Bulk and boundary factorization in rational CFT with topological defects

Fix the two cuts and all exterior ribbon coupons. Write $e_A={\mathrm{\Delta }}_A{\eta }_A$, $S=U{\otimes }^{+}A{\otimes }^{-}V$ and $W=U^{\vee }{\otimes }^{+}X{\otimes }^{-}{V}^{\vee }$. Let R denote the right average of [F6]. For $\varphi :U\otimes V\to X$ and $\psi :W\to 1$ put $F(\varphi )={\rho }_X^l({\mathrm{id}}_A\otimes \varphi )(c_{A,U}^{-1}\otimes {\mathrm{id}}_V)$, $H(\psi )=({\mathrm{id}}_A\otimes \psi )({\mathrm{id}}_A\otimes {\rho }_W^l)(e_A\otimes {\mathrm{id}}_W)$, $J_1=R_{S,X}F$, $J_2=R_{W,A}H$, $K_1(\alpha )=\alpha ({\mathrm{id}}_U\otimes {\eta }_A\otimes {\mathrm{id}}_V)$, $K_2(\beta )={ϵ}_A\beta$. Centrality of $e_A$ and module associativity make F,H left-linear. The two averages commute, so J1,J2 are bimodule morphisms. The unit gives F K1=id; the inverse Frobenius-pairing identity gives H K2=id on bimodule morphisms. R fixes these morphisms, hence J1 K1=id and J2 K2=id. Thus K1 J1 and K2 J2 are actual idempotents on the ordinary coupon spaces and their images identify with the two field spaces. This argument and [F6]'s trace-duals apply to zero spaces without total-dimension divisions.

Fix the actual FRS simple cap coupons lambda and bar-lambda, with $\lambda \bar{\lambda }=1$. Retain their order, ribbon core orientations and turns. The source's turned bulk cup is $({\theta }_U\otimes {\mathrm{id}}_{U^{\vee }})c_{U^{\vee },U}{\bar{\lambda }}_{\bar{U},U}$; its opposite cap is the composition dual obtained by reversing that word. This is a morphism, not a scalar set equal to 1. To calibrate it against [F3]'s pivotal dualities, the Drinfeld comparison gives $({\mathrm{id}}_{U^{\vee }}\otimes u_U^{-1})b_{U^{\vee }}=c_{U^{\vee },U}^{-1}{b}_U$. Since $j_U=u_U{\theta }_U$, $b_U^R=({\mathrm{id}}_{U^{\vee }}\otimes {\theta }_U^{-1})c_{U^{\vee },U}^{-1}{b}_U$ and therefore $({\theta }_U\otimes {\mathrm{id}}_{U^{\vee }})c_{U^{\vee },U}{b}_U^R=b_U$. Scaling the source cup scales its opposite cap inversely. This calculation is valid before any self-dual-label identification and supplies the actual turn used in FFS (2.45). It does not impose two independent gauges on a single self-dual Hom space.

The following trace contraction will remove every total-dimension normalization. Put $Q={\mathrm{id}}_U\otimes ({\eta }_A{ϵ}_A)\otimes {\mathrm{id}}_V$. Transport the induced left module $S=U\otimes A\otimes V$ to the free left module A tensor U tensor V using $\chi =c_{A,U}^{-1}\otimes {\mathrm{id}}_V$, with inverse $c_{A,U}\otimes {\mathrm{id}}_V$. The inverse braiding has source U tensor A and target A tensor U. The inverse hexagon and naturality with m give $\chi {\rho }_S^l=(m_A\otimes {\mathrm{id}}_{UV})({\mathrm{id}}_A\otimes \chi )$; naturality with eta epsilon gives $\chi Q{\chi }^{-1}=({\eta }_A{ϵ}_A)\otimes {\mathrm{id}}_{UV}$. The left average of eta epsilon on the regular A module is $m({\mathrm{id}}_A\otimes \eta ϵ)({\mathrm{id}}_A\otimes m)(e_A\otimes {\mathrm{id}}_A)=({\mathrm{id}}_A\otimes ϵm)(e_A\otimes {\mathrm{id}}_A)={\mathrm{id}}_A$. Consequently $L_{S,S}(Q)={\mathrm{id}}_S$. Trace adjointness of the left average, proved in [F6], gives $\mathrm{tr}(fQ)=\mathrm{tr}(f)$ for every left-linear $f:S\to S$. In particular the closed coupon graph ${ϵ}_A\beta ({\mathrm{id}}_{U^{\vee }}\otimes \alpha \otimes {\mathrm{id}}_{V^{\vee }})({\tilde{b}}_U\otimes {\eta }_A\otimes b_V)$ equals ${\mathrm{tr}}_X(\alpha T(\beta ))$: cutting its two simple loops reads it as ${\mathrm{tr}}_S(QT(\beta )\alpha )$, and then apply the identity just proved and cyclicity. This computes the unit/counit coupon graph as a trace, not as $d_X$ times an undefined normalized trace.

