Typed algebraic derivations of the Frobenius sewing moves

For each $a,b$, write $d_{ab}(1)={\sum }_j{p}_j\otimes q_j$ and take $x\in A_{ab}$. Apply the first typed Frobenius identity with labels $(a,b,a,b)$ to ${\eta }_a(1)\otimes x$. The left unit equation gives ${\mathrm{\Delta }}_{abb}(x)$; the right-hand composite gives ${\sum }_j{p}_j\otimes {\mu }_{bab}(q_j,x)$. Applying $1_{A_{ab}}\otimes {ϵ}_b$ and the right counit to this equality gives ${\sum }_j{p}_j{b}_{ba}(q_j,x)=x\mathrm{.}$ This is the typed snake ${\rho }_{A_{ab}}(1_{A_{ab}}\otimes b_{ba}){\alpha }_{A_{ab},A_{ba},A_{ab}}(d_{ab}\otimes 1_{A_{ab}}){\lambda }_{A_{ab}}^{-1}=1_{A_{ab}},$ where the associator maps $(A_{ab}\otimes A_{ba})\otimes A_{ab}$ to $A_{ab}\otimes (A_{ba}\otimes A_{ab})$. Apply the second Frobenius identity with labels $(b,a,b,a)$ to $y\otimes {\eta }_a(1)$, then apply ${ϵ}_b\otimes 1_{A_{ba}}$ and the left counit. For $y\in A_{ba}$ and the same $d_{ab}(1)={\sum }_j{p}_j\otimes q_j$, this gives ${\sum }_j{b}_{ba}(y,p_j)q_j=y$, which is the other typed snake ${\lambda }_{A_{ba}}(b_{ba}\otimes 1_{A_{ba}}){\alpha }_{A_{ba},A_{ab},A_{ba}}^{-1}(1_{A_{ba}}\otimes d_{ab}){\rho }_{A_{ba}}^{-1}=1_{A_{ba}}\mathrm{.}$ The closed object $C$ has the same two snakes with $b_C$ and $d_C$: apply its uncolored Frobenius identities to ${\eta }_C\otimes x$ and $x\otimes {\eta }_C$ and then use its two counits. Writing $d_{ab}(1)={\sum }_j{p}_j\otimes q_j$, the first snake says $x\mapsto (y\mapsto b_{ab}(x,y))$ has inverse $\varphi \mapsto {\sum }_j{p}_j\varphi (q_j)$, because $b_{ab}(x,q_j)=b_{ba}(q_j,x)$. The second says $y\mapsto (x\mapsto b_{ba}(y,x))$ has inverse $\psi \mapsto {\sum }_j\psi (p_j)q_j$. Thus both induced maps to the opposite algebraic duals are isomorphisms.

The trace equation says $b_{ab}=b_{ba}{\tau }_{A_{ab},A_{ba}}$. To prove copairing symmetry explicitly, put $d_{ab}(1)={\sum }_j{p}_j\otimes q_j$ and $d_{ba}(1)={\sum }_k{r}_k\otimes s_k$. For arbitrary $x\in A_{ab}$ and $y\in A_{ba}$, contract each proposed tensor in $A_{ab}\otimes A_{ba}$ against ${\varphi }_y\otimes {\psi }_x$, where ${\varphi }_y(t)=b_{ba}(y,t)$ and ${\psi }_x(t)=b_{ab}(x,t)$. The two contractions are respectively ${\sum }_j{b}_{ba}(y,p_j)b_{ab}(x,q_j)=b_{ab}(x,y),$ ${\sum }_k{b}_{ba}(y,s_k)b_{ab}(x,r_k)={\sum }_k{b}_{ab}(s_k,y)b_{ab}(x,r_k)=b_{ab}(x,y)\mathrm{.}$ The first equality uses the second snake for $d_{ab}$; the second uses trace symmetry followed by the first snake for $d_{ba}$. By step 1.1 the maps $y\mapsto {\varphi }_y$ and $x\mapsto {\psi }_x$ exhaust the two dual spaces. Their tensor products separate tensors, proving the typed equality $d_{ab}={\tau }_{A_{ba},A_{ab}}{d}_{ba}:k\to A_{ab}\otimes A_{ba}$. The same proof gives $d_C={\tau }_{C,C}{d}_C$.

