De Rham integration is a cochain map

The definition of the cochain differential, followed by integration and [F2], gives $(\delta I_M^k(\omega ))(c)=I_M^k(\omega )(\mathrm{\partial }c)={\int }_{\mathrm{\partial }c}\omega ={\int }_{c}d\omega =I_M^{k+1}(d\omega )(c)\mathrm{.}$ The same coefficients and alternating boundary signs occur in [F2] and [F3], so no sign adjustment is required.

Since step 1.1 holds for every chain, the two linear functionals are equal. Together with the degreewise real linearity in [F1], this is precisely the cochain-map identity. For $k=0$ it is the fundamental-theorem formula for values at the two endpoints of each smooth one-simplex, including constant paths. For $k=\dim M$, the form $d\omega$ is zero, and the same equality proves $\delta I_M^k(\omega )=0$ without assuming there are no higher-dimensional singular simplices.

If $k>\dim M$ or $k<0$, the form and all relevant source differentials are zero, so the identity is an equality of zero cochains; this includes the map out of degree minus one. If $M$ is empty, both complexes are zero. Zero chains and degenerate simplices were included in [F2], so they create no exception to step 1.1. Every evaluation uses a finite chain and a previously well-defined integral; no choice axiom enters.