Additive singular cohomology cross product

Let $R$ be a commutative unital ring, $X,Y$ spaces and $p,q\ge 0$. Use the positive coboundary convention of Singular cochain complex with coefficients. For homogeneous cochains $\phi \in C^p(X;R)$ and $\psi \in C^q(Y;R)$, define a functional $J(\phi ,\psi )$ on the tensor chain complex by $J(\phi ,\psi )(x\otimes y)=\{\begin{array}{ll}\phi (x)\psi (y),& \mathrm{\mid }x\mathrm{\mid }=p,\mathrm{\mid }y\mathrm{\mid }=q,\\ 0,& \mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }=p+q\text{ and }\mathrm{\mid }x\mathrm{\mid }\mathrm{\ne }p\mathrm{.}\end{array}$ Extend linearly. The formula is $R$-balanced, since multiplying either input chain by $r$ multiplies its value by $r$; commutativity of $R$ is used here. Each total degree is a finite direct sum.

Let $T:C(X\times Y;R)\to C(X;R){\otimes }_{R}C(Y;R)$ be the natural chain homotopy inverse of shuffle supplied by Singular product chain equivalence by simplex models. For cocycles define the additive singular cohomology cross product by $[\phi ]\times [\psi ]=[J(\phi ,\psi )T]\in H^{p+q}(X\times Y;R),$ using Singular cohomology with coefficients. The next lemma The additive singular cohomology cross product is well-defined ↗ proves that the cochain is closed, that both changes of representatives yield coboundaries, that another chain homotopy inverse gives the same class, and that the resulting $R$-bilinear map is natural. These verifications precede every use of the quotient product.

This convention puts no extra $(-1{)}^{pq}$ in $J$: its differential obeys $\delta J(\phi ,\psi )=J(\delta \phi ,\psi )+(-1{)}^{p}J(\phi ,\delta \psi )$ with positive coboundaries. It is an additive cross product only; cup products and cohomology-ring assertions are not part of this definition. Negative-degree inputs, empty factors or the zero ring give zero products. In degree zero, the vertex formula gives $(\phi \times \psi )(x,y)=\phi (x)\psi (y)$. No AC is required.