Khovanov homology and Jones categorification

The endpoint realizations and the common cap. Regard a planar diagram as a diagram on the projection sphere, and choose a small closed disk $U$ about its mark meeting the diagram in one crossing-free arc only. The complementary disk contains a proper interval diagram; its two endpoints can be put at any prescribed two boundary points by an orientation-preserving disk coordinate change, preserving the interval orientation. Realize its crossings at small distinct positive or negative heights in a thickening of that disk, and straighten short end collars. Round the thickening to a standard three-ball $B$. Closing the interval by a fixed unknotted arc in the complementary ball $B_{\ast }$ gives exactly the original diagram on the sphere: the two projected disks glue along their boundary and the cap is the single deleted arc. This operation changes no crossing, sign, or resolution; it replaces the marked circle in every resolution by one interval. Make these choices for both diagrams with the same $B,B_{\ast }$, endpoints and cap $J$. The cap and slightly longer knot collars may be chosen identical as parametrized embeddings at both ends. Indeed an orientation-preserving change of circle parametrization can prescribe its parametrization on an interval; it extends over the remaining interval by a strictly increasing smooth map. Such circle reparametrizations are isotopic to the identity, as seen by interpolating their increasing lifts $f:\mathbb{R}\to \mathbb{R}$, $f(x+1)=f(x)+1$, with the identity lift. Thus there is a smooth isotopy $F_t$ of the closed parametrized knots with these endpoint embeddings and this common source interval $J$. Reparametrize time to make it stationary near $0,1$. Here the existence of $F_t$ is exactly the hypothesis that the two closed diagrams represent the same oriented knot.

Trivialize a moving ball, including its endpoint charts. Along the track of the slightly enlarged source interval $J$, choose a smooth pair of normal vectors to the nonzero knot tangent, equal to the same standard pair near $t=0,1$. Such a pair exists on the rectangle $J\times I$: start with local transverse frames, extend them along the interval and then in time, and correct their terminal angle by a smooth rotation, equal to zero initially and to that angle terminally. The terminal angle has a smooth real lift on an interval. Regarding $S^3$ as the unit sphere in ${\mathbb{R}}^4$, use the explicit tube $(F_t(s)+u{n}_1(s,t)+v{n}_2(s,t))\mathrm{/}\sqrt{1+u^2+v^2}$, with the two normals tangent to $S^3$. Its derivative is injective along the zero section. For uniformly small transverse coordinates it is an embedded tube; compactness gives a common width and excludes intersections with the rest of the knot outside the enlarged interval. Rounded end caps give a smooth family of embedded three-balls $b_t:B_{\ast }\to S^3$ about the cap, intersecting the knot in the chosen cap and its standard end collars. At the two stationary ends this family is the same specified ball chart: $b_0=b_1$. Apply [F1] to $b_t$ and, if necessary, a slightly larger ball chart, to get $H_t$ with $H_t{b}_0=b_t$. Then $G_t=H_t^{-1}{F}_t$ fixes the cap and its collar throughout. Its restriction to the complementary source interval is a proper-arc isotopy in the fixed $B=S^3\setminus \mathrm{Int}{B}_{\ast }$. At $t=0$ it is the first tangle; at $t=1$ it is $h$ applied to the second, where $h=H_1^{-1}{\mathrm{\mid }}_B$ is a diffeomorphism equal to the identity on a boundary collar, because $b_1=b_0$ on the slightly larger ball. No identification of the terminal ball chart with an arbitrary initial chart has been assumed.

