mathrm{/}{C}_{\ast }(U;R)$ is a chain map. Since $E$ preserves both $N$ and $C_{\ast }(U;R)$, it descends to each quotient; the identity $1-P=dE+Ed$ shows that $\bar{P}q$ and $q\bar{P}$ are homotopic to the respective identities. Applying ${\mathrm{Hom}}_R(-,R)$ preserves these explicit homotopy identities. Hence $q^{\ast }:H^{\ast }(X,U;R)⟶H^{\ast }({\mathrm{Hom}}_R(C_{\ast }(X;R)\mathrm{/}N,R))$ is an isomorphism. Define the relative cup product by composing the preceding product with $(q^{\ast }{)}^{-1}$. This inverse on cohomology is unique; auxiliary choices in an alternative small-chain comparison cannot affect it.

The same definition applies whenever a specified neighborhood or subcomplex replacement has separately proved this quotient-cochain comparison and its compatibility with the pair maps. Such a replacement is an additional hypothesis, not an assertion for every triad. In general a simplex in $A\cup B$ can meet both members without lying in either, so vanishing on $N$ alone does not mean vanishing on $C_{\ast }(A\cup B)$.

If either relative class is zero, its product is zero by the displayed primitive. If $A=B=\mathrm{\varnothing }$, then $N=0$ and $q$ is the identity, recovering absolute cup. If $A=X$ or $B=X$, one source group is zero. Empty $X$, zero coefficients and point spaces satisfy the same formulas. Degree-zero factors evaluate at a vertex; negative-degree groups and their possible primitives are zero. All singular simplices, including degenerate ones, are retained. The small-chain operators use least subdivision counts, and the extension of $D$ is prescribed on simplex generators. No AC is required.