Cup Cap Cross Products and Cohomology Rings

Let $X,Y$ be spaces and $R$ a commutative unital ring, allowing $R=0$. Use ordinary, unnormalized singular chains, with the coefficient convention in Singular cochain complex with coefficients. Put $F_n(X,Y;R)={⨁}_{p+q=n}{C}_p(X;R){\otimes }_R{C}_q(Y;R),d(a\otimes b)=\mathrm{\partial }a\otimes b+(-1{)}^{p}a\otimes \mathrm{\partial }b(a\in C_p(X;R))\mathrm{.}$ Negative chain degrees and the boundary of a vertex are zero. For a simplex $\sigma :{\mathrm{\Delta }}^n\to X\times Y$, write $x={\mathrm{pr}}_X\sigma$ and $y={\mathrm{pr}}_Y\sigma$. The Alexander–Whitney map is the linear map ${\mathrm{AW}}_n:C_n(X\times Y;R)⟶F_n(X,Y;R),{\mathrm{AW}}_n(\sigma )={\sum }_{p=0}^{n}x[0,\dots ,p]\otimes y[p,\dots ,n]\mathrm{.}$ Here $x[i_0,\dots ,i_k]$ means restriction along the affine face with that ordered vertex list. The formula gives a finite chain for each generator and extends uniquely linearly; it does not require choosing representatives or fillings. Projections and the diagonal are continuous in the binary product topology. With the postcomposition maps of The induced singular chain map of a continuous map, the diagonal approximation is $D_X=\mathrm{AW}\circ {\mathrm{\Delta }}_{\mathrm{\#}}:C_{\ast }(X;R)⟶F_{\ast }(X,X;R),\mathrm{\Delta }(x)=(x,x)\mathrm{.}$

For completeness, the formula is a chain map. In $d\mathrm{AW}(\sigma )$, the first-factor faces at cut $p\ge 1$ have signs $(-1{)}^i$, $0\le i\le p$; the second-factor faces at cut $p\le n-1$ have signs $(-1{)}^j$, $p\le j\le n$. The first-factor term deleting its last vertex at cut $p$ is $(-1{)}^{p}x[0,\dots ,p-1]\otimes y[p,\dots ,n]$. It cancels the second-factor term deleting its first vertex at cut $p-1$, whose sign is $(-1{)}^{p-1}$. All remaining first-factor terms have $i<p$; they are precisely the cuts of the face of $\sigma$ omitting $i$ in which the omitted vertex lies before the cut. All remaining second-factor terms have $j>p$ and are precisely those with the omitted vertex after the cut. Each has the sign of that face in $\mathrm{\partial }\sigma$. This exhausts $\mathrm{AW}(\mathrm{\partial }\sigma )$ and proves $d\mathrm{AW}=\mathrm{AW}\mathrm{\partial }$. In degree zero both sides are zero, as required. The same face calculation shows that postcomposition commutes with $\mathrm{\partial }$; hence $D_X$ is also a chain map.

For continuous $f:X\to X^{\mathrm{\prime }}$ and $g:Y\to Y^{\mathrm{\prime }}$, face restriction commutes with postcomposition, term by term, giving $\mathrm{AW}(f\times g{)}_{\mathrm{\#}}=(f_{\mathrm{\#}}\otimes g_{\mathrm{\#}})\mathrm{AW},D_{X^{\mathrm{\prime }}}{f}_{\mathrm{\#}}=(f_{\mathrm{\#}}\otimes f_{\mathrm{\#}})D_X\mathrm{.}$ In degree zero AW sends $(x,y)$ to $x\otimes y$, preserving the augmentation that assigns $1$ to each vertex. If a factor is empty, both complexes are zero; the same holds over the zero ring. For one-point factors the formula still uses all higher singular simplices: degenerate simplices have not been discarded. The cancellation also applies to these simplices because it is an identity of face maps. Only finite sums and specified maps were used. No AC is assumed, and the arbitrary-product nonemptiness clause of the product definition is not used.