For $a,b,c\in S$, the full copairing-invariance identity is the following equality of maps $A_{ab}\to A_{ac}\otimes A_{cb}$: $\begin{array}{rl}{\displaystyle L_{ab}^c}& {\displaystyle :=({\mu }_{abc}\otimes 1_{A_{cb}}){\alpha }_{A_{ab},A_{bc},A_{cb}}^{-1}(1_{A_{ab}}\otimes d_{bc}){\rho }_{A_{ab}}^{-1}}\\ {\displaystyle }& {\displaystyle ={\mathrm{\Delta }}_{acb}}\\ {\displaystyle }& {\displaystyle =(1_{A_{ac}}\otimes {\mu }_{cab}){\alpha }_{A_{ac},A_{ca},A_{ab}}(d_{ac}\otimes 1_{A_{ab}}){\lambda }_{A_{ab}}^{-1}=:R_{ab}^c\mathrm{.}}\end{array}$ The first composite goes through $A_{ab}\otimes k$, $A_{ab}\otimes (A_{bc}\otimes A_{cb})$, $(A_{ab}\otimes A_{bc})\otimes A_{cb}$. The last goes through $k\otimes A_{ab}$, $(A_{ac}\otimes A_{ca})\otimes A_{ab}$, $A_{ac}\otimes (A_{ca}\otimes A_{ab})$. To prove the first equality, precompose the second typed Frobenius law with labels $(a,b,c,b)$ by $(1_{A_{ab}}\otimes {\eta }_b){\rho }_{A_{ab}}^{-1}$. Its left side becomes ${\mathrm{\Delta }}_{acb}$ by the right unit, and its right side becomes $L_{ab}^c$ by $d_{bc}={\mathrm{\Delta }}_{bcb}{\eta }_b$. For the last equality precompose the first typed Frobenius law with labels $(a,c,a,b)$ by $({\eta }_a\otimes 1_{A_{ab}}){\lambda }_{A_{ab}}^{-1}$. Its left side becomes ${\mathrm{\Delta }}_{acb}$ by the left unit, and its right side becomes $R_{ab}^c$ by $d_{ac}={\mathrm{\Delta }}_{aca}{\eta }_a$. Thus every displayed map is well typed and the equality follows directly from defining relations, without an implicit transpose. Replacing the three open spaces by $C$ gives both closed copairing invariances with the same associators/unitors.

