Lee deformation and Rasmussen’s s-invariant

The monic relations make $\mathbb{A}$ free over $\mathbb{Z}$ and $\mathbb{R}$ free over $\mathbb{Z}[h]$, each with basis $1,X$. Both relations are homogeneous for the displayed degrees. For the odd maps, the merge quotient is well-defined because $p(C_1-C_2)=0$. The split formula is independent of whether $j(C)=C_1$ or $C_2$, since $(C_1-C_2)\wedge C_1=(C_1-C_2)\wedge C_2=C_1\wedge C_2$.

In $\mathbb{R}$, associativity, commutativity and the unit law follow from quotient-ring multiplication. On the basis $1,X$, the counit composites $(ϵ\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }$ and $(\mathrm{i}\mathrm{d}\otimes ϵ)\mathrm{\Delta }$ both send $1\mapsto 1$ and $X\mapsto X$. Cocommutativity follows from the symmetric formulas for $\mathrm{\Delta }(1)$ and $\mathrm{\Delta }(X)$. For coassociativity, both iterated comultiplications send $X$ to $X\otimes X\otimes X$ and send $1$ to $1\otimes X\otimes X+X\otimes 1\otimes X+X\otimes X\otimes 1-h(1\otimes 1\otimes X+1\otimes X\otimes 1+X\otimes 1\otimes 1)+h^{2}1\otimes 1\otimes 1.$ Thus the unit, counit and coassociativity axioms hold.

On the ordered basis $1\otimes 1,1\otimes X,X\otimes 1,X\otimes X$, each of $\mathrm{\Delta }m$, $(m\otimes \mathrm{i}\mathrm{d})(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })$, and $(\mathrm{i}\mathrm{d}\otimes m)(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})$ has respective outputs $1\otimes X+X\otimes 1-h1\otimes 1,X\otimes X,X\otimes X,hX\otimes X\mathrm{.}$ Therefore the Frobenius identities hold. On every cube square, the two unsigned composites agree by tensor functoriality on disjoint circles, associativity for two merges, coassociativity for two splits, or the checked Frobenius identities for a merge-split pair. Opposite cube signs make the two signed composites cancel, so the Bar–Natan differential squares to zero. A based odd cube with the square-compatible sign data in [A1] is also a cochain complex: the coefficient of each height-two state in $d^2$ is the sum of its two signed composites, zero by the defining condition. The right marked action commutes with merges because the quotient carries $Z_p$ to the marked output circle. It commutes with splits because $((C_1-C_2)\wedge j(z))\wedge Z_p=(C_1-C_2)\wedge j(z\wedge Z_p)$, including a split of the marked circle: either lift of the marked output differs by $C_1-C_2$, whose extra wedge vanishes. Also $X^2$ acts by zero. Every state module is free over $\mathbb{A}$: choose the exterior products of unmarked circle generators as module basis and append $Z_p$ on the right for their $X$ multiples. Modulo $2$ this action is precisely the even marked action, and reduction of $\mathbb{R}$ modulo $(2,h)$ retains that action through $\mathbb{B}$, rather than setting $X=0$. Existence of such odd sign assignments, the mod-2 comparison as a homotopy equivalence, and the localized rank-one condition are not claimed by this Definition; they are separate assertions of the foundations lemma.

Straighten only a crossing-free marked germ. Translate the mark to $z=0$ and rotate its oriented tangent to the positive real axis, by paths of positive planar similarities. Since the mark is a regular non-double point, there is a small disk meeting the diagram in just one arc. The inverse-function theorem expresses that arc as $y=f(x)$, where $f(0)=f^{\mathrm{\prime }}(0)=0$. It can be straightened by the supported shear $(x,y)\mapsto (x,y-t\chi (x,y)f(x))$, $0\le t\le 1$: choose a cutoff equal to one near the smaller graph segment and supported in a rectangle of width and height comparable to a sufficiently small $r$. There $f(x)=O(r^2)$, whereas ${\mathrm{\partial }}_y\chi =O(r^{-1})$. Thus $1-t({\mathrm{\partial }}_y\chi )f>0$ after shrinking $r$. Each vertical coordinate map is strictly increasing and the identity outside a compact interval, so each shear is a global orientation-preserving diffeomorphism; its inverse is smooth by its positive vertical derivative. Choose nested neighborhoods so that after this shear the diagram in $\mathrm{\mid }z\mathrm{\mid }<2r$ is the real diameter and no other strand meets that disk. Shrink $r<1$ if necessary. The part not in the real interval $\mathrm{\mid }x\mathrm{\mid }<r$ consequently satisfies $\mathrm{\mid }z\mathrm{\mid }\ge r$. All these isotopies merely transport the mark and crossing decorations; their local supports meet no crossing.