For arbitrary spaces $X,Y$ and every commutative unital ring $R$, the Alexander–Whitney map $A$ and signed shuffle map $S$ are natural chain maps Both preserve degree-zero augmentation, and $AS$ and $SA$ are naturally chain homotopic to the respective identity maps. Moreover, for the specified natural inverse $T$ constructed below there is a natural chain homotopy $A\simeq T$. This holds for ordinary unnormalized chains and requires no AC.

Let $X$ be a space, let $R$ be a commutative unital ring, and use the positive coboundary $\delta \phi =\phi \mathrm{\partial }$ of Singular cochain complex with coefficients. For $p,q\ge 0$ and cochains $\phi \in C^p(X;R)$, $\psi \in C^q(X;R)$, their cup product is $(\phi ⌣\psi )(\sigma )=\phi (\sigma [0,\dots ,p])\psi (\sigma [p,\dots ,p+q])(\sigma :{\mathrm{\Delta }}^{p+q}\to X)\mathrm{.}$ Extend from simplex generators $R$-linearly. The product is $R$-bilinear in the cochains by distributivity and commutativity in $R$. Zero or negative-degree inputs give zero.

With $J(\phi ,\psi )$ the tensor functional of Additive singular cohomology cross product and $A=\mathrm{AW}$ from Alexander–Whitney map and diagonal approximation, this is exactly $J(\phi ,\psi )A{\mathrm{\Delta }}_{\mathrm{\#}}$. Indeed only the cut of bidegree $(p,q)$ survives. Equivalently, define the AW external cochain by $J(\phi ,\psi )A$ and pull it back along the diagonal. There is no extra cochain sign.

The earlier additive cohomology cross product was expressed through a specified shuffle inverse $T$. The comparison with it is an equality of classes: Alexander--Whitney and shuffle are natural chain-homotopy inverses constructs $K$ with $A-T=dK+Kd$. For cocycles, The additive singular cohomology cross product is well-defined gives $Jd=0$, hence $J(A-T){\mathrm{\Delta }}_{\mathrm{\#}}=JKd{\mathrm{\Delta }}_{\mathrm{\#}}=\delta (JK{\mathrm{\Delta }}_{\mathrm{\#}})\mathrm{.}$ Here postcomposition with the diagonal commutes with boundary, as proved in the diagonal definition. Thus the AW cup class equals diagonal pullback of the earlier external product; equality of the two chosen external cochains is not required. At total degree zero the displayed primitive is zero and the identity is literal. The same tensor-functional identity gives $\delta (JA{\mathrm{\Delta }}_{\mathrm{\#}})=0$ for cocycles, so the compared classes exist.

For degree-zero cochains the formula on a vertex is ordinary multiplication of their values. A constant cochain of value $1$ multiplies any cochain on either side without changing it, including on disconnected spaces. On empty $X$ or over $R=0$ all cochains and products are zero. Face restrictions make sense for every singular simplex, including degenerate ones. A bare abelian coefficient group has no specified multiplication to insert in this formula. No AC or selection of representatives is used in this definition or comparison.

For cochains $\phi \in C^p(X;R)$ and $\psi \in C^q(X;R)$, $p,q\ge 0$, over a commutative unital ring with positive coboundary, $\delta (\phi ⌣\psi )=\delta \phi ⌣\psi +(-1{)}^p\phi ⌣\delta \psi \mathrm{.}$ Thus the product of cocycles is a cocycle, and changing either cocycle representative by a coboundary changes their product by a coboundary.

For a space $X$ and commutative unital coefficient ring $R$, put $H^{\ast }(X;R)={⨁}_{n\ge 0}{H}^n(X;R),[\phi ]⌣[\psi ]=[\phi ⌣\psi ]\mathrm{.}$ The groups and equality of representatives are those of Singular cohomology with coefficients. The multiplication descends through both quotient maps by Cup product Leibniz identity, including simultaneous changes. Extend to finite sums of homogeneous classes by distributivity; only finitely many summands occur, so each product belongs to the displayed direct sum.

