Functoriality of even integral Khovanov homology up to sign

A local integral comparison category. Use oriented blue facets of label $1$ and red facets of label $2$; two blue facets meet a red facet along each seam, with the usual merge orientation. Decorations on blue facets are powers of a dot. One can define their closed evaluation by cutting at the circular seams and applying the two Frobenius algebras $B=\mathbb{Z}[X]\mathrm{/}(X^2)$, ${ϵ}_B(1)=0$, ${ϵ}_B(X)=1$, ${\mathrm{\Delta }}_B(1)=1\otimes X+X\otimes 1$, ${\mathrm{\Delta }}_B(X)=X\otimes X$, and $R=\mathbb{Z}$ with counit $-1$ and comultiplication $-1\otimes 1$. Contract a seam by ${ϵ}_B(x\bar{y})$, where the positive blue side is selected by right-hand cyclic order and $\bar{X}=-X$. Cut-independence is the unit, counit, associativity and Frobenius identities of these displayed algebras. Equivalently, this is the following direct specialization of the two-pigment evaluation. With roots $x_1,x_2$, put $d=x_1-x_2$, $h=x_1+x_2$, $t=-x_1{x}_2$, and use $B_h=\mathbb{Z}[h,t][X]\mathrm{/}(X^2-hX-t)$. Its primitive idempotents have counits ${\beta }_1=1\mathrm{/}d$, ${\beta }_2=-1\mathrm{/}d$, and conjugation interchanges them. A connected blue facet of genus $g$ with $b$ seam circles contributes ${\beta }_i^{1-g-b}$; a red facet contributes $(-1{)}^{1-g-b}$; the seam contributes ${\beta }_i$ for its positive blue pigment. Sum over the two colors, which must be different at each seam. This is a finite tensor contraction in $B_h$ and $R$, hence is integral before any localization. Define states of a closed web as foams from the empty web modulo the common radical of every closing evaluation; define all other morphisms by bending to a closed web and taking that radical. Gluing preserves the radical because every closing after gluing is a closing before gluing. These definitions provide a local category, with no assertion of full faithfulness for a previously defined homology-valued functor.

Reduction to an almost braid closure is relative. A finite one-to-one tangle diagram, possibly with additional closed components, can be braided while preserving a prescribed short crossing-free endpoint arc. Here is the relative finite braiding construction needed, rather than an invocation of an AC-stated existence theorem. Close that arc on the left, choose a generic height away from its small fixed rectangle, and subdivide the remaining finitely many projected arcs at extrema and between changes of over/under status. A backward arc can be replaced by two forward arcs extending past the diagram and the exterior return arc: lift its small triangular detour above the rest when it is over, or below when under, then straighten the two free legs to the height direction. Its attaching intervals stay fixed and its chosen side has no strand between the old and new arc, so sliding across this triangle is an isotopy of the embedded tangle. Subdivision makes every such detour contain at most one old crossing of a fixed status. The detours introduce only forward legs and are chosen away from the fixed endpoint rectangle. Repeating on the finite backward list gives a braid whose rightmost $n$ strands are closed, and whose remaining strand retains the prescribed open endpoints. This is an almost braid closure. All moves occur in compact balls away from the fixed arc; smoothing the finite corners and extending the local velocities gives a relative smooth isotopy and a finite Euclidean movie. By [F2], conjugating a sweep by this isotopy conjugates its projective map. Therefore it suffices to prove the sweep identity for an almost braid closure.

