The Khovanov to Floer spectral sequence

The filtration has only the levels $0,\dots ,n$, so it is finite, exhaustive and separated; [F1] says each level is a subcomplex.

Apply [F2] to this bounded filtered complex. Its pages stabilize and the limiting page is the associated graded of $H(C)$ with the filtration induced by the inclusions $F_{p}C\hookrightarrow C$.

This argument identifies the abutment for the fixed filtered complex only. No map between filtered complexes for two different link diagrams occurs in [F1] or [F2], so diagram-independent filtered quasi-isomorphism type is not a consequence of this convergence result.

Compose the Hurewicz isomorphism in [F2], the Alexander-duality isomorphism in [F1], and coordinate-sum modulo two on ${\mathbb{Z}}^r$. The local intersection description in [F1] sends a meridian of $K_i$ to $\pm e_i$, whose coordinate sum is $1$ modulo two. This defines ${ϵ}_L$ and proves it is surjective because $r\ge 1$; hence its kernel has index two.

An ambient isotopy from $L$ to $L^{\mathrm{\prime }}$ restricts to an orientation-preserving diffeomorphism of their complements and carries each meridian class to a meridian class. It therefore carries $\ker {ϵ}_L$ to $\ker {ϵ}_{L^{\mathrm{\prime }}}$.

By [F3] and [F4], the index-two kernel determines a connected two-sheeted cover. On a punctured tube around $K_i$, its two monodromies are the meridian value $1$ and a longitude value ${\eta }_i\in {\mathbb{F}}_2$. The local model [F5] has exactly these monodromies, so covering classification identifies its punctured part with the restricted cover. Attach its core to obtain a smooth extension in the chosen charts. This constructs an extension; independence of the smooth total-space diffeomorphism type requires the normalization below, not merely equality of boundary monodromies.

Compare two oriented tubular charts ${\phi }_0,{\phi }_1$ around the same component. Express them over a common oriented core parameter $s$; a change of the original core parameter only replaces the unit phase $e^{\mathrm{i}{\eta }_{j}s}$ by a smooth unit loop $a_j(s)$ of the same winding number ${\eta }_j$. Let $A(s)\in G{L}^{+}(2,\mathbb{R})$ be the unique vertical linear change for which ${\phi }_1\circ A$ and ${\phi }_0$ induce the same normal-quotient identification. Write $A=RP$ by smooth polar decomposition, with $R(s)\in SO(2)$ and $P(s)$ positive definite. The compatible-tube assertion [F6] gives an ambient isotopy fixing the core and carrying ${\phi }_0$ to ${\phi }_1\circ A$ near it. The fiber-linear isotopy $R((1-t)P+tI)$ then connects $A$ to $R$ through invertible maps. On a common smaller tube its composition with ${\phi }_1$ is an isotopy of open embeddings, with open track, fixing the core. The relative open-set clause of [F6] extends it near the compact core. Thus an ambient diffeomorphism $g$, isotopic to the identity through maps fixing the whole link, satisfies $g\circ {\phi }_0={\phi }_1\circ R$ near this component. Choose the supports in disjoint component tubes and compose the finitely many isotopies.

Identify $R(s)$ with a unit complex number, of winding $k$. The image under $g$ of the first chart longitude has slope ${\lambda }_1+k\mu$ in the second chart, so preservation of the meridian character gives ${\eta }_0\equiv {\eta }_1+k(\mathrm{m}\mathrm{o}\mathrm{d}2)$. The unit loop $R(s)a_0(s)\mathrm{/}{a}_1(s)$ therefore has even winding. It has a smooth periodic square root $r(s)$: lift its argument on $[0,2\pi ]$, halve it, and use even winding to match the endpoints and derivatives. In the two branch models $b_j(s,z)=(s,a_j(s)z^2)$ the map $(s,z)\mapsto (s,r(s)z)$ consequently covers $g$, since $a_1{r}^2=R{a}_0$. It is an orientation-preserving smooth diffeomorphism including $z=0$. The two signs of $r$ differ by the deck map. Choose the sign matching the complement-cover isomorphism at one point of each connected punctured tube; covering uniqueness makes them agree throughout that tube. Hence the local lifts glue to the complement lift. This proves independence of the oriented smooth total-space diffeomorphism type from arbitrary oriented tubular charts. It does not assert a smooth lift over the identity base map between arbitrary unnormalized charts.

