Frobenius Algebras and Oriented Two-Dimensional Sewing

Finite smooth cutoffs and partitions on the compact corner charts are obtained from [F7]: restrict ambient Euclidean bumps to each half-space or quadrant chart, using a strict support margin to extend by zero. Choose a finite family whose smaller chart neighborhoods cover the compact set, and divide its bumps by their positive sum there. Thus the interpolation below uses actual finite smooth partitions without applying a boundaryless manifold theorem to corners. A corner in a compact surface with faces is isolated, so compactness gives finitely many corners. Choose pairwise disjoint quadrant charts about them; their two coordinate directions give product collars of the incident faces that agree on the quadrant. Delete smaller quadrant neighborhoods and round the new boundary junctions inside those neighborhoods. The remaining ambient surface is a smooth manifold with boundary, so [F4] gives collars of its remaining original boundary arcs. Restrict those collars to the black and free arcs. On each overlap with a corner chart, interpolate their inward transverse vector fields with the coordinate normal field by a smooth partition of unity, taking the coordinate field on a smaller corner neighborhood. Positive inward normal components remain positive under this interpolation. Near a corner use the two coordinate fields for the two incident faces; they commute there and are tangent to the other face. Their smooth ambient local flows from [F8], restricted to inward times and a common sufficiently small width on each compact face, give face collars equal to the fixed product collars in the smaller quadrant. Away from corners the two faces do not meet, so no further compatibility is required. This finite construction gives the compatible face collars in the definition. A black face glues to an oppositely oriented black face of the same type, and the interval endpoint colors match by the stated rule; therefore the quotient retains an oriented surface with corners and the unglued free faces retain their labels.

Arbitrary compatible black collars can also be compared. Each collar is the flow collar of its normal-coordinate vector field near the face. These fields are inward transverse to that face and tangent to its incident free faces; interpolate the two fields linearly on a common neighborhood. Transversality stays positive, and tangency is preserved. Their flows give a smooth family of collars $c_t$ of a common smaller width, fixed on the face and preserving the free faces. On this neighborhood the velocity $W_t=({\mathrm{\partial }}_t{c}_t)\circ c_t^{-1}$ vanishes on the black face and is tangent to the free faces. Multiply it by a cutoff equal to one on a still smaller collar and zero outside the common neighborhood. The ambient time-dependent flow (obtained from [F8] on the product with the time coordinate) is a diffeomorphism carrying $c_0$ to $c_t$ on that smaller collar, fixing all black parameterizations and preserving the free faces. At the corners use ambient extensions of the product charts; tangency makes their restricted flows preserve each free face. Thus the resulting collar comparison descends across a glued seam. Shortening or increasing reparameterization of a collar is handled by the same construction. A diffeomorphism of either input transports its collar to a compatible collar on the other representative, so this comparison also proves independence of representatives.

For three composable representatives, both parenthesized composites identify the same three collared surfaces along the same two black boundary parameterizations. In product coordinates, both quotients are obtained by replacing the two seam collars by the same concatenated interval; a smooth increasing reparameterization of that interval gives a diffeomorphism fixed on the exterior black boundary. Hence composition is associative on equivalence classes. Gluing an identity cylinder merely lengthens a collar and is equivalent to the original surface by step 2.1.

Ordered disjoint union is associative with the empty surface as strict unit at the level of words and diffeomorphism classes, and interchange makes it a bifunctor. The cylinder that permutes components gives the symmetry. Its double application is the identity permutation. For any morphisms $f:X\to X^{\mathrm{\prime }}$ and $g:Y\to Y^{\mathrm{\prime }}$, the two composites in the symmetry-naturality square are the same disjoint union of $f$ and $g$ with the same block permutation of boundary components, so they represent the same relative surface, with the component identification equal to the identity in every corresponding black collar. Each of the two hexagon composites for three words is a cylinder for the same permutation of the three blocks, so those composites also agree; a permutation with an empty block is the identity. The unit and associator are identities in this strict category, so the monoidal pentagon and triangle also hold. Thus the component-permutation cylinders satisfy the braiding and symmetry axioms directly.