An actual rotation recharting. Choose $0<ϵ<\min (r\mathrm{/}4,1\mathrm{/}4)$ and set $a=iϵ$. On the oriented stereographic sphere use the fractional linear coordinate $w=-\frac{1+\bar{a}z}{z-a}\mathrm{.}$ The matrix $\left(\begin{array}{cc}-\bar{a}& -1\\ 1& -a\end{array}\right)\mathrm{/}\sqrt{1+\mathrm{\mid }a{\mathrm{\mid }}^2}$ is unitary with determinant one: its columns have norm one and scalar product zero, and its unnormalized determinant is $1+\mathrm{\mid }a{\mathrm{\mid }}^2$. Hence its fractional linear action is the action of an orientation-preserving rotation of the round two-sphere. Its omitted old point is $a$, which lies off the real germ in the marked disk, hence in a face adjacent to the mark and outside all crossing disks. The old omitted point $\mathrm{\infty }$ is sent to $-\bar{a}$, which is an ordinary point of the new chart. The map is smooth on the entire spherical diagram and preserves the oriented local crossing charts and all transported decorations.

A quantitative exterior closing arc. For every old diagram point outside the interval $\mathrm{\mid }x\mathrm{\mid }<r$, step 1.1 and the triangle inequality give $\mathrm{\mid }w\mathrm{\mid }\le \frac{1+ϵ\mathrm{\mid }z\mathrm{\mid }}{\mathrm{\mid }z\mathrm{\mid }-ϵ}\le B:=\frac{1+ϵr}{r-ϵ}\mathrm{.}$ The second inequality holds because the rational function has derivative $-(1+{ϵ}^2)\mathrm{/}(s-ϵ{)}^2<0$ for $s\ge r$. The inequality $B<1\mathrm{/}ϵ$ is equivalent to $2ϵ<r(1-{ϵ}^2)$, which follows from $ϵ<r\mathrm{/}4$ and $ϵ<1\mathrm{/}4$. Choose a real $R$ strictly between $B$ and $1\mathrm{/}ϵ$. On the real germ $\mathrm{\mid }w(x){\mathrm{\mid }}^2=\frac{1+{ϵ}^2{x}^2}{x^2+{ϵ}^2},$ whose derivative with respect to $x^2$ is $({ϵ}^4-1)\mathrm{/}(x^2+{ϵ}^2{)}^2<0$. Since the value at $x=0$ is $1\mathrm{/}{ϵ}^2>R^2$ and at $\mathrm{\mid }x\mathrm{\mid }=r$ is at most $B^2<R^2$, the circle $\mathrm{\mid }w\mathrm{\mid }=R$ meets that germ at exactly $x=\pm c$, with $c^2=\frac{1-R^2{ϵ}^2}{R^2-{ϵ}^2},0<c<r\mathrm{.}$ Both intersections are transverse, since $c\mathrm{\ne }0$ and the derivative just computed is nonzero there. The only diagram points outside the closed disk $\mathrm{\mid }w\mathrm{\mid }\le R$ form the single open arc parametrized by $-c<x<c$; its closure contains the mark and has no crossing. The image is a round circular arc, because a fractional linear map sends a circle on the sphere to a circle and its pole $iϵ$ is not on the real circle. Every other strand and every crossing is strictly inside the disk. Thus this is the claimed exterior closing arc, with uniform crossing-free neighborhoods of it and of its two cut points.

Record spatial coordinate identifications. A planar shear in step 1.1 lifts on a spatial projection chart by $(x,y,h)\mapsto ({\mathrm{\Phi }}_t(x,y),h)$. Its compact planar support can also be cut off in height on a larger box containing the link: use its compactly supported time-dependent planar velocity multiplied by a height cutoff equal to one on that box, with zero height velocity. The flow is complete, agrees with the displayed lift near the link, and extends by the identity to $S^3$. The initial translations and rotations have their usual orientation-preserving similarity lifts and smooth conformal extensions at the point at infinity. For the rechart, realize the projection sphere as the equator of $S^3\subset {\mathbb{R}}^3\oplus \mathbb{R}$. If $Q\in SO(3)$ is the rotation from step 2.1, then $(v,s)\mapsto (Qv,s)$ is an orthogonal determinant-one map of ${\mathbb{R}}^4$ preserving $S^3$ and preserving its orientation. On the equatorial collar it transports the projection direction, the signed height, and the over/under datum. The Euclidean projection and the equatorial product projection require an explicit chart identification. Write $q=(0,0,1,0)$ for the spatial stereographic point, $z=x+iy$, $X(z)=\frac{(2x,2y,\mathrm{\mid }z{\mathrm{\mid }}^2-1)}{1+\mathrm{\mid }z{\mathrm{\mid }}^2},d(z)=\frac{2}{1+\mathrm{\mid }z{\mathrm{\mid }}^2}\mathrm{.}$ Standard inverse spatial stereographic projection is $E(z,h)=\frac{(2x,2y,\mathrm{\mid }z{\mathrm{\mid }}^2+h^2-1,2h)}{1+\mathrm{\mid }z{\mathrm{\mid }}^2+h^2}\mathrm{.}$ The product-collar chart with the same first normal derivative is $T(z,h)=(\sqrt{1-d(z{)}^2{h}^2}X(z),d(z)h)\mathrm{.}$ On $h=0$ these charts agree and have identical full first derivatives: both normal derivatives are $d(z)e_4$, while their tangential derivatives are those of $X$. On a compact planar neighborhood containing the diagram, the coordinate transition $\mathrm{\Psi }=E^{-1}T$ therefore satisfies $\mathrm{\Psi }-\mathrm{i}\mathrm{d}=O(h^2)$ and $D\mathrm{\Psi }-I=O(\mathrm{\mid }h\mathrm{\mid })$ uniformly. Choose a slightly larger compact planar box and cut off this difference there and in a height slab $\mathrm{\mid }h\mathrm{\mid }<\eta$. A height cutoff has derivative $O({\eta }^{-1})$, multiplying an error $O({\eta }^2)$; thus the resulting global compactly supported displacement $V$ has $\mathrm{\parallel }DV\mathrm{\parallel }<1\mathrm{/}2$ for small $\eta$. Every $\mathrm{i}\mathrm{d}+tV$ is injective by the Lipschitz bound, is a local orientation-preserving diffeomorphism, and is onto: it is proper, its image is closed and open, and the plane-height space is connected. Its inverse is smooth by the inverse-function theorem. This is an actual ambient coordinate isotopy, extending by the identity to $S^3$. First shrink any existing Euclidean height realization by a positive constant so it lies where both cutoffs equal one. Height scaling is an embedding isotopy: at every double point both branch heights are multiplied by the same positive factor. The new spatial representative is then exactly $T(z,h)$, with product-collar base coordinate $z$ and signed height $d(z)h$. Because $d(z)>0$, at a double point it preserves the strict over/under ordering. Hence the displayed Euclidean diagram is identified with the product-collar diagram by this explicit spatial map; they are not assumed literally equal off the equator. These are actual marked link isotopies and coordinate identifications with specified endpoints. Composing these coordinate maps and height isotopies gives the asserted spatial identification; no claim that the rotation is a planar isotopy is used.