Correct the terminal diffeomorphism on the arc. Identify $B$ with the unit ball and choose the coordinate origin off the second compact arc $T$. The chart can have this property by a small interior change fixing its boundary collar. Since $h$ is the identity near the boundary, for $0<r\le 1$ the maps $h_r(x)=\{\begin{array}{ll}rh(x\mathrm{/}r),& \mathrm{\mid }x\mathrm{\mid }\le r,\\ x,& \mathrm{\mid }x\mathrm{\mid }\ge r\end{array}$ are smooth diffeomorphisms, agreeing smoothly at $\mathrm{\mid }x\mathrm{\mid }=r$, and equal to the identity on a fixed boundary collar. Put $h_0=1$. For $r$ less than the positive distance from the origin to $T$, $h_r{\mathrm{\mid }}_T$ is exactly the identity. Thus $(s,r)\mapsto h_r(T(s))$ is a smooth isotopy of proper arcs even at $r=0$, although the ambient Alexander formula need not be smooth at $(x,r)=(0,0)$. Each slice is embedded, its endpoints and collars are fixed, and its final slice is $h(T)$. Reverse this arc isotopy and concatenate it with step 2.1, using stationary time changes at the join. The resulting smooth arc isotopy runs from the first tangle to the second relative to the same boundary endpoints and collars. This argument uses no claim that the diffeomorphism group of a three-ball is smoothly contractible.

Projection and relative perturbations. Choose the standard ball model as a rounded thickening of the projection disk, with the two fixed endpoint collars radial in that disk and with constant height. The rest of the compact arc track is in the interior of the ball. Shorten the protected collars if needed. Compactness then separates their projected endpoint neighbourhoods from every other branch: off a slightly larger source collar the radial coordinate in the disk is uniformly bounded away from its boundary. On the remaining transition collars the projection is still a fixed regular embedded arc. Reparametrize the isotopy in time to be stationary near its two regular end diagrams. In the finite-dimensional family of [F2], replace the time cutoff $\lambda (t)$ by $\lambda (t)\rho (s)$, where $\rho$ vanishes on the protected source collars and is positive on their complement. Embed the compact source rectangle in Euclidean space; multiplying coordinate polynomials of degree at most $43$ by this cutoff independently prescribes jets through order three at any at most six distinct free source points. Explicitly, at points $X_i$, the polynomials ${\prod }_{j\mathrm{\ne }i}\mathrm{\mid }X-X_j{\mathrm{\mid }}^8$ times suitable cubic Taylor polynomials prescribe the required jets, since multiplication by the nonzero cutoff is invertible on jets. Keep the spatial height coordinate unchanged. Small parameters preserve slice embeddings by compact local immersion and distant-point separation, fix the time-end diagrams, and fix the protected endpoint collars. The collars meet no singular stratum, and the transition region is regular with no other projected branch, so absence of perturbations on the collars loses no needed transversality.

All singularities remain interior and finite. Apply [F2] to that family on the free source configurations. The equations $g_s=0$ have codimension two in $(s,t)$; the extra degeneracy $\det (g_{ss},g_{sss})=0$ has total codimension three and is avoided. Distinct-point equality has codimension two in $(s_1,s_2,t)$; equality plus tangent dependence has codimension three, giving isolated quadratic double tangencies with nonzero relative normal velocity. Triple equality has codimension four in $(s_1,s_2,s_3,t)$; adding tangent dependence or a vanishing derivative increases codimension and is avoided. Quadruple equality has codimension six in dimension five and is absent. Simultaneous distinct events have combined codimension larger than the shared-time configuration dimension; independent jets at up to six points exclude them. Countably many coordinate charts and AC${}_{\omega }$ allow a parameter outside the resulting countable null union. As in [F2], near a regular source point a projected coordinate is strictly monotone, excluding accumulation along deleted diagonals; near an ordinary cusp its Taylor form $(b{s}^2,e(t-t_0)s+c{s}^3)$, $bec\mathrm{\ne }0$, has a single local double arc with no accumulation of other event types. Elsewhere the relevant configurations are compact, and isolated transverse events are finite. There are no endpoint accumulations because the protected endpoint neighbourhoods are fixed and isolated, and no end-time accumulations because those diagrams are stationary and regular. These arguments prove the required relative version directly; they do not apply a closed-source statement to an interval without modification.

