Zero traces and the Conway nonsliceness theorem

The two isotopy classes are fixed by the two named diagrams; no equivalence of knots is inferred from their polynomial or homological invariants.

The orientations are auxiliary choices for the later oriented invariants; the underlying knot types in this definition are fixed by the diagrams and do not depend on a choice of a different representative in the same ambient isotopy class.

The attaching framing determines a parallel curve on the boundary torus of the chosen tubular neighborhood. By [F1], the zero framing is the one whose parallel represents the preferred longitude, which is the stated attaching convention.

The construction uses the supplied preferred-longitude representative as attaching data; it makes no uniqueness claim about such representatives and does not change the chosen attachment.

The definition specifies $R$ as dotted and $B,G$ as zero-framed attaching circles. Thus $R$ contributes one $1$-handle, while $B$ and $G$ contribute the two $2$-handles in the stated handlebody.

The two meridional-sublink conditions are invariant under ambient isotopy, and each says that the indicated $2$-handle goes geometrically once over the $1$-handle. The separately stated mixed-crossing condition is a link invariant by [F4], while [F3] fixes the crossing-sign convention; it is not inferred from either meridional-sublink condition.

The solid torus $N(K)=S^1\times D^2$ deformation retracts to its core circle, so by [F4] it has $H_1(N(K);\mathbb{Z})\cong \mathbb{Z}$ and $H_2(N(K);\mathbb{Z})=0$. The product torus has $H_1\cong {\mathbb{Z}}^2$ by [F2]; inclusion sends the product section to the core generator and the meridian to zero because the meridian bounds a fiber disk. Hence $H_1(\mathrm{\partial }N(K))\to H_1(N(K))$ is surjective with kernel exactly $\mathbb{Z}[{\mu }_K]$. Exactness in [F3] identifies $H_2(N(K),\mathrm{\partial }N(K);\mathbb{Z})$ with $\mathbb{Z}$, and its boundary map sends a relative normal disk generator to $[{\mu }_K]$.

Enlarge $N(K)$ by a product collar and apply excision [F3] to the pair $(S^3,E_K)$. It identifies $H_2(S^3,E_K;\mathbb{Z})$ with $H_2(N(K),\mathrm{\partial }N(K);\mathbb{Z})$, with the normal disk generator preserved. By the long exact sequence of $(S^3,E_K)$ and [F4], the connecting map $H_2(S^3,E_K)\to H_1(E_K)$ is an isomorphism, because the adjacent groups $H_2(S^3)$ and $H_1(S^3)$ vanish. Naturality sends the relative normal disk to its boundary positive meridian. Thus $H_1(E_K;\mathbb{Z})$ is infinite cyclic on $[{\mu }_K]$.

Let $q$ be the unique point of $B\cap \mathrm{int}{D}_R$, and put the current attaching circle $G$ transverse to $D_R$. Its intersection set is finite: it is a compact zero-dimensional transverse submanifold, disjoint from $\mathrm{\partial }{D}_R=R$. If it is empty there is nothing to do. Otherwise choose $p\in G\cap \mathrm{int}{D}_R$. A sufficiently close framed parallel $B^{\mathrm{\prime }}$ of $B$ is disjoint from $B\cup G\cup R$ and meets $D_R$ once, at a point $q^{\mathrm{\prime }}$ close to $q$. Indeed near $q$ the transverse arc $B$ has product coordinates in which $D_R$ is $z=0$ and $B$ is a vertical interval; choose the parallel's displacement there tangent to the disk. Away from this coordinate ball the compact sets $B$ and $D_R$ are disjoint, so a sufficiently small parallel creates no other intersection. The displacement can be interpolated in the framed normal disk bundle without changing the prescribed parallel's framing.

Orient the signed parallel $B^{\mathrm{\prime }}$ so that its intersection sign with $D_R$ at $q^{\mathrm{\prime }}$ is opposite to the sign of the intersection of $G$ with $D_R$ at $p$. Reversing the orientation of the parallel changes this sign without changing its underlying framed curve, and [F2] permits either signed parallel. Join $p$ to $q^{\mathrm{\prime }}$ by a smooth embedded arc $a$ in $\mathrm{int}{D}_R$ whose interior misses $G\cap D_R$, $q$, and $q^{\mathrm{\prime }}$. Such an arc is obtained by a polygonal path in the disk minus the finitely many forbidden points, removing loops in its finite subdivision and smoothing its corners in small disjoint disks. A thin strip about $a$ misses the remaining intersections and $B$. Choose the strip's normal product neighborhood so that the short branches of $G$ at $p$ and the chosen oriented $B^{\mathrm{\prime }}$ at $q^{\mathrm{\prime }}$ are vertical, while its interior misses all three retained components. In its positive side, join short intervals of these vertical branches by an embedded band running just above $a$. Give the band its product framing and choose the band-end coordinates by the full seam convention in [F2], so its orientation extends the orientation of $G$ over the selected signed parallel. Band summing $G$ with this $B^{\mathrm{\prime }}$ is therefore an allowed slide over $B$.