This is the singular cohomology ring. To check the ring assertion before using it, let $\phi ,\psi ,\eta$ have degrees $p,q,r$. On an arbitrary $(p+q+r)$-simplex both parenthesizations of their cochain cup product evaluate to $\phi (\sigma [0,\dots ,p])\psi (\sigma [p,\dots ,p+q])\eta (\sigma [p+q,\dots ,p+q+r])\mathrm{.}$ Restriction to a face and then a face of that face is restriction to the listed vertex block, and multiplication in $R$ is associative. Thus cochain associativity holds exactly and descends to classes. Bilinearity gives the two distributive laws. Addition, additive inverses and zero come from the abelian quotient groups. Products of degrees $p,q$ have degree $p+q$.

Its multiplicative unit is the class of the cochain $\mathbf{1}\in C^0(X;R)$ assigning $1_R$ to every singular vertex. On any edge, $\delta \mathbf{1}=1_R-1_R=0$. The front/back formula gives $\mathbf{1}⌣\phi =\phi =\phi ⌣\mathbf{1}$ on every simplex. This includes all components at once: degree-zero cochains need not have finite support on the set of vertices or components.

If $X$ is empty or $R=0$, the ring is the zero ring and its unit equals zero, as allowed by our unital-ring convention. The formula for three factors works also when any degree is zero and for degenerate simplices; for one-point $X$ it has the same unit. Nothing here chooses a representative for every class: the quotient operation is defined by representative independence. No AC is used.

For continuous $f:X\to Y$ and a commutative unital ring $R$, pullback on singular cohomology satisfies $f^{\ast }(a⌣b)=f^{\ast }a⌣f^{\ast }b,f^{\ast }(1)=1.$ Cup product is associative already on cochains. Consequently $f^{\ast }:H^{\ast }(Y;R)\to H^{\ast }(X;R)$ is a unital graded-ring homomorphism. Homotopic maps give the same homomorphism.

Let $X$ be a space and $R$ a commutative unital ring. Put $A=\mathrm{AW}:C_{\ast }(X\times X;R)\to C_{\ast }(X;R){\otimes }_R{C}_{\ast }(X;R)$ and let $\tau (x,y)=(y,x)$. On homogeneous tensors put $W(a\otimes b)=(-1{)}^{pq}b\otimes a(\mathrm{\mid }a\mathrm{\mid }=p,\mathrm{\mid }b\mathrm{\mid }=q)\mathrm{.}$ Then $WA{\tau }_{\mathrm{\#}}$ and $A$ are naturally chain homotopic. In particular $W{D}_X$ and $D_X$, with $D_X=A{\mathrm{\Delta }}_{\mathrm{\#}}$, are naturally chain homotopic. No AC is required.

Let $X$ be a space, $R$ a commutative unital ring, $a\in H^p(X;R)$ and $b\in H^q(X;R)$, with $p,q\ge 0$. Then $a⌣b=(-1{)}^{pq}b⌣a\mathrm{.}$ The defining cochain operation need not be graded commutative. No AC is required.

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.

Relative cup products are natural for maps of excisive triples: if $f:X^{\mathrm{\prime }}\to X$ takes $A^{\mathrm{\prime }}$ into $A$ and $B^{\mathrm{\prime }}$ into $B$, then $f^{\ast }(u⌣v)=f^{\ast }u⌣f^{\ast }v\in H^{p+q}(X^{\mathrm{\prime }},A^{\mathrm{\prime }}\cup B^{\mathrm{\prime }};R)\mathrm{.}$ Use the excisive comparisons in the relative-product definition for both triples. In particular these exist when each pair of subspaces is open in its union.