The relative movie. By [F3], each exceptional time of step 5.1 gives one oriented R1, R2 or R3 in an interior disk disjoint from the protected endpoint collars. Between events, continue crossings by the inverse function theorem and extend the motions of the intervening embedded edges by normal-coordinate cutoffs, as in [F2]. The cutoffs can vanish on the protected collars, since those arcs are stationary and isolated. Hence the intervening planar isotopies fix them too. A finite subdivision of time and composition produces the asserted finite relative movie. The construction starts with either pair of marks, including two different marks on one diagram, so it proves that case as well. AC${}_{\omega }$ has been used only in [F1], [F2] and their countable chart unions.

The quotient by the monic polynomial $X^2$ is the free abelian group with basis $1,X$. The displayed multiplication is the quotient-ring multiplication, so it is associative, commutative, and has unit $1$. On this basis, $ϵ$ and $\mathrm{\Delta }$ satisfy both counit identities; explicitly, $(ϵ\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(1)=1$, $(\mathrm{i}\mathrm{d}\otimes ϵ)\mathrm{\Delta }(1)=1$, and the same composites send $X$ to $X$.

Cocommutativity follows by swapping tensor factors in the two displayed values of $\mathrm{\Delta }$. For coassociativity, both $(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(1)$ and $(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })\mathrm{\Delta }(1)$ equal $1\otimes X\otimes X+X\otimes 1\otimes X+X\otimes X\otimes 1$, and both composites send $X$ to $X\otimes X\otimes X$. Hence $(\mathcal{A},m,\eta ,\mathrm{\Delta },ϵ)$ is a commutative Frobenius algebra once the Frobenius compatibility is checked.

On the ordered basis $1\otimes 1,1\otimes X,X\otimes 1,X\otimes X$ of $\mathcal{A}\otimes \mathcal{A}$, each of $(m\otimes \mathrm{i}\mathrm{d})(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })$, $\mathrm{\Delta }m$, and $(\mathrm{i}\mathrm{d}\otimes m)(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})$ has respective outputs $1\otimes X+X\otimes 1$, $X\otimes X$, $X\otimes X$, and $0$. Thus the Frobenius identities hold. The degree shifts in the Definition give each edge map total quantum degree $0$; the quotient-ring construction and all checks are choice-free.

Fix a state $v$ and two distinct coordinates $i<j$ with $v(c_i)=v(c_j)=0$. The two length-two paths from $v$ to $v+e_i+e_j$ induce equal unsigned maps: saddles on disjoint circles commute as tensor maps; when the two saddles meet one component, the three possible merge/split patterns give respectively associativity, coassociativity, or one of the two equal Frobenius composites in [F1]. These are all possibilities because each edge changes one smoothing by one saddle.

The sign on the path that changes $i$ then $j$ has exponent ${\sum }_{r<i}v(c_r)+{\sum }_{r<j}v(c_r)+1$, while the path that changes $j$ then $i$ has exponent ${\sum }_{r<j}v(c_r)+{\sum }_{r<i}v(c_r)$. Their signs are opposite, so the two equal unsigned composites from step 1.1 cancel in the sum defining the component of $d^2$ from $v$ to $v+e_i+e_j$. If there are fewer than two crossings, no length-two paths occur and $d^2$ is the empty sum; otherwise every term of $d^2$ lies in exactly one such pair. Hence $d^2=0$.

If an edge does not meet the basepoint circle, its map is the identity on the basepoint factor and hence commutes with $X_p$. If the edge merges the basepoint circle with another circle, associativity gives $m(Xa\otimes b)=Xm(a\otimes b)$. If it splits the basepoint circle, the Frobenius identity gives $\mathrm{\Delta }(Xa)=(X\otimes 1)\mathrm{\Delta }(a)=(1\otimes X)\mathrm{\Delta }(a)$; on $a=1,X$ these are respectively $X\otimes X$ and $0$ on all three sides. Thus every edge map commutes with $X_p$, so $\mathrm{im}(X_p)$ is a subcomplex.

By [F3], the unreduced differential already squares to zero, and by step 1.1 it preserves $\mathrm{im}(X_p)$; thus the induced quotient differential also squares to zero. In each state the based factor is $\mathcal{A}=\mathbb{Z}\{1,X\}$, and $X_p$ has image $\mathbb{Z}\{X\}$. Therefore the quotient factor is $\mathbb{Z}\{1\}$, with raw quantum degree $+1$; subtracting one gives reduced degree $0$.

