Zero th de rham cohomology is locally constant functions

Given: A smooth real function $f$ on $M$; $B^0=0$.

De rham cohomology: The real de Rham cohomology is $H_{\mathrm{d}\mathrm{R}}^k(M)=Z^k(M)\mathrm{/}{B}^k(M)$, with $Z^k,B^k$ as in def-closed-and-exact-differential-forms. This is def-cohomology-object-of-a-cochain-complex in real vector spaces. Only a closed form $\omega$ represents a class $[\omega ]$. For closed forms $\omega ,\omega \mathrm{\prime }$, equality $[\omega ]=[\omega \mathrm{\prime }]$ means precisely $\omega \mathrm{\prime }-\omega =d\eta$ for some $(k-1)$-form $\eta$. Addition and real scalar multiplication are induced by those of forms. All groups on the empty manifold are zero.

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.