Let $F$ be the terminal diffeomorphism of the given ambient isotopy from $L$ to $L^{\mathrm{\prime }}$. First choose the target tubular charts by transporting the source charts through $F$. Corresponding longitudes have the same character, and in these transported branch coordinates a lift is $(s,z)\mapsto (s,z)$, or its deck translate. By step 1.2 and [F4] the complement covers are isomorphic over $F$; on every punctured tube that isomorphism is one of these two local lifts, so it extends smoothly across the branch core. The resulting map is an orientation-preserving diffeomorphism of the closed total spaces. Finally steps 3.1–4.1 compare the transported target charts with any originally chosen target charts. Thus $\mathrm{\Sigma }(L^{\mathrm{\prime }})\cong \mathrm{\Sigma }(L)$ as oriented manifolds, with no unjustified smooth extension of an arbitrary normal germ.

Regard $D^2$ as the unit disk in $\mathbb{C}$. On the Riemann sphere define $q(w)=(a_2{w}^2+a_1)\mathrm{/}(w^2+1)$, with its usual values at the poles and at infinity. This is a degree-two map branched precisely over $a_1,a_2$: $q(w)-a_1=(a_2-a_1)w^2\mathrm{/}(1+w^2)$ near $w=0$, and $q(1\mathrm{/}v)-a_2=(a_1-a_2)v^2\mathrm{/}(1+v^2)$ near $v=0$. The nonzero factors have local square roots, giving exactly the square charts of [F2]. Thus $S=q^{-1}(D^2)$ is a compact oriented surface with smooth boundary and a simple branched map to $D^2$. Its punctured restriction is connected: lifting a meridian about either branch point interchanges the two sheets. By [F1] it is the prescribed disk cover, since the two puncture meridians generate its fundamental group. This constructs the extension rather than assuming it from a link-only definition.

Triangulate the disk with both branch points as vertices. A fan triangulation of an $m$-gon has $V=m$, $E=2m-3$, $F=m-2$ and Euler characteristic $1$; inserting interior vertices preserves this count. Each open cell lifts twice except the two branch vertices, which lift once. Consequently $\chi (S)=2\cdot 1-2=0$. The boundary monodromy is $1+1=0$, so there are two boundary circles. Capping gives a closed connected orientable surface with Euler characteristic $2$, hence a sphere by [F3]. To apply [F4] with smooth circles in the standard sphere, use the explicit model of step 1.1: the closed exterior disk $\widehat{\mathbb{C}}\setminus \mathrm{int}{D}^2$ contains no branch value and is simply connected, so its inverse image consists of two disks by [F1]. Their boundaries are disjoint smooth circles, because $q$ is unramified over $\mathrm{\partial }{D}^2$. Thus $S$ is exactly the closed region between these circles in the standard Riemann sphere, and [F4] identifies it as an annulus.

Take $q{\mathrm{\mid }}_S\times {\mathrm{i}\mathrm{d}}_{[0,1]}$. Its local models along the two arcs, including their boundary endpoints, are $(z,t)\mapsto (z^2,t)$ by step 1.1. Away from the arcs it is the connected product cover with the required meridians. Projection of the punctured product to its disk factor induces an isomorphism on fundamental groups, so [F1] identifies this restriction with the specified cover of $B\setminus A$. Transporting the constructed extension along that covering isomorphism proves the required extension. Its total space is $S\times [0,1]\cong S^1\times [0,1{]}^2$, a solid torus after the usual rounding of product corners.

An explicit oriented branched annulus. Fix $c>0$ and put $A=(\mathbb{R}\mathrm{/}2\pi \mathbb{Z})\times [-c,c]$, oriented by $d\theta \wedge dt$. Its complex coordinate is $z=e^{-t+i\theta }$, whose complex orientation is this same orientation. The holomorphic map $q(z)=z+z^{-1}=(2\cosh t\cos \theta ,-2\sinh t\sin \theta )$ identifies $A$ with a two-fold branched cover of the closed ellipse $D_c$ having semiaxes $2\cosh c,2\sinh c$. Its deck map is $\mathrm{ȷ}(\theta ,t)=(-\theta ,-t)$, and its two fixed points map to $2,-2$. To verify its image and degree, every point outside the real segment $[-2,2]$ lies on a unique confocal ellipse: the expression $x^2\mathrm{/}(4{\cosh }^{2}t)+y^2\mathrm{/}(4{\sinh }^{2}t)$ decreases strictly from a value greater than one to at most one as $t$ increases from zero to $c$. Its two preimages have coordinates $(\theta ,t)$ and $(-\theta ,-t)$. Points inside that segment have the two preimages on $t=0$, except its endpoints. The derivative $1-z^{-2}$ vanishes precisely at $z=\pm 1$; locally $q(z)-2=(z-1{)}^2\mathrm{/}z$ and $q(z)+2=(z+1{)}^2\mathrm{/}z$, so each is a simple square branch in smooth complex coordinates. Both boundary components map diffeomorphically to the ellipse boundary.