After cutting the based circle, every state has one distinguished interval factor $M$. The quotient identification sends the surviving based label $1$ to $1\in M$. For a merge into that interval, $m(1\otimes 1)=1$ and $m(X\otimes 1)=0$ in the quotient, exactly the displayed $m_M$; for a split, $\mathrm{\Delta }(1)=1\otimes X+X\otimes 1$ and the term with $X$ on the distinguished interval is zero in $M$, leaving $X\otimes 1$, exactly ${\mathrm{\Delta }}_M(1)$. Edges away from the interval retain $m$ and $\mathrm{\Delta }$. Hence this statewise identification commutes with every edge and gives the stated based tangle cube isomorphism.

If $n=\mathrm{\mid }C(D)\mathrm{\mid }$, there are exactly $2^n$ states; each complete smoothing of a finite closed diagram has finitely many circles. Since $\mathcal{A}$ is free of rank two with basis $1,X$, each state module is finite-rank free, and the finite sum of these modules is finite-rank free. The procedure visits each of finitely many states and at most $n$ outgoing edges per state, so it terminates.

Each edge changes $\mathrm{\mid }v\mathrm{\mid }$ and $i$ by $+1$ and changes the algebraic degree by $-1$ through $m$ or $\mathrm{\Delta }$, so its total quantum-degree change is $+1-1=0$. By [F2] the returned differential squares to zero. The state set is finite, so only finitely many homological degrees occur; this is a bounded bigraded cochain complex.

Fix a quantum degree and tensor the finite free complex with $\mathbb{Q}$. For an integer matrix $d$, the kernel after tensoring is $(\ker d)\otimes \mathbb{Q}$: if $d(a\mathrm{/}r)=0$ with $a$ integral and $r\mathrm{\ne }0$, then $d(a)=0$ because the target is torsion-free. The image after tensoring is $\mathrm{im}(d)\otimes \mathbb{Q}$ by clearing denominators, so rationalized homology is $H^i(C)\otimes \mathbb{Q}$ and has dimension ${\mathrm{rank}}_{\mathbb{Z}}{H}^i(C)$. Rank-nullity in this bounded rational complex gives $\dim C^i=\dim H^i+\dim \mathrm{im}{d}^i+\dim \mathrm{im}{d}^{i-1}$; alternating over $i$ cancels the image dimensions. The same argument applies to the reduced complex, whose terms are finite-rank free.

On a state $v$, summing $q^{\sum \mathrm{deg}(a)}$ over its $k(v)$ independent labels gives $(q+q^{-1}{)}^{k(v)}$, so the unreduced chain Euler sum is ${\sum }_v(-1{)}^{\mathrm{\mid }v\mathrm{\mid }-n_{-}}{q}^{\mathrm{\mid }v\mathrm{\mid }+n_{+}-2n_{-}}(q+q^{-1}{)}^{k(v)}$. With $q=-A^{-2}$, one has $q+q^{-1}=\delta$ and, term by term, $(-1{)}^{\mathrm{\mid }v\mathrm{\mid }-n_{-}}{q}^{\mathrm{\mid }v\mathrm{\mid }+n_{+}-2n_{-}}{A}^{-(n-2\mathrm{\mid }v\mathrm{\mid })}=(-A^3{)}^{-w}$. The state contribution is consequently $(q+q^{-1})(-A^3{)}^{-w}{A}^{n-2\mathrm{\mid }v\mathrm{\mid }}{\delta }^{k(v)-1}$. For the reduced cube, the based label contributes degree zero by [F4], leaving $(q+q^{-1}{)}^{k(v)-1}$ in place of $(q+q^{-1}{)}^{k(v)}$, so its state contribution is $(-A^3{)}^{-w}{A}^{n-2\mathrm{\mid }v\mathrm{\mid }}{\delta }^{k(v)-1}$.

