The ring form of de Rham’s theorem needs the singular cup product

A linear bijection does not by itself preserve multiplication: $T:\mathbb{R}\to \mathbb{R}$, $T(x)=2x$, is linear with inverse $x\mapsto x\mathrm{/}2$, but $T(1\cdot 1)=2$ while $T(1)T(1)=4$. It also fails to preserve the unit. This calculation identifies the logical information missing from bare vector-space bijectivity, without claiming that the actual integration map has this defect.

For the actual comparison, [F2] supplies the missing product equation by a specific coboundary. Its front/back cut has no extra cochain sign; the signed shuffle integral and simplex Stokes produce that equation. In [F1] restriction preserves this same cup formula, so injectivity of restriction transports the equation to continuous singular cohomology. The vertex integral separately supplies the unit. Thus the product and unit information used in the ring assertion is explicit.

The hypotheses of [F1] include neither compactness nor connectedness. Its empty-manifold case is the zero unital algebra; the point and degree-zero unit are covered by vertex evaluation. [F2] includes degree-zero and degree-one endpoints, boundary targets and degenerate simplices, and needs no choice. The countable-choice assumption of [F1] is confined to the global comparison isomorphisms; the distinction in step 1.1 does not supply or remove that assumption.