The complete Kirby-color band-slide identity with open colored graph coupons

For each $X$ choose actual $u_{i,a}:i\to X$, $v^{i,a}:X\to i$ with $v^{i,a}{u}_{j,b}={\delta }_{ij}{\delta }_{ab}{1}_i$ and ${\sum }_{i,a}{u}_{i,a}{v}^{i,a}=1_X$. Define ${\iota }_X={\sum }_{i,a}{\iota }_i(u_{i,a}^{\ast }\otimes v^{i,a}):X^{\ast }\otimes X\to B$. Changing a multiplicity-space basis multiplies the two entries by inverse matrices, leaving this sum unchanged. For $f:X\to Y$, expand its scalar simple blocks $v_{Y}f{u}_X$ in the two decompositions. This gives the exact dinaturality equation ${\iota }_Y(1_{Y^{\ast }}\otimes f)={\iota }_X(f^{\ast }\otimes 1_X)$. If another family $t_X:X^{\ast }\otimes X\to Z$ obeys that equation, its finite decomposition is $t_X={\sum }_{i,a}{t}_i(u_{i,a}^{\ast }\otimes v^{i,a})$. Thus the unique map $t:B\to Z$ with $t{\iota }_i=t_i$ satisfies $t{\iota }_X=t_X$. This proves the coend property explicitly, including uniqueness, without an existence or Fubini theorem.

Define the required structural maps by $\mathrm{\Delta }{\iota }_X=({\iota }_X\otimes {\iota }_X)(1_{X^{\ast }}\otimes b_X\otimes 1_X)$, $\epsilon {\iota }_X=e_X$, and $m({\iota }_X\otimes {\iota }_Y)={\iota }_{Y\otimes X}(1_{X^{\ast }}\otimes c_{X,Y^{\ast }\otimes Y})$. The last target uses $(Y\otimes X{)}^{\ast }=X^{\ast }\otimes Y^{\ast }$, or its canonical dual comparison. These formulas define maps first on the finitely many simple summands. Dinaturality, braiding naturality and the explicit dual-mate identities in [F1] show that their stated words hold for arbitrary $X,Y$, proving well-definedness under every choice of decomposition. Substituting the first two words gives $(1_B\otimes \epsilon )\mathrm{\Delta }{\iota }_X=({\iota }_X\otimes e_X)(1_{X^{\ast }}\otimes b_X\otimes 1_X)={\iota }_X$ by a snake; the opposite snake gives $(\epsilon \otimes 1_B)\mathrm{\Delta }=1_B$ too. No other Hopf algebra axiom is needed.

Fix simple $j,k$ and decompose each $i\otimes j$ using $u_{k,a}:k\to i\otimes j$ and $v^{k,a}:i\otimes j\to k$, with $v^{k,b}{u}_{k,a}={\delta }_{ab}{1}_k$. Set $F_{i,a}=(v^{k,a}\otimes 1_{j^{\ast }})(1_i\otimes b_j):i\to k\otimes j^{\ast }$ and $G_{i,a}=(1_i\otimes r_j)(u_{k,a}\otimes 1_{j^{\ast }}):k\otimes j^{\ast }\to i$. Their inverses as adjunction correspondences insert $e_j$ or $b_j^R$ and retract a snake, so the $F$ and $G$ families span the respective simple multiplicity spaces. Define ${\mathrm{ptr}}_j(a)=(1_i\otimes r_j)(a\otimes 1_{j^{\ast }})(1_i\otimes b_j)$ on $a\in \mathrm{End}(i\otimes j)$. The displayed words give $G_{i,b}{F}_{i,a}={\mathrm{ptr}}_j(u_{k,b}{v}^{k,a})$. Closing in stages and then cyclically gives ${\mathrm{tr}}_i(G_{i,b}{F}_{i,a})={\mathrm{tr}}_{i\otimes j}(u_{k,b}{v}^{k,a})={\mathrm{tr}}_k(v^{k,a}{u}_{k,b})={\delta }_{ab}{d}_k$. Since $i$ is simple, $G_{i,b}{F}_{i,a}={\delta }_{ab}(d_k\mathrm{/}{d}_i)1_i$; compositions with different simple labels vanish. Consequently the actual complete trace-dual resolution on $k\otimes j^{\ast }$ is ${\sum }_{i,a}{d}_i{F}_{i,a}{G}_{i,a}=d_k{1}_{k\otimes j^{\ast }}$. This includes every fusion multiplicity and uses no positivity.

