Local coordinate cup products generate top relative cohomology

For $v\mathrm{\ne }0$ put $\rho (v)={\max }_i\mathrm{\mid }{v}_i\mathrm{\mid }$. The homotopy $h_t(v)=((1-t)+t\mathrm{/}\rho (v))v$ is continuous on ${\mathbb{R}}^r\setminus 0$, stays nonzero, is the identity at $t=0$, and at $t=1$ lands on $\mathrm{\partial }{C}_r$. It fixes $\mathrm{\partial }{C}_r$ for every $t$. Thus $\mathrm{\partial }{C}_r\hookrightarrow {\mathbb{R}}^r\setminus 0$ is a strong deformation-retract inclusion. The cube and Euclidean space are both nonempty and contractible by $v\mapsto (1-t)v$, and their inclusion induces a cohomology isomorphism by [F5]. A point has cohomology $R$ in degree zero and zero above: its positive unnormalized cochain complex alternates zero and identity differentials.

By [F6], $H_k(C_r,\mathrm{\partial }{C}_r;R)$ is $R$ at $k=r$ and zero elsewhere. With integral chains the same assertion gives a free integral homology group in just degree $r$. Apply [F7] with coefficients $R$: all its Ext terms vanish, since either the zero resolution or the length-zero identity resolution of $\mathbb{Z}$ computes them. Thus $H^k(C_r,\mathrm{\partial }{C}_r;R)$ is $R$ for $k=r$ and zero elsewhere, and evaluation identifies its top group with ${\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z},R)$. Let $v_r$ be the class taking value $1_R$ on the positively oriented cube generator.

We spell out the needed pair-sequence comparison. For a nonempty contractible space $V$ and a nonempty subspace $T$, the initial restriction $H^0(V;R)=R\to H^0(T;R)$ includes the constant functions and is injective. The sequence of [F4] therefore gives $H^0(V,T)=0$, $H^1(V,T)=H^0(T)\mathrm{/}R$, and $H^k(V,T)\cong H^{k-1}(T)$ for $k\ge 2$, with the isomorphism given by the connector. Consequently a map of pairs between two such contractible ambient spaces, inducing a cohomology isomorphism on their subspaces, induces isomorphisms in all relative degrees: in degree one it respects the same constant subgroup, and in higher degrees use the natural connector square of [F4]. Apply this to step 1.1. The inclusion $j_r:(C_r,\mathrm{\partial }{C}_r)\to ({\mathbb{R}}^r,{\mathbb{R}}^r\setminus 0)$ gives the claimed relative-cohomology isomorphism in every degree. When $r=1$, the subspace has two components and $H^0(T)\mathrm{/}R=(R\oplus R)\mathrm{/}R(1,1)$, so this argument includes that endpoint.

For the sign, represent the positively oriented relative cube class by its oriented simplex decomposition. One explicit decomposition, after rescaling to $[0,1{]}^r$, has a simplex for each permutation $\pi$: its vertices are $0,e_{\pi (1)},e_{\pi (1)}+e_{\pi (2)},\dots ,(1,\dots ,1)$, with coefficient $\mathrm{sgn}\pi$. These simplices cover the cube, since the decreasing order of a point's coordinates gives its containing simplex and their differences give its nonnegative barycentric coordinates. Ties describe their common faces. Their signed boundaries cancel on shared interior faces and leave the oriented outer faces. The resulting relative cycle $c_r$ is the positive generator in [F6]: on the characteristic disk's interior each simplex has determinant sign $\mathrm{sgn}\pi$, compensated by its coefficient, so the decomposition subdivides that oriented disk with multiplicity one. Equivalently, restriction at a point with strictly ordered coordinates sees one oriented simplex with coefficient one. The characteristic orientation selects exactly this relative disk class in [F6]. This check concerns relative singular chains with their affine parameterizations, not a cellular cochain substituted into the cup formula.

Projection $W\to {\mathbb{R}}^p$ identifies the pair $(W,A)$ up to deformation with $({\mathbb{R}}^p,{\mathbb{R}}^p\setminus 0)$: contract the second coordinate to zero, preserving $A$ throughout. On absolute spaces and subspaces these are homotopy equivalences by [F5], so step 2.1 and [F4] give the same identification in relative cohomology. Similarly for $(W,B)$ and the second factor. Inside the cube product $K=C_p\times C_q=C_{p+q}$ put $A_0=\mathrm{\partial }{C}_p\times C_q$ and $B_0=C_p\times \mathrm{\partial }{C}_q$. Contracting the unused cube coordinate gives the analogous two pair identifications. The inclusions $A_0\hookrightarrow A$ and $B_0\hookrightarrow B$ induce cohomology isomorphisms: contract their unused coordinates and use the radial deformation of step 1.1 on the other coordinate. Thus restriction identifies $H^p(W,A)$ with $H^p(K,A_0)$, taking ${\mathrm{pr}}_1^{\ast }{u}_p$ to ${\mathrm{pr}}_1^{\ast }{v}_p$ when $j_p^{\ast }{u}_p=v_p$, and similarly for the other factor. This last assertion follows from literal commutation of projections with the inclusions and the pair naturality in [F4].