A commuting twist. Choose a smooth nondecreasing $\rho :[-c,c]\to [0,1]$, zero near $-c$, one near $c$, equal to $1\mathrm{/}2$ on $[-ϵ,ϵ]$ for some $0<ϵ<c$, and satisfying $\rho (-t)=1-\rho (t)$. For example choose its upper half from a smooth step and reflect. Set $T(\theta ,t)=(\theta +2\pi \rho (t),t)$. The symmetry makes $T\mathrm{ȷ}=\mathrm{ȷ}T$ modulo $2\pi$, since $2\pi \rho (-t)=2\pi -2\pi \rho (t)$. It therefore descends to a smooth disk map $h$ fixing the ellipse boundary. Near either branch it is $q\mapsto -q$, because $T$ adds $\pi$ in the central strip, so descent is smooth there as well. In particular $h$ interchanges the branch points, and $T$ fixes both annular boundaries pointwise. By [F1], $T$ is exactly a positive annular twist. It remains to prove the handedness of its quotient, rather than infer it merely from the interchange.

An orientation-preserving disk coordinate. On the quotient of $t\ge ϵ$ write $\phi =-\theta$. Then $q=(2\cosh t\cos \phi ,2\sinh t\sin \phi )$. Choose $r_0\in (0,1)$ and a smooth increasing $t(r)$ from $ϵ$ to $c$ on $[r_0,1]$. Define $a(r)=2\cosh t(r)$ and $b(r)=2\sinh t(r)$ there. Extend $a,b$ to $[0,1]$ so $a(0)=b(0)=0$, both derivatives are strictly positive, and both are linear near zero. Here is a finite smooth construction: retain their positive derivative germs near $r_0$, on an interval so short that each derivative integral is less than half its prescribed value at $r_0$. Extend each derivative positively and constantly near zero, initially with total integral smaller than that value. Add a nonnegative bump in the intervening interval, with its unique scalar chosen to make the integral equal to that value. Integrating now matches every derivative and the value at $r_0$. Thus $F(r,\phi )=(a(r)\cos \phi ,b(r)\sin \phi )$ is a smooth odd diffeomorphism from the round disk to $D_c$. Smoothness and invertibility at zero follow from its positive diagonal linear formula there. Away from zero its Jacobian is $a^{\mathrm{\prime }}b{\cos }^2\phi +a{b}^{\mathrm{\prime }}{\sin }^2\phi >0$. The ellipses are strictly nested, so the equation $x^2\mathrm{/}a(r{)}^2+y^2\mathrm{/}b(r{)}^2=1$ proves global injectivity and surjectivity. Thus these are actual orientation-preserving disk coordinates, with both branch points inside $r<r_0$.

The quotient is the positive half-twist, with its sign. On the central ellipse $h(w)=-w$, and oddness of $F$ makes this the polar rotation $\phi \mapsto \phi +\pi$. On an outer confocal ellipse, $T$ changes $\phi$ by $-2\pi \rho (t)$, which modulo $2\pi$ is the angle $2\pi (1-\rho (t))$. Consequently in the single disk coordinate of step 2.2, $F^{-1}hF(r,\phi )=(r,\phi +\pi \chi (r)),$ where $\chi =1$ for $r\le r_0$ and $\chi (r)=2(1-\rho (t(r)))$ outside. This is smooth, one near the branch points, and zero near the boundary. The isotopy replacing the angle by $s\pi \chi (r)$ moves the two points counterclockwise through $+\pi$ in an orientation-preserving coordinate. Thus it is the standard positive half-twist, not its inverse. This explicitly fixes the relative sign with the positive annular twist in [F1].