For the two-$\mathrm{\Omega }$ slide, cut its participating components near the attaching band and place those endpoint pairs adjacently below the remaining graph. Evaluating that exterior with variable cut colors gives $T_{X,Y}:X^{\ast }\otimes X\otimes Y^{\ast }\otimes Y\otimes U\to V$, retaining arbitrary other boundary factors and coupons. It is dinatural in the two surgery variables: a morphism inserted on their paths passes crossings by [F1]'s naturality, twists by twist naturality and bends by its explicit inverse mate formulas; no coupon lies on a surgery component itself. Define $\mathrm{\Phi }:B\otimes B\otimes U\to V$ by its simple restrictions $\mathrm{\Phi }({\iota }_i\otimes {\iota }_j\otimes 1_U)=T_{i,j}$. step 1.1's finite decomposition proves $T_{X,Y}=\mathrm{\Phi }({\iota }_X\otimes {\iota }_Y\otimes 1_U)$ for all $X,Y$. This is the full exterior factorization, including arbitrary external coupons, not a claim that those coupons are natural as functions of their own graph colors.

For $a\in \mathrm{End}(k\otimes j^{\ast })$ define the literal composite $T_{kj}(a)=(1_{k^{\ast }}\otimes 1_k\otimes e_j)(1_{k^{\ast }}\otimes a\otimes 1_j)(b_k^R\otimes 1_{j^{\ast }}\otimes 1_j):j^{\ast }\otimes j\to k^{\ast }\otimes k$. It sends the identity to $b_k^R{e}_j$. Substituting $F,G$ gives $T_{kj}(F_{i,a}{G}_{i,a})=(u_{k,a}^{\ast }\otimes v^{k,a})(1_{j^{\ast }}\otimes b_i^R\otimes 1_j)$. To check every bend, first expand $F$: its $b_j$ and the final $e_j$ retract by the $j^{\ast }$ snake, leaving $v^{k,a}$ on $i\otimes j$. In the other half use $(1_{k^{\ast }}\otimes u_{k,a})b_k^R=(u_{k,a}^{\ast }\otimes 1_{i\otimes j})b_{i\otimes j}^R$, the dual-mate coevaluation identity. Expand $b_{i\otimes j}^R=(1_{j^{\ast }}\otimes b_i^R\otimes 1_j)b_j^R$; the $r_j$ in $G$ retracts the outer right snake. The remaining word is exactly the stated $u^{\ast },b_i^R,v$ composite, with domain $j^{\ast }\otimes j$ and target $k^{\ast }\otimes k$. Applying $T$ to step 2.2's resolution thus proves the complete morphism identity ${\sum }_{i,a}{d}_i(u_{k,a}^{\ast }\otimes v^{k,a})(1_{j^{\ast }}\otimes b_i^R\otimes 1_j)=d_k{b}_k^R{e}_j$.