Compatibility with protected long ends. The exterior arc and two slightly longer joining collars have injective projection and are disjoint in projection from the remaining compact part, except at their adjacent ends. On a narrow planar neighborhood of that arc its spatial height is a smooth function $h(x,y)$, extended from the arc in tubular coordinates. A supported vertical translation $(x,y,z)\mapsto (x,y,z-t\eta (x,y)h(x,y))$ sets its height to zero on a smaller neighborhood; this is a diffeomorphism for every $t$, since its derivative in $z$ is one. It cannot reverse an over/under ordering, and there are no double points in its smaller marked support. Choose the joining neighborhoods slightly longer before cutting, so the resulting cut collars are fixed smooth product collars. The remaining diagram is a compact tangle in the round disk, and the single planar exterior arc is its marked closure. These are precisely endpoint data that may be retained as protected long endpoints in a relative movie construction. Every operation has preserved the transported mark, and none passes it through a crossing. Finite composition proves the assertion, including its recorded spatial coordinate identification. No algebraic comparison of the two charts, no arbitrary mark-slide identity, and no purely planar comparison have been assumed.

The two-saddle calculation. Quotients commute with quotients. Two split factors anticommute because $u\wedge v=-v\wedge u$. A quotient and a split commute when the quotient carries the split difference to the corresponding output difference. These give $\kappa =1,-1,1$, respectively, for two merges, two splits and a mixed pair whose saddle types are unchanged by interchange. The remaining connected nonzero case starts with two circles and two bands between them: either order first merges and then splits. Both composites factor through the quotient identifying the two input circles. If the two band arrows point towards the same input circle, their final split differences are opposites, giving $\kappa =-1$; if they point towards different circles, the differences agree, giving $\kappa =1$. Finally, alternating endpoints on a single input circle give a split followed by a merge in either order. The quotient kills the split difference, so both composites are zero. These exhaust the cases: discard components missing both arcs, and distinguish whether each arc joins equal or distinct components; if both are self-arcs, their endpoints either alternate or do not. A spectator generator is carried identically throughout, so the computation on the incident generators proves the full exterior-map equation.

Reversing an arrow is a gauge change. Reverse the arrow at coordinate $i$. The unsigned edge map changes by $r(e)=-1$ on its split edges and by $r(e)=1$ on its merge edges; other edges have $r(e)=1$. The face coefficient changes by the product of these four $r$'s. For nonzero composites this follows by multiplying the two equations for the edge maps. For a zero face, the $i$-saddle is a split in one ordering and a merge in the other. Reversing its arrow reverses the left/right test defining $\kappa$, so the coefficient changes by $-1$, again the product of the $r$'s. Consequently the product of the six $\kappa$'s on a three-cube is unchanged by reversing any arrow: every edge belongs to exactly two of its faces. It suffices to check one choice of arrows for every unoriented three-saddle configuration.