Use torus coordinates $(s,t)\in (\mathbb{R}\mathrm{/}\mathbb{Z}{)}^2$ with the two marked X arcs centered at (0,3/4) and (0,1/4), tangent to the s direction and signed as in the Statement. Cut along $t=0$ and $t=1\mathrm{/}2$, keeping the p and q seam order. The two normalized RT cut maps in [F3] identify the state space with the finite direct sum of ordinary coupon tensor spaces $\mathrm{Hom}(U_p\otimes U_q,X)\otimes \mathrm{Hom}(U_{\bar{p}}\otimes X\otimes U_{\bar{q}},1)$. The pants are the two resulting cylinders with one marked arc, and rigid bending reads precisely those two Hom spaces. Define $K_{pq,ab}$ by putting $K_1({\alpha }_a)$ in the first coupon ball and $K_2({\beta }_b)$ in the second and reversing the cuts with their actual FRS cap coupons, including the turn in step 1.2. The s circle bounds a disk in this solid torus, so this is the cap cobordism occurring in the factorized collar, rather than a new abstract basis vector. Indeed the factorized connecting collar is two product half-collars; its two inserted simple cap balls are exactly these inverse-cut balls. Their remaining t-direction ribbons and field coupons have the stated order, with no permutation of the exterior marked X arcs.