Summing the state contributions in step 2.1 gives the two displayed formulas for the chain Euler sums. Step 1.1 identifies those sums with the Euler characteristics of integral cohomology. When $D$ is the crossingless unknot, $n_{+}=n_{-}=0$ and $k=1$, so the unreduced value is $q+q^{-1}$ and the reduced value is $1$.

Meaning of every local map. Use rectangles with top boundary points $1,\dots ,n$ and bottom boundary points $1^{\mathrm{\prime }},\dots ,n^{\mathrm{\prime }}$, both listed left to right. Let $I$ join $a$ to $a^{\mathrm{\prime }}$. Let $E_i$ join top $i$ to $i+1$, bottom $i^{\mathrm{\prime }}$ to $(i+1{)}^{\mathrm{\prime }}$, and all other $a$ to $a^{\mathrm{\prime }}$. In a product $UV$, $U$ is above $V$ and equal middle labels are joined. Thus $E_i{E}_i$ is $E_i$ with one disjoint middle circle, and $E_1{E}_2{E}_1$ is boundary-fixing planar isotopic to $E_1$; likewise with $1,2$ interchanged. For the latter assertion, trace top $1,2$ to one cap, bottom $1^{\mathrm{\prime }},2^{\mathrm{\prime }}$ to one cup, and top $3$ to bottom $3^{\mathrm{\prime }}$ along the remaining arc; there is no closed component. Contracting the intervening bends gives the stated isotopy. Close any of these matchings by a fixed exterior resolution. Order the resulting circle factors by the least arc label on each circle, using a fixed total order on all local and exterior arcs; factor permutations have no Koszul signs. A dot on an arc means multiplication by $X$ on the circle containing that arc. A saddle replacing an occurrence of $I$ by $E_i$, or conversely, means precisely $m$ on the two ordered circle factors if they merge, and $\mathrm{\Delta }$ if they split, followed by the indicated factor permutation. All unmentioned factors have the identity map. This defines the maps below on every tensor basis $\{1,X{\}}^{\otimes k}$, even when several local arcs belong to the same global circle. In the based case use $M$ for the distinguished factor and [F2] for a saddle meeting it.

Circle coordinates and neck cutting. For a disjoint middle circle put its factor first and write $w=1\otimes w_1+X\otimes w_X$. If a saddle splits that circle from an arc carrying dot operator $x$, its map in these coordinates is $u\mapsto (xu,u)$; if it merges the circle into an arc carrying dot operator $y$, its map is $(w_1,w_X)\mapsto w_1+y{w}_X$. These follow from $\mathrm{\Delta }(a)=1\otimes Xa+X\otimes a$ and $m(a\otimes b)=ab$. Two successive opposite saddles between the same two local arcs equal the sum of a dot on each arc. If the arcs are on distinct circles, this is the identity $\mathrm{\Delta }m(a\otimes b)=Xa\otimes b+a\otimes Xb$: on $1\otimes 1,1\otimes X,X\otimes 1,X\otimes X$, both sides are $X\otimes 1+1\otimes X,X\otimes X,X\otimes X,0$. If they are on one circle it is $m\mathrm{\Delta }(a)=2Xa$. The same formulas with the distinguished factor reduced modulo $X$ prove the assertion in that case too. These are the exact tensor interpretations of the local neck-cutting identities, including when two dot positions are on the same circle.

An explicit cancellation rule. Suppose a differential between consecutive degrees has block matrix $\left(\begin{array}{cc}\varphi & \beta \\ \gamma & \delta \end{array}\right):A_0\oplus R\to B_0\oplus S$, with $\varphi$ an isomorphism. Set $L(a,r)=(a+{\varphi }^{-1}\beta r,r)$ and $T(b,s)=(b,s-\gamma {\varphi }^{-1}b)$. These invertible integral changes of coordinates give $Td{L}^{-1}=\mathrm{diag}(\varphi ,\delta -\gamma {\varphi }^{-1}\beta )$. Conjugate the adjacent differentials too. The equation $d^2=0$ and invertibility of $\varphi$ imply that the incoming differential has zero $A_0$ coordinate and the outgoing differential vanishes on $B_0$. Thus there is a direct summand $A_0\stackrel{\varphi }{\to }{B}_0$, contracted by $h{\mathrm{\mid }}_{B_0}={\varphi }^{-1}$, $h=0$ elsewhere. Inclusion and projection of the remaining coordinates, conjugated by $L,T$, are chain maps $i,p$ with $pi=1$ and $1-ip=dh+hd$. This supplies actual chain maps and a homotopy for every cancellation used below, rather than merely an acyclicity assertion. If the blocks are homogeneous, these maps have quantum degree zero and $h$ has cohomological degree $-1$.