Separating a configuration. Collapse each saddle disk to a four-valent vertex and retain the regular arcs joining its four ports. Discard vertex-free circles. This gives a planar four-valent multigraph with three vertices. If it is disconnected, one component contains just one vertex. If it has a two-edge cut, cut those edges and cap the resulting pair of ports in each piece by an arc. The original diagram is then obtained by joining the two capped marked arcs; one piece again has one vertex and the other two. The caps can be chosen in a thin neighbourhood of the cut, so they preserve the cyclic orders and inside/outside intersection test for every two-saddle face within a piece. Every resolution of the joined diagram has the circle count $k_1+k_2-1$; in the disconnected case it has count $k_1+k_2$. Thus the merge/split status of a saddle is its status in its own capped piece. A face using one saddle from each piece is an ordinary interchange with coefficient $(-1{)}^{b_i{b}_j}$ by step 1.1. The two faces using both saddles of the two-vertex piece have equal coefficients and cancel in the boundary product. If the one-vertex piece has split indicator $b_c$, the product from the four remaining faces is $(-1{)}^{b_{c}N}$, where $N$ is the sum of the split counts in the two orders of the two saddles in the other piece. Each order has the same split count, namely $(k_{\mathrm{o}\mathrm{u}\mathrm{t}}-k_{\mathrm{i}\mathrm{n}}+2)\mathrm{/}2$. Hence $N$ is even and the boundary product is $1$. This also proves the disconnected case, using separate exterior generators rather than the marked connected sum.

The irreducible list. Suppose the graph is connected and has no two-edge cut. It has no loop: a loop uses two of the four ports at its vertex, and the remaining two edges would be a cut. Write $m_{ij}$ for the number of edges between vertices $i,j$. The three degree equations give $m_{12}=m_{23}=m_{31}=2$. At each vertex the two parallel edges bound a lens on one side; the remaining edges must lie on the other side, since they must reach the third vertex without crossing. Thus the planar graph has three lens faces and two triangular faces, with its cyclic orders determined up to reflection and permutation of the vertices. Resolving its three vertices gives eight choices. Permuting the vertices identifies choices having the same number of switches from the resolution into three lens circles. The four representatives, obtained by zero, one, two and three switches, have respectively these circle-and-band descriptions: three circles with three bands forming a triangle; two circles joined by two bands, with a self-band on one circle separating their attachment points; one circle with two parallel inside bands and one outside band alternating with each of them; and two circles joined by three parallel bands. This enumeration follows also by doing the switches successively: the first merges two lens circles, the second merges the remaining two circles, and the third splits their common circle. It accounts for all eight choices, without an assumption about the arrows.

Triangle and three parallel bands. In the triangle representative direct the three bands cyclically. A face with the third band absent is two successive merges, of coefficient $1$. With the third band performed first, the two remaining arrows point towards different circles, again giving coefficient $1$ by step 1.1. Thus all six coefficients are $1$. In the three-parallel-band representative direct all three arrows from the first circle to the second. The three bottom faces have two arrows towards the same circle and coefficient $-1$. After the third band merges the two circles, the two remaining bands are nonalternating self-bands; their two split factors anticommute, so each top face also has coefficient $-1$. In both representatives the six-face product is $1$.

Digon with a self-band. Place the two circles to the left and right, draw the two joining bands above and below, and draw the self-band across the left circle, separating the upper and lower attachment points. Direct both joining bands towards the right circle. The face of the two joining bands before the self-band is a same-head interchange and has coefficient $-1$. After the self-band, those bands form a tree on three circles and the corresponding face has coefficient $1$. The two faces involving the self-band before the other joining band are ordinary mixed interchanges and have coefficient $1$. After a joining band is performed, the other joining band and the self-band have alternating endpoints. The two resulting zero faces are reflections across the horizontal midline, with the self-band arrow retained and the two equally directed joining arrows exchanged. Reflection reverses the inside/outside oriented-intersection test, so their $\kappa$'s are opposite. Their product is $-1$, cancelling the single nonzero coefficient $-1$. The total product is $1$, for either direction of the self-band arrow.

Two inside bands and an outside band. List the endpoints clockwise as $c,a,b,c,b,a$, where $a,b$ are the two parallel inside bands and $c$ is outside. Direct $a,b$ upwards. Their bottom face consists of two splits and has coefficient $-1$. After performing $c$, their heads lie on the same one of the two resulting circles; their top face is a same-head interchange and has coefficient $-1$. The faces $(a,c)$ before $b$ and $(b,c)$ before $a$ are zero faces with the same inside-arrow direction and the same outside arrow, hence have equal coefficients. Their product is $1$. In the two remaining faces, performing the other inside band has separated the endpoints of $c$; the two events are an ordinary merge and split with coefficient $1$. Thus the six-face product is again $1$, for either direction of $c$. Steps 3.1–4.3 and arrow reversal in step 2.1 prove that every three-cube has boundary product $\prod \kappa =1$. Consequently $\varphi =-\kappa$ also has boundary product $1$, because there are six faces.

Constructing the signs. Induct on cube dimension $n$. For $n=0$ there are no signs; for $n=1$ choose the only edge sign $1$. For the induction step construct signs on the lower face $v_n=0$, set every edge in direction $n$ to $1$, and define an upper-face edge in direction $i<n$ by ${ϵ}_i(v,1)={\varphi }_{i,n}(v){ϵ}_i(v,0)$. This enforces each vertical face product. For an upper face in directions $i,j<n$, multiply the six face equations for its three-cube with direction $n$: the lower face and four vertical faces already have their prescribed products, while the six prescribed products multiply to $1$ by step 4.3. Therefore the upper face has its prescribed product too. The resulting differential has square zero: on a nonzero face the two unsigned composites differ by $\kappa$, while their signed coefficient ratio is $-\kappa$; on a zero face both composites vanish. Every operation in this induction is a finite multiplication of signs.