Transport to the given cover and lift uniqueness. Apply the preceding model in an oriented coordinate disk supporting the given half-twist. Its branched cover is identified with the model without a covering-classification choice axiom: slit the disk along two disjoint intervals joining the branch points to its outer boundary. Over the resulting simply connected slit disk the two sheets are obtained by unique path lifting from one specified sheet point; endpoints are independent of path by homotopy lifting in [F2]. Crosswise gluing of the two slit banks is forced by the nontrivial monodromy of a small branch meridian, as the local square map shows. This is exactly the gluing of the model. The identifications are smooth locally. At the slit endpoints use the compatible base and source square charts of the chosen representative; the lifted identifications are then linear in the local square-root coordinates and fill smoothly. Outside the support disk its two exterior sheet collars are retained. The two possible lifts differ by $\mathrm{ȷ}$; this deck map interchanges the two boundary components, while $T$ fixes them. Thus precisely one lift fixes the annular boundary pointwise, and it is the transported $T$. Only one finite initial sheet choice was made.

Representatives and orientations. Any change of smooth half-twist cutoff is an isotopy through the displayed angle functions, constant near the two points and the outer boundary. Lift it by [F2], using the boundary-fixed sheet. On the complement of the branch points, ordinary smooth local covering coordinates make the lift smooth. Near the branch points the isotopy is the fixed half-turn representative; it therefore extends in the square coordinates and is stationary there. More generally the same argument transports a supplied supported coordinate isotopy defining another standard representative of this half-twist class, chosen standard in the branch charts throughout. It gives a smooth relative-boundary isotopy of lifts. If the annular orientation is reversed, change $\theta$ to $-\theta$ while retaining $t$; the actual formula for $T$ becomes $(\theta ,t)\mapsto (\theta -2\pi \rho (t),t)$. It is the inverse twist in the positively oriented annular convention. Conversely reversing the half-twist motion reverses this angle. These calculations prove the full orientation and standard-representative assertions and also explain why the other, boundary-swapping lift was excluded. The assertion concerns this smooth mapping-class lift; it does not assert that an arbitrary unnormalized smooth base germ has a smooth square-root lift.

Parametrize the splitting torus by ${\mathbb{R}}^2\mathrm{/}{\mathbb{Z}}^2$ with ${\lambda }_U=(1,0)$ and ${\mu }_U=(0,1)$. Thus the two compression meridians are $(1,0)$ and $(1,3)$ by [F3]. They give exactly the positive-sign diagram in [F1]. Reversing the oriented splitting gives the negative sign diagram; equivalently conjugate the surface coordinate, retaining the same attaching sets. Both signs have three intersections and zero normalized periodic lattice. They therefore satisfy the corrected full-all-sector admissibility condition, including every empty sector; admissibility is not being inferred from a finite set of realized generators.

Take actual allowed generic hat data for either of these literal diagrams. This need not be the flat equation: its prescribed generator charts may be retained. A nonconstant strip would compactify to a relative disk. Distinct endpoints are impossible by [F1]. For equal endpoints the explicit plane contraction of [F1] is a boundary-preserving relative homotopy to the constant map. For each region point q its degree is invariant under this homotopy: the boundary never meets q, so the boundary winding of its small disk, or equivalently the pullback divisor degree, is constant. Every region degree is consequently zero. The ACTUAL positivity and energy inequality in [F2] force zero energy, hence a constant strip. Thus there is no index-one strip for any of these allowed data. At a constant generator its operator is the exact transverse operator of real index zero, as in [F1]; it does not enter the index-one count. This argument does not assume that merely seeing quadrilateral regions excludes arbitrary multiply covered disk domains.

The generator module is therefore ${\mathbb{F}}_2^3$ with zero differential. Its homology has total rank three. The actual closed-manifold invariance in [F3] identifies this diagram computation with $\widehat{HF}(L(3,1);{\mathbb{F}}_2)$; the negative-sign calculation proves the same for its reversed orientation. No individual Spin-c grading, absolute normalization or correction term has been inferred from this total rank calculation.

The cube module and its increasing filtration are fixed by [F1]. The candidate total differential sums over ordered immediate-successor paths and their polygon-count maps, with diagonal disk differential. Each component from $I$ to $J\ge I$ lowers $n-\mathrm{\mid }I\mathrm{\mid }$ to $n-\mathrm{\mid }J\mathrm{\mid }$, so it preserves every increasing filtration subspace whenever the count is defined. Square-zero, count finiteness and identification of total homology still require proof. No filtered comparison between different presentations is a premise of this fixed-data assertion.