The condition ${\mathrm{\partial }}_{1}M=\mathrm{\varnothing }$ is preserved by gluing and disjoint union, so the circle-only objects and no-free-boundary morphisms form the stated closed symmetric monoidal subcategory. If $S\mathrm{\ne }\mathrm{\varnothing }$, choose $a\in S$: a genus-zero circle-to-circle surface with an additional free circular face colored $a$ lies in the full circle-object subcategory but not in the closed category. If $S=\mathrm{\varnothing }$, a coloring of a nonempty free boundary would require an element of the empty set, so every morphism has no free boundary and the full subcategory equals the closed category.

For the strong symmetric monoidal functors of [F5] and [F9], identity natural transformations satisfy the binary and unit compatibility equations of [F6]; if two transformations satisfy them, substitution in those equations shows their componentwise vertical composite also satisfies them. Their identity and associativity laws hold componentwise in the target category [F1]. Thus the morphisms between the functors in the definition form the stated category.

Choose a basis $(u_1,\dots ,u_n)$ of $A$. The map $\mathrm{\Theta }:A\to A^{\ast }$ given by $\mathrm{\Theta }(v)(a)=ϵ(av)$ is an isomorphism by [F1]. Define $v_i:={\mathrm{\Theta }}^{-1}(u_i^{\ast })$. Then $ϵ(u_j{v}_i)={\delta }_{ji}$, and [F2] gives $a={\sum }_iϵ(a{v}_i)u_i,a={\sum }_i{v}_iϵ(u_{i}a)\mathrm{.}$ For the second identity, apply $z\mapsto ϵ(u_{j}z)$ to its proposed right side: ${\sum }_iϵ(u_j{v}_i)ϵ(u_{i}a)=ϵ(u_{j}a)$. These functionals separate elements by nondegeneracy in the second variable, so the right side equals $a$. Set $e:={\sum }_i{v}_i\otimes u_i$.

Conversely, suppose $(A,\mu ,\eta ,\mathrm{\Delta },ϵ)$ satisfies [F3], and put $d:=\mathrm{\Delta }\circ \eta :k\to A\otimes A$. With the canonical associators and unitors, the two snake composites are $S_L:={\rho }_A\circ (1_A\otimes \beta )\circ {\alpha }_{A,A,A}\circ (d\otimes 1_A)\circ {\lambda }_A^{-1}:A\to A$ and $S_R:={\lambda }_A\circ (\beta \otimes 1_A)\circ {\alpha }_{A,A,A}^{-1}\circ (1_A\otimes d)\circ {\rho }_A^{-1}:A\to A\mathrm{.}$ For $S_L$, replace $\beta$ by $ϵ\mu$ and use the Frobenius identity $(1_A\otimes \mu )\alpha (\mathrm{\Delta }\otimes 1_A)=\mathrm{\Delta }\mu$; naturality of $\alpha$ moves $\eta$ to the first input. The left unit equation reduces the composite to ${\rho }_A(1_A\otimes ϵ)\mathrm{\Delta }$, which is $1_A$ by the right counit. For $S_R$, use $(\mu \otimes 1_A){\alpha }^{-1}(1_A\otimes \mathrm{\Delta })=\mathrm{\Delta }\mu$; the right unit equation reduces it to ${\lambda }_A(ϵ\otimes 1_A)\mathrm{\Delta }$, which is $1_A$ by the left counit. Thus both snake composites are identities.

The first displayed dual-basis identity gives ${\sum }_iϵ(a{v}_i)u_i=a$, and the second gives ${\sum }_i{v}_iϵ(u_{i}a)=a$, so this $e$ has the required two contraction properties. If another tensor has the first property, contracting its first factor against every functional $x\mapsto ϵ(ax)$ gives the same vector for every fixed $a$; these functionals range over all of $A^{\ast }$ by nondegeneracy. Hence the tensors are equal. The contraction map $A\otimes A\to {\mathrm{Hom}}_k(A^{\ast },A)$ is an isomorphism in finite dimension, so the tensor is intrinsically defined and independent of the basis.

To prove $(a\otimes 1)e=e(1\otimes a)$, test the two tensors against arbitrary $\varphi \otimes \psi \in A^{\ast }\otimes A^{\ast }$. Nondegeneracy supplies $c,d\in A$ with $\varphi (z)=ϵ(cz)$ and $\psi (z)=ϵ(zd)$. The first contraction is ${\sum }_iϵ(ca{v}_i)ϵ(u_{i}d)=ϵ(cad)$ by the second dual-basis identity applied to $d$. The second is ${\sum }_iϵ(c{v}_i)ϵ(u_{i}ad)=ϵ(cad)$ by the same identity applied to $ad$. Since such product functionals separate tensors, the two tensors agree.