All actual rank-two foams are covered. In the positive-label raw foam category a facet of label at least three has no two-pigment coloring. Every closure of a foam containing it therefore evaluates to zero, so that foam is in the radical. Each allowed tetrahedral vertex has such a facet: its first positive-label model contains $a+b+c\ge 3$; its second has $a>c\ge 1$, $b\ge 1$, and contains $a+b\ge 3$. Thus every nonzero rank-two foam is vertex-free, with precisely the facets of step 1.1; no zero-facet removal assertion is required. To compare raw left-hand Robert–Wagner signs with the right-hand evaluation above, let $B_i$ be the blue pigment-$i$ surfaces, $R$ the red surface, $M$ the seam count, and $t_i$ the number of seams positive on blue pigment $i$ in right-hand order. Their two sign exponents are $\chi (B_1\cup R)\mathrm{/}2+t_2$ and $\chi (B_2\cup R)\mathrm{/}2+t_1$. Their difference modulo two is $\chi (B)\mathrm{/}2+\chi (R)+M=\chi (B)\mathrm{/}2$, since $\chi (R)\equiv M(\mathrm{m}\mathrm{o}\mathrm{d}2)$. Their common denominator is $d^{\chi (B)\mathrm{/}2}$. Consequently raw evaluation differs by $(-1{)}^{\chi (B)\mathrm{/}2}$. At $h=t=0$ a nonzero scalar evaluation with $D$ blue dots has $2D=\chi (B)$; replacing each raw blue dot by its negative cancels this sign. Positive-degree red decorations specialize to zero. Thus every closed evaluation, and hence every bent radical, agrees after this explicit dot normalization. In particular the normalized dotted sphere is $+1$; it is not identified with the unrenormalized raw dotted sphere, which is $-1$.

Pairing bases, without red compression. A closed web with $k$ underlying blue circles has exactly $2^k$ boundary pigment states: a pigment changes across every red vertex on its blue circle, and there are an even number of these vertices because blue orientation changes there. Localized closed evaluation factors through these states, so its pairing rank is at most $2^k$. For a pure-blue web, dotted disk cups and complementary dotted caps have a unimodular tensor-product pairing: the one-circle Gram matrix is $\left(\begin{array}{cc}0& 1\\ 1& h\end{array}\right)$, of determinant $-1$. Hence its universal module is exactly the free module with those $2^k$ cups. For a general web $W$, there is a foam $I_W$ with underlying blue cylinders from standardly oriented blue circles to $W$. To construct it, checkerboard-color the complement of the underlying blue cylinders; complete every wrongly oriented region boundary by a red copy, push its interior into that region and assign the opposite color in the new thin region. The boundary red arcs are completed in the same way. This finite construction uses only parallel copies in collars, so produces the prescribed boundary and no singular vertices. The reflected composite $I_W^{\ast }{I}_W$ commutes with dots: crossing a seam changes $X$ to $h-X$, and each blue cylinder crosses an even number of orientation changes. It is degree zero, so on the pure-blue module it is multiplication by an integer. For each fixed boundary pigment, its color contraction is a single nonzero term of absolute value one; hence that integer is $\pm 1$. Thus $I_W$ has a left inverse. Its cokernel is a direct summand of the universal module, is torsion-free because all its pairing evaluations lie in the domain $\mathbb{Z}[h,t]$, and has rank zero by the localized $2^k$ bound. It is zero, and $I_W$ is invertible. Bending now identifies every local morphism space with the corresponding blue cup module. In particular a $k$-arc pairing has basis degrees $2\mathrm{\mid }L\mathrm{\mid }$ in its degree-zero endomorphism calculation, so ${\mathrm{End}}^0=\mathbb{Z}1$. This argument does not compress a possibly knotted red surface in a ball.