Comparison. If $ϵ,{ϵ}^{\mathrm{\prime }}$ realize the same face products, their quotient $\rho (e)={ϵ}^{\mathrm{\prime }}(e)\mathrm{/}ϵ(e)$ has product $1$ around every face. Give the lowest vertex sign $1$, and give any other vertex the product of $\rho$ along a monotone path to it. Two such paths differ by interchanging adjacent coordinate changes; each interchange is a square and leaves the product unchanged. Therefore these vertex signs are well-defined and their ratio across an edge is $\rho (e)$. Multiplying every state module by its vertex sign gives a chain isomorphism. These scalar maps preserve both normalized gradings and, when a mark is supplied, commute with its marked action by [F2]. This proves all the assertions.

A spatial chart and small embedding perturbations. If $S$ is empty take the empty movie. Extend its spatial projection constantly across appended product collars to a boundaryless surface. Its differential has rank at most two, so [F5] makes its image in $S^3$ null; in particular the original compact spatial image is null. Choose a point outside that compact image and use a spatial stereographic chart, so the entire surface lies in ${\mathbb{R}}^3\times I$. If endpoint diagrams use the fixed point at infinity, choose this missing point near infinity, outside the track, and move it to infinity by an ambient point motion in a small ball disjoint from the end links; this fixes smaller end collars, exactly as in the compact-track argument for link isotopies. Write its embedding as $e(x)=(a(x),h(x))$. Small $C^1$ perturbations of $e$ on a fixed compact support in the interior remain embeddings: injectivity of each local differential has a uniform lower bound on finitely many smaller charts, while images of source pairs outside those chart neighbourhoods have positive minimum separation. Interpolating a sufficiently small perturbation with $e$ gives an embedding isotopy fixed on the collars. Its velocity extends by [F5], working in the boundaryless collar extension $S^3\times \mathbb{R}$ and cutting it off away from the fixed end collars. Thus these perturbations preserve the actual collared surface up to ambient isotopy.

Morse height realized on the embedded surface. Cover the complement of a smaller product collar by finitely many chart cores and choose supported bumps equal to one there by [F1]. In each surface chart with coordinates $x_1,x_2$, perturb the height $h$ by finite linear combinations of $\chi x_1,\chi x_2$, extended by zero. At a zero of its differential, the parameter derivatives span the cotangent fiber on at least one core. Thus the evaluation of the differential is transverse to the zero section. [F1] gives arbitrarily small parameters with transverse differential, hence nonsingular critical Hessians. Choose them so small that step 1.1 applies and the height stays strictly between zero and one off the collars. On the collars $h=t$ is unchanged and has no critical point. By [F2] there are finitely many critical points; separate their heights by the supported small perturbation in [F2], still within the embedding neighbourhood. The embedding is now $(a,h^{\mathrm{\prime }})$ with excellent Morse height $h^{\mathrm{\prime }}$, and is ambient isotopic to the original surface relative to the collars.

Isolating the Morse pictures. Let $p$ be one of the finitely many critical points, of height $c$. Since $d{h}_p^{\mathrm{\prime }}=0$ and the surface embedding is immersive, the spatial derivative $d{a}_p$ is injective. Choose planar projection transverse to this tangent plane. In a small source disk the spatial projection is therefore a diffeomorphism onto a planar disk, and [F2] writes the height there as $c+x^2+y^2$, $c-x^2-y^2$, or $c-x^2+y^2$, after a change of its planar disk coordinates. These give respectively one circle born, one circle dying, or the two smoothings of a saddle band. The orientations are the boundary orientations induced from the same oriented surface. We may arrange that the projected critical point misses every other branch at height $c$: equality with a distinct regular-level branch imposes two planar equations on its one source coordinate, so a transverse perturbation has no solution. Likewise regular immersion, transverse double points and absence of triple points on the rest of that slice follow from the one-slice jet conditions in [F4] or [F5]. Here the finite-dimensional families are supported away from the critical source disk and the end collars; on the disk only its projection direction or an arbitrarily small spatial linear perturbation is needed to make its tangent projection invertible. Finitely many disjoint source disks permit these choices simultaneously, with no change in $h^{\mathrm{\prime }}$. Transversality on the remaining slice pieces uses independent chart bumps exactly as in [F4]. All conditions are open away from the selected local critical model. Compactness then gives a small time interval about each $c$, disjoint from the others, in which the critical picture occupies a disk missing all other projected strands, and the complementary diagrams are regular and vary by planar isotopy. Thus there is one isolated birth, saddle or death in each such interval.

