De rham homotopy formula on a product

Given: Write $\omega ={\alpha }_t+dt\wedge {\beta }_t$ with tangential families.

The interval homotopy operator is coordinate independent: The interval operator $K:{\mathrm{\Omega }}^k(M\times [0,1])\to {\mathrm{\Omega }}^{k-1}(M)$ is coordinate independent and maps smooth forms to smooth forms.

The local coordinate formula for the exterior derivative: Let $(U,x^1,\dots ,x^n)$ be a smooth chart on a smooth manifold and $\omega$ a smooth $k$-form on $U$, with $k\ge 0$. Summing over increasing $k$-tuples $I$, and writing $d{x}^I=d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_k}$, if $\omega ={\sum }_I{\omega }_{I}d{x}^I$, then $d\omega ={\sum }_{I}d{\omega }_I\wedge d{x}^I\mathrm{.}$

Newton–Leibniz needs only continuity on $[a,b]$, differentiability on $(a,b)$, and a Riemann-integrable extension of the interior derivative: Let $a<b$. Suppose $G:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f:[a,b]\to \mathbb{R}$ is Riemann integrable and $f(x)=G^{\mathrm{\prime }}(x)(a<x<b),$ then ${\int }_a^{b}f=G(b)-G(a)\mathrm{.}$ No derivative of $G$ at either endpoint is assumed, and the two endpoint values assigned to the integrable extension $f$ do not enter the conclusion.

Integration along the unit interval for a differential form: For $\omega ={\alpha }_t+dt\wedge {\beta }_t$ smooth up to the endpoints, $K\omega ={\int }_0^1{\beta }_{t}dt$ in positive degree, while $K=0$ in degree zero and on zero terms.

Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral: On a compact rectangle, a continuous parameter derivative may be passed through the integral when represented by a continuous function.