Here is the full reversed-factor transpose behind that identity. Set $U=A_{ab}$, $V=A_{bc}$, $W=A_{ac}$, $U^{\vee }=A_{ba}$, $V^{\vee }=A_{cb}$, $W^{\vee }=A_{ca}$. Let $P:(U\otimes V)\otimes (V^{\vee }\otimes U^{\vee })\to (U\otimes U^{\vee })\otimes (V\otimes V^{\vee })$ be the following specified composite of associators and flips: $\begin{array}{rl}{\displaystyle (U\otimes V)\otimes (V^{\vee }\otimes U^{\vee })}& {\displaystyle \stackrel{{\alpha }_{U,V,V^{\vee }\otimes U^{\vee }}}{\to }U\otimes (V\otimes (V^{\vee }\otimes U^{\vee }))}\\ {\displaystyle }& {\displaystyle \stackrel{1_U\otimes {\alpha }_{V,V^{\vee },U^{\vee }}^{-1}}{\to }U\otimes ((V\otimes V^{\vee })\otimes U^{\vee })}\\ {\displaystyle }& {\displaystyle \stackrel{1_U\otimes {\tau }_{V\otimes V^{\vee },U^{\vee }}}{\to }U\otimes (U^{\vee }\otimes (V\otimes V^{\vee }))}\\ {\displaystyle }& {\displaystyle \stackrel{{\alpha }_{U,U^{\vee },V\otimes V^{\vee }}^{-1}}{\to }(U\otimes U^{\vee })\otimes (V\otimes V^{\vee })\mathrm{.}}\end{array}$ Thus $P((x\otimes y)\otimes (r\otimes s))=(x\otimes s)\otimes (y\otimes r)$. Define $B={\lambda }_k(b_{ab}\otimes b_{bc})P$, a perfect pairing between $U\otimes V$ and $V^{\vee }\otimes U^{\vee }$. Its perfection follows from the two finite-dimensional perfect pairings of step 1.1: in dual bases its matrix is a permutation of the tensor product of two invertible pairing matrices, hence invertible. The explicit transpose equality of maps $(U\otimes V)\otimes W^{\vee }\to k$ is $b_{ac}({\mu }_{abc}\otimes 1_{A_{ca}})=B(1_{U\otimes V}\otimes {\mathrm{\Delta }}_{cba})\mathrm{.}$ For its verification write $d_{ab}(1)={\sum }_j{p}_j\otimes q_j$. Step 2.2 with labels $(c,a,b)$ gives ${\mathrm{\Delta }}_{cba}(z)={\sum }_j{\mu }_{cab}(z,p_j)\otimes q_j$. Therefore the right side at $(x\otimes y)\otimes z$ equals ${\sum }_j{b}_{ab}(x,q_j)b_{bc}(y,z{p}_j)={\sum }_j{b}_{ab}(x,q_j)b_{ba}(yz,p_j)=b_{ab}(x,yz)=b_{ac}(xy,z)\mathrm{.}$ The first intervening equality is indexed associativity under ${ϵ}_b$; the next is the second snake for $d_{ab}$; the last is indexed associativity under ${ϵ}_a$. This proves the reversed-factor transpose including its exact flip. The pairing invariance itself is the equality $b_{ac}({\mu }_{abc}\otimes 1)=b_{ab}(1\otimes {\mu }_{bca}){\alpha }_{A_{ab},A_{bc},A_{ca}}$ on $(A_{ab}\otimes A_{bc})\otimes A_{ca}$. Hence both the pairing and its coproduct transpose, for every triple of labels, have been expanded as actual typed composites.

For arbitrary $a,b,t\in S$ put $O_{ab}^t:={\mu }_{atb}{\mathrm{\Delta }}_{atb}:A_{ab}\to A_{ab},e_a^t:=O_{aa}^t{\eta }_a(1)\in A_{aa}\mathrm{.}$ For all $a,b,c,t$, the full multiplication slides, as maps $A_{ab}\otimes A_{bc}\to A_{ac}$, are $O_{ac}^t{\mu }_{abc}={\mu }_{abc}(O_{ab}^t\otimes 1_{A_{bc}})={\mu }_{abc}(1_{A_{ab}}\otimes O_{bc}^t)\mathrm{.}$ To prove the first, the first Frobenius law with labels $(a,t,b,c)$ gives ${\mu }_{atc}{\mathrm{\Delta }}_{atc}{\mu }_{abc}={\mu }_{atc}(1_{A_{at}}\otimes {\mu }_{tbc}){\alpha }_{A_{at},A_{tb},A_{bc}}({\mathrm{\Delta }}_{atb}\otimes 1_{A_{bc}})={\mu }_{abc}({\mu }_{atb}{\mathrm{\Delta }}_{atb}\otimes 1_{A_{bc}})\mathrm{.}$ The last equality is indexed associativity. For the second use the second Frobenius law with labels $(a,b,t,c)$: ${\mu }_{atc}{\mathrm{\Delta }}_{atc}{\mu }_{abc}={\mu }_{atc}({\mu }_{abt}\otimes 1_{A_{tc}}){\alpha }_{A_{ab},A_{bt},A_{tc}}^{-1}(1_{A_{ab}}\otimes {\mathrm{\Delta }}_{btc})={\mu }_{abc}(1_{A_{ab}}\otimes {\mu }_{btc}{\mathrm{\Delta }}_{btc})\mathrm{.}$ Set the first or second input to the appropriate unit to obtain $O_{ab}^t(x)=e_a^{t}x=x{e}_b^t$. Consequently for $x\in A_{ab},y\in A_{ba}$, $b_{ab}(O_{ab}^{t}x,y)={ϵ}_a(e_a^{t}xy)={ϵ}_a(x{e}_b^{t}y)=b_{ab}(x,O_{ba}^{t}y)\mathrm{.}$ Here $e_a^{t}x=x{e}_b^t$ is the preceding unit specialization, so this step needs no unproved centrality or extra symmetry.