The square-zero substitution has a precise additional finite hypothesis. Denote polygon operations by ${\mu }_k$ and the chosen immediate top inputs by ${\theta }_{IJ}$. For $A<B$, form the finite top-input sum $M_{AB}={\sum }_{A=I_0<\cdots <I_r=B}{\mu }_r({\theta }_{I_0{I}_1},\dots ,{\theta }_{I_{r-1}{I}_r}),$ where every step changes one coordinate; ${\mu }_1$ is the differential on a single top input. Suppose the actual surgery data satisfy $M_{AB}=0$ for every such interval, in their actual auxiliary complexes and compatible labels. Expand the component $(D^2{)}_{IJ}$ as the sum over two consecutive input-containing polygon operations. The boundary identity of [F2], summed over all paths from $I$ to $J$, pairs these with exactly the terms obtained by inserting one $M_{AB}$ among the top inputs of an input-containing polygon. Every consecutive subpath is counted once in this insertion sum, including length-one differential terms. Thus $M_{AB}=0$ implies $D^2=0$. For $I=J$ this is the vertex differential identity. This is a finite substitution in the actual polygon identities; it does not calculate the top-input sums.

There are only finitely many vertex generators, coordinate paths and required arities. Once the actual admissible subordinate data and top inputs are supplied, [F2] gives finite counts for each term and hence for every differential coefficient. The filtration calculation in step 1.1 then applies literally. In the $n=1$ case, the matrix has diagonal vertex differentials and sole off-diagonal triangle map $f_{01}$; the resulting two-summand complex is its mapping cone with the usual homological shift. This is an actual algebraic identification whenever those data exist, not yet a comparison with the third filling. For $n=0$ it is exactly the given hat complex of $Y$.

Apply [F4] to the actual surgery data: its top cycles and vanishing $M_{AB}$ supply exactly the extra hypothesis of step 2.1. Together with the retained local count identities [F2], steps 1.1, 2.1, 3.1 give an actual finite differential preserving the stated filtration. In particular the branched-cover definition later uses a genuine complex, not just a candidate module.

For one coordinate, [F5] supplies the actual comparison of this cone with the third filling, so the cone's short exact sequence gives the surgery long exact sequence. The cyclic triad ordering is the one with slope labels $\lambda ,\mu +\lambda ,\mu$, as fixed in the Statement; the source's Section2, p.6, describes its cyclic symmetry. For zero coordinates the identity complex gives the asserted result.

For several coordinates use [F6], rather than composing unspecified vertexwise comparison maps. Filter the enlarged $\{0,1{\}}^{n-1}\times \{0,1,\mathrm{\infty }\}$ complex by the first $n-1$ coordinates. Its successive quotients are the acyclic single-coordinate three-term complexes from [F5], including the quadrilateral entry. A finite filtered complex with acyclic successive quotients is acyclic, by induction through the short exact sequences of filtration pieces. The subcomplex with last coordinate $\mathrm{\infty }$ and the quotient with last coordinate $0$ or $1$ therefore have isomorphic homology via the connecting map, with the usual cone shift. Induction identifies the latter cube's total homology with the all-$\mathrm{\infty }$ hat homology. The source provides the higher entries of this enlarged complex and their square-zero in [F4], [F5], [F6]; no new canonical comparison map is claimed.

Closed-manifold invariance [F3] identifies this final group with $\widehat{HF}(Y;{\mathbb{F}}_2)$. The construction and every count used here are the imported compatible surgery construction, combined with the retained AComega analytic package. Its geometric data and arities for the fixed link are finite; the cyclic Hamiltonian-copy sequence used in the single-coordinate argument is countable, so the existing countable-choice branch suffices. The source itself does not formulate a choice hierarchy; this accounting uses the retained analytic CC interface and the finite/countable auxiliary selections just specified. No full-AC classification, presentation-independent filtered quasi-isomorphism, or restriction to $S^3$ is introduced.