RII: all four modules and signs. Take two strands and order the crossing with states $(I,E)$ first, the opposite crossing with states $(E,I)$ second. The four states $00,01,10,11$ are $U=E$, $V=I$, $W=E$ with a middle circle, and $Z=E$, respectively. The local differential is $d^0(u)=(su,xu,u),d^1(v,w_1,w_X)=tv-w_1-y{w}_X,$ where $s:E\to I$ and $t:I\to E$ are the two saddles, $x$ is the dot on the arc from which the middle circle splits, and $y$ is the dot on the arc into which it merges. Here $C^1=V\oplus W_1\oplus W_X$; the minus signs on the $10\to 11$ edge are exactly the cube sign for the second crossing. Step 2.1 gives $ts=x+y$, so these displayed differentials square to zero. Define $i(v)=(v,tv,0),p(v,w_1,w_X)=v-s{w}_X,$ with $i,p$ zero in other local degrees, and define $h^1(v,w_1,w_X)=w_X,h^2(z)=(0,-z,0)\mathrm{.}$ Then $di=0$, $pd=0$, and $pi=1$. In degree zero $hd=1$, in degree two $dh=1$, and in degree one $dh+hd$ sends $(v,w_1,w_X)$ to $(s{w}_X,w_1-tv+(x+y)w_X,w_X)$, which is exactly $1-ip$. The surviving $V$ has local degree one and quantum grading increased by one; in the shift conventions of the question this is $[-1]\{-1\}$. Thus both acyclic pairs and the signed quotient are explicitly identified. The other crossing order follows by swapping the two cube coordinates with state sign $(-1{)}^{v_1{v}_2}$.

RIII: ordered states and the first cancellation. Let each crossing have states $(I,E_i)$, and form the word $121$ from top to bottom, with crossings ordered in that order. Write $P=E_1{E}_2$ and $Q=E_2{E}_1$. In degrees $0,1,2,3$ the ordered states are $I;(E_1^{100},E_2^{010},E_1^{001});(P^{110},(E_1\bigsqcup ◯{)}^{101},Q^{011});E_1^{111}\mathrm{.}$ In the last term use the boundary-fixing isotopy $E_1{E}_2{E}_1\cong E_1$ of step 1.1. Denote a degree-one element by $(a,b,c)$ and the circle coordinates in state $101$ by $(w_1,w_X)$. Introduce four precisely located saddles: $u:E_1\to P$ inserts $E_2$ below $E_1$, $v:E_2\to P$ inserts $E_1$ above $E_2$, $w:E_1\to Q$ inserts $E_2$ above $E_1$, and $z:E_2\to Q$ inserts $E_1$ below $E_2$. Let $x$ and $y$ be the dot operators on the bottom cup and top cap of $E_1$, respectively. The degree-one differential is $d^1(a,b,c)=(-ua+vb,-xa+yc,-a+c,wc-zb),$ with output coordinates $(P,W_1,W_X,Q)$. Indeed the six cube edges are $100\to 110$ with sign $-$, $010\to 110$ with sign $+$, $100\to 101$ with sign $-$, $001\to 101$ with sign $+$, $010\to 011$ with sign $-$, and $001\to 011$ with sign $+$. The two edges to $101$ split the middle circle from the bottom cup or top cap, giving exactly the two displayed circle-coordinate pairs by step 2.1. The $101\to 111$ edge has sign $-$ and merges that middle circle into the remaining through arc of $E_1{E}_2{E}_1$. Its $W_1\to E_1^{111}$ component is therefore $-1$, with its $W_X$ component minus the dot on that through arc. Cancel this $-1$ component by step 2.2. There are no remaining degree-three terms; the differential into $(P,W_X,Q)$ is the corresponding projection of the displayed $d^1$.