We have $A\cup B=W\setminus 0$ and $A_0\cup B_0=\mathrm{\partial }{C}_{p+q}$. Using $\rho (x,y)=\max \{{\max }_i\mathrm{\mid }{x}_i\mathrm{\mid },{\max }_j\mathrm{\mid }{y}_j\mathrm{\mid }\}$ in step 1.1 gives a deformation retraction of $W\setminus 0$ onto this product boundary. Step 2.1 therefore makes the bottom restriction $H^{\ast }(W,W\setminus 0;R)\to H^{\ast }(K,\mathrm{\partial }K;R)$ an isomorphism as well. The upper triple has open subspaces $A,B$ and so has the open relative product of [F3]. The lower triple has the proved CW product comparison of [F2]. Although that lower triple is not open, the needed comparison with the upper product is direct: restriction from $W$ to $K$ carries cochains vanishing on $A$ and $B$ to cochains vanishing on $A_0$ and $B_0$, and it commutes literally with the Alexander--Whitney front/back formula. It also commutes with the quotient maps from the quotient by the sum of the two relative subcomplexes to the quotient by their union. The upper quotient comparison of [F3] and the lower quotient comparison of [F2] are isomorphisms, so their inverses commute with restriction as well. Thus the relative-cup square commutes, with all three restriction maps just proved isomorphisms.

The relative shuffle of $c_p\otimes c_q$ represents the positive generator $c_{p+q}$ of the product cube. In the interiors of the two chosen factor simplices, a shuffle interleaves their ordered edge directions; the determinant of the resulting ordered product directions is its shuffle permutation sign times the two factor determinant signs. The coefficients in $c_p,c_q$ and in [F8]'s shuffle cancel exactly these signs. The product decomposition therefore has multiplicity one and positive orientation on every interior simplex, and its remaining boundary lies in $A_0\cup B_0$ by the boundary rule. By the characteristic orientation argument of step 2.2, its relative class is $[c_{p+q}]$. This is why the ordered factors give positive sign, rather than merely an unspecified unit.

The finite CW pairs $(C_p,\mathrm{\partial }{C}_p)$ and $(C_q,\mathrm{\partial }{C}_q)$ have the degreewise finite-free $R$-homology calculated in step 1.2. The coefficient rings $\mathbb{Z}$ and ${\mathbb{F}}_2$ are PIDs. Thus [F1] applies. In total degree $p+q$ its source has just the single summand $R{v}_p{\otimes }_{R}R{v}_q$, and it sends $v_p\otimes v_q$ to the product of their two projection pullbacks by [F2]. This is a generator of $H^{p+q}(K,\mathrm{\partial }K;R)$. The isomorphism square in step 3.2 transports this to the product in the statement. It already proves the generator assertion without an orientation-sign convention.

Choose relative cocycles $\phi ,\psi$ representing $v_p,v_q$, with $\phi (c_p)=\psi (c_q)=1_R$. In the tensor of relative chain complexes, $c_p,c_q$ are cycles and $J=J(\phi ,\psi )$ satisfies $Jd=0$. Let $a,b$ be the quotient AW and shuffle of [F2], with $ab-1=dL+Ld$ from [F8]. The relative product is the unique class whose pullback to the smaller quotient is $Ja$. Evaluating that class on the actual relative shuffle cycle therefore gives $Jab(c_p\otimes c_q)=J(c_p\otimes c_q)+J(dL+Ld)(c_p\otimes c_q)=1_R\mathrm{.}$ Both homotopy terms vanish by the two cycle equations. Step 3.3 identifies this cycle with the positive cube generator, so the product is $v_{p+q}$, not its negative over $\mathbb{Z}$. The commuting square of step 3.2 and the defining normalization through $j_{p+q}$ prove the asserted positive local normalization.

The case $p=1$ or $q=1$ uses the explicit degree-one quotient by constants in step 2.1; the interval relative cycle has boundary $[1]-[-1]$ and the chosen dual evaluates to one. The integer normalization fixes the sign, while over ${\mathbb{F}}_2$ the two signs agree. Any other integral generator is the negative of the normalized generator or that generator itself, and the only nonzero field generator over ${\mathbb{F}}_2$ is the normalized one; bilinearity therefore proves the generator assertion for every allowed choice of the two generators. Zero classes have zero products by bilinearity. There are no empty Euclidean factors here, since $p,q\ge 1$; zero dimensions and the zero ring are outside this lemma's hypotheses. Both radial homotopy endpoints and their nonzero domains were checked in step 1.1. Shared or degenerate singular faces are retained, and all affine shuffle sums are finite. AC is used in [F9] only for [F1]'s PID splittings and homology sections/bases and [F7]'s integral cycle projections; the radial comparisons, naturality and orientation determinants introduce none.