The actual crossing and tensor conventions. Let $I$ be the two-strand pairing and $A_i$ the thick web at adjacent positions $i,i+1$. In the over-southwest-to-northeast source chart, $I$ is the A-smoothing and the underlying blue pairing of $A_i$ is the B-smoothing. Use auxiliary crossing factors $(A_i\{-1\}\to I)$ in degrees $(-1,0)$ and $(I\to A_i\{1\})$ in degrees $(0,1)$ for the two inverse physical braid generators. These are the two rank-two foam factors with all-parallel term in degree zero; the braces here are the formal shift convention, opposite to their printed $q$ exponent. Surface-time cochain duality, with intrinsic grading changed from $-\chi$ to $\chi$, gives $(I\to E_i\{1\})$ and $(E_i\{-1\}\to I)$, respectively. Adding the common writhe shifts $+1$ and $-1$ yields exactly $(I\{1\}\to E_i\{2\})$ and $(E_i\{-2\}\to I\{-1\})$, the normalized ordinary factors. For an arbitrary balanced boundary orientation, cut an annulus at its basepoint and apply the same region completion as in step 3.1 to blue radial curtains. Its red facets convert the standard alternating boundary orientation to the prescribed one, while its blue curtains and cochain degree remain unchanged. Reversing an arrow is handled by bending this same boundary curtain; the underlying smoothing and the differential are unchanged, and the normalization is still $\mathrm{\mid }v\mathrm{\mid }-n_{-}$ and $\mathrm{\mid }v\mathrm{\mid }+n_{+}-2n_{-}$. At a curl this is also checked directly by the two kink retractions in [F4]. A zip or unzip transports to a primitive signed saddle, since its bent minimal-degree morphism space has one cup basis vector and its two-pigment contraction has coefficient a unit. Square commutation makes these edge signs a cube cocycle, removed by vertex signs from a fixed vertex; all two-saddle composites here are nonzero, as their blue Frobenius values show. Two blue-curtain completion choices differ on each state by a dot-linear degree-zero unit of the pure-blue module, hence by $\pm 1$. After correcting the cube edges, connectedness of the cube forces these state signs to be one common sign. Therefore local insertion and the comparison of whole complexes are coherent up to one overall sign, which is the precise coherence required here. Finally prefix and suffix totalization conventions differ by the vertex sign $(-1{)}^{{\sum }_{i<j}{r}_i{r}_j}$ for actual factor degrees $r_i$. Across the $i$th differential its ratio is $(-1{)}^{{\sum }_{j\mathrm{\ne }i}{r}_j}$, exactly the difference of the two tensor signs. Thus this is a local comparison of actual complexes with their orders and duals, not only an equality of closed graded ranks.

An explicit finite verification model. The next matrices are defined independently in the local category just constructed. Their equations can be checked faithfully after adjoining the distinct pigments. Write a three-strand boundary coloring as $v\in \{0,1{\}}^3$, where the two pigment values are $0,1$ after scaling their difference. A word in $A=A_1$ and $C=A_2$ has basis the paths $(v_0,\dots ,v_m)$ such that at its $i$th-letter step the relevant two colors are distinct and $v_j$ is either $v_{j-1}$ or their interchange. With ${\delta }_i(v)=v_i-v_{i+1}$, the elementary operators have the following complete rules: ${\eta }_i:I\to A_i$ inserts the unchanged color path with coefficient ${\delta }_i(v)$; ${\epsilon }_i:A_i\to I$ deletes a step only when its two endpoints agree, with coefficient $1$; ${\mathrm{\Delta }}_i:A_i\to A_i{A}_i$ sums all allowed middle color orders with coefficient $1$; and $m_i:A_i{A}_i\to A_i$ removes the middle color $u$ with coefficient $1\mathrm{/}{\delta }_i(u)$. Indeed the circle birth is the sum of its two idempotents and its cap has values ${\beta }_i$; zip followed by unzip has the values $0,d,-d,0$ on equal/equal, first/second, second/first, and second/second colors. Normalizing unzip to $1$ gives exactly these rules. The unit and counit identities determine the displayed digon coefficients as well: ${\delta }_i(u){\beta }_i=1$. All four are actual integral foams; only their color coordinates use inverses of $d$. Put $U_i={\mathrm{\Delta }}_i{\eta }_i$ and $V_i={\epsilon }_i{m}_i$. Faithfulness of this color test follows from the defining radical: a morphism whose every pigment coefficient is zero pairs to zero with every closure. Torsion-freeness from step 3.1 then reflects equality back before localization.