The topology assertion has a direct finite construction. Put the planar $k$-component unlink, $k\ge 1$, in its standard $k$-bridge position: the upper and lower balls each contain $k$ boundary-parallel arcs with the same noncrossing matching of their $2k$ endpoints. Choose their disjoint bridge disks. Cutting either ball along those disks trivializes the branch construction. Its double cover is two balls joined by $k$ one-handles; the first joins the two zero-handles, and each remaining handle adds one genus. Thus it is a genus-$k-1$ handlebody. The two bridge matchings and disk systems agree in the projection sphere, so the cover of the unlink is the double of this handlebody with matching meridian disks. Doubling each of its $k-1$ co-core disks gives disjoint nonseparating two-spheres. Cutting them removes those one-handles and leaves a sphere with paired ball holes; regluing the pairs gives ${\mathrm{\#}}^{k-1}(S^2\times S^1)$. For $k=1$ it is the double of a ball, hence $S^3$. The cover is constructed by crosswise gluing two labeled copies of the cut balls along the finitely many bridge disks, using $(s,z)\mapsto (s,z^2)$ along the branch arcs and the matching endpoint charts. These charts give its orientation and smooth structure directly. Thus this is a finite realization of the local branch convention in [F1], without invoking the generic AC-dependent covering classification or any Floer map.

Here is the complete finite algebra to which the external Floer map computation [F4]–[F5], applied below is applied. Order the unmarked circles of a based $k$-circle state, and assign to them generators $x_c$. The reduced state space is ${\mathrm{\Lambda }}_{{\mathbb{F}}_2}⟨x_c⟩$: the basis monomial on a subset labels those circles by $X$, the others by $1$, and the marked circle by the quotient label $1$. Exterior multiplication is commutative over ${\mathbb{F}}_2$ and each $x_c^2=0$. Thus this state space has dimension $2^{k-1}$, once an actual Floer vertex identification with it is supplied. This is a computation of the required state algebra, not yet of $\widehat{HF}$.

Write $A={\mathbb{F}}_2[X]\mathrm{/}(X^2)$. For an unmarked merge the required map is $m(1,1)=1$, $m(1,X)=m(X,1)=X$, $m(X,X)=0$. For an unmarked split it is $\mathrm{\Delta }(1)=1\otimes X+X\otimes 1$, $\mathrm{\Delta }(X)=X\otimes X$. If a marked circle merges with an unmarked one, reducing the output kills $X$ on the marked factor, so $1\otimes 1\mapsto 1$ and $1\otimes X\mapsto 0$. If a marked circle splits, reduce the child containing the original point: $\mathrm{\Delta }(1)$ then becomes $1_{\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{k}\mathrm{e}\mathrm{d}}\otimes X_{\mathrm{n}\mathrm{e}\mathrm{w}}$ (with factors exchanged when necessary). Hence marked split creates the unmarked label $X$, not $1$. All untouched factors are carried identically. These are literal quotient calculations and include both marked cases.

Equivalently, unmarked merge is the exterior-algebra quotient identifying $x_c,x_d$ with $x_e$. Unmarked split is $(x_d+x_e)\wedge \mathrm{\Lambda }\mathrm{ȷ}$, where $\mathrm{ȷ}(x_c)=x_d$ and other generators are unchanged; choosing $x_e$ instead gives the same map, since $(x_d+x_e{)}^2=0$. Marked merge is quotient by the disappearing $x_c$, and marked split is wedge with the newly created $x_e$. Thus these four precise quotient/wedge operations, in a common state basis, are the finite map targets for the analytic surgery-edge calculation.

Apply [F4] to the actual cover constructed in step1.1. Its rank-one exterior module gives the rank $2^{k-1}$, with generators represented by the arc lifts specified in [F5]. That same input supplies the actual Floer triangle maps and their common state bases. Under these bases the quotient/wedge operations of [F4], [F5] are precisely the four finite reduced tables proved in steps 1.2, 2.1, 3.1. In particular $[{\gamma }_0]=0$ kills the unwanted marked-circle factor and forces marked split to create the new $X$ label.

The vertex and map computations in the preceding step are externally imported, not inferred from analytic regularity [F3] or from a chosen common-top transfer. The topology and quotient algebra remain the local finite arguments above. All coefficients are ${\mathbb{F}}_2$, so no integral sign or instanton-saddle formula is used. The actual finite diagram has finitely many states and edges, with the analytic data in the stated AComega branch. This proves the full fixed-resolution Floer TQFT assertion with its compatible marked bases.