Here are explicit domains for connector compatibility. Write $i:A\hookrightarrow X$, $U=A\cup B$, and let ${\mathrm{\partial }}_A:H^k(A;R)\to H^{k+1}(X,A;R)$ be the pair connector. Let ${\mathrm{\partial }}_{A,B}:H^k(A,A\cap B;R)⟶H^{k+1}(X,U;R)$ be the connector obtained by extending to a cochain on $X$ vanishing on $B$, taking its coboundary, and using the small-chain quotient comparison. Then ${\mathrm{\partial }}_{A,B}(x⌣i^{\ast }y)={\mathrm{\partial }}_{A}x⌣y(x\in H^p(A;R),y\in H^q(X,B;R)),$ and ${\mathrm{\partial }}_{A,B}(i^{\ast }x⌣y)=(-1{)}^{p}x⌣{\mathrm{\partial }}_{A}y(x\in H^p(X,B;R),y\in H^q(A;R))\mathrm{.}$ For $B=\mathrm{\varnothing }$ these are the ordinary pair identities. All coboundaries have the positive sign convention; $R$ is commutative unital. No AC is needed.

Let $X$ be a space and $R$ a commutative unital ring. For $p,n\ge 0$, $\phi \in C^p(X;R)$, and a singular $n$-simplex $\sigma$, define $\phi \cap \sigma =\{\begin{array}{ll}\phi (\sigma [0,\dots ,p])\sigma [p,\dots ,n],& n\ge p,\\ 0,& n<p\mathrm{.}\end{array}$ Extend $R$-bilinearly in $\phi$ and the finite chains of Singular simplices and singular chain groups with coefficients. An $R$-linear cochain is a function on simplex generators; each value multiplies one specified back-face generator. Thus the formula is well-defined on finite formal chains and is $R$-balanced in its two inputs. It gives $C^p(X;R){\otimes }_R{C}_n(X;R)⟶C_{n-p}(X;R),$ with negative chain groups zero. This is the cap product with cohomology first: evaluate on the front face and retain the back face.

The face convention is the same as Singular cup product on cochains. In terms of its Alexander–Whitney diagonal $D_X$, cap is the bidegree-$(p,n-p)$ part of $D_X$, followed by evaluation of the first factor by $\phi$. There is no additional sign in this evaluation. Subsequent boundary and projection identities use this order.

For $p=0$, this multiplies a simplex by the value of $\phi$ at its first vertex; in particular the constant value-one cochain acts as the identity on chains. For $p=n$, it returns $\phi (\sigma )$ times the last vertex, a degree-zero chain, whose boundary is zero. For $p>n$ it is zero by definition. On a degenerate simplex the same face formula applies; unnormalized chains retain these generators. Empty $X$, zero inputs and the zero ring give zero maps. The construction also applies to the higher singular simplices of a point and requires no AC.

For $\phi \in C^p(X;R)$, $c\in C_n(X;R)$, $p,n\ge 0$, over a commutative unital ring, $\mathrm{\partial }(\phi \cap c)=(-1{)}^p(\phi \cap \mathrm{\partial }c-\delta \phi \cap c)\mathrm{.}$ Consequently cap induces an $R$-bilinear map $H^p(X;R){\otimes }_R{H}_n(X;R)⟶H_{n-p}(X;R)\mathrm{.}$ Negative chain groups are zero, and no AC is assumed.

For continuous $f:X\to Y$, $\alpha \in H^p(Y;R)$ and $x\in H_n(X;R)$, $f_{\ast }(f^{\ast }\alpha \cap x)=\alpha \cap f_{\ast }x\mathrm{.}$ For $\alpha \in H^p(X;R)$, $\beta \in H^q(X;R)$ and $x\in H_n(X;R)$, $(\alpha ⌣\beta )\cap x=\beta \cap (\alpha \cap x)\mathrm{.}$ Here $R$ is commutative unital and cap evaluates the front face, retaining the back face. Both identities hold already on cochains and chains. No AC is needed.