Four complete third-move matrices. Products below compose rightmost first, juxtaposed operators act on the indicated tensor factors, and omitted entries are zero. At each three-crossing cube order vertices by masks $(0,1,2,4,3,5,6,7)$; hence its four consecutive terms have sizes $1,3,3,1$. A positive factor is $A_i\to I$ in degrees $(-1,0)$ and a negative factor is $I\to A_i$ in degrees $(0,1)$, with the prefix tensor signs. The four source/target sign patterns are $(+++;+++)$, $(--+;+--)$, $(++-;-++)$ and $(---;---)$ on the words $ACA,CAC$. The source and target term lists, respectively, are $(ACA;(CA,AA,AC);(A,C,A);I)$ and $(CAC;(AC,CC,CA);(C,A,C);I)$ in case I; $(A;(AA,CA,I);(ACA,A,C);AC)$ and $(C;(I,CA,CC);(A,C,CAC);AC)$ in case II; $(AC;(C,A,ACA);(I,CA,AA);A)$ and $(AC;(CAC,C,A);(CC,CA,I);C)$ in case III; and $(I;(A,C,A);(AC,AA,CA);ACA)$ and $(I;(C,A,C);(CA,CC,AC);CAC)$ in case IV. Their maps on the four consecutive terms are $0,M_1,M_2,1$ in I, $0,N_1,N_2,1$ in II, $1,P_1,P_2,0$ in III, and $1,Q_1,Q_2,0$ in IV, where $M_1=\left(\begin{array}{ccc}0& -(1_A{\eta }_C)m_A& 1_{AC}\\ -{\mathrm{\Delta }}_C(1_C{\epsilon }_A)& U_C{V}_A& 0\\ 1_{CA}& -({\eta }_C{1}_A)m_A& 0\end{array}\right),M_2=\left(\begin{array}{ccc}0& 1_C& 0\\ 1_A& 0& 1_A\\ 0& 0& 0\end{array}\right),$ $N_1=\left(\begin{array}{ccc}-V_A& 0& 1\\ -({\eta }_C{1}_A)m_A& 1_{CA}& 0\\ U_C{V}_A& -{\mathrm{\Delta }}_C(1_C{\epsilon }_A)& 0\end{array}\right),N_2=\left(\begin{array}{ccc}{m}_A(1_A{\epsilon }_C{1}_A)& 1_A& 0\\ 0& 0& 1_C\\ 0& 0& 0\end{array}\right),$ $P_1=\left(\begin{array}{ccc}-(1_C{\eta }_A{1}_C){\mathrm{\Delta }}_C& 0& 0\\ 1_C& 0& 0\\ 0& 1_A& 0\end{array}\right),P_2=\left(\begin{array}{ccc}{U}_C& -{\mathrm{\Delta }}_C(1_C{\epsilon }_A)& U_C{V}_A\\ 0& 1_{CA}& -({\eta }_C{1}_A)m_A\\ 1& 0& 0\end{array}\right),$ and $Q_1=\left(\begin{array}{ccc}0& 1_C& 0\\ 1_A& 0& 0\\ 0& 1_C& 0\end{array}\right),Q_2=\left(\begin{array}{ccc}0& -({\eta }_C{1}_A)m_A& 1_{CA}\\ -{\mathrm{\Delta }}_C({\epsilon }_A{1}_C)& U_C{V}_A& -{\mathrm{\Delta }}_C(1_C{\epsilon }_A)\\ 1_{AC}& 0& 0\end{array}\right)\mathrm{.}$ These are rank-two representatives of the four local slides: the moving strand goes from first top endpoint to third bottom endpoint, so its two crossings are the first two source factors and last two target factors. The remaining third source/first target crossing is the fixed internal crossing, positive in I–II and negative in III–IV. This specifies the physical variants without relying on an unlabelled picture or an omitted six-valent foam.

