Type-A top support layers are controlled by reflection localization

The two natural maps are injective. A flagged module is right-free, and after passing to the fraction field its graph summands separate. A section supported on two disjoint sets of graphs is zero, since it is zero in this localization. This proves ${\mathrm{\Gamma }}_{y}M\cap {\mathrm{\Gamma }}_{<y}M=0$ and ${\mathrm{\Gamma }}_{\ge y}M\cap {\mathrm{\Gamma }}_{\mathrm{\ne }y}M={\mathrm{\Gamma }}_{>y}M$, giving the two injections. Support cuts and quotients commute with idempotents: write a word sum as $M\oplus M^{\mathrm{\prime }}$ and take each cut componentwise. Its canonical flag layers therefore split into those of $M,M^{\mathrm{\prime }}$. A graded projective summand $P$ of a finite free graded polynomial module is graded free: lift a homogeneous basis of $P\mathrm{/}P{R}_{+}$ to obtain a free graded module mapping onto $P$, using its lower degree bound to prove surjectivity by successive degree reduction. Projectivity splits this surjection; its finite graded kernel has zero reduction and hence is zero by the same lower-degree argument. Thus the two canonical layers in [F3] remain free for all summands.

Evaluation at $1$ identifies $\mathrm{Hom}(R_y,B)$ with ${\mathrm{\Gamma }}_{y}B$ for a word $B$: in its fraction-field graph decomposition every section supported on $y$ satisfies the $y$ graph relations, and right torsion-freeness gives them before localization. By the special Hom formula this module is free. The quotient ${\mathrm{\Gamma }}^{y}B$ is free as well: twist the right action by $y^{-1}$, identifying this quotient with the identity-graph quotient of $B{R}_{y^{-1}}$. This twisted word has a $\mathrm{\Delta }$ flag, since it is obtained by successive left generator tensors from the single flagged graph $R_{y^{-1}}$ using [F4]. Its identity quotient is its length-zero layer and is free. Evaluation, cuts, and these identifications commute with summand idempotents, so the graded-projective argument in step 1.1 gives freeness of all four layers for $M$. The sequence $0\to {\mathrm{\Gamma }}_{\mathrm{\ne }y}M\to M\to {\mathrm{\Gamma }}^{y}M\to 0$ consequently splits on the right. The same argument gives right freeness of $M\mathrm{/}{\mathrm{\Gamma }}_{y}M$: the twisted word $B{R}_{y^{-1}}$ also has a $\mathrm{\nabla }$ flag by the same left-generator closure from $R_{y^{-1}}$. Its identity support is its length-zero submodule, and the quotient has a flag of the remaining right-free layers. Thus its identity submodule, and hence ${\mathrm{\Gamma }}_{y}B$, splits on the right; applying the summand idempotent retains a right-free quotient and a right splitting. In particular the inclusion ${\mathrm{\Gamma }}_{y}M\subset M$ is right-saturated: if $mr\in {\mathrm{\Gamma }}_{y}M$ with $r\mathrm{\ne }0$, then the graph relations kill $mr$, and right torsion-freeness implies that they kill $m$, so $m\in {\mathrm{\Gamma }}_{y}M$. This saturation is the consequence needed for reduction in a minimal generator degree.

Fix a reflection $t$ and localize the right polynomial ring at the prime hyperplane equation $c_t$; all homogeneous polynomials not divisible by $c_t$ become invertible. A word becomes a direct sum of shifted graph modules $R_a$ with $a<at$, and two-graph union modules $R_{\{a,at\}}$, with $a<at$. We prove this by induction on the word. It holds for the unit, since $e<t$. Tensoring a graph with $B_s$ has the two graph layers $a,sa$ from rank-one calculus; if $sa\mathrm{\ne }at$, their extension splits in this localization, and both singleton labels still have positive $t$ direction. Indeed $a<at$ means $a({\beta }_t)$ is a positive coordinate root; a simple permutation $s$ preserves positivity of every positive root except its own, and the exception is exactly $sa=at$. In that exceptional case the tensor is the two-graph union module by the rank-one presentation. For a two-graph union module the same rank-one exact sequence gives the old pair $\{a,at\}$ and the translated pair $\{sa,sat\}$. When these pairs differ, every cross extension splits by [F4], since its hyperplane is invertible; hence the tensor splits into these two pair modules. When the pairs coincide, $sa=at$, the rank-one invariant decomposition $R=R^s\oplus {\alpha }_s{R}^s$ gives two shifted copies of the pair module directly. This proves the claimed decomposition for words. For a summand, retain the description as a summand of that direct sum; no localized classification theorem is required.