For the boundary cut put $N=X{\otimes }_B{M}_{\mathrm{\ell }}$, $M=M_r$, and $Y=M^{\vee }\otimes N$. In [F4], take the left algebra to be $A_0=\mathbf{1}$, with multiplication $m_0={\lambda }_{\mathbf{1}}={\rho }_{\mathbf{1}}:\mathbf{1}\otimes \mathbf{1}\to \mathbf{1}$, unit and counit ${\eta }_0={ϵ}_0={\mathrm{id}}_{\mathbf{1}}$, and comultiplication ${\mathrm{\Delta }}_0=m_0^{-1}$. The unitors and the canonical duality of $\mathbf{1}$ give the Frobenius and symmetry identities; moreover $m_0{\mathrm{\Delta }}_0={\mathrm{id}}_{\mathbf{1}}$ and ${ϵ}_0{\eta }_0={\mathrm{id}}_{\mathbf{1}}=d_{\mathbf{1}}{\mathrm{id}}_{\mathbf{1}}$, with $d_{\mathbf{1}}={\mathrm{tr}}_j({\mathrm{id}}_{\mathbf{1}})=1$. Thus $A_0$ is symmetric special Frobenius under [L1]. By [L1], $M^{\vee }$ has right action ${\rho }_{M^{\vee }}^r:M^{\vee }\otimes A\to M^{\vee }$ characterized by ${\mathrm{ev}}_M({\mathrm{id}}_{M^{\vee }}\otimes {\rho }_M)={\mathrm{ev}}_M({\rho }_{M^{\vee }}^r\otimes {\mathrm{id}}_M)$ as maps $M^{\vee }\otimes A\otimes M\to \mathbf{1}$. Give it the canonical left $A_0$-action ${\lambda }_{M^{\vee }}:\mathbf{1}\otimes M^{\vee }\to M^{\vee }$. The unit coherence axioms make these actions commute, so $M^{\vee }$ is an $A_0$-$A$-bimodule. Since $N$ is a left $A$-module, [F4] applies with right algebra $B=A$ and gives the projector $P=P_{M^{\vee },N}$ on $Y$. With ${\psi }_{+}:N\otimes U_i\to M$ and ${\psi }_{-}:M\otimes U_{\bar{i}}\to N$, use the ACTUAL source cup ${\bar{\lambda }}_{i,\bar{i}}$ in $e_{\lambda }({\psi }_{+})=({\mathrm{ev}}_M\otimes {\mathrm{id}}_{U_i^{\vee }})({\mathrm{id}}_{M^{\vee }}\otimes {\psi }_{+}\otimes {\mathrm{id}}_{U_i^{\vee }})({\mathrm{id}}_Y\otimes {\bar{\lambda }}_{i,\bar{i}}),f({\psi }_{-})=({\mathrm{id}}_{M^{\vee }}\otimes {\psi }_{-})({\tilde{b}}_M\otimes {\mathrm{id}}_{U_i^{\vee }})\mathrm{.}$ The source cap dual to $\bar{\lambda }$ is composition dual, not a snake evaluation. If $\bar{\lambda }=a_i{b}_{(U_i)}$ in the fixed ordered duality coordinate, $e_{\lambda }=a_i{e}_{\mathrm{b}\mathrm{a}\mathrm{r}\mathrm{e}}$; its inverse is $a_i^{-1}$ times the ordinary snake inverse. Equivalently the inverse using the actual source $\lambda$ includes the factor $d_i$, since its bent cup/cap is $\mathrm{id}\mathrm{/}{d}_i$. The second map $f$ has the usual module-evaluation inverse. Moving the transposed $M$ action across its evaluation and using centrality of ${\mathrm{\Delta }}_A{\eta }_A$ gives $e_{\lambda }P=e_{\lambda }$ and $Pf=f$; reversing this readback gives the module intertwining equations. Thus these are isomorphisms onto the fixed Hom spaces, not just maps into them.

Take the actual FRS cylinder P on this torus: two product X ribbons and the two A bands along the s cycle, attached by the combined left/right actions and the coproduct/counit pieces of FFS (2.48). Push the two bands through the product collars into the two coupon balls without moving a marked arc. The word read on the first ball is $E_1(\varphi )={\rho }_X^r({\rho }_X^l\otimes {\mathrm{id}}_A)({\mathrm{id}}_A\otimes \varphi \otimes {\mathrm{id}}_A)(c_{A,U}^{-1}\otimes {\mathrm{id}}_V\otimes {\mathrm{id}}_A)({\mathrm{id}}_U\otimes c_{A,V}^{-1}\otimes {\mathrm{id}}_A)({\mathrm{id}}_{U\otimes V}\otimes e_A)\mathrm{.}$ Here append $e_A=(a_1,a_2)$, move a1 left past V and then U, and act left by a1 and right by a2 on the coupon output. This is exactly K1 R F: expanding ${\rho }_S^r$ moves a1 past V and then $m({\eta }_A,a_1)=a_1$; F moves it past U. Both inverse crossings are forced by this readback. The opposite X orientation in the other ball reads $E_2(\psi )={ϵ}_A{m}_A(H(\psi )\otimes {\mathrm{id}}_A)({\rho }_W^r\otimes {\mathrm{id}}_A)({\mathrm{id}}_W\otimes e_A),H(\psi )=({\mathrm{id}}_A\otimes \psi )({\mathrm{id}}_A\otimes {\rho }_W^l)(e_A\otimes {\mathrm{id}}_W)\mathrm{.}$ It is K2 R H. Its two copairings reduce by the inverse Frobenius pairing; the opposite orientation uses the pivotal cyclic invariance of $e_A$ proved from symmetry in [F6], not an untwisted interchange of its two outputs. These are the two actual cut cylinder words. Thus under the exact cutting isomorphism Z(P) is the direct sum of E1 tensor E2, and its image has basis the $K_{pq,ab}$ from step 2.2. Gluing P to the original collar solid torus absorbs each additional A band into its annular A ribbon by $\mathrm{\Delta }m=(m\otimes \mathrm{id})(\mathrm{id}\otimes \mathrm{\Delta })$, module associativity and $m\mathrm{\Delta }=\mathrm{id}$. The opposite band uses the same pivotal cyclic identity. The original state is consequently fixed by P.