Regular slabs retain the actual isotopy. Cut at regular levels on the two sides of each critical interval. Extend the height over the product collars when a boundaryless carrier is needed. On every complementary closed slab the normalized-gradient flow [F3] identifies the surface with $C\times [a,b]$ and its embedding with $(F_t(x),t)$, for a finite union of circles $C$. Compactness makes $C$ finite, and each $F_t$ is a spatial embedding. If a slab is empty there is nothing to do. Use slightly extended regular bands at the cuts. Choose a smooth function $\rho$ on each such band, equal to the cut time on a smaller central strip and equal to $t$ near the band's two outer faces. Replace $F_t$ by $F_{\rho (t)}$. This replacement is an isotopy of the traced surfaces, not an assertion that $\rho$ is an ambient time diffeomorphism: the interpolation $(F_{(1-s)t+s\rho (t)}(x),t)$ is an embedding for every $s$, because its time coordinate is still $t$ and its spatial slices are embeddings. It is fixed near the band's outer faces, so it extends by the identity over the rest of the surface and ambiently by [F5]. In particular the traces on both sides of a cut now agree with the same stationary product near it. Bands are chosen disjoint from the critical pictures and the prescribed end collars. The regular slabs, with these smaller product strips at their ends, therefore have stationary regular endpoint projections, while the intervening critical pieces remain unchanged on smaller neighbourhoods of their critical levels. The endpoint diagrams at the cuts are regular by step 3.1.

Finite projection events on each regular slab. Apply [F4] to each stationary-ended $F$ of step 4.1. It gives an arbitrarily small smooth perturbation $F^{\mathrm{\prime }}$ fixed near both slab ends, with finitely many mutually distinct cusp, quadratic double-tangency and transverse triple times. Interpolate linearly in the spatial ${\mathbb{R}}^3$ chart between $F$ and $F^{\mathrm{\prime }}$; take the perturbation small enough that all intermediate slices are embeddings, using compact local immersion and distant-pair separation as in step 1.1. The traces $(F_{s,t}(x),t)$ are therefore embedding isotopies fixed on the slab ends, and extend ambiently by [F5]. Thus the chosen generic isotopy represents the original traced regular slab up to ambient isotopy relative to its cuts. By [F4] its projected events are precisely oriented R1, R2 and R3 and the intervals between them are planar isotopies. This uses the generic-isotopy supplier on the actual traced family; replacing it merely by an arbitrary Reidemeister sequence between the same endpoint links would not establish the trace assertion.

Gluing and boundary cases. There are finitely many critical intervals and regular slabs. All slab perturbations are fixed near their cuts, so their traced surfaces glue smoothly to the unchanged local critical pictures. Their ambient isotopies glue using collars, or compose after extending each by the identity off its support. The resulting finite ordered movie retains all boundary parametrizations and smaller end collars. Critical intervals contain only their one Morse event; the generic projection events are in the complementary slabs, so all exceptional times are distinct. Empty incoming or outgoing links merely omit the respective end cylinders; closed surface components have their own finitely many births and deaths, and are included by the same construction. The empty surface has already been treated. Prescribed regular end diagrams were never perturbed. The orientations and crossing height orders throughout are those of the actual embedded surface. Countable choice has been used only in the named suppliers and countable-chart transversality, not full AC. This proves the stated existence and trace comparison without any movie-move classification.

Move a genuine ambient ball by translations. Choose an interior point $p$ of $F$ and a radius $r>0$ with $\mathrm{\mid }p\mathrm{\mid }+r<1$. The embeddings of the closed radius-$r$ ball $D$ centered at $p$ given by $x\mapsto x-tp$, $0\le t\le 1$, stay in the interior of $B^4$: the norm of a point in their image is at most $(1-t)\mathrm{\mid }p\mathrm{\mid }+r<1$. Their compact track has a neighborhood with closure in the interior of $B^4$. By [F1] this isotopy extends ambiently in ${\mathbb{R}}^4$ with support in that neighborhood. Its time-one map fixes an ambient neighborhood of the outer boundary, preserves orientation and takes $p$ to $0$, while agreeing with translation on the whole ball $D$. Replace $F$ by its image. No conclusion about any nonround ball complement has been drawn.

Flatten the surface on an actual neighborhood of zero. Let $P$ be the tangent plane of the new surface at $0$. Orthogonal projection to $P$ has invertible derivative on that tangent plane, so [F3] writes the surface as $(u,h(u))$ in $P\oplus P^{\perp }$, with $h(0)=0$ and $Dh(0)=0$. Choose one sufficiently small closed disk in the $u$ coordinates. Its graph is a compact embedded disk in the interior of $B^4$. The maps $u\mapsto (u,(1-t)h(u))$ are an isotopy of that compact disk, since projection to $u$ remains injective with injective differential. Choose the disk small enough that its entire compact track lies in the interior of $B^4$. Apply [F1] again there, obtaining an ambient orientation-preserving diffeomorphism that fixes an ambient neighborhood of the outer boundary and maps the full graph disk to the flat disk. The surface is now exactly $P$ near $0$: the disk interior is an open neighborhood of $0$ within the embedded surface. Its complement outside a smaller such graph disk is compact and misses $0$, so it has positive distance from $0$. Choose $\epsilon >0$ smaller than that distance and the flat disk radius, and less than $1$. Then the surface meets the actual round ball $B_{\epsilon }^4$ precisely in $P\cap B_{\epsilon }^4$. Compose the two ambient diffeomorphisms to obtain $\mathrm{\Psi }$. They fix a common ambient neighborhood $N$ of the outer boundary. For the supplied collar $c:K\times [0,\eta )\to F$, the open set $c^{-1}(N)$ contains $K\times \{0\}$. A finite cover of the compact circle by product neighborhoods gives one $0<\delta <\eta$ with $c(K\times [0,\delta ])\subset N$. Thus $\mathrm{\Psi }$ fixes this sufficiently small restriction of the supplied collar; no fixedness of its entire image is asserted.