Define $\mathrm{\Delta }(a)=(a\otimes 1)e$. Then $(ϵ\otimes 1)\mathrm{\Delta }(a)={\sum }_iϵ(a{v}_i)u_i=a$. Also $(1\otimes ϵ)\mathrm{\Delta }(a)=a{\sum }_i{v}_iϵ(u_i)=a$ by the second identity in step 1.1 at $a=1$. The central relation of step 2.2 gives $\mathrm{\Delta }(ab)=(a\otimes 1)\mathrm{\Delta }(b)=\mathrm{\Delta }(a)(1\otimes b)\mathrm{.}$ Expanding these two equalities with the associator is exactly the pair of Frobenius identities in [F3].

The defining contraction also yields, for all $x,y,a\in A$, $(\beta \otimes \beta )(x\otimes y,\mathrm{\Delta }(a))={\sum }_iϵ(xa{v}_i)ϵ(y{u}_i)=ϵ(yxa)=\beta (yx,a)\mathrm{.}$ This establishes the contraction identity.

If ${\mathrm{\Delta }}^{\mathrm{\prime }}$ is another coproduct with counit $ϵ$ and the two Frobenius identities, write $e^{\mathrm{\prime }}={\mathrm{\Delta }}^{\mathrm{\prime }}(1)={\sum }_j{x}_j^{\mathrm{\prime }}\otimes y_j^{\mathrm{\prime }}$. Applying the counit identities to ${\mathrm{\Delta }}^{\mathrm{\prime }}(a)$ and using each Frobenius identity gives ${\sum }_jϵ(a{x}_j^{\mathrm{\prime }})y_j^{\mathrm{\prime }}=a,{\sum }_j{x}_j^{\mathrm{\prime }}ϵ(y_j^{\mathrm{\prime }}a)=a\mathrm{.}$ Step 2.1 therefore gives $e^{\mathrm{\prime }}=e$. The Frobenius identity then gives ${\mathrm{\Delta }}^{\mathrm{\prime }}(a)=(a\otimes 1)e^{\mathrm{\prime }}=(a\otimes 1)e=\mathrm{\Delta }(a)$, so the coproduct is unique.

Use the tensor-product pairing ${\beta }^{\otimes 3}$ to test coassociativity. For any $x,y,z,a\in A$, step 4.1 gives $⟨x\otimes y\otimes z,(\mathrm{\Delta }\otimes 1)\mathrm{\Delta }(a)⟩=\beta (zyx,a)=⟨x\otimes y\otimes z,(1\otimes \mathrm{\Delta })\mathrm{\Delta }(a)⟩\mathrm{.}$ The pairing ${\beta }^{\otimes 3}$ is nondegenerate: in any basis its matrix is the Kronecker product of three invertible matrices for $\beta$. Thus the two tensors are equal, proving coassociativity.

The two snake identities exhibit inverse maps between $A$ and $A^{\ast }$ for the left and right pairings, respectively; hence $\beta$ is nondegenerate in both variables. It follows that $ϵ$ is a Frobenius functional by [F1], and step 4.2 gives uniqueness of $\mathrm{\Delta }$. Symmetry of $\beta$ and commutativity of $\mu$ remain the separate properties defined in [F3].

If $M=\mathrm{\varnothing }$, use the empty function and empty generator word; all assertions are vacuous. For nonempty $M$, the finite bump and partition constructions used below also work in its corner charts: extend a chart to an ambient Euclidean open set, use [F8] on nested balls there, restrict to the half-space or quadrant, and extend by zero outside the chart. A strict support margin makes that extension smooth, including on boundary faces. Smaller restricted balls cover each compact set in question; choose a finite subcover. The corresponding nonnegative bumps $b_j$ have positive sum near that compact set, so $b_j\mathrm{/}{\sum }_i{b}_i$ form a smooth finite subordinate partition there. Each selected bump is one on a smaller chart neighborhood. For a cutoff equal to one near an entire compact set and supported in a prescribed open neighborhood use $1-{\prod }_j(1-b_j)$; its support lies in the finite union of the strict chart supports. This proves the corner-chart partition and cutoff assertions directly from [F8], rather than applying a boundaryless partition theorem to corners. There are finitely many corners: the depth-two stratum of a compact surface with faces is a closed discrete zero-manifold, so its singleton neighborhoods have a finite subcover. Choose disjoint quadrant charts at these corners. The two coordinate directions in each quadrant give product collars of the two incident faces. Delete smaller quadrant neighborhoods, round their new junctions inside those neighborhoods, and apply [F2] to the remaining ambient smooth surface with boundary. Restrict its collars to the remaining original face arcs. Interpolate the associated inward transverse fields with the coordinate normal fields on the finite overlaps, keeping the latter fields fixed in smaller quadrant charts. Positivity of the inward component is preserved by the partition-of-unity interpolation. The two fields commute and are tangent to the opposite face in those smaller charts, so their local ambient flows from [F10], restricted to inward times and a sufficiently small common collar width, give exactly the product corner collars; elsewhere the faces are disjoint. Thus the black and free faces have compatible collars. On a black collar prescribe $f=t$ at an incoming face and $f=1-t$ at an outgoing face.