In a two-graph union ring the explicit presentation is the pullback of its two graph rings with equal values on their common hyperplane. A section supported on the high label $y$ has value divisible by $c_t$ on that graph, while the lower-layer quotient is the unrestricted value there. If $yt<y$, a singleton $y$ cannot occur in step 2.2, so in the localized word the image of ${\mathrm{\Gamma }}_{y}B\to {\underset{‾}{\mathrm{\Gamma }}}_{\le y}B$ is exactly $c_t$ times the lower layer. Dually the image of the upper layer into ${\mathrm{\Gamma }}^{y}B$ is $c_t$ times that quotient. These image descriptions commute with idempotents and hence hold for $M$. In a homogeneous basis of either free target, the image coordinates are therefore divisible by $c_t$ after localization, and hence in $R$ itself: $R$ is a polynomial UFD and no factor coprime to $c_t$ can remove this divisibility. Doing this for every distinct reflection with $yt<y$ shows that both images are contained in $p_y$ times their targets.

The special Hom formula computes $\mathrm{rk}{\mathrm{\Gamma }}_{y}M$ first for words by evaluation in step 2.1; for a summand its Hom is the corresponding graded summand, and its $\mathrm{\nabla }$ layer is the matching idempotent component of the free layer. The same localization image containment and the word's equality descend to that component, so it suffices to compare the word first. In the positive-degree rank convention of this page, the word's special Hom calculation gives $\mathrm{rk}{\mathrm{\Gamma }}_{y}B={\sum }_{\nu }(B:{\mathrm{\nabla }}_y(\nu ))v^{\mathrm{\ell }(y)-\nu }$. The lower layer has rank ${\sum }_{\nu }(B:{\mathrm{\nabla }}_y(\nu ))v^{-\mathrm{\ell }(y)-\nu }$; multiplying by $p_y$ raises degrees by $2\mathrm{\ell }(y)$ and gives exactly the former polynomial. Since the injection of step 3.1 is a comparison of free graded modules with equal graded rank polynomials, their Hilbert series agree; in each degree it is an injection of finite-dimensional spaces of equal dimension, hence surjective. Thus ${\mathrm{\Gamma }}_{y}B={\underset{‾}{\mathrm{\Gamma }}}_{\le y}B{p}_y$, and applying its commuting idempotent gives the first equality for $M$. The dual special Hom formula applied to graph targets gives the identical dimension comparison for the second image, proving ${\underset{‾}{\mathrm{\Gamma }}}_{\ge y}M={\mathrm{\Gamma }}^{y}M{p}_y$.

Inverting $p_y$ makes the two image equalities isomorphisms. The first equality and the lower-layer rank just computed give $\mathrm{rk}{\mathrm{\Gamma }}_{y}M={\sum }_{\nu }(M:{\mathrm{\nabla }}_y(\nu ))v^{\mathrm{\ell }(y)-\nu }$; the second and the upper-layer rank ${\sum }_{\nu }(M:{\mathrm{\Delta }}_y(\nu ))v^{\mathrm{\ell }(y)-\nu }$ give $\mathrm{rk}{\mathrm{\Gamma }}^{y}M={\sum }_{\nu }(M:{\mathrm{\Delta }}_y(\nu ))v^{-\mathrm{\ell }(y)-\nu }$. This proves the original two image assertions and both rank formulas without assuming those assertions as facts.