We justify formal use by faithful bending. Suppose $V$ has the dual $V^{\vee }$, evaluation $b_V:V\otimes V^{\vee }\to \mathbf{1}$ and coevaluation $d_{V^{\vee }}:\mathbf{1}\to V^{\vee }\otimes V$, with both snakes proved from Frobenius and units as in this lemma step 1.1. For $f:U\to V$ define $D(f):=b_V(f\otimes 1_{V^{\vee }}):U\otimes V^{\vee }\to \mathbf{1.}$ For $h:U\otimes V^{\vee }\to \mathbf{1}$ define $J(h):={\lambda }_V(h\otimes 1_V){\alpha }_{U,V^{\vee },V}^{-1}(1_U\otimes d_{V^{\vee }}){\rho }_U^{-1}:U\to V\mathrm{.}$ Naturality of associators/unitors and the snake on $V$ give $J(D(f))=f$. In detail the composite is the map obtained by tensoring $f$ with $d_{V^{\vee }}$ and then evaluating $b_V$ on its first two factors; the snake is ${\lambda }_V(b_V\otimes 1_V){\alpha }_{V,V^{\vee },V}^{-1}(1_V\otimes d_{V^{\vee }}){\rho }_V^{-1}=1_V\mathrm{.}$ Precomposing this map by $f$ proves the identity. Thus D is injective on Hom sets with no vector-space assumption. For tensor outputs use $V=V_1\otimes V_2$, dual $V_2^{\vee }\otimes V_1^{\vee }$, evaluation exactly the reversed-factor P-composite in step 3.1, and coevaluation formed by inserting $d_{V_1^{\vee }}$ between the two factors of $d_{V_2^{\vee }}$, using canonical unitors and reassociating to $(V_2^{\vee }\otimes V_1^{\vee })\otimes (V_1\otimes V_2)$. In flattened tensor notation this is $(1_{V_2^{\vee }}\otimes d_{V_1^{\vee }}\otimes 1_{V_2})d_{V_2^{\vee }}$. The two constituent snakes cancel successively, so the tensor version of D is also injective. Every pairing calculation below is an equality of maps with external tensor factors of the indicated types. For example the two sides of the module equation in the later module calculation are maps $C\otimes A_{aa}\to C$; applying D adds one external C factor and gives the three-composite equality displayed there. The product-transpose in step 3.1 uses the external factors $A_{ab},A_{bc},A_{ca}$; after inserting the coproduct expression from step 2.2, its contraction is precisely the second snake composed with indexed associativity. The tensor-output tests in the later transport, coalgebra and window calculations use P with both external dual factors, so injectivity of D applies. In all such displays, input letters name these external factors, juxtaposition is the displayed indexed product, and evaluation is the displayed pairing. The equalities use only the defining maps, their tensor products and structural equations; they do not assume scalar-valued points separate morphisms in a general category. Thus the displayed identities hold in the formal presentation. In the vector-space specialization, pure-tensor evaluations give the same formulas, and the additional finite-dimensional nondegeneracy assertion follows from step 1.1.

The coproduct slides, maps $A_{ac}\to A_{ab}\otimes A_{bc}$, are ${\mathrm{\Delta }}_{abc}{O}_{ac}^t=(O_{ab}^t\otimes 1_{A_{bc}}){\mathrm{\Delta }}_{abc}=(1_{A_{ab}}\otimes O_{bc}^t){\mathrm{\Delta }}_{abc}\mathrm{.}$ Test against $u\in A_{ba}$ and $v\in A_{cb}$ by the perfect output pairing $⟨r\otimes s,u\otimes v⟩=b_{ab}(r,u)b_{bc}(s,v)$. The scalar for the left composite is ${ϵ}_a((O_{ac}^{t}x)vu)={ϵ}_a(x{O}_{ca}^t(vu))$ by the transpose identity above. By the product slides this is both ${ϵ}_a(x(O_{cb}^{t}v)u)$ and ${ϵ}_a(xv(O_{ba}^{t}u))$. Transposing $O_{bc}^t$ or $O_{ab}^t$ under its opposite pairing gives respectively the rightmost and middle composite. The general coproduct-contraction identity used here is the reversed-factor transpose from step 3.1 with labels renamed: $⟨{\mathrm{\Delta }}_{abc}x,u\otimes v⟩={ϵ}_a(xvu)$. Since the product pairing is perfect, these scalar equalities prove both typed map equalities.