Each free boundary component is either a circle or an interval between corners. On a circle use $g(s)=\frac{1}{2}+\delta \cos (2\pi s)$ with $0<\delta <1\mathrm{/}4$. For an interval, its endpoint values are $0$ or $1$. If they differ, use the monotone profile $g(s)=s$ or $1-s$. If both are $0$, take a smooth nonnegative bump ${\rho }_0$ symmetric about $1\mathrm{/}2$, supported in $(1\mathrm{/}4,3\mathrm{/}4)$ and positive at $1\mathrm{/}2$; such a bump is obtained from [F8] using a ball centered at $1\mathrm{/}2$. Normalize it by its positive finite integral to obtain $\rho$ with ${\int }_0^1\rho =1$. Define $\theta (s)=1-2{\int }_0^s\rho (u)du$ and $g(s)={\int }_0^s\theta (u)du$. Then $\theta$ is nonincreasing, equals $1$ and $-1$ near the respective endpoints, satisfies $\theta (1-s)=-\theta (s)$, and has its unique zero at $1\mathrm{/}2$ with ${\theta }^{\mathrm{\prime }}(1\mathrm{/}2)=-2\rho (1\mathrm{/}2)<0$. Thus $g^{\mathrm{\prime }}(s)=\theta (s)$ has one transverse zero, while $g(0)=g(1)=0$, $g(s)>0$ for $0<s<1$, and $g(s)\le 1\mathrm{/}2$; its endpoint derivative germs are $+1$ and $-1$. If both endpoints are $1$, use one minus that profile. The endpoint germs match the black-corner collar coordinates. A smooth inward extension $g(s)+th(s,t)$ is chosen with $h=0$ near the endpoints, so it agrees with the black-corner collars, and $h>0$ at each free-face critical point, so the full differential is nonzero there. Every critical value lies strictly between $0$ and $1$. Do not scale a profile after its endpoint germs have been fixed. Instead choose pairwise disjoint free-face coordinate neighborhoods of its finitely many critical points, with closures away from corners. By [F8] choose a cutoff ${\beta }_j$ supported in each neighborhood and identically one on a smaller neighborhood of the critical point. Replace $g$ there by $g+c_j{\beta }_j$. On the transition annuli $\mathrm{\mid }{g}^{\mathrm{\prime }}\mathrm{\mid }$ has a positive minimum; choose $\mathrm{\mid }{c}_j\mathrm{\mid }\mathrm{\parallel }{\beta }_j^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$ smaller than that minimum. Hence no new free-face critical point occurs, while the derivative and Hessian near the old point are unchanged and its value shifts by exactly $c_j$. Choose the finitely many constants successively, avoiding the finitely many previously selected critical values. The supports are away from corners, so every endpoint germ remains fixed. Extend each perturbation constantly in the inward variable on a smaller collar and cut it off farther in using [F8]; its inward derivative on the face is zero, so the prescribed nonzero normal derivative at the critical point is unchanged. The compact perturbation supports have positive distance in value from $0,1$, so sufficiently small constants retain the range and the prescribed extreme levels.

