The vector-space de Rham comparison is automatically a ring isomorphism

A degreewise linear isomorphism need not preserve multiplication even when it preserves the unit. Let $A=\mathbb{R}[u]\mathrm{/}(u^3)$ with $\mathrm{\mid }u\mathrm{\mid }=2$, and define the degree-preserving linear bijection $T:A\to A$ by $T(1)=1,T(u)=2u,T(u^2)=2u^2\mathrm{.}$ It is invertible on the basis $(1,u,u^2)$, but $T(u^2)=2u^2$ whereas $T(u)T(u)=4u^2$. Hence linear bijectivity alone cannot imply multiplicativity.

In the de Rham comparison the source product is wedge and the target product is the cup product of [F1]. The missing assertion is therefore the equality $I_{\ast }([\alpha ]\wedge [\beta ])=I_{\ast }[\alpha ]⌣I_{\ast }[\beta ]$. This does not follow from degreewise bijectivity; [F2] proves it by constructing a specific cochain homotopy between $I(\alpha \wedge \beta )$ and $I\alpha ⌣I\beta$. Only after adjoining that result may a bijective de Rham comparison be called a ring isomorphism.

The zero ring and empty-space cases satisfy multiplicativity vacuously, and in degree zero on a point both products are ordinary scalar multiplication; neither special case establishes the general product law. The unit alone is insufficient, as step 1.1 already preserves it. No endpoint or representative-choice issue occurs, and [F2]'s compatibility is choice-free. The false claim is the word "automatically," not the ring theorem obtained after proving [F2].