Pairing adjunction in [F1] is the equality of maps $C\otimes A_{aa}\to k$ $b_C(1_C\otimes i_a^{\ast })=b_{aa}(i_a\otimes 1_{A_{aa}})\mathrm{.}$ Precomposing with the flip and using the symmetries of $b_C$ and $b_{aa}$ gives its reflected equality $b_C(i_a^{\ast }\otimes 1_C)=b_{aa}(1_{A_{aa}}\otimes i_a)$ on $A_{aa}\otimes C$. Faithful bending from step 4.1 and the two snakes from step 1.1 give both copairing transports: $(i_a\otimes 1_C)d_C=(1_{A_{aa}}\otimes i_a^{\ast })d_{aa}:k\to A_{aa}\otimes C,$ $(1_C\otimes i_a)d_C=(i_a^{\ast }\otimes 1_{A_{aa}})d_{aa}:k\to C\otimes A_{aa}\mathrm{.}$ For the first equality let $d_C(1)={\sum }_j{p}_j\otimes q_j$ and $d_{aa}(1)={\sum }_k{r}_k\otimes s_k$ in vector notation. The complete contraction against external factors $u\in A_{aa}$ and $c\in C$ is $\begin{array}{rl}{\displaystyle \sum _j{b}_{aa}(u,i_a(p_j))b_C(c,q_j)}& {\displaystyle =\sum _j{b}_C(i_a^{\ast }u,p_j)b_C(c,q_j)=b_C(c,i_a^{\ast }u),}\\ {\displaystyle \sum _k{b}_{aa}(u,r_k)b_C(c,i_a^{\ast }{s}_k)}& {\displaystyle =\sum _k{b}_{aa}(u,r_k)b_{aa}(i_a(c),s_k)=b_{aa}(i_a(c),u)=b_C(c,i_a^{\ast }u)\mathrm{.}}\end{array}$ Each first equality is adjunction or its reflection; each contraction cancellation is a snake. As composites these are tensor-output applications of D from step 4.1, so their equality proves the first transport in the formal presentation too. Applying the flip to that equality and using $\tau d_C=d_C$ and $\tau d_{aa}=d_{aa}$ proves the second, with source and target exchanged exactly as displayed. This gives both transports, including the source's exercise.

Pair the proposed cozipper coalgebra identity against arbitrary $c,d\in C$ using the nondegenerate pairing $b_C\otimes b_C$. For either symmetric Frobenius object $X=C$ or $X=A_{aa}$, the snake and Frobenius identities imply $(b_X\otimes b_X)(u\otimes v,{\mathrm{\Delta }}_X(x))=b_X(vu,x)$ with the multiplication and tensor order shown. Hence the left side of ${\mathrm{\Delta }}_C{i}_a^{\ast }=(i_a^{\ast }\otimes i_a^{\ast }){\mathrm{\Delta }}_{aaa}$ pairs to $b_C(dc,i_a^{\ast }x)$. The right side pairs to $b_{aa}(i_a(d)i_a(c),x)=b_{aa}(i_a(dc),x)$, because $i_a$ preserves multiplication and $C$ is commutative. Pairing adjunction makes this $b_C(dc,i_a^{\ast }x)$ as well. Nondegeneracy proves the coproduct equality. Setting the first argument of pairing adjunction to $1_C$ gives ${ϵ}_C(i_a^{\ast }x)={ϵ}_a(x)$ for every $x$, proving counit preservation.