Use only the concentric round shell. The explicit map $(z,t)\mapsto (\epsilon +(1-\epsilon )t)z$ is a diffeomorphism $S^3\times [0,1]\to B^4\setminus \mathrm{int}{B}_{\epsilon }^4$. The inner boundary curve is the great circle $P\cap S_{\epsilon }^3$, hence a standard unknot after radial rescaling; an orthonormal basis of $P$ extends by finite linear algebra to an oriented orthonormal basis of ${\mathbb{R}}^4$, giving an orientation-preserving linear rotation identifying it with the usual equator. Near the inner sphere the surface is a radial product in $P$, so this radial shell coordinate already supplies its product collar.

Respect a restriction of the given outer surface collar. For completeness the original outer surface collar need not be radial. In its smooth proper half-space chart the radial coordinate has nonzero inward derivative along a transverse collar direction. In a supplied parametrization of that collar, [F3] and compactness of its boundary circle therefore reparametrize it, for $r$ near $1$, as $(r{f}_r(\theta ))$ with $f_r:S^1\hookrightarrow S^3$ a smooth embedding family and $f_1$ the original parametrized $K$. By [F1] on $S^3$, choose a smooth ambient family $G_r$ with $G_1=\mathrm{i}\mathrm{d}$ and $G_r{f}_1=f_r$. Choose a smooth function $h(r)$ equal to $r$ near $1$ and to $1$ near the inner edge of this thin outer collar, with values in the parameter interval where $G$ is defined. The fiberwise diffeomorphism $(z,r)\mapsto (G_{h(r)}^{-1}z,r)$ is the identity at $r=1$ and near that inner edge, so extends as the identity over the rest of the shell; near the outer end it takes the surface to the literal product $K\times [1-\delta ,1]$. Thus adjust the shell identification by this map. It leaves the inner sphere and the boundary knot fixed, and changes neither genus nor the fixedness of a sufficiently small restriction of the supplied outer collar by $\mathrm{\Psi }$ from steps 1.1–2.1. Removing one interior disk from a connected surface leaves it connected, retains genus and adds one boundary circle; its Euler characteristic decreases from $1-2g$ to $-2g$. Orient the resulting shell surface as a cobordism from the inner unknot to $K$. Reversing the interval and the surface orientation gives the reverse direction if desired; a half-turn reverses the orientation of the standard unknot. This proves the full genus, orientation and collared-end assertions under countable choice, using only an explicitly concentric shell.

The quotient $A_L=\mathbb{Q}[X]\mathrm{/}(X^2-1)$ has basis $1,X$, and its multiplication table is the displayed $m_L$. Hence multiplication is associative and commutative with unit $1$. The counit identities follow on $1,X$: $({ϵ}_L\otimes \mathrm{i}\mathrm{d}){\mathrm{\Delta }}_L(1)=1$, $({ϵ}_L\otimes \mathrm{i}\mathrm{d}){\mathrm{\Delta }}_L(X)=X$, and the other counit composite gives the same values.

For input $1$, both composites $({\mathrm{\Delta }}_L\otimes \mathrm{i}\mathrm{d}){\mathrm{\Delta }}_L$ and $(\mathrm{i}\mathrm{d}\otimes {\mathrm{\Delta }}_L){\mathrm{\Delta }}_L$ equal $1\otimes X\otimes X+X\otimes 1\otimes X+1\otimes 1\otimes 1+X\otimes X\otimes 1$. For input $X$, both equal $1\otimes 1\otimes X+1\otimes X\otimes 1+X\otimes 1\otimes 1+X\otimes X\otimes X$. Thus ${\mathrm{\Delta }}_L$ is coassociative. Cocommutativity follows by swapping tensor factors in its two displayed values.

On the ordered basis $1\otimes 1,1\otimes X,X\otimes 1,X\otimes X$, each of $(m_L\otimes \mathrm{i}\mathrm{d})(\mathrm{i}\mathrm{d}\otimes {\mathrm{\Delta }}_L)$, ${\mathrm{\Delta }}_L{m}_L$, and $(\mathrm{i}\mathrm{d}\otimes m_L)({\mathrm{\Delta }}_L\otimes \mathrm{i}\mathrm{d})$ has outputs $1\otimes X+X\otimes 1$, $1\otimes 1+X\otimes X$, $1\otimes 1+X\otimes X$, and $1\otimes X+X\otimes 1$. The Frobenius identities therefore hold. On a cube square, two changes on disjoint circles commute by tensor functoriality; two merges give the same composite by associativity of $m_L$; two splits give the same composite by coassociativity of ${\mathrm{\Delta }}_L$; and a merge-split pair gives the same composite by the Frobenius identities just checked. In each case the cube signs on the two routes are opposite, so the square contribution to $d_L^2$ is zero. Every length-two path in the cube lies in one such square, so summing over squares proves $d_L^2=0$.