One of the new band-sum arcs runs along one edge of the band between $p$ and $q^{\mathrm{\prime }}$. Together with a parallel of $a$ in $D_R$, it bounds an embedded product disk in the positive half-neighborhood. Of the two endpoints, one has positive intersection sign and the other negative sign. Orienting this arc from the positive-sign crossing to the negative-sign crossing, it enters the positive side of $D_R$ at the first endpoint and exits at the second; thus the entire arc lies on the positive side. Its interior misses the slid circle and the retained components: the other edge of the band is chosen on the opposite transverse side of this disk, and the rest of $B^{\mathrm{\prime }}$ and $G$ misses the neighborhood. Push this arc across the disk and slightly into the negative side of $D_R$, rounding its ends. In strip coordinates this replaces the arch above $z=0$ joining the two opposite-sign intersections by an arc below $z=0$; extend the motion to its thin tubular neighborhood. This is a link isotopy supported away from $B\cup R$. It removes precisely $p,q^{\mathrm{\prime }}$ and creates none, while all old intersections away from $p$ persist. Thus a slide followed by this isotopy changes $\mathrm{\mid }G\cap D_R\mathrm{\mid }$ from $N$ to $N+1-2=N-1$. Transport the attaching framing by this isotopy.

Repeat steps 1.1–3.1 for the current attaching circle whenever an intersection remains. Each new parallel and strip can be chosen sufficiently thin for that finite stage, and the fixed $B,R,D_R$ are retained. The nonnegative integer intersection count drops by one at every stage, so after exactly the original number of intersections the current attaching circle is disjoint from $D_R$. The resulting finite sequence consists solely of the permitted framed slides and link isotopies and proves the full statement.

The portion of $\mathrm{\partial }X$ added by deleting $\nu (D)$ is the normal-circle bundle $D^2\times S^1$, and it meets $E_K$ in $S^1\times S^1$. By excision [F2], the relative pair $(B^4,X)$ has the homology of $(D^2\times D^2,D^2\times S^1)$. The latter solid-torus subspace has $H_1\cong \mathbb{Z}$ generated by a normal circle and $H_2=0$, while $D^2\times D^2$ is contractible by [F3]. Its pair sequence [F2] therefore identifies $H_2(B^4,X;\mathbb{Z})$ with $\mathbb{Z}$, generated by a normal disk whose boundary is that normal circle.

In the pair sequence of $(B^4,X)$, the segment $H_2(B^4)⟶H_2(B^4,X)\stackrel{\mathrm{\partial }}{\to }{H}_1(X)⟶H_1(B^4)$ has zero end groups by [F3]. Hence $\mathrm{\partial }$ is an isomorphism. The generator from step 1.1 maps to the normal circle on $D^2\times S^1$. A path in the base disk $D^2$ from an interior point to a boundary point gives a homotopy of these normal circles to the boundary meridian of $K$, with orientation fixed by [F4]. Thus $H_1(X;\mathbb{Z})=\mathbb{Z}[{\mu }_X]$.

The inclusion $E_K\hookrightarrow X$ carries the boundary meridian ${\mu }_K$ to the same normal circle ${\mu }_X$. It is therefore an isomorphism between the infinite cyclic groups in [F1] and step 2.1.

Fix disjoint oriented knots $K,J\subset S^3$ with a regular oriented diagram $D$, and write $[J]=m[{\mu }_K]$ in $H_1(E_K;\mathbb{Z})$ using [F7]. The inclusion $E_K\hookrightarrow S^3\setminus K$ is a homotopy equivalence: in the punctured tube $N(K)\setminus K$, radial motion in each normal disk retracts onto its boundary torus. Thus the same coefficient $m$ is obtained in the open complement. Let $\mathrm{\ell }(D)$ be the half-sum of mixed crossing signs. At a mixed crossing where $K$ passes over $J$ with sign $ϵ$, switching the crossing changes $\mathrm{\ell }$ by $-ϵ$. In an oriented crossing ball straighten $K$ to the positive $x$-axis. When $ϵ=+1$, the old underpassing arc of $J$ runs in the positive $y$ direction, and the new arc above $K$ minus the old arc represents the negative positive-meridian class; when $ϵ=-1$, reversing that tangent reverses the class. Hence the same switch changes $[J]$ by $-ϵ[{\mu }_K]$, so it changes $m$ by $-ϵ$. After switching all mixed crossings, $J$ passes over $K$ at every projected intersection. Raise $J$ in the viewing-normal direction and move its projection into a disk disjoint from the projection of $K$; this is an isotopy in $S^3\setminus K$. The final $J$ lies in a $3$-ball disjoint from $K$, so its homology class is zero, and a regular diagram with separated projections has half-sum zero. By [F6] the half-sum is unchanged by this isotopy. Since every crossing switch changed the half-sum and meridian coefficient by the same integer, their initial values agree: $\mathrm{\ell }(D)=m$. In particular, for any framed parallel $L_f$ of a knot $L$, its framing integer measured by this half-sum is its meridian coefficient in $H_1(E_L)$; it is zero exactly when $[L_f]=0$. A degree-one parallel with zero class is a preferred longitude by [F8].