The relation $i_a^{\ast }(i_a(c)x)=c{i}_a^{\ast }(x)$ holds for $c\in C$, $x\in A_{aa}$. To prove it, pair both sides against arbitrary $d\in C$. Pairing adjunction, the unital algebra equation for $i_a$, centrality, and associativity give $b_C(d,i_a^{\ast }(i_a(c)x))=b_{aa}(i_a(d),i_a(c)x)=b_{aa}(i_a(dc),x)=b_C(d,c{i}_a^{\ast }(x))\mathrm{.}$ The middle equality uses $i_a(d)i_a(c)=i_a(dc)$ and $C$-commutativity; the last uses associativity of $b_C$. Since $b_C$ is nondegenerate by step 1.1 with the closed object, equality against every $d$ proves the claim. Thus $W_a=i_a^{\ast }{i}_a$ is multiplication by $w_a:=W_a(1_C)$ and is a $C$-bimodule map. The Frobenius identities for $C$ then slide $W_a$ across either leg of ${\mu }_C$ and ${\mathrm{\Delta }}_C$; for example $W_a{\mu }_C={\mu }_C(W_a\otimes 1_C)={\mu }_C(1_C\otimes W_a)$ and $(W_a\otimes 1_C){\mathrm{\Delta }}_C=(1_C\otimes W_a){\mathrm{\Delta }}_C={\mathrm{\Delta }}_C{W}_a$.

Finally the colored open-window reduction is, for arbitrary $a,t$, $i_a^{\ast }{O}_{aa}^t=W_t{i}_a^{\ast }:A_{aa}\to C,W_t=i_t^{\ast }{i}_t\mathrm{.}$ For $x\in A_{aa}$ write ${\mathrm{\Delta }}_{ata}x={\sum }_j{r}_j\otimes s_j$ with $r_j\in A_{at},s_j\in A_{ta}$. For arbitrary $c\in C$ the scalar pairing of the left side against $c$ is $\begin{array}{rl}{\displaystyle \sum _j{ϵ}_a(i_a(c)r_j{s}_j)}& {\displaystyle =\sum _j{ϵ}_t(s_j{i}_a(c)r_j)}\\ {\displaystyle }& {\displaystyle =\sum _j{ϵ}_t(i_t(c)s_j{r}_j)}\\ {\displaystyle }& {\displaystyle ={ϵ}_t(i_t(c)i_t(i_a^{\ast }x))}\\ {\displaystyle }& {\displaystyle =b_C(c,i_t^{\ast }{i}_t{i}_a^{\ast }x)\mathrm{.}}\end{array}$ The first equality is the indexed trace on $A_{at}\otimes A_{ta}$, after associating; the second is centrality on $A_{ta}$; the third is the colored Cardy composite $i_t{i}_a^{\ast }={\mu }_{tat}{\tau }_{A_{at},A_{ta}}{\mathrm{\Delta }}_{ata}:A_{aa}\to A_{tt}$; the last is pairing adjunction for $i_t$. Perfectness of $b_C$ proves the reduction. This proves a window of every free-boundary color t, not just t=a. In the singleton case it specializes to the existing $i_a^{\ast }{O}_a=W_a{i}_a^{\ast }$ calculation. All color sums here are finite tensor decompositions; none is a sum over S.

Let $H={\mu }_C{\mathrm{\Delta }}_C$ and $O_a={\mu }_{aaa}{\mathrm{\Delta }}_{aaa}$. For any Frobenius object $X$ and $x,y\in X$, the first and second Frobenius identities respectively give $\mu \mathrm{\Delta }(xy)=x\mu \mathrm{\Delta }(y),\mu \mathrm{\Delta }(xy)=\mu \mathrm{\Delta }(x)y\mathrm{.}$ For example, the first formula expands $\mathrm{\Delta }(xy)$ as $(x\otimes 1)\mathrm{\Delta }(y)$ and then multiplies its two factors; the second expands it as $\mathrm{\Delta }(x)(1\otimes y)$. Thus $H$ is a $C$-bimodule map and $O_a$ an $A_{aa}$-bimodule map. Their coproduct slides follow by composing the two typed Frobenius identities with $\mathrm{\Delta }$: $\mathrm{\Delta }\mu \mathrm{\Delta }=(\mu \otimes 1)(1\otimes \mathrm{\Delta })\mathrm{\Delta }=(1\otimes \mu )(\mathrm{\Delta }\otimes 1)\mathrm{\Delta }\mathrm{.}$ By coassociativity, the outer expressions are $(O\otimes 1)\mathrm{\Delta }$ and $(1\otimes O)\mathrm{\Delta }$. Since $W_a$ is multiplication by $w_a$ and $H$ is a $C$-bimodule map, $W_{a}H=H{W}_a$. These are the required closed-window and handle slides.