Use a local boundary strip $x\in [-1,1]$, $y\in [-3,3]$, $z=0$, with boundary module ribbons near $x=\pm 1$ in the solid cylinder $x^2+z^2\le 2$. Refine its trivalent network to have the cut edge $y=0$. Adding this edge is implemented by expanding a special A bubble and moving one leg to the other boundary segment: Frobenius moves the coproduct through the multiplication, module associativity moves its action, and $m\mathrm{\Delta }=\mathrm{id}$ removes the bubble. Hence the state is unchanged. Reading the newly inserted edge with the M ribbon bent to M-dual gives exactly $P=({\rho }_{M^{\vee }}^r\otimes {\rho }_N^l)({\mathrm{id}}_{M^{\vee }}\otimes {\mathrm{\Delta }}_A{\eta }_A\otimes {\mathrm{id}}_N)\mathrm{.}$ Before fusion the same local refinement in the B region gives $P_{X,M_l}$. It is therefore that actual projector which is replaced by its split inclusion and restriction, with $ir=P_{X,M_l}$ and $ri={\mathrm{id}}_N$; the descended A action is [F4]'s. No arbitrary projector is inserted into a previously unrelated network. The replacement transports both field coupons and the disk graph.

Choose ordinary trace-duals ${\alpha }^a:X\to S$ to ${\alpha }_a:S\to X$ and ${\beta }^b:A\to W$ to ${\beta }_b:W\to A$, so both traces are delta; [F6] proves their existence without $d_X$ division. Reflect the cap K and replace its field coupons by these duals, with the opposite simple cap coupons composition dual to those retained in step 2.2. Call its unscaled functional $L_{\mathrm{r}\mathrm{a}\mathrm{w}}$. Gluing it to K gives S2 times S1 with the product coordinate t. The four coupon-free layers between the two field coupon balls and the two simple cap balls contain two opposed simple strands: two layers have p/r and two q/s. The RT map of a two-point-sphere cylinder vanishes for different simple labels; for equal label i the insertion on the two strands is ${\mathrm{coev}}_i{\mathrm{ev}}_i\mathrm{/}{d}_i$. This follows from the one-dimensional Hom space and the equality ${\mathrm{ev}}_i{\mathrm{coev}}_i=d_i$. Apply it at the four layers. The total factor is ${\delta }_{pr}{\delta }_{qs}\mathrm{/}(d_p{d}_q{)}^2$. The two remaining closed coupon components are the unit/counit trace contractions of step 2.1, one on X and one on A, and are ${\delta }_{ac}$ and ${\delta }_{bd}$. All lambda and opposite-lambda pairs are composition dual, including the turned pair of step 1.2, so no extra cap scalar is dropped. The empty S2 times S1 has value 1, the trace of the sphere identity. Hence $L_{pq,ab}^{\mathrm{r}\mathrm{a}\mathrm{w}}(K_{rs,cd})=\frac{{\delta }_{pr}{\delta }_{qs}{\delta }_{ac}{\delta }_{bd}}{(d_p{d}_q{)}^2}\mathrm{.}$ This is the actual geometric Gram calculation, with only nonzero simple dimensions in denominators. Thus $L=(d_p{d}_q{)}^2{L}_{\mathrm{r}\mathrm{a}\mathrm{w}}$ is the true dual cap, including for $d_X=0$.