Check of the matrix equations and the two required features. Here is the full finite checking rule. Enumerate the eight input and eight output colorings, then the paths of each listed word by the two choices at each permitted step in step 4.2. Apply a matrix entry by successively replacing the local subpath by the stated operator's outputs, multiply coefficients, and add coincident paths. A differential at coordinate $r$ has coefficient $(-1{)}^{{\sum }_{l<r}(b_l-{\mathbf{1}}_{\text{factor }l\text{ positive}})}$ and applies $\epsilon$ or $\eta$ according as that factor is positive or negative. Direct substitution gives $d^{\mathrm{\prime }}F=Fd$ on all $72$ state-path columns in each case. For each nonzero component, the internal degree $(b_1-{\mathbf{1}}_{\text{internal positive}})$ in the target is at least $(b_3-{\mathbf{1}}_{\text{internal positive}})$ in the source; there are $80$ nonzero color coefficients in each case, all satisfying this inequality. Finally compare I with II and III with IV by complementing the first two source bits and last two target bits. Their path words are then identical. On pairs of states of equal internal degree their coefficients agree: $144$ coefficient tests in each comparison. This enumeration is also a short exact verification algorithm for all the asserted equations: every basic operator coefficient is $0,1$ or $-1$, sums use exact integers, and the only division is by a color difference $\pm 1$. For general difference $d$, every entry of $F$ has common power $d^{({\mathrm{\ell }}_{\mathrm{t}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{t}}-{\mathrm{\ell }}_{\mathrm{s}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{c}\mathrm{e}})\mathrm{/}2}$, where $\mathrm{\ell }$ counts thick letters; every coefficient in $d^{\mathrm{\prime }}F-Fd$ has common power $d^{({\mathrm{\ell }}_{\mathrm{t}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{t}}-{\mathrm{\ell }}_{\mathrm{s}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{c}\mathrm{e}}+1)\mathrm{/}2}$. Thus the checks at a unit difference prove the rational identities for arbitrary distinct pigments, and step 4.2 proves the integral foam identities. The complete executable rule, including all four matrices, is retained in the calculation record below. Our representatives are internally nondecreasing, the dual filtration direction to the convention in the cited sweep calculation; this is sufficient because both sweep endpoints have external degree zero.

These are actual local equivalences. The dotted pairing calculation in step 3.1 and the boundary-crossing inverse equivalences in [F4] identify the full degree-zero endomorphism ring of each local braid complex with $\mathbb{Z}$. Let $G$ be any of the known equivalences in [F4], transported through step 4.1, and let $F$ be one of step 5.1's chain maps. Then $G^{-1}F$ is an integer scalar in the homotopy category. To determine that scalar, give all three strands the same pigment. Every state with a thick letter vanishes, and only the parallel term remains, in degree zero. Its coefficient is a homotopy invariant and is multiplicative, since all other states are zero in this channel. The parallel coefficient of an integral equivalence and its inverse multiply to $1$, so each is $\pm 1$; the parallel coefficient of every $F$ above is $1$. Consequently $F=\pm G$ in the homotopy category and is itself an equivalence. This proves comparison with the fixed local maps without requiring an unverified equality between a cone representative and a chosen parameter value in a published family.

The first and second moves used by the sweep. In the normalized ordinary Frobenius algebra, let $u$ create the small circle and $e$ cap it. The four relevant kink maps are $u$, $(Y-X)u$, $e$, and $e(Y-X)$, where $X$ is the continuing-arc dot and $Y$ the small-circle dot. Indeed $v-\mathrm{\Delta }=(Y-X)u$ and $w-m=e(Y-X)$: on $1,X$, $v=2Yu$, $\mathrm{\Delta }(1)=1\otimes X+X\otimes 1$, $\mathrm{\Delta }(X)=X\otimes X$, and the reverse calculation is the same counit computation. These are precisely [F4]'s two kink retractions. In the auxiliary picture label the two fronts so $R{1}_{+}=u$, $R{1}_{-}=(Y-X)u$, $R{1}_{+}^{-1}=e(Y-X)$ and $R{1}_{-}^{-1}=e$; reversing all labels changes no conclusion. Their only nonzero components have external degree zero. For one pair of inverse external crossings, order the middle summands as $I,A_i{A}_i$. The two physical orders have inclusions $(1,-U_i{)}^T,(1,U_i{)}^T$ and projections $(1,V_i),(1,-V_i)$. The unit/counit rules in step 4.2 give the chain-map equations, and $V_i{U}_i=0$, so their projection-inclusion composites are $1$. Their homotopy equivalences are also supplied by the actual inverse-crossing retractions in [F4], or by the same scalar test as step 7.1. Thus a succession of $n$ such pairs is supported only on palindromic thick/parallel states, preserves external degree, and the two fronts differ on a palindrome with $2k$ thick letters by $(-1{)}^k$, for either the inclusion or the projection. This is one sign from each double-thick pair; the all-parallel component is always $1$.