Use the finite corner-chart cutoff construction of step 1.1 to choose $\chi =1$ near the compact boundary with support in the neighborhood carrying the prescribed smooth collar function $f_{\mathrm{c}\mathrm{o}\mathrm{l}}$. Shrink that neighborhood so $f_{\mathrm{c}\mathrm{o}\mathrm{l}}$ lies in $[0,1]$, with equality only on the prescribed black faces. Set $f_0=\chi f_{\mathrm{c}\mathrm{o}\mathrm{l}}+(1-\chi )\mathrm{/}2$, interpreting the first term as zero off that neighborhood. The strict support margin makes it smooth, it agrees with the prescribed function near every face, and its range and extreme levels are correct. On the resulting boundary neighborhood the full differential is nonzero: on black faces the collar derivative is nonzero; on free faces either the tangential derivative or the chosen normal derivative is nonzero; at corners the tangential germ is the nonzero linear endpoint germ. Compactness therefore leaves a compact set $K$ in the interior outside which no perturbation is needed. Cover $K$ by finitely many coordinate charts $U_i$ and smaller open sets $V_i$ whose union contains $K$. Choose the $V_i$ with compact closures in $U_i$ and apply [F9] on the boundaryless interior to obtain bumps ${\chi }_i$ supported in $U_i$ and equal to one near ${\bar{V}}_i$. The differentials of the finitely many functions ${\chi }_i{x}_i^j$ span every cotangent fiber on a neighborhood of $K$.

If $K=\mathrm{\varnothing }$, the constructed differential is already nonzero everywhere in the interior, so there are no interior critical points and no parameter choice or density argument is needed. Retain $f_0$ in this case. If $K\mathrm{\ne }\mathrm{\varnothing }$, at least one two-dimensional chart is present and the parameter space below has positive dimension. For parameters $a_{ij}$ in an open ball about $0$, set $f_a=f_0+{\sum }_{i,j}{a}_{ij}{\chi }_i{x}_i^j$. Put $W={\bigcup }_i{V}_i$ and $Q={\bigcup }_{i,j}\mathrm{supp}({\chi }_i{x}_i^j)$, extending each chart-supported function by zero outside its chart. Then $W$ is an open neighborhood of $K$, $Q$ is compact in the interior, and $W\subseteq Q$: on each $V_i$, the differentials $d({\chi }_i{x}_i^j)=d{x}_i^j$ span the cotangent fiber, so at least one of those functions has nonzero differential at each point and that point lies in its support. The parameter derivatives therefore span every cotangent fiber on $W$. On $Q\setminus W$, $d{f}_0$ is nonzero because $K\subseteq W$ and $d{f}_0$ has no zeros outside $K$. For each point of this compact set, continuity of $(p,a)\mapsto d(f_a{)}_p$ gives a product neighborhood on which the differential remains nonzero; a finite subcover yields a common parameter neighborhood of $0$ preserving nonvanishing throughout $Q\setminus W$. Outside $Q$ the perturbation vanishes, so the differential remains $d{f}_0\mathrm{\ne }0$. Restrict the parameter ball to that neighborhood and small enough that the range stays in $[0,1]$. On $W$ the evaluation map $(p,a)\mapsto d(f_a{)}_p$ is transverse to the zero section because its parameter derivatives span each cotangent fiber. Apply [F3] to this family on the positive-dimensional manifold $W$: its bad parameter set is null. The parameter space is positive-dimensional, so [F4] makes the good parameters dense; in particular the restricted open ball contains one. For this choice, all critical points lie in $W$ and are nondegenerate. By [F6] the interior restriction is Morse. Its critical set is closed and contained in the compact set $Q$, so it is compact; because each nondegenerate critical point is isolated, it is finite.

If there are no interior critical points, skip these interior perturbations. Otherwise choose pairwise disjoint small interior chart neighborhoods of the finitely many critical points and use [F9] to obtain bumps supported there and equal to one on smaller closed neighborhoods. Choose a smooth Riemannian metric on $M$ by covering the compact surface with finitely many corner charts and patching their Euclidean metrics with the finite smooth partition construction of step 1.1. For its induced cotangent norm, $\mathrm{\mid }df\mathrm{\mid }$ is continuous and positive on the compact complement of those smaller neighborhoods, so it has a positive minimum. Choose the bump supports and then the constants small enough that the norm of the perturbation differential is smaller than this minimum on the transition annuli; no new critical point is created there. Near each critical point the bump is constant, so its differential and Hessian are unchanged while its critical value shifts by that constant. Choose the constants successively to avoid the finite set of free-face and already fixed interior values. The boundary collars stay fixed, and the constants can be small enough that the range remains in $[0,1]$. Thus all critical values are distinct. On a component with no incoming or outgoing black face, first place the free/interior function in a compact subinterval of $(0,1)$; its absent endpoint level is then empty.