The Cardy relation says $i_a{i}_a^{\ast }={\mu }_{aaa}\tau {\mathrm{\Delta }}_{aaa}$. For $x\in A_{aa}$ and $c\in C$, cyclicity of the symmetric open trace and centrality of $i_a(c)$ give $b_C(c,i_a^{\ast }{O}_{a}x)={ϵ}_a(i_a(c){\mu }_{aaa}{\mathrm{\Delta }}_{aaa}x)={ϵ}_a(i_a(c){\mu }_{aaa}\tau {\mathrm{\Delta }}_{aaa}x)\mathrm{.}$ For the middle equality, write $\mathrm{\Delta }x={\sum }_j{x}_j\otimes y_j$; the two scalar expressions are ${\sum }_j{ϵ}_a(i_a(c)x_j{y}_j)$ and ${\sum }_j{ϵ}_a(i_a(c)y_j{x}_j)$. Cyclicity moves $y_j$ to the front in the first expression, and centrality moves $i_a(c)$ back to the front, giving the second. Cardy changes the final term to ${ϵ}_a(i_a(c)i_a(i_a^{\ast }x))$. Pairing adjunction makes this $b_C(c,W_a{i}_a^{\ast }x)$. Nondegeneracy of $b_C$ proves the typed equality $i_a^{\ast }{O}_a=W_a{i}_a^{\ast }$. This is the open-handle reduction used when a cozipper follows an open handle.

For $q\ge 1$, let ${\mu }_{a_0\dots a_q}^{(q)}:A_{a_0{a}_1}\otimes \cdots \otimes A_{a_{q-1}{a}_q}\to A_{a_0{a}_q}$ be the iterated product, parenthesized to the left; indexed associativity makes it independent of parenthesization after the canonical associators. When $a_q=a_0$, the map $i_{a_0}^{\ast }{\mu }^{(q)}$ is invariant under cyclic rotation of the factors together with their endpoint labels. The claim is immediate for $q=1$. For $q\ge 2$, put $X=i_{a_0}(c)x_1\in A_{a_0{a}_1}$ and $Y=x_2\cdots x_q\in A_{a_1{a}_0}$. Pair the original value against $c\in C$; pairing adjunction gives ${ϵ}_{a_0}(XY)$. The indexed trace equation gives ${ϵ}_{a_0}(XY)={ϵ}_{a_1}(YX)$. Centrality says $Y{i}_{a_0}(c)=i_{a_1}(c)Y$, so the latter is ${ϵ}_{a_1}(i_{a_1}(c)Y{x}_1)$, which is the pairing against $c$ of the cyclically rotated cozipper output. Since $b_C$ is nondegenerate, those outputs in $C$ agree. In the singleton case this is $i^{\ast }{\mu }_A^{(q)}(x_1,\dots ,x_q)=i^{\ast }{\mu }_A^{(q)}(x_2,\dots ,x_q,x_1)\mathrm{.}$ This proves the cyclic cozipper rotation without assuming that $A_{aa}$ is commutative.

Every operation just derived is local to the labels occurring on its factors. The snakes pair $A_{ab}$ only with $A_{ba}$; the products and coproducts preserve the concatenated endpoint sequence; $i_a$ and $i_a^{\ast }$ meet only at the matching color $a$; $W_a$ and $O_a$ retain their color; and the Cardy equation uses exactly $A_{ba}\otimes A_{ab}$ before the flip. Consequently inserting any derived move into a larger composite changes no free-boundary label or black parameterization. This checks the color side condition required when the identities are used in surface normal-form reductions.