The additive singular cohomology cross product is well-defined

On a tensor of bidegree $(p+1,q)$, $J(\phi ,\psi )d$ has only the term $\phi (dx)\psi (y)=J(\delta \phi ,\psi )(x\otimes y)$. On bidegree $(p,q+1)$, only $(-1{)}^p\phi (x)\psi (dy)$ remains, equal to $(-1{)}^{p}J(\phi ,\delta \psi )(x\otimes y)$. On every other bidegree of total degree $p+q+1$, all terms vanish by the support definition of $J$. Homogeneous tensors generate the total complex, proving the asserted identity. In particular $J(\phi ,\psi )$ is closed when both inputs are closed; its composite with $T$ is closed because $T$ is a chain map.

If $\phi$ changes to $\phi +\delta u$ with $\mathrm{\mid }u\mathrm{\mid }=p-1$ while $\psi$ is closed, step 1.1 gives $J(\delta u,\psi )=\delta J(u,\psi )$. After composing with $T$ the change is $\delta (J(u,\psi )T)$. If $\psi$ changes by $\delta v$ while $\phi$ is closed, step 1.1 gives $J(\phi ,\delta v)=(-1{)}^p\delta J(\phi ,v)$, yielding the coboundary $\delta ((-1{)}^{p}J(\phi ,v)T)$. A changed cocycle is still closed because ${\delta }^2=0$, so applying the two calculations successively handles simultaneous changes. When an input degree is zero, its negative-degree cochain is zero and the corresponding change is absent. This proves descent through both cohomology quotients.

Let $T^{\mathrm{\prime }}:D\to F$ be another chain homotopy inverse of $S$. Choose the supplied homotopies $dL+Ld=1_D-ST$ and $dP+Pd=T^{\mathrm{\prime }}S-1_F$ (negating a homotopy if needed). Then $K=T^{\mathrm{\prime }}L+PT$ has $dK+Kd=T^{\mathrm{\prime }}(1-ST)+(T^{\mathrm{\prime }}S-1)T=T^{\mathrm{\prime }}-T$, using the chain-map identities. For a closed functional $\chi =J(\phi ,\psi )$, $\chi (T^{\mathrm{\prime }}-T)=\chi dK+\chi Kd=\delta (\chi K)$ since $\chi d=0$. Thus both inverse choices give the same class. In total degree zero, the potential primitive has negative degree and is zero; the same identity gives equality directly.

On cochains, $J$ is additive and $R$-linear in each variable by its evaluation formula and commutativity of $R$. Precomposition with $T$ is linear, so step 2.1 descends this bilinearity to cohomology and hence to the tensor product of cohomology modules. For maps $f:X^{\mathrm{\prime }}\to X$ and $g:Y^{\mathrm{\prime }}\to Y$, naturality of the specified $T$ gives $T_{X,Y}(f\times g{)}_{\mathrm{\#}}=(f_{\mathrm{\#}}\otimes g_{\mathrm{\#}})T_{X^{\mathrm{\prime }},Y^{\mathrm{\prime }}}$. Evaluation on homogeneous tensors gives $J(\phi ,\psi )(f_{\mathrm{\#}}\otimes g_{\mathrm{\#}})=J(\phi f_{\mathrm{\#}},\psi g_{\mathrm{\#}})$. Combining these equalities and passing to classes proves $(f\times g{)}^{\ast }(\alpha \times \beta )=f^{\ast }\alpha \times g^{\ast }\beta$. Independence in step 2.2 makes this naturality independent of the particular inverse used to express the classes.

If either factor space is empty or either cochain class is zero, the product is zero by step 2.1 and bilinearity; the zero ring likewise gives zero modules. At $p=q=0$, $T$ is the inverse vertex-pair identification, so the product value is $\phi (x)\psi (y)$ and the two unit values on points multiply to $1$. Steps 1.1 and 2.1 separately cover $p=0$ or $q=0$, with no missing negative cochains. All chosen homotopies in step 2.2 are supplied as part of the inverse data, and the canonical inverse in [F2] is explicit, so no family of arbitrary choices or AC is used.