Finite-type based tensor bars can lack finite global corrections

Actual based factors and the raw seed. Take two copies of the module in [F1]. These are actual finite-Cartan-type integrable modules satisfying all four based axioms, with the specified stable free integral lattices and crystal charts. The raw seed is $e=b_{0,0}\otimes b_{0,2}$, of factor weights $2,-2$ and total weight zero. In standard coordinates its two factors are $u_{0,0}+u_{1,1}\mathrm{/}(q+q^{-1})$ and $u_{0,2}-u_{1,3}\mathrm{/}(q+q^{-1})$, so both factors carry genuine deformations. The constituent lengths grow without bound, and the module is outside the globally bounded highest/lowest classes. Every vector still lies in finitely many finite standard constituents.

The full tensor bar is finite, integral and triangular. On each input vector, only finitely many standard constituent pairs occur. Their finite strings bound all nonzero divided $E^{(r)}$ and $F^{(r)}$ actions, so [F3] evaluates as a finite sum. Each scalar $a_r$ is Laurent integral and each divided action preserves the actual integral factor lattices by [F1]; hence $\mathrm{\Psi }$ preserves their algebraic tensor lattice. Unitarity and the standard compatible factor bars make ${\mathrm{\Psi }}^2=1$. Its raw matrix is $(C^{-1}\otimes C^{-1})\mathrm{\Theta }(C\otimes C)$ followed by coefficient bar. The degree-zero term is exactly the identity, and every other term strictly raises the first weight because the shears preserve weights. All columns are finite. Thus this example retains the full integral-action and field-bar assertions.

Exact projection to a finite cut. In raw coordinates let ${\pi }_J$ kill tuples with a factor index exceeding $J$ and identify the remaining raw indices with those of $M^{(J)}\otimes M^{(J)}$. Each shear raises a constituent index; standard divided actions preserve constituent indices. Therefore every path in the raw bar expansion raises indices or preserves them, and deleting indices above $J$ gives exactly the bar expansion with the last outgoing shear omitted. Equivalently standard constituent projection commutes with $\mathrm{\Theta }$, and its compositions with $C$ and $C^{-1}$ are their finite-cut versions. Consequently ${\pi }_J\mathrm{\Psi }={\mathrm{\Psi }}^{(J)}{\pi }_J$. This is an identity of the complete finite-column raw matrices; it does not assume that intersecting a cyclic field submodule gives a balanced restriction.

No finite normalized fixed lift. Suppose $g$ were a finite-support fixed vector with $g-e$ having coefficients in $q\mathbb{Z}[q]$. Let $K$ be the largest factor index in its support and take $J\ge K+2$. Its projection is the same finite coefficient vector, is fixed by step 2.1 and has the required seed residue. It equals $G^{(J)}$: indeed the difference is a finite fixed vector whose every coefficient is in $q\mathbb{Z}[q]$; at its least first weight, triangularity forces each coefficient to equal its own bar, impossible for a nonzero strictly positive polynomial. Thus the difference vanishes. But the projected candidate has zero coefficient at $b_{J,0}\otimes b_{J,n_J}$, whereas [F2] proves that coefficient nonzero in $G^{(J)}$. This contradiction disproves existence of $g$, even without imposing a strict-transfer condition on its proposed corrections.

Failure of the integral balanced map. Put $\mathcal{M}=M_A{\otimes }_A{M}_A$, $L=L_0{\otimes }_{A_0}{L}_0$ and $L^{-}=\mathrm{\Psi }(L)$, the actual infinity chart. Their raw coefficient rings give $\mathcal{M}\cap L=⨁\mathbb{Z}[q](b\otimes b^{\mathrm{\prime }})$. There is no nonzero $\mathrm{\Psi }$-anti-invariant vector in this intersection: at its least first weight an anti-invariant finite vector would have a polynomial coefficient $c$ with $\bar{c}=-c$, forcing $c=0$, a contradiction. Every vector $v$ in $\mathcal{M}\cap L\cap L^{-}$ is therefore fixed: both $v$ and $\mathrm{\Psi }v$ belong to $\mathcal{M}\cap L$, and $\mathrm{\Psi }v-v$ is anti-invariant. If reduction from that triple intersection to $(\mathcal{M}\cap L)\mathrm{/}(\mathcal{M}\cap qL)$ were onto the raw residue class of $e$, its preimage would be such a fixed finite vector congruent to $e$ with positive-polynomial corrections, contradicting step 3.1. Hence the actual integral tensor reduction map fails surjectivity.

Conclusion and boundary. Two genuine four-axiom based integrable $A_1$ modules thus have a finite-column integral involutive triangular quasi-$R$ tensor bar but no algebraic normalized global correction for the specified raw seed. In particular no tensor global basis with the promised finite transitions can exist, and the integral tensor triple fails the balanced reduction property explicitly exhibited in step 4.1. Finite-dimensional products, bounded highest/lowest constructions and separately compatible finite-cut direct limits are not refuted; the counterexample concerns the unqualified arbitrary based finite-type tensor assertion. Every index, module and map is explicit and all arguments use finite arithmetic or induction, with no choice assumption.