Proposition: F0 fails

The core package is incomplete. Not reproduced in the record: the generators and coefficient class of ${\mathcal{A}}^m(V)$ and the module action (D0.1–D0.4); the topology of ${\mathcal{K}}_{\mathrm{int}}$ (D0.3); $E_{\mathrm{res}}$ (D0.5); a set equation for $O=U_{\mathrm{can}}$, refused at three places (D0.8); the display of $m^{\circ }$ (D0.10); the identity of the blocks in ${\mathcal{C}}_{\mathrm{cross}}$ (D0.11); the normalization of $T^{\mu \nu }(\varphi ,\psi )$ and the remaining weights (D0.12); $\mathsf{A}$ beyond its role, ${\sigma }_{\epsilon }$, $f$, ${\dot{a}}_2$, $\mathsf{D}$ (D0.13); the ${q^{L,I,J}}_{\alpha \beta }$ beyond their order bound, $\mathsf{P}$, $\mathfrak{b}$, $g$, ${E^{\mathrm{A}}}_m$ (D0.14); the display of the candidate class (D0.16); and the parametrization and presentation-level form of the correction class (D0.17). Hence no preliminary lift $F_0:{\mathcal{P}}_{\mathrm{int}}\to {\mathcal{D}}^{\prime }(W;E_{\mathrm{res}})$ with $r_O{F}_0=m^{\circ }\mathsf{Rec}$ can be stated, and no space $\mathcal{A}\subseteq \mathrm{Hom}({\mathcal{P}}_{\mathrm{int}},Y)$ is available against which a null defect could be tested.

Proof. A linear map requires a domain, a codomain, and a value on each element of the domain. The domain is known through its chart and its reconstruction rule but not its generators or coefficient class (D0.1, T/P); the codomain is undefined (D0.5). The value on a presentation of any of the six entries is its composition with ${\mathcal{C}}_{\mathrm{cross}}$, the vertex Hessian, and weights according to $m^{\circ }$ (D0.10, P); the internal kernels exist for every completion of the undefined data (Theorem thm:exist) but $\mathsf{A}$, $f$, ${\sigma }_{\epsilon }$, $\mathsf{D}$ and the ingredients of $K_8$, $K_9$ are not fixed, so no particular distribution is singled out. Proposition prop:classstatus shows that ${\mathcal{C}}_{\mathrm{part}}$ does not yield $\mathcal{A}$. Each cited item is in status P, R, U, C, or T${}^{\ast }$\ in the manifest.