The covered disk pages and their binding fillings. Use the standard disk open book of $S^3$ about the braid axis $U$: its complement of a small axis neighborhood is $D^2\times S^1$, and the closed two-strand braid meets each disk in two points. This description is explicit in $S^3\subset {\mathbb{C}}^2$: the axis is $\{z_1=0\}$ and the page parameter is $u=\arg z_1\mathrm{/}(2\pi )$; the disk page is parametrized by $z_2$, with $z_1$ having the remaining modulus and chosen argument. Cut the page parameter at zero. The two moving points over the resulting interval are a braid isotopy of the twice-marked disk. Its double branched cover is the corresponding annular product by the same square-coordinate and sheet-gluing construction as [F2]. On gluing the ends, the annular monodromy is the boundary-fixed lift of the braid mapping class, hence $T^k$ by [F3]. There is no extra deck map: the page-parameter loop near the axis bounds a small meridian disk in its neighborhood disjoint from the branch link, and has trivial meridian character. Moreover the axis itself links the two-strand braid twice, so its mod-two character is zero as well. Thus the cover of its solid-torus neighborhood consists of two ordinary copies, one for each annular boundary. In each copy the meridian is the loop $u$ at a fixed boundary-page point. Consequently the whole branched cover is exactly the annular mapping torus with both boundary tori filled along these page-parameter meridians. The local branched models retain the smooth structures.

Untwist the mapping torus, retaining both meridians. Write $A=(\mathbb{R}\mathrm{/}2\pi \mathbb{Z}{)}_{\theta }\times [-1,1{]}_t$, with a positive twist $T(\theta ,t)=(\theta +2\pi \rho (t),t)$, where $\rho =0$ near $-1$ and $\rho =1$ near $1$. The mapping torus has the equivalence $(\theta ,t,1)\sim (\theta +2\pi k\rho (t),t,0)$. The function ${\theta }^{\mathrm{\prime }}=\theta +2\pi k\rho (t)u(\mathrm{m}\mathrm{o}\mathrm{d}2\pi )$ is well-defined across that equivalence and gives an actual diffeomorphism of the mapping torus with $S_{{\theta }^{\mathrm{\prime }}}^1\times [-1,1{]}_t\times S_u^1$. Its inverse on the cut parameter interval is $\theta ={\theta }^{\mathrm{\prime }}-2\pi k\rho (t)u$, with the same endpoint compatibility. At the lower boundary the filling meridian $u\mapsto ({\theta }_0,-1,u)$ has class $(0,1)$ in the $({\theta }^{\mathrm{\prime }},u)$ basis. At the upper boundary the corresponding meridian has class $(k,1)$. This computes their actual slopes, not only their intersection number.

The two solid tori. Split the product at any fixed interior $t$. Fill its lower end along $(0,1)$. This caps the $u$ circle in the annulus $[-1,t]\times S_u^1$ by a disk, so that piece is the solid torus $V_{-}=S_{{\theta }^{\mathrm{\prime }}}^1\times D^2$. Its meridian on the common torus is $m=(0,1)$ and its core longitude is $l=(1,0)$. The upper filled piece is also a solid torus: the integral shear $({\theta }^{\mathrm{\prime }},u)\mapsto ({\theta }^{\mathrm{\prime }}-2\pi ku,u)$ extends over its whole product collar, is single-valued because $k$ is an integer, and takes its filling slope $(k,1)$ to $(0,1)$. After that shear it is the same disk-capped product as the lower piece. Before undoing the shear, its meridian on the common torus is precisely $kl+m$. Hence this is a genus-one gluing with that meridian slope.

Identify the fixed unknot peripheral basis. Identify $V_{-}$ with the standard unknot exterior, using the same $S^3\subset {\mathbb{C}}^2$ decomposition: $E_U=\{\mathrm{\mid }{z}_1\mathrm{\mid }\ge \epsilon \}$ for a small standard axis neighborhood $\{\mathrm{\mid }{z}_1\mathrm{\mid }\le \epsilon \}$. On its boundary the unknot meridian ${\mu }_U$ is the $\arg z_1$ circle, which generates the core circle of $E_U$, and the preferred longitude ${\lambda }_U$ is the $\arg z_2$ circle, which bounds its meridian disk. Thus take $l={\mu }_U$, $m={\lambda }_U$ by an actual product-coordinate identification. Up to the global orientation choice allowed in the statement, the other solid torus is attached with meridian $k{\mu }_U+{\lambda }_U$. For $k=3$ this is exactly [F4]. For $k=-3$, the reflection $(z_1,z_2)\mapsto ({\bar{z}}_1,z_2)$ reverses the orientation of $S^3$, preserves the unknot, and sends ${\mu }_U$ to $-{\mu }_U$ and ${\lambda }_U$ to ${\lambda }_U$. It carries the slope $-3{\mu }_U+{\lambda }_U$ to $3{\mu }_U+{\lambda }_U$ and extends over the filled solid torus by the same meridian identification. Hence the other handedness has the same $L(3,1)$ description up to orientation reversal. Reversing the orientation of the branch knot does not change its mod-two meridian character.