Choose a spanning disk $D_R$ for the unknotted dotted circle $R$ so that $B$ meets it once, as allowed by [F1]. Apply [F5] to obtain a finite sequence of slides of $G$ over $B$ and isotopies after which the remaining attaching circle is disjoint from $D_R$. The supplier clears actual intersections one at a time; it does not infer geometric disjointness from a zero signed count.

At each slide of $G$ over $B$, the oriented band sum represents $[G]\pm [B_f]$ in $H_1(E_B)$, where $B_f$ is the signed framed parallel of $B$. By step 1.1, the meridian coordinate of $[B_f]$ is its framing integer $f(B)$. Thus the new linking number is $\mathrm{lk}(B,G)\pm f(B)$. The initial framing pushoff of $B$ is its preferred longitude by [F1], so [F8] makes its class zero and step 1.1 gives $f(B)=0$. Therefore $\mathrm{lk}(B,G)$ remains zero after every slide.

For the same band sum, expanding the linking of the new attaching circle with its transported pushoff gives $f(G\mathrm{\#}(\pm B))=f(G)+f(B)\pm 2\mathrm{lk}(B,G)\mathrm{.}$ The two self-pairings are the old framings and the two cross-pairings are equal by the symmetry in [F4]. The expansion is valid because the band and its pushoff are carried in disjoint product strips, so the four old pairings are unchanged outside the strips. Each framing integer is the meridian coordinate of its pushoff by step 1.1. All framing terms are zero initially by [F1], [F8] and step 1.1, and the mutual linking remains zero by step 2.1. Induction over the finite sequence in [F5] shows the final attaching framing is zero; isotopies preserve it.

The circle $B$ still goes geometrically once over the $1$-handle $R$, while the remaining attaching circle is disjoint from the cancellation disk. Therefore the attaching sphere of the $2$-handle $B$ meets the belt sphere of the $1$-handle once, and [F3] cancels this pair. Since the remaining attaching circle and its framing lie outside the cancellation neighborhood, the transported remaining handle is a single $0$-framed $2$-handle on a $4$-ball. By [F1] this is $X_0(K_G)$. Slides preserve the handlebody by [F2], so $X(B,G,R)\cong X_0(K_G)$.

Interchange $B$ and $G$. The other meridional condition supplies the starting intersection for [F5], and the linking number remains zero by [F4]. The reversed slide sequence preserves both the zero mutual linking and zero framing by the calculations in steps 2.1 and 3.1. Cancel $(R,G)$ by the same one-intersection argument as in step 4.1. This gives $X(B,G,R)\cong X_0(K_B)$; combining with step 4.1 yields both trace descriptions.

Suppose $D\subset B_1$ is a smooth proper slice disk for $K$. Its oriented normal $2$-plane bundle is trivial because $D$ is a disk. Choose a nowhere-zero normal section and push $D$ to a parallel disk $D^{\mathrm{\prime }}$ in the exterior $X$ of a smaller product tube of $D$. Its boundary $K^{\mathrm{\prime }}$ is a degree-one parallel of $K$, identified radially with a curve on the fixed boundary torus. It bounds $D^{\mathrm{\prime }}$ in $X$, so [F3] makes its class zero in $H_1(E_K)$. It is therefore a preferred longitude in the sense of [F2]. To compare with the fixed framing, let $s_0$ be its section pushoff representing ${\lambda }_K$. In product coordinates on the tube, the classes of $s_0$ and ${\mu }_K$ form a basis of the torus homology: its two coordinate circles generate $H_1(S^1\times S^1;\mathbb{Z})$. The disk-induced section has class $[K^{\mathrm{\prime }}]=[s_0]+n[{\mu }_K]$, where $n$ is the winding number of its normal direction relative to $s_0$. Both section classes vanish in $H_1(E_K)$, so $n=0$ by [F6]. The relative direction map $S^1\to S^1$ consequently has a periodic real-valued angle lift; scaling that lift to zero homotopes the two normal framings. Fibre rotations along this homotopy give isotopic attaching tubes. Thus the disk tube induces the fixed preferred-longitude framing, rather than merely some curve satisfying [F2].