An interior critical point of index zero or two has a disk birth or death neighborhood by the Morse lemma, giving ${\eta }_C$ or ${ϵ}_C$. An interior saddle is a rectangular band attached to two disjoint subintervals of the preceding regular level. This level is a compact one-manifold, so each affected component is a circle or an interval. The band is untwisted, because its orientation extends that of the surface. If both affected level components are circular, or both feet lie on one circular component, the usual two circle configurations give ${\mu }_C$ or ${\mathrm{\Delta }}_C$. If an interval occurs, the only possibilities are two distinct intervals, two feet on one interval, or one interval and one circle. Indeed these exhaust whether the two feet belong to the same or different level components and whether those components have endpoints. Two feet on a single interval split it into an interval and a circle; reversing the slab gives the interval/circle merge. Two different intervals are reconnected into two intervals. A twisted reconnection would give a nonorientable band and is excluded.

The six mixed representatives and their types are the following. A permutation of the listed input or output slots is implemented by the symmetry cylinders; it never changes an interval's own endpoint labels. In rows 1–2 the canonical type is $I_{ab}\otimes I_{cd}\to I_{ad}\otimes I_{cb}$. These are the two choices of which incoming interval is split in a two-interval reconnection. Rows 3–6 have the types shown. The letter at a newly exposed free edge is determined by the free edge already meeting that end of the slab, so no arbitrary color is introduced.

At a free critical point the tangential restriction has a nondegenerate extremum and the inward normal derivative is nonzero by construction. In coordinates $t\ge 0$ inward and $s$ tangential its chart can be taken as $f=f(p)\pm s^2\pm t$: apply [F7] to the map $(s,t)\mapsto (s,f(s,t)-f(s,0))$, whose Jacobian determinant is the nonzero inward normal derivative, followed by the one-dimensional Morse coordinate on the boundary. A free minimum with positive inward normal derivative births an interval and gives ${\eta }_a$; a free maximum with negative inward derivative kills one and gives ${ϵ}_a$. A free maximum with positive inward derivative joins the two local level ends: if they close the one interval into a circle the slab is $i_a^{\ast }$ (two corners), and otherwise they join two intervals into one, giving ${\mu }_{abc}$ (six corners). A free minimum with negative inward derivative is the reverse: $i_a$ has two corners and ${\mathrm{\Delta }}_{abc}$ has six. The label at the extremum is the label of the single free face through that point; all other endpoint labels are carried by product collars. There is no zero-normal-derivative case in the function constructed in the preceding boundary-relative construction, and no assertion about a full boundary Hessian is needed.

Here is a boundary-relative verification of the cuts in that table. Write $X^{-}$ for an incoming black interval read in induced surface-boundary orientation, $U^{+}$ for an outgoing interval, and $F_s$ for an intervening free arc of color $s$. In rows 1–2, call the inputs $X=I_{ab},Y=I_{cd}$ and outputs $U=I_{ad},V=I_{cb}$. The one-saddle slab is two interval product rectangles joined by the untwisted rectangular band along one edge in each rectangle, hence is a disk: successively attaching each rectangle along a full band edge leaves a disk. Its boundary word is $U^{+}{F}_d{Y}^{-}{F}_c{V}^{+}{F}_b{X}^{-}{F}_a\mathrm{.}$ For row 1 cut the ${\mathrm{\Delta }}_{acb}$/${\mu }_{acd}$ seam $I_{ac}$; for row 2 cut the ${\mathrm{\Delta }}_{cbd}$/${\mu }_{abd}$ seam $I_{bd}$. Each composite consists of two disks joined along that seam. Reading the remaining edges gives precisely the displayed word. The apparent crossing in each formula is a permutation of black slots; it does not identify any free edges. In rows 3–4 the slab is an annulus with outer boundary word $U^{+}{F}_b{X}^{-}{F}_a$ and inner boundary the incoming closed circle $Z^{-}$. This can be seen directly from the untwisted band on a circle and an interval: slit the circle at a point away from the band foot, then cut through the intervening product strip to the indicated outer free arc; the band and the slit level collars unfold to a disk. For the composite, the zipper is an annulus with inner boundary $Z^{-}$; attaching the product disk along its interval seam replaces an interval of the outer boundary by the complementary boundary arc of that disk. This disk attachment preserves the annulus and gives exactly $U^{+}{F}_b{X}^{-}{F}_a$. Choose an auxiliary arc in the zipper from $Z$ to the surviving part of its free arc, of color $a$ in row 3 and $b$ in row 4. Its interior avoids the product seam. Cutting along this auxiliary root-to-hole arc gives one disk. It is not the interval gluing seam: cutting that seam instead separates the original disk and annulus. In either the slab or composite split the outer cyclic word at the auxiliary arc's endpoint, writing it $E_1{E}_2$; the cut word is $E_{1}s{Z}^{-}{s}^{-1}{E}_2$. Choose corresponding endpoints on the indicated free arc, so the two cut words and their paired seam sides agree, including all black parameterized subarcs and free labels. Rows 5–6 are the height reversals of these annular cut models, with the original interval parameterizations retained. Matching these disk words gives diffeomorphisms relative to black collars: after rounding their nonblack corners, an orientation-preserving circle map extends across the disk by interpolating its lifted angular map to the identity near the disk center; on a prescribed boundary collar use its product extension. Regluing each pair $s,s^{-1}$ recovers the annular diffeomorphism. Thus these are actual local factorizations, including the two interval reconnections and all four interval/circle orientations.