Put $\mathrm{\Lambda }={\sum }_i{d}_i{\iota }_i{b}_i^R$. Braiding naturality at $b_i^R$ says $c_{j,i^{\ast }\otimes i}(1_j\otimes b_i^R)=b_i^R\otimes 1_j$. Hence the multiplication word gives $m({\iota }_j\otimes \mathrm{\Lambda })={\sum }_i{d}_i{\iota }_{i\otimes j}(1_{j^{\ast }}\otimes b_i^R\otimes 1_j)$. Expand ${\iota }_{i\otimes j}$ using step 1.1 and apply step 3.1 separately for each output $k$. The result is ${\sum }_k{d}_k{\iota }_k{b}_k^R{e}_j=\mathrm{\Lambda }{e}_j$. The ${\iota }_j$ are all summand inclusions, so this is the full identity $m(1_B\otimes \mathrm{\Lambda })=\mathrm{\Lambda }\epsilon$. step 2.1 consequently gives $\mathrm{\Gamma }(1_B\otimes \mathrm{\Lambda })=(1_B\otimes \mathrm{\Lambda }\epsilon )\mathrm{\Delta }=1_B\otimes \mathrm{\Lambda }$, and precomposing with $\mathrm{\Lambda }$ gives $\mathrm{\Gamma }(\mathrm{\Lambda }\otimes \mathrm{\Lambda })=\mathrm{\Lambda }\otimes \mathrm{\Lambda }$. This is a full-channel calculation and does not use a simple-meridian projector, modular inversion, Gauss sums or surgery normalization.

Inserting $\mathrm{\Lambda }$ at a component cut is exactly the finite formal coloring $\mathrm{\Omega }$: $b_i^R$ closes its two oriented cut endpoints and its coefficient is $d_i$. Reversing that component's orientation replaces $i$ by $i^{\ast }$; dual-mate compatibility of $j$, ribbon twist duality, and sphericality $d_i=d_{i^{\ast }}$ give the same evaluated graph after reindexing the sum. Thus no independent antipode or integral theorem is needed for orientation reversal. Moving a pair of $\mathrm{\Lambda }$ cuts past each other also does not change their value, since naturality gives $c_{B,B}(\mathrm{\Lambda }\otimes \mathrm{\Lambda })=\mathrm{\Lambda }\otimes \mathrm{\Lambda }$ from $c_{\mathbf{1},\mathbf{1}}=1$.

The complete endpoint word created by the framed band sum on those two cuts is $({\iota }_X\otimes {\iota }_{Y\otimes X})(1_{X^{\ast }}\otimes 1_X\otimes 1_{X^{\ast }}\otimes c_{X,Y^{\ast }\otimes Y})(1_{X^{\ast }}\otimes b_X\otimes 1_X\otimes 1_{Y^{\ast }}\otimes 1_Y)$, exactly $\mathrm{\Gamma }({\iota }_X\otimes {\iota }_Y)$. To read it geometrically, follow the rectangle of the attaching band: the slid strand turns around one end, making $b_X$ and its $X^{\ast }$ return; its new parallel follows the stationary strand in its specified framing; at the other end $X$ passes the ordered pair $Y^{\ast },Y$, giving $c_{X,Y^{\ast }\otimes Y}$ with its hexagon expansion. The two parallels have cable color $Y\otimes X$ and reversed dual cable $X^{\ast }\otimes Y^{\ast }$. The new parallel follows the entire stationary component outside the little band ball: applying the same exterior $\mathrm{\Phi }$ to ${\iota }_{Y\otimes X}$ accounts for that whole cable. Every exterior crossing cables by the hexagons and every exterior twist by balancing. Thus no false assertion that the global band sum is supported in the attaching-band ball is made. Its full evaluation is $\mathrm{\Phi }(\mathrm{\Gamma }(\mathrm{\Lambda }\otimes \mathrm{\Lambda })\otimes 1_U)$, equal to $\mathrm{\Phi }(\mathrm{\Lambda }\otimes \mathrm{\Lambda }\otimes 1_U)$ by step 4.1. The rectangle's framing is retained: any twist in it or in the stationary parallel is read on both sides by the same cable twist word, never discarded by an unframed move.

