Naturality of the de Rham map

Let $\sigma$ be a smooth singular $k$-simplex in $M$. By [F2], $F\circ \sigma$ is a smooth simplex in $N$: composing $F$ with one target-valued neighbourhood extension of $\sigma$ supplies its extension. Functoriality in [F3] on that domain gives ${\sigma }^{\ast }(F^{\ast }\omega )=(F\circ \sigma {)}^{\ast }\omega$ along the simplex. Their affine coefficients, and therefore their integrals in [F4], are identical. Hence $I_M^k(F^{\ast }\omega )(\sigma )={\int }_{\sigma }{F}^{\ast }\omega ={\int }_{F\circ \sigma }\omega =I_N^k(\omega )(F_{\mathrm{\#}}\sigma )=(F^{\ast }{I}_N^k(\omega ))(\sigma )\mathrm{.}$

Every chain is a finite linear combination of simplices, so step 1.1 gives equality of the two cochains. If $d\omega =0$, both sides represent cohomology classes by [F1]–[F3]. Passing to those classes yields $I_M([F^{\ast }\omega ])=[F^{\ast }{I}_N^k(\omega )]=F^{\ast }{I}_N([\omega ])\mathrm{.}$ Changing $\omega$ by $d\eta$ changes its pullback by $d{F}^{\ast }\eta$ and its integration cochain by $\delta I_N^{k-1}(\eta )$; precomposition carries the latter to $\delta F^{\ast }{I}_N^{k-1}(\eta )$. Thus the identity is independent of representatives on both routes.

In degree zero step 1.1 is the equality $(\omega \circ F)(\sigma (v_0))=\omega (F(\sigma (v_0)))$. In degree one it equates the pullback integrals on the entire closed parameter interval, with no requirement on distinct endpoints. Constant and degenerate simplices, maps wholly into a boundary, and zero forms satisfy the same coefficient identity. Negative degrees have zero source, and an empty source manifold has zero target cochain groups; an empty target admits such a map only from an empty source. All formulas are defined on each supplied simplex and finite chain, so no choices of representatives or extensions are made simultaneously.