Choose regular values $a<b$ surrounding each critical value, and first consider a compact slab $Q=f^{-1}([a,b])$ containing no interior or free-face critical value. Extend it a little through its two regular black level faces using the product level charts supplied by [F7] from the nonzero tangential or interior differential; at an actual incoming or outgoing black face use the given collar to add a short exterior strip. This gives a neighborhood of $Q$ whose only boundary near $Q$ is the free boundary, with no remaining corner at the extended black levels. On a free face define the tangent field $X=({\mathrm{\partial }}_{s}f{)}^{-1}{\mathrm{\partial }}_s$ and extend it smoothly to a smaller collar; its denominator is nonzero throughout the regular slab. Near an interior point choose a coordinate direction with nonzero $df$ and divide its coordinate field by that derivative. The finite chart partitions from step 1.1 glue these fields while preserving $df(X)=1$ and tangency along the free faces. The extended neighborhood is a smooth manifold with boundary, and its entire boundary near $Q$ is the free face to which $X$ is tangent; thus [F11], not a theorem requiring tangency to the transverse level faces, gives its boundary-preserving local flows. In interior charts the same flow is supplied by [F10]; uniqueness makes the charts agree. At every point of compact $Q$ the open flow domain contains $(-{\epsilon }_j,{\epsilon }_j)\times U_j$ for some neighborhood. A finite subcover gives a common positive lower time bound. Since $df(X)=1$, a trajectory stays in $Q$ until its level reaches an endpoint; it cannot cross a free face by [F11]. Restarting in this finite cover with increments below the common bound constructs the trajectory through every time needed between $a$ and $b$. This proves finite-time existence on the slab without an unjustified completeness assumption. The map $f^{-1}(a)\times [a,b]\to Q$, $(p,t)\mapsto {\mathrm{\Phi }}_{t-a}(p)$, has inverse $q\mapsto ({\mathrm{\Phi }}_{a-f(q)}(q),f(q))$. These maps are smooth by [F10]–[F11] and preserve each free component, hence every color; they identify the slab with its product, with the level collars retained. Each critical slab is one of the interior, free-boundary or mixed local models already proved. Concatenating the finitely many slabs gives the generator word and identity cylinders. Apply the finite construction to each connected component and tensor the words; the empty component family gives the empty word.