The disk connecting manifold is a ball. In its minimal three-edge A network move the lower A ribbon upward along the M module action, merge it by module associativity, and remove the resulting bubble by specialness. Move the remaining A ribbon through ${\psi }_{+}$ by its intertwining equation and remove it by the unit. The remaining annular M/N ribbon with its two coupons closes to exactly $\mathrm{tr}(e_{\lambda }f)$ with the same actual bar-lambda as in step 2.3. The outgoing ball coefficient is 1; the dual two-point sphere test has its actual cap correction D, cancelling the closed S3 factor $D^{-1}$. Thus the geometric disk matrix is $c_{\mathrm{b}\mathrm{n}\mathrm{d}}=(\mathrm{tr}(e_{\lambda ,a}{f}_b){)}_{ab}$. It is nondegenerate by [F5], since the maps in step 2.3 identify the field spaces with the fixed Hom spaces. The complete resolution from [F5] is consequently $P={\sum }_{i,a,b}{d}_i(c^{\mathrm{b}\mathrm{n}\mathrm{d}}{}^{-1}{)}_{ba}{f}_b{e}_{\lambda ,a}\mathrm{.}$ The local sewn boundary graph is precisely $f_b{e}_{\lambda ,a}$: the double cutting tube is normalized by [F3], and its inserted source cap is the same bar-lambda already included in $e_{\lambda }$. This proves the defect-free boundary factorization coefficient with its inverse ordering. The local ir/ri fusion of step 3.2 gives the identical disk matrix and sewn graph for the boundary-defect fields, proving the stated boundary-defect formula. Rescaling the actual source cup scales both $e_{\lambda }$ and its disk matrix; the inverse matrix and the sewn graph change reciprocally. No bare-cap normalization or nonzero total module dimension is assumed.

Keep the same actual lambda coupons in the sphere two-point graph and in $L_{\mathrm{r}\mathrm{a}\mathrm{w}}$. The graph gives the following finite inverse-matrix computation. For the cap labels (p,q) of step 2.2 the first sphere field is $\beta :U_{\bar{p}}\otimes X\otimes U_{\bar{q}}\to A$. Apply the following bending with $U=U_{\bar{p}},V=U_{\bar{q}}$; its dual-label output is $X\to U_p\otimes A\otimes U_q$, using the fixed pivotal dual comparisons. Thus the resulting sphere matrix is $c_{X,\bar{p},\bar{q}}$ and $d_U{d}_V=d_p{d}_q$. Bend a field $\varphi :U\otimes X\otimes V\to A$ to $X\to U^{\vee }\otimes A\otimes V^{\vee }$ with the two simple cup vertices appearing in FFS (A.13), and call this map $f_{\lambda }$; $g_{\lambda }$ is the same oriented word with its two simple cap vertices. In unrescaled pivotal coordinates the first word is $({\mathrm{id}}_{U^{\vee }}\otimes \varphi \otimes {\mathrm{id}}_{V^{\vee }})({\tilde{b}}_U\otimes {\mathrm{id}}_X\otimes b_V)$ times the two actual nonzero cup scalars. The inverse word is $({\widetilde{ev}}_U\otimes {\mathrm{id}}_A\otimes {\mathrm{ev}}_V)({\mathrm{id}}_U\otimes \gamma \otimes {\mathrm{id}}_V)$ times the corresponding cap scalars. For each simple strand the bent cup/cap composite is a scalar times ${\mathrm{id}}_U$; its trace is $\lambda \bar{\lambda }=1$, so that scalar is $1\mathrm{/}{d}_U$. Thus $f_{\lambda }{g}_{\lambda }=g_{\lambda }{f}_{\lambda }=(d_p{d}_q{)}^{-1}\mathrm{id}$, including the actual source turns. Expand them in the ordinary trace-dual bases: $f_{\lambda }({\varphi }_a)={\sum }_c{\mathrm{\Delta }}_{ac}{\varphi }^c$ and $g_{\lambda }({\varphi }^b)={\sum }_d{\mathrm{\Omega }}_{bd}{\varphi }_d$. Then $\mathrm{\Delta }\mathrm{\Omega }=\mathrm{\Omega }\mathrm{\Delta }=(d_p{d}_q{)}^{-1}I$. The RT sphere connecting manifold has two outgoing sphere components and no surgery components; its exact coefficient is $D^{-1+2}=D=S_{00}^{-1}$. Closing it with the source two-point block tests therefore gives $c=S_{00}^{-1}$ times its actual two-coupon sphere graph. Define $H_{\lambda }^{\mathrm{d}\mathrm{u}\mathrm{a}\mathrm{l}}$ to be the closed opposite graph in $L_{\mathrm{r}\mathrm{a}\mathrm{w}}$ after replacing the collar annulus by the meridian-exchanged filling. Composing $H_{\lambda }^{\mathrm{d}\mathrm{u}\mathrm{a}\mathrm{l}}$ with the sphere graph, bending one coupon by $f_{\lambda }$ and the other by $g_{\lambda }$, contracts the two ordinary trace-dual pairs to deltas, by step 2.1 and [F6]. The remaining matrix is $\mathrm{\Omega }\mathrm{\Delta }$. Consequently $\left(S_{00}{d}_p{d}_q{H}_{\lambda }^{\mathrm{d}\mathrm{u}\mathrm{a}\mathrm{l}}\right)c=I=c\left(S_{00}{d}_p{d}_q{H}_{\lambda }^{\mathrm{d}\mathrm{u}\mathrm{a}\mathrm{l}}\right)\mathrm{.}$ This derives both nondegeneracy of the geometric sphere matrix and its inverse. It is the source A.18–19 contraction written with ordinary trace-duals; no $d_A{d}_X$ normalization or extension by cancelling a zero denominator is used. The label reflection and cap order put c at (bar p,bar q), exactly as in the Statement.

