De Rham integration respects wedge and cup in cohomology

For a smooth $n$-simplex $\sigma$, [F9] supplies an extension $\bar{\sigma }:O\to M$, where $O$ is open in the affine span and contains ${\mathrm{\Delta }}^n$. By [F8], ${\bar{\sigma }}^{\ast }\alpha$ and ${\bar{\sigma }}^{\ast }\beta$ are smooth forms on the boundaryless open set $O$, even when $\sigma$ meets $\mathrm{\partial }M$. Define the degree-$N$ form on the Euclidean-open product $O\times O$ by ${\mathrm{\Theta }}_{\sigma }(\alpha ,\beta )={\mathrm{pr}}_1^{\ast }({\bar{\sigma }}^{\ast }\alpha )\wedge {\mathrm{pr}}_2^{\ast }({\bar{\sigma }}^{\ast }\beta )\mathrm{.}$ Only its germ along $Q_n$ will be integrated. Changing the extension does not change that germ's restriction or any derivatives along $Q_n$: the pulled-back forms agree on the interior of ${\mathrm{\Delta }}^n$ and hence with all derivatives on its closure. Consequently all the affine-chain integrals below are independent of the extension, by [F5] and [F6]. No product manifold $M\times M$ with corners is used.

For $N\ge 1$ and a smooth $(N-1)$-simplex define $K(\alpha ,\beta )(\sigma )={\int }_{h_{N-1}}{\mathrm{\Theta }}_{\sigma }(\alpha ,\beta ),$ extending from simplex generators linearly to chains. For $N=0$, put $K=0$ in the zero group $C_{\mathrm{\infty }}^{-1}$. Each integral is over a specified finite chain; step 1.1 gives its unique value without choosing extensions simultaneously for all simplices. The wedge and integral are bilinear, so this defines a real-bilinear cochain operator. For a smooth $f:L\to M$, the equality $(f\bar{\sigma }{)}^{\ast }={\bar{\sigma }}^{\ast }{f}^{\ast }$ in [F8] makes the integrands for $K_L(f^{\ast }\alpha ,f^{\ast }\beta )(\sigma )$ and $K_M(\alpha ,\beta )(f\sigma )$ identical. Thus $K_L(f^{\ast }\alpha ,f^{\ast }\beta )=f^{\ast }{K}_M(\alpha ,\beta )$.

Let $N\ge 1$ and evaluate on a smooth $N$-simplex $\sigma$. Pulling ${\mathrm{\Theta }}_{\sigma }$ back by the diagonal affine simplex $a_N$ gives ${\sigma }^{\ast }(\alpha \wedge \beta )$ by [F8]. In the sum $b_N$ of [F5], the cut with dimensions $(r,N-r)$ integrates to zero by [F6] unless $(r,N-r)=(p,q)$. That remaining cut has integral $({\int }_{\sigma [0,\dots ,p]}\alpha )({\int }_{\sigma [p,\dots ,N]}\beta )$, again by [F6]. By the actual cup formula [F1], therefore, $(I(\alpha \wedge \beta )-I\alpha ⌣I\beta )(\sigma )={\int }_{a_N-b_N}{\mathrm{\Theta }}_{\sigma }\mathrm{.}$ This uses the signed shuffle with its orientation signs already calculated, not an assumed multiplicative comparison theorem.

Substitute the affine-chain identity [F5] into step 2.2. On the $i$th face model, pullback by ${\delta }_i\times {\delta }_i$ changes ${\mathrm{\Theta }}_{\sigma }$ into ${\mathrm{\Theta }}_{\sigma {\delta }_i}$; the restricted extension is admissible on a neighbourhood of the face. Thus the face sum is exactly $K(\alpha ,\beta )(\mathrm{\partial }\sigma )=\delta K(\alpha ,\beta )(\sigma )$. For the other term, apply [F7] to every affine simplex summand of $h_N$ to get ${\int }_{\mathrm{\partial }{h}_N}{\mathrm{\Theta }}_{\sigma }={\int }_{h_N}d{\mathrm{\Theta }}_{\sigma }$. All these pullbacks have smooth neighbourhood extensions by step 1.1. Equation [F8] gives $d{\mathrm{\Theta }}_{\sigma }(\alpha ,\beta )={\mathrm{\Theta }}_{\sigma }(d\alpha ,\beta )+(-1{)}^p{\mathrm{\Theta }}_{\sigma }(\alpha ,d\beta )\mathrm{.}$ By step 2.1 their integrals over $h_N$ are precisely $K(d\alpha ,\beta )(\sigma )$ and $(-1{)}^{p}K(\alpha ,d\beta )(\sigma )$. This proves the asserted cochain identity in total degrees $N\ge 1$.

If $N=0$, both forms are functions and $I(\alpha \beta )$ and $I\alpha ⌣I\beta$ agree on every vertex as the product of their values. Here $\delta K=0$, and both derivative terms evaluate $h_0=0$, so the right side is zero too. For $N=1$, $K(\alpha ,\beta )$ itself uses $h_0=0$, whereas its derivative terms use $h_1$; step 3.1 includes precisely this first endpoint case. At top form degree or above it, a zero form is treated as zero, but smooth chains in those degrees remain present and the same chain identity still applies.

Now let $d\alpha =d\beta =0$. By [F8] their wedge is closed, and by [F4] its image under $I$ and both individual images are cocycles. By [F2] the cup of the latter is a cocycle as well. The two derivative terms in step 3.1 vanish by bilinearity, leaving the claimed coboundary. To check the form quotient explicitly, if $\alpha$ changes to $\alpha +d\xi$ and $\beta$ to $\beta +d\eta$, the wedge changes by $d(\xi \wedge \beta +(-1{)}^p\alpha \wedge \eta +\xi \wedge d\eta ),$ by the signed derivative rule and the two closure equations; omit negative-degree terms when $p=0$ or $q=0$. Its image is exact by [F4], and the cup representatives descend by [F2]. Hence the equality is an equality of well-defined products of cohomology classes. Naturality is the operator identity proved in step 2.1.

Empty manifolds have zero cochains and zero form spaces; zero inputs give zero throughout. On a point the total-degree-zero product is the product of values and positive-degree forms vanish. Constant simplices, repeated vertices and all other degenerate simplices remain covered by the finite affine-chain identities. Boundary targets are handled by the actual extension in step 1.1, not by pushing extensions outside $M$. The recursion, finite integrals and unique extension-independent values require no AC. Neither the countable-choice global de Rham isomorphism nor any unsupplied continuous-chain smoothing is used.