The two sweep movies and their grading. For an almost closure of a braid word on $n+1$ strands, move the left-hand closing arc to the right first in front of and then behind the diagram. Reading the strand's track gives a kink, $n$ inverse-crossing pairs, one of the four slides in step 5.1 at each braid crossing, the inverse $n$ pairs, and the inverse kink. Their composite maps are ${\mathrm{s}\mathrm{w}}_{+}$ and ${\mathrm{s}\mathrm{w}}_{-}$ between the same two endpoint complexes, and the full sweep is ${\mathrm{s}\mathrm{w}}_{-}^{-1}{\mathrm{s}\mathrm{w}}_{+}$. Split cohomological degree into internal degree from the fixed braid crossings and external degree from its moving strand. The endpoint complexes have external degree zero. Steps 6.1 and 8.1 show that each middle slide is externally nonincreasing, while all kink and pair maps preserve external degree. Hence only the external-degree-preserving parts of the slides can contribute between the endpoints. By step 6.1 those parts agree for the two fronts. When a local three-factor order is inserted into the whole diagram, tensor permutations add their Koszul vertex signs. Corresponding parallel factors have degree zero and corresponding thick factors have degrees $1$ or $-1$, of the same parity on the two fronts. Therefore the permutation phases agree and do not alter the diagonal comparison.

Cancellation of the pair and kink discrepancies. The $n$ external crossings on the remote closing arcs lie outside every slide disk. Their thick/parallel bits are unchanged during the middle slides. Each nonzero pair inclusion and projection has a palindromic state by step 8.1; its half on those remote arcs therefore determines the same integer $k$ at entry and exit. The front/back discrepancies are $(-1{)}^k$ at both ends, whose product is $1$. After the pair and slide maps have been compared, the only difference is the kink factor $p=Y-X$ at the beginning and its factor $p^{\mathrm{\prime }}$ at the end. Its two dot positions can be carried on the two closing-arc pieces outside the braid box and outside every pair or slide disk. Dot multiplication there commutes with each constituent map, since the remaining factor is an identity curtain; thus the middle map $B$ satisfies $Bp=p^{\mathrm{\prime }}B$. It follows directly that $eBpu=e{p}^{\mathrm{\prime }}Bu$: the back sweep $eBpu$ equals the front sweep $e{p}^{\mathrm{\prime }}Bu$. Hence ${\mathrm{s}\mathrm{w}}_{+}={\mathrm{s}\mathrm{w}}_{-}$ in the auxiliary homotopy category, and the sweep is the identity there. Each replacement by a fixed local map in [F1] changes at most one overall scalar $\pm 1$ by step 7.1 and the analogous pair/kink comparison. Planar insertion preserves that scalar and the homotopies. Consequently the sweep for the actual normalized ordinary maps is $\pm 1$, with one sign on the whole complex, not signs varying with states or inputs.

By [F3], connect the two spherical movies by finitely many Euclidean movie relations, distant interchanges and sweeps. Each Euclidean relation or interchange preserves its map up to one sign by [F2]; each sweep does so by step 10.1. Composing these finitely many equations gives one overall sign on the full integral homology group. A birth, death or saddle contributes quantum degree $1,1,-1$ by [F1], so the composite has degree $b+d-s=\chi (\mathrm{\Sigma })$ and cohomological degree zero. Identities and concatenation are the identity map and composition for fixed movies; quotienting every morphism by its overall negative makes the resulting assignment a functor. This proves every assertion of the Statement under exactly [A1].