Let \(X,C,Y\) be sets and let \(c:X\to C\) and \(r:X\to Y\) be functions. There is a function \(F:c(X)\to Y\) with \(r=F\circ c\), viewing \(c\) as a map to its image, if and only if \[c(x)=c(x')\ \Longrightarrow \ r(x)=r(x')\] for all \(x,x'\in X\). When it exists, \(F\) is unique on \(c(X)\). This is the set-level form of the fiber-constancy argument for finite-map pullback; no finiteness is required here.

Mathlib's Function.FactorsThrough is the displayed fiber-constancy condition. Its Function.factorsThrough_iff extends to the entire codomain \(C\) when \(Y\) is nonempty; the image formulation above needs no such extra assumption.

Apply this to any set of representatives of the rational enriched objects, first taking \(c(\mathcal B)\) to be the rational quadratic-isometry class of \(q\) and \(r(\mathcal B)=R(\mathcal B)\). Both are unchanged by the declared arrows. A pair of isometric forms with unequal trace ratios excludes every set-level recovery rule from that isometry class on a domain containing the pair. It consequently excludes recovery from any invariant determined solely by that class, including the Clifford algebra's isomorphism class and the quadratic center character \(\chi _q\). In particular, a fixed function of \(L(2,\chi _D)\), or of any fixed collection of values determined solely by the same character, cannot recover \(R\) there. The two rationally isometric fibers provide such a pair.

The criterion asserts neither regularity nor continuity of \(F\). Even fiber constancy would not supply those properties, a choice of a physical observable, or a physical parameter law. The obstruction concerns invariants of \(q\) alone and this independently defined trace ratio. It makes no assertion about arithmetic data that also retain the reference pairing and mass.