We first give the elementary cut procedure used in every certificate below. Let $P(B_0;B_1,\dots ,B_h)$ be a round disk with $h$ disjoint smaller round disks removed. Each $B_j$ is the cyclic word of marked black intervals and colored free arcs on that component, or a single marked black circle. The orientation on $B_0$ is opposite to that on the interior holes, as dictated by the surface orientation. A word is always read in the induced surface-boundary direction; $X^{-}$ and $U^{+}$ denote an incoming and outgoing black slot. To cut $P$ into a disk, pick a nonblack basepoint on each component with such a point, or a basepoint on a black circle otherwise, and draw disjoint arcs $s_j$ from $B_0$ to $B_j$. Arrange the arcs at the root in the order $1,\dots ,h$. After cutting them the disk boundary word is $B_0{s}_1{B}_1{s}_1^{-1}{s}_2{B}_2{s}_2^{-1}\cdots s_h{B}_h{s}_h^{-1}\mathrm{.}$ Black circle basepoints are auxiliary, not extra markings. A desired circle parameterization is respected on its whole collar, including this point. If two cut patches have this same word, match the corresponding black collars by their parameterizations and match each colored free arc by an increasing smooth map. Match the two sides of every $s_j$ by identical product maps. This gives a disk-boundary map extending to the collars. After rounding the nonblack corners, lift the remaining angular map $h$ to a strictly increasing map of $\mathbb{R}$ with $h(\theta +2\pi )=h(\theta )+2\pi$. The interpolation $(1-u)\theta +uh(\theta )$ has positive derivative for $0\le u\le 1$. Use $u=0$ near the center and $u=1$ near the boundary; its polar extension is smooth and orientation preserving, and agrees with the prescribed collar map after the collar is attached. Regluing the paired $s_j$ edges gives the desired diffeomorphism. This proves the cut criterion directly for these planar models; no surface normal-form theorem is used.

Here are the three gluing rules for computing the certificates. A disk generator with cyclic word $EX$ glued to another disk with word $X^{-1}D$ along an interval seam $X$ becomes the disk with word $ED$; this follows by placing the two collars along the two sides of the same straight segment in a disk. If either disk has holes, carry their disjoint removed disks through this construction; the holes are simply retained. Gluing a boundary circle of one planar patch to one of a different planar patch joins their annular collars and gives a planar patch with the unglued boundary components; this can be seen by placing one punctured disk in the deleted disk of the other, using inversion of the annular radial coordinate on the collar. Finally, two disks glued along two disjoint interval seams give an annulus when the boundary splicing yields two boundary cycles. To check this without a classification theorem, first glue one seam to obtain a disk; the second seam identifies two boundary intervals with opposite induced orientations. Their standard rectangular collar is a strip joining those intervals outside the disk. The two complementary boundary arcs are distinct, and a cut through that strip returns a disk. Its regluing is precisely $P(B_0;B_1)$. All these constructions respect the marked portions of the boundary. The rule for one interval seam can be applied repeatedly to any tree of disk/annulus patches.

A product generator is a disk with cyclic boundary word $U_{ac}^{+}{F}_c{Y}_{bc}^{-}{F}_b{X}_{ab}^{-}{F}_a$. A coproduct generator is the same disk read with reversed heights, with word $U_{ab}^{+}{F}_b{V}_{bc}^{+}{F}_c{X}_{ac}^{-}{F}_a$. The unit and counit are disks with a single black interval and its complementary free arc. A zipper is an annulus with words $(U_{aa}^{+}{F}_a;X_C^{-})$, and a cozipper is an annulus with words $(X_{aa}^{-}{F}_a;U_C^{+})$. The closed product and coproduct are punctured disks with three marked circle boundaries; the closed unit and counit have one. These are concrete choices of the elementary patches. The gluing rules in step 2.1 give the following exhaustive table for the defining equations of the data definition. In the table each displayed word belongs to an individual full boundary component; a semicolon separates components. $F_a$ means one free arc of color $a$, including the case of the complementary arc of a single marked interval.

Every row is a finite application of the gluing rules to specified disks or punctured disks. The rules prove planarity by construction; planarity is not inferred merely from an Euler characteristic or an appeal to the classification under proof. For an open row with one boundary component the displayed word is already the cut disk. For the other rows cut the indicated root-to-hole arcs, and use step 1.1. The identical words and paired seam sides give orientation-preserving diffeomorphisms fixed on every external black collar and preserving each free-edge label. In particular the Cardy row retains separate $a$- and $b$-labeled boundary components, and the open Frobenius rows retain their different domains and codomains. The cyclic trace row allows rotation of the disk boundary word but fixes the actual black slots after their specified symmetry; it gives no open commutativity assertion.

The rows give the closed Frobenius, open indexed Frobenius, cyclic trace, zipper algebra, centrality, pairing adjunction and Cardy equations of the data definition. The strong monoidal coherence maps of a functor carry the product-cylinder associators and unitors coherently, and its symmetry-preservation equation from [F3] carries every symmetry cylinder to the target symmetry. Applying it to these actual surface equalities therefore gives exactly the typed equations in the data definition. No relation-completeness assertion has been used.