In $E_U$ the group is generated by ${\mu }_U$ and ${\lambda }_U$ is trivial, by its product $S^1\times D^2$ coordinates. Gluing the second solid torus kills the slope $k{\mu }_U+{\lambda }_U$. By [F5] there are no further relations: a primitive complementary boundary longitude identifies the core generator of the attached solid torus with the corresponding word in the exterior generator, so that generator is eliminated from the pushout presentation. The group is therefore $⟨{\mu }_U\mid {\mu }_U^k=1⟩$, cyclic of order $\mathrm{\mid }k\mathrm{\mid }=3$. All new coordinate and slope calculations were finite; the only inherited choice use is the already declared cover and peripheral-convention branch. This proves both assertions for either oriented trefoil without replacing a meridian-slope calculation by the determinant alone.

If an edge avoids the marked circle, its map acts as the identity on that factor and commutes with $X_p$. For a merge involving the marked circle, associativity gives $m(Xa\otimes b)=Xm(a\otimes b)$, with the marked factor placed on the appropriate input. For a split of that circle, the Frobenius identity gives $\mathrm{\Delta }(Xa)=(X\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(a)=(\mathrm{i}\mathrm{d}\otimes X)\mathrm{\Delta }(a)$; for $a=1$ both sides equal $X\otimes X$, and for $a=X$ both sides vanish. Thus every edge map commutes with $X_p$, so $\mathrm{im}{X}_p$ is a subcomplex.

The quotient differential is well-defined by step 1.1 and squares to zero by [F3]. In each state the marked-circle factor has basis $1,X$, and multiplication by $X$ has image the one-dimensional span of $X$. The quotient therefore replaces that factor by ${\mathbb{F}}_2\{1\}$; the shift $j-1$ assigns the surviving label degree zero. This gives the stated reduced cochain complex and its cohomology.

On the associated graded cube the diagonal differential is the ordinary vertex differential. Therefore its homology is the direct sum of the resolution Floer groups. The next differential is induced by the components changing one coordinate: higher polygon components lower the increasing filtration by at least two and cannot contribute to this first differential. Each surviving component is the actual surgery triangle map of that edge.

Use the SAME vertex state identification simultaneously for all incident edges. By the exact Floer supplier [F2], an unmarked merge has table $1\otimes 1\mapsto 1$, $1\otimes X,X\otimes 1\mapsto X$, $X\otimes X\mapsto 0$; an unmarked split has $1\mapsto 1\otimes X+X\otimes 1$ and $X\mapsto X\otimes X$. Marked merge kills the input with the other circle labelled $X$ and sends its $1$ input to $1_{\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{k}\mathrm{e}\mathrm{d}}$. Marked split sends $1_{\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{k}\mathrm{e}\mathrm{d}}$ to $1_{\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{k}\mathrm{e}\mathrm{d}}\otimes X_{\mathrm{n}\mathrm{e}\mathrm{w}}$, with the tensor order determined by which child contains the point. These last two formulas are obtained by quotienting the marked factor, not by keeping an extra unreduced factor. Tensor the identity on every untouched circle. These are exactly the reduced Frobenius saddle maps of [F3].

For an explicit grading binding, let $n_{+}(D),n_{-}(D)$ be the signed crossing counts. By [F4] the surgery bit vector $I$ is the $0\mathrm{/}1$ vector of $\bar{D}$, so the normalized cochain degrees are $i=\mathrm{\mid }I\mathrm{\mid }-n_{+}(D),\tilde{j}={\sum }_{c\mathrm{\ne }p}{\mathrm{deg}}_q(a_c)+\mathrm{\mid }I\mathrm{\mid }+n_{-}(D)-2n_{+}(D),$ where $a_c\in \{1,X\}$, with degrees $1,-1$, and the marked surviving label has reduced degree zero. Every saddle table in step 2.1 lowers the sum of label degrees by one, including marked merge and split. Increasing $\mathrm{\mid }I\mathrm{\mid }$ by one therefore raises $i$ by one and preserves $\tilde{j}$. The spectral-sequence filtration index $p=n-\mathrm{\mid }I\mathrm{\mid }$ satisfies $i=n_{-}(D)-p$; it is not an independent Floer cochain grading.