Convert the source inward-normal convention uniformly to the pinned outward-normal convention: reverse each boundary surface and its directed end arcs and dualize its effective labels, while leaving the oriented three-manifold and physical interior ribbon graph unchanged. This is exactly the opposite-object operation in [F3]. Thus the source upper $(X,+)$ and lower $(X,-)$ become pinned upper $X^{\vee }$ and lower $X$, and every incoming end is changed by the same operation. Both normal and end-arc direction reverse, so the oriented ribbon plane $(n_{\mathrm{i}\mathrm{n}},\gamma )$ is the same plane as $(-n_{\mathrm{i}\mathrm{n}},-\gamma )$; its preferred white/black faces do not interchange. Use source torus coordinates $(s,t)$ of period $2\pi$, source orientation $ds\wedge dt$, horizontal $s$ along source arrow 1 and vertical $t$ along upward arrow 2. Put the upper mark at $(0,3\pi \mathrm{/}2)$, lower at $(0,\pi \mathrm{/}2)$, with source arc tangents $+{\mathrm{\partial }}_s,-{\mathrm{\partial }}_s$. The middle p seam is $t=\pi$ and the outer q seam is $t=0$. The source normal conversion gives pinned outgoing orientation $-ds\wedge dt$ and arc tangents $-{\mathrm{\partial }}_s,+{\mathrm{\partial }}_s$. These are the signed marks used on the common exterior and on both fillings, not a newly transported set of marks. Let $r_0=2^{-1\mathrm{/}2}$ and parametrize their actual common boundary once by $q(s,t)=(r_0{e}^{is},r_0{e}^{it})\subset S^3\subset {\mathbb{C}}^2\mathrm{.}$ In the standard oriented $S^3$ let $T_1=\{\mathrm{\mid }{z}_1\mathrm{\mid }\ge \mathrm{\mid }{z}_2\mathrm{\mid }\}$ and $T_2=\{\mathrm{\mid }{z}_2\mathrm{\mid }\ge \mathrm{\mid }{z}_1\mathrm{\mid }\}$. The original filling is $T_X=+T_1$, the factorized cap is $K=-T_2$, and the opposite cap is $L=+T_2$. The explicit wedge maps are ${\mathrm{\Phi }}_X(r,s,t)=(\sqrt{1-r^2}{e}^{is},r{e}^{it}),{\mathrm{\Phi }}_K(\rho ,s,t)=(\rho e^{is},\sqrt{1-{\rho }^2}{e}^{it}),0\le r,\rho \le r_0\mathrm{.}$ The first collapses the t circle at $r=0$, the second the s circle at $\rho =0$, precisely the vertical and horizontal contractible cycles of the original source wedges. In the source inward-normal convention their orientations are $-dr\wedge ds\wedge dt$ and $-d\rho \wedge ds\wedge dt$. Direct differentiation shows these maps preserve orientation into $+T_1$ and $-T_2$. Indeed with $z_2=x+iy$, positive $S^3$ has volume $ds\wedge dx\wedge dy=-rdr\wedge ds\wedge dt$, whereas in the $\rho$ chart it has volume $+\rho d\rho \wedge ds\wedge dt$. Consequently the outgoing boundaries of K and TX are both $-ds\wedge dt$, L has boundary $+ds\wedge dt$ as an incoming cap, and their boundary parametrizations are the identical q, pointwise. The closure TX union L is therefore literally the positively oriented $T_1\cup T_2=S^3$; no S versus inverse-S choice or central minus-identity is inferred from slopes.