The second cancellation and the common residual matrix. In the resulting complex, the $c\to W_X$ component is $+1$. Cancel it by step 2.2: its row is $-a+c$, so the residual coordinate substitution is $c=a$. The retained degree-one coordinates are $(a,b)$, the retained degree-two coordinates are $(P,Q)$, and the differentials become $I\stackrel{\left(\genfrac{}{}{0px}{}{s_1}{s_2}\right)}{\to }{E}_1\oplus E_2\stackrel{\left(\begin{array}{cc}-u& v\\ w& -z\end{array}\right)}{\to }P\oplus Q,$ where $s_i:I\to E_i$ is the saddle inserting $E_i$. The first map is obtained from the original $d^0=(s_1,s_2,s_1)$ by deleting the canceled coordinate; it is compatible with $c=a$. Repeat both cancellations for $212$ with $1$ and $2$ interchanged. Its retained coordinates before reordering are $(E_2,E_1)$ and $(Q,P)$, with matrix $\left(\begin{array}{cc}-z& w\\ v& -u\end{array}\right)$. Swapping both pairs, without a sign, gives exactly the preceding matrix and the preceding $d^0$. The four entries refer to the same saddles on the same labelled boundary matchings: for instance $E_2\to P$ inserts $E_1$ above, and $E_1\to P$ inserts $E_2$ below, on both sides. Hence these are actual equal tensor maps by step 1.1, not identifications of unlabelled spaces. Their kernels, composites and signs need no further $\tau$ identities. Composing the two explicit cancellation maps and their inverses gives the RIII equivalence; the homotopy for a composite of retractions $(i_1,p_1,h_1)$ and $(i_2,p_2,h_2)$ is $h_1+i_1{h}_2{p}_1$.

Quantum degrees. A state of local height $r$ has quantum degree increased by $r$. In RII, $W_1$ has the underlying grading of $E$ increased by two, $W_X$ has the grading of $E$, and $Z$ has the grading of $E$ increased by two. Thus both canceled isomorphisms and the maps of step 3.1 preserve quantum degree. In RIII, $W_1^{101}$ and $E_1^{111}$ both have the grading of $E_1$ increased by three; $W_X^{101}$ and $E_1^{001}$ both have the grading of $E_1$ increased by one. The common residual complex has quantum increases $0,1,2$ in its three degrees. Every change of coordinates in step 2.2 is therefore homogeneous.

Exterior differentials and the marked quotient. Dots commute with every exterior saddle: $m(Xa,b)=Xm(a,b)=m(a,Xb)$ and $\mathrm{\Delta }(Xa)=(X\otimes 1)\mathrm{\Delta }(a)=(1\otimes X)\mathrm{\Delta }(a)$. Two saddles in disjoint local disks commute before cube signs, including when their participating arcs lie on a common global circle, by associativity, coassociativity and the Frobenius identity in [F1]. A middle circle being delooped is contained in the local disk, so no exterior saddle meets it. Boundary-fixing isotopies of the matchings preserve the labelled exterior arcs. All matrix changes, inverses of $\pm 1$ blocks, projections, inclusions and contractions above therefore commute with unsigned exterior saddles and with $X$ at a fixed exterior marked point. The total differential, with local crossings first, is $d_{\mathrm{l}\mathrm{o}\mathrm{c}}+(-1{)}^r{d}_{\mathrm{e}\mathrm{x}\mathrm{t}}$ on local degree $r$. A local homotopy lowers $r$ by one, so its two exterior terms in $dh+hd$ cancel; local degree-zero maps commute with the signed total differential. These identities prove the same retractions for the total cube. The identities descend modulo $\mathrm{im}{X}_p$ because every map and every homotopy commutes with $X_p$; this argument does not assume that every local saddle acts trivially on the distinguished interval. It proves all assertions, integrally and choice-free.