Theorem: Fixed-domain obstruction

If a retained pair satisfies both actual orderings and ${\mathcal{C}}_{f,g}\ne 0$, no symmetric causally factorizing map ${\mathsf{T}}_2$ exists on the fixed $D_2$.

Proof. Factorization and symmetry would force

contradicting the nonzero order-$\hslash$ coefficient. Higher orders or a contact added to the proposed time-ordered value cannot satisfy two unequal prescribed equalities.