The ribbon ends also match pointwise, including their frames. At the upper and lower marks write $g={\mathrm{\partial }}_s$, $h={\mathrm{\partial }}_t$, using the common radial coordinate $r=\mathrm{\mid }{z}_2\mathrm{\mid }$. K and L have inward geometric normal $+{\mathrm{\partial }}_r$, TX has inward normal $-{\mathrm{\partial }}_r$. Source upper/lower signs are $(+,-)$ for K and TX and $(-,+)$ for L; source arc tangents are $(g,-g)$ for K and TX and $(-g,g)$ for L. Thus their pinned boundary arcs are $(-g,g)$ for K and TX and $(g,-g)$ for the incoming L boundary. The physical core directions are $(+{\mathrm{\partial }}_r,-{\mathrm{\partial }}_r)$ for K and $(-{\mathrm{\partial }}_r,+{\mathrm{\partial }}_r)$ for both L and TX. Preferred white normals are $(h,-h)$ for all three, with black normals $(-h,h)$. These face signs follow from the oriented ribbon plane and the three-volume, not just from the core arrows: at TX upper the plane is $(-{\mathrm{\partial }}_r,{\mathrm{\partial }}_s)$ in volume $-dr\wedge ds\wedge dt$, giving white normal $+{\mathrm{\partial }}_t$; at TX lower it is $(-{\mathrm{\partial }}_r,-{\mathrm{\partial }}_s)$, giving white normal $-{\mathrm{\partial }}_t$. Reversing both cap volume and end arc gives the same white normals for L. In particular TX and L glue with identical core vectors and white-face vectors at the same q points. An explicit complete X strip is $F(s,y)=(\sqrt{1-y^2}{e}^{is},iy),\mathrm{\mid }s\mathrm{\mid }<\delta ,-r_0\le y\le r_0,$ with core $+{\mathrm{\partial }}_y$ from upper $y=-r_0$ to lower $y=+r_0$, ribbon orientation $-ds\wedge dy$, and preferred white normal $+{\mathrm{\partial }}_x$. At its two ends these are precisely the two TX/L entries above. Hence the closed X strip has one continuous preferred-face field. The incoming object reversal is a change of boundary encoding, not a rotation of the physical strip.