The terms of $m_L$ and ${\mathrm{\Delta }}_L$ inherited from $\mathbb{Q}[X]\mathrm{/}(X^2)$ have total quantum degree $0$ after the state shift. The new multiplication term $m_L(X\otimes X)=1$ changes the label degree from $-2$ to $1$, and the edge state shift adds $1$, so it raises $q$ by $4$. The new term $1\otimes 1$ in ${\mathrm{\Delta }}_L(X)$ changes label degree from $-1$ to $2$, and its edge state shift again adds $1$, so it raises $q$ by $4$. Thus every differential term preserves or raises $q$, making each $F^p{C}_L$ a subcomplex. The state set and each state module are finite, so only finitely many $q$-degrees occur and the filtration is finite. All algebra and cube checks use finite bases and no choice axiom.

A cancellation identity over any ring. Suppose a cochain differential $d$ on $M=A\oplus B\oplus R$ has a degree-one block $b:A\to B$ which is an isomorphism. Set $h{\mathrm{\mid }}_B=b^{-1}$ and $h{\mathrm{\mid }}_{A\oplus R}=0$. Then $h^2=0$ and $hdh=h$. Consequently $E=1-dh-hd$ is a projection commuting with $d$: expansion using $d^2=h^2=0$ and $hdh=h$ gives $E^2=E$ and $dE=Ed$. Writing $j:R\to M$ and $q:M\to R$ for the direct-sum maps, set $i=Ej=j-hdj$, $p=qE=q-qdh$, and $d_R=q(d-dhd)j$. The $A$-coordinate of a vector in $\mathrm{im}E$ is minus $hd$ of its $R$-coordinate, and its $B$-coordinate is zero; hence $q:\mathrm{im}E\to R$ and $Ej$ are inverse. Thus $pi=1$, $ip=E$, $di=i{d}_R$, $pd=d_{R}p$, and $1-ip=dh+hd$. These identities also give $d_R^2=0$. If the blocks and their inverse are right-module maps, so are all these maps. This calculation applies to entire finite exterior cubes at once; their differentials are included in $d$, rather than omitted as spectators.

The RII four-state model. Call the straight matching $I$ and the cup-cap matching $U$. Rotate the disk and order the crossing whose smoothing pair is $(I,U)$ first, followed by the crossing whose smoothing pair is $(U,I)$. This order refers to the geometric smoothings; for either orientation of the two strands the pair contains one oriented positive and one oriented negative crossing. The local states $00,01,10,11$ are $U,I,U\bigsqcup z,U$. In state $10$ label the circle through the upper arc $a$, the small circle $z$, and the circle through the lower arc $c$; the exterior closure may identify $a$ and $c$. These labels therefore denote actual generators, with that equality imposed when applicable. At $00\to 10$ the map is the split $(a-z)\wedge (-)$ after choosing its lift; at $10\to 11$ the quotient identifies $z$ with $c$. Each edge has its prescribed sign, which is $\pm 1$ and may depend on $v$. In $10$ set $Y=z-c$ and write $L=\mathrm{\Lambda }W$, $K=Y\wedge \mathrm{\Lambda }W$, where $W$ is the circle group of $U$. The quotient $10\to 11$ is the identity on $L$ and zero on $K$. The $K$-coefficient of $00\to 10$ is minus the identity, because $a-z=(a-c)-Y$. This remains true when $a=c$. Thus both indicated blocks are invertible right-module blocks for every exterior closure.

The merge curl. Write $W_v$ for the old circle group and $c_v$ for the circle through the curl's continuing arc. The two local states are $\mathrm{\Lambda }(W_v\oplus \mathbb{Z}z)$ and $\mathrm{\Lambda }{W}_v$, with merge quotient $m(z)=m(c_v)=c_v$. Put $Y=z-c_v$ and decompose the first state as $L_v\oplus K_v$, where $L_v=\mathrm{\Lambda }{W}_v$ and $K_v=Y\wedge \mathrm{\Lambda }{W}_v$. The merge is the identity $L_v\to \mathrm{\Lambda }{W}_v$ and kills $K_v$. External edges preserve both summands. Indeed a merge takes $Y$ to its new $Y$, while a split takes $Y$ to the new $Y$ up to a multiple of its own split difference; that extra term vanishes when wedged with the same difference. On $L$ the external maps are the old maps. On $K$, identifying $Y\wedge w$ with $w$ negates each external split and leaves each merge unchanged. Put $\sigma (v)=(k(v)-k(0)+\mathrm{\mid }v\mathrm{\mid })\mathrm{/}2$. It is an integer, unchanged by a merge and increased by one by a split. Thus $w\mapsto (-1{)}^{\sigma (v)}Y\wedge w$ identifies $K$ with the old complex. The other two summands cancel by step 1.1, using the signed merge pivot and its inverse. With the ordinary cube ordering its pivot coefficient can be taken $(-1{)}^{\mathrm{\mid }v\mathrm{\mid }}$. The normalization makes the survivor's cohomological and quantum degrees those of the old diagram: the extra small circle and extra exterior generator change the local quantum degree by $-1$, cancelling the positive-curl shift $+1$.