The de Rham map is an isomorphism on convex coordinate domains

Set $H(x,t)=c+t(x-c)$. Convexity makes this target-valued for $0\le t\le 1$; its polynomial coordinate expression gives all required local extensions. Its endpoint maps are $ip$ and the identity. For a closed form $\omega$ of positive degree, the constant-map pullback is zero because its derivative is zero. Thus [F2] gives $\omega =d(L_H\omega )$. For a closed zero-form $f$, the same identity has $L_{H}f=0$ and $df=0$, so $f-f(c)=0$. Conversely constant functions are closed. Hence de Rham cohomology is zero in positive degrees and is $\mathbb{R}$ in degree zero, with evaluation at $c$ inverse to the constant-function map.

On the point $\{c\}$ there is exactly one simplex $s_j$ in every degree $j\ge 0$. For $j\ge 1$, its boundary is $({\sum }_{a=0}^j(-1{)}^a)s_{j-1}$, equal to $s_{j-1}$ when $j$ is even and zero when $j$ is odd. Therefore the cochain group is $\mathbb{R}$ in each nonnegative degree, with ${\delta }^k=0$ for even $k$ and ${\delta }^k=\mathrm{id}$ for odd $k$. This unnormalized complex has $H^0=\mathbb{R}$ and $H^k=0$ for $k>0$: in positive odd degree the kernel is zero, and in positive even degree the entire kernel is the preceding image.

The unmodified radial homotopy need not extend into a boundary target beyond the time endpoints. Use the time-flattened prism guaranteed by [F3] instead. It has the same endpoint maps and gives a degree-one chain operator $P$ with ${\mathrm{id}}_{\mathrm{\#}}-(ip{)}_{\mathrm{\#}}=\mathrm{\partial }P+P\mathrm{\partial }\mathrm{.}$ For a cochain $\phi$ of degree $k\ge 1$ set $D\phi =\phi P_{k-1}$, and set $D=0$ in degree zero. Direct evaluation gives $\delta D\phi +D\delta \phi =\phi -(ip{)}^{\ast }\phi \mathrm{.}$ In degree zero the first term is zero and the second is $\phi \mathrm{\partial }{P}_0$, so this identity still holds. No dual exactness theorem or selected cochain extension is used.

Since $pi={\mathrm{id}}_{\{c\}}$, [F6] gives $i^{\ast }{p}^{\ast }=\mathrm{id}$ on cochains of the point. Step 2.1 gives $p^{\ast }{i}^{\ast }=\mathrm{id}$ on cohomology of $V$, because the difference on any cocycle is the coboundary $\delta D\phi$. Thus $i^{\ast },p^{\ast }$ are inverse cohomology maps. More explicitly in positive degree, for a cocycle $\phi$ let $\beta =i^{\ast }\phi$. In odd degree $\beta =0$ by step 1.2, so take $\gamma =0$; in positive even degree take the preceding-degree point cochain with the same scalar value as $\beta$, so $\delta \gamma =\beta$. Then $\phi =\delta (D\phi +p^{\ast }\gamma )\mathrm{.}$ This proves positive-degree vanishing without a representative-selection principle. In degree zero, evaluation at $c$ and constant cochains give the inverse identifications with $\mathbb{R}$.

By [F5], integration sends the constant function $a$ to the cochain with value $a$ on every point simplex. Thus the degree-zero map is the identity under the two identifications with $\mathbb{R}$ in steps 1.1 and 3.1. In every positive degree both groups vanish, so their unique linear map is an isomorphism; negative degrees are zero by the complex conventions. A coordinate diffeomorphism and its inverse give inverse pullback maps in both theories, and [F1] transports these conclusions to the coordinate domain.

Nonemptiness is used only to fix one contraction centre; the empty domain instead has zero groups on both sides and still a comparison isomorphism, but not the asserted degree-zero identification with $\mathbb{R}$. For $n=0$ the nonempty domain is a point, already calculated in step 1.2, and all positive-degree forms vanish. Degree one is covered by the zero odd-degree kernel at the point and the explicit primitive formulas. Constant and degenerate simplices are retained throughout. The endpoint flattening in step 2.1 preserves $c$ and $x$, and [F2] handles both endpoints for forms. A single centre and explicit operators suffice; no choices over families of domains or cohomology classes are made.