For an open strand keep its fixed color $Y$ and its entire factor $Y^{\ast }\otimes Y$, rather than imposing coend dinaturality in a coupon-dependent graph variable. The displayed operator ${\widehat{\mathrm{\Gamma }}}_Y$ first sends $Y^{\ast }\otimes Y$ to $(Y^{\ast }\otimes Y)\otimes (Y^{\ast }\otimes Y)$ by $1_{Y^{\ast }}\otimes b_Y\otimes 1_Y$, then sends only the second pair through ${\iota }_Y$ before multiplication with the surgery variable. step 4.1 implies $m({\iota }_Y\otimes \mathrm{\Lambda })=\mathrm{\Lambda }{e}_Y$, by step 1.1's finite decomposition if $Y$ is not simple. Substitution into this exact lifted word gives ${\widehat{\mathrm{\Gamma }}}_Y(1_{Y^{\ast }\otimes Y}\otimes \mathrm{\Lambda })=(1_{Y^{\ast }\otimes Y}\otimes \mathrm{\Lambda }{e}_Y)(1_{Y^{\ast }}\otimes b_Y\otimes 1_Y)=1_{Y^{\ast }\otimes Y}\otimes \mathrm{\Lambda }$ by the left snake. For simple $j$ the same identity also follows by applying the actual projection $B\to j^{\ast }\otimes j$ to $\mathrm{\Gamma }({\iota }_j\otimes \mathrm{\Lambda })={\iota }_j\otimes \mathrm{\Lambda }$. Hence it is an identity on the complete open-color factor, not only after a categorical trace.

Fix the actual colored graph and its coupons, and cut only a short interior segment of the strand to be slid. After bending that segment the exterior has type $(Y^{\ast }\otimes Y)\otimes B\otimes U\to V$. Its factorization through $B$ uses dinaturality only along the $\mathrm{\Omega }$ surgery component, exactly as in step 2.3; the first object $Y$ and every graph coupon remain fixed. The band reading of step 5.2 now uses ${\widehat{\mathrm{\Gamma }}}_Y$ on that fixed first factor, so step 5.3 equates its before/after composites with the entire fixed exterior. To make the finite-cut method fully explicit, write $1_Y={\sum }_{j,a}{u}_{j,a}{v}^{j,a}$ and insert that resolution on the short segment before bending. On each term absorb the actual $u_{j,a},v^{j,a}$ into its two adjacent exterior graph portions. Its local segment is now the fixed simple $j$ and step 5.3 applies to that term with those very coupons held constant. Summing restores exactly the original graph; basis independence is the matrix identity $\sum uv=1_Y$. No family of arbitrary coupons indexed by variable $X$ is moved through a universal dinaturality argument.

The bending used in step 6.1 is injective on the relevant open-graph Hom space: for $f:A\otimes Y\to C$, its mate is $(f\otimes 1_{Y^{\ast }})(1_A\otimes b_Y)$, with inverse $(1_C\otimes e_Y)(-\otimes 1_Y)$; their composites are the two snakes. Other incident orientations use $b_Y^R,r_Y$ with their inverse mates. Thus equality of the bent expressions proves equality of the original open-strand morphisms before any closure. If the segment is part of a more complicated coupon-bearing graph, all portions outside its short cut, including junction coupons, are precisely the fixed exterior of step 6.1, so the same equality proves their slide as well. Opposite slide directions read the equality backwards; other orientations and either ordering of the two participating components use step 5.1's explicit dual reindexing and cut interchange. This covers all framed colored-strand and $\mathrm{\Omega }$-component cases in the Statement.

All calculations are identities of finite categorical composites, and hence survive arbitrary exterior pre/post-composition and tensor factors. No $D$, Gauss sum, normalized surgery invariant or topological presentation theorem was used. In a surgery application a band slide preserves the component count and changes the linking matrix by $B\mapsto P^{t}BP$, with $P$ the integral elementary basis change adding the stationary attaching vector to the slid one, so signature is preserved even when singular; that observation adds no assumption to the ribbon identity just proved. Countable choice enters only through [F1]; every fusion basis, cut expansion and coend component sum is finite. The complete local ribbon claims are therefore proved independently of all later Gauss/FR/TQFT consumers.