Embed the actual lower alpha and upper beta coupon balls of K and L by the same wedge coordinates, keeping their source preferred sides and their actual lambda cap words. The p ribbons occupy the middle t interval between the two balls; the q ribbons occupy the complementary interval through $t=0$ and the collapsed s disk. Thus the maps in step 5.2 preserve the actual source routing and the marked collars, not only the meridian homology classes. The annular A ribbon of TX runs along the noncontractible s cycle. Cutting it at the coupon balls reads the already computed E1/E2 separability-action words of step 3.1 with the handedness fixed by step 6.1. Move its actions through the dual field coupons by the bimodule equations. The adjacent Frobenius contraction is exactly $L_S(Q)={\mathrm{id}}_S$ from step 2.1; it removes this annulus without a total-dimension factor. Cut the remaining graph at the X trace and the two actual simple cap balls. Its lower coupon is ${\alpha }^a:X\to U_p\otimes A\otimes U_q$, its upper coupon is ${\beta }^b:A\to U_{\bar{p}}\otimes X\otimes U_{\bar{q}}$, and the intervening cap words bend the latter into $R_{\lambda }({\beta }^b):U_p\otimes A\otimes U_q\to X$. Thus its closed word is exactly ${\mathrm{tr}}_X(R_{\lambda }({\beta }^b){\alpha }^a)$ with the actual source cap scalars: the graph $H_{\lambda }^{\mathrm{d}\mathrm{u}\mathrm{a}\mathrm{l}}$ of step 5.1. To check framing rather than assuming it from a planar picture, divide this closed graph into its two field rectangles, actual cap balls, and intervening product strips. The rectangles and cap balls are the very same framed pieces used in the inverse-pairing computation. Their endpoint strip parametrizations are not reset. On the remaining strips in the hemisphere disk $D=\{z_1\text{ real and nonnegative}\}$, the normalized s-direction is the constant nonzero vector $(i,0)$, extending through $z_1=0$. The wedge product framing has zero angular monodromy even through the cap tip. The X trace strip has the identical endpoint trivializations explicitly checked in step 6.1. The p/q intervals in D can be separated without crossings. Hence all framed pieces and their product identifications agree with the inverse word: an extra full twist would have to occur on one of these product strips, where the constant trivialization excludes it. This identifies the actual closed ribbon operator, retaining the quarter-turns already present in lambda and the pivotal coupons, and not just its ambient S3.

The pinned invariant of this positive closed S3 is $S_{00}$ times its ribbon operator. Thus $L_{\mathrm{r}\mathrm{a}\mathrm{w}}$ applied to the original filling is $S_{00}{H}_{\lambda }^{\mathrm{d}\mathrm{u}\mathrm{a}\mathrm{l}}$, and the genuine dual cap from step 4.1 extracts $S_{00}(d_p{d}_q{)}^2{H}_{\lambda }^{\mathrm{d}\mathrm{u}\mathrm{a}\mathrm{l}}=d_p{d}_q(c_{X,\bar{p},\bar{q}}^{\mathrm{d}\mathrm{e}\mathrm{f}}{)}^{-1}\mathrm{.}$ Applying the common exterior cobordism to the finite cap expansion proves the stated bulk formula, with the exact source inverse ordering. No individual decorated term is identified with the original one merely because its weight agrees.

Perform the local torus comparisons in the reference Lagrangian spanned by the s meridian of K. For the K/L Gram closure both filling kernels equal that meridian. For the L/TX closure the kernels are the s and t meridians, while the intermediate reference agrees with the L kernel; the Maslov triple has a repeated pair. Thus these closures have zero Maslov correction, with closed weights zero, and the closed values used above are exactly $Z(S^2\times S^1)=1$ and $Z(S^3)=D^{-1}$. Sphere cuts and boundary balls have zero first homology and no reference correction. For arbitrary object Lagrangians, use the actual reference-change weighted cylinders of [F3]; its composition/reference identity transports the same finite graph equalities. Give each global gluing map the Statement's unique weight $n_G+n_{\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{z}\mathrm{e}\mathrm{d}}+{\mu }_W=n_{\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}}$. This aligns the weights of the finite expansion with the common exterior and supplies neither anomaly cancellation nor equality of individual coupon terms. Full AC and countable choice are used only through [F3]; all bases and sums here are finite. The derived geometric pairings apply also to zero spaces: $X=0$ gives empty field bases and zero local network, and $d_X=0$ for nonzero X causes no denominator anywhere. These are abstract oriented FRS correlators, not analytic VOA sewing or a path-integral measure.