Conversely, suppose $i:X_0(K)\hookrightarrow S^4$ is a smooth codimension-zero embedding. In the $0$-handle of $X_0(K)$ take the cone on $K$ from an interior point $p$, and join that cone disk to the core disk $D^2\times \{0\}$ of the $2$-handle along their common boundary $K$. After smoothing the join, this is a piecewise-smooth embedded $2$-sphere, smooth away from the cone point $p$.

Use oriented coordinates at $p$ and standard stereographic coordinates centred at $i(p)$. The derivative of $i$ is invertible by [F4], because its differential is injective and both dimensions are four. By [F5], $i$ is a local diffeomorphism near $p$. If its local orientation sign is negative, compose it with a fixed orientation-reversing reflection of the standard $S^4$; this preserves the existence of the embedding and makes the local sign positive. Thus its derivative $A$ lies in ${\mathrm{G}\mathrm{L}}^{+}(4,\mathbb{R})$.

A tubular neighborhood of $D$ is $D^2\times D^2$ and meets the other hemisphere $B_2$ in the attaching region $S^1\times D^2$. Consequently $B_2\cup \nu (D)$ is obtained from a $4$-ball by attaching one $0$-framed $2$-handle along $K$, hence is diffeomorphic to $X_0(K)$ by [F1]. This union is a smooth codimension-zero submanifold of $S^4$.

Identify the cone near $p$ with the radial cone on $K\subset S^3$. Along each direction $u\in K$, the derivative at the cone point of the target distance is $\parallel Au\parallel >0$. Compactness of $K$ makes this derivative uniformly positive, so [F5] gives a smooth radial intersection parameter for all sufficiently small target spheres. Taylor expansion in $C^1$ along $K$, followed by radial normalization, makes the resulting link ${\mathrm{\Gamma }}_{ϵ}$ $C^1$-close to ${\varphi }_A(K)$, where ${\varphi }_A(u)=Au\mathrm{/}\parallel Au\parallel$. The polar decomposition $A=QP$ joins $A$ to the identity through ${\mathrm{G}\mathrm{L}}^{+}(4,\mathbb{R})$: join the positive-definite $P$ to $I$ linearly and join $Q\in \mathrm{S}\mathrm{O}(4)$ to $I$ by a path in the connected group $\mathrm{S}\mathrm{O}(4)$. Hence ${\varphi }_A(K)$ is isotopic to $K$. The chosen normal-disk tube of $K$, transported by ${\varphi }_A$, is a tube of ${\varphi }_A(K)$. For sufficiently small $ϵ$, the tubular projection restricts to a local diffeomorphism from ${\mathrm{\Gamma }}_{ϵ}$ to ${\varphi }_A(K)$. It is homotopic to the identity because the curves are $C^1$-close, so it has degree one; a local diffeomorphism of circles with degree one is a diffeomorphism. Thus ${\mathrm{\Gamma }}_{ϵ}$ is the graph of a small normal section. Scaling that section to zero gives a smooth knot isotopy to ${\varphi }_A(K)$, so ${\mathrm{\Gamma }}_{ϵ}$ is isotopic to $K$.

Choose the round ball bounded by that small sphere in the fixed stereographic chart. Its complement $W=S^4\setminus \mathrm{int}(B_{ϵ})$ is a standard $4$-ball, as seen by inversion in the chart with its point at infinity. Shrink $ϵ$ so that the ball misses the compact part of the image sphere outside the cone chart. The image sphere is smooth away from the cone point; transversality makes its remainder after deleting the cone cap a smooth properly embedded disk in $W$ with product boundary collar and boundary ${\mathrm{\Gamma }}_{ϵ}$. Append the trace annulus of the isotopy in step 2.2 inside a boundary collar of $W$, with the isotopy stationary near its ends. The result is a smooth slice disk with boundary exactly $K$.

Steps 1.1, 2.1 construct an embedding from a slice disk, and steps 1.2, 1.3, 2.2, 3.1 construct a slice disk from an embedding. These are the two implications in the statement.

By [A1] and [A2], every countable-choice premise of the retained local diagram, linking, and handle arguments is available. The published zero-trace equality [F8] is used in the adopted classical full-AC scope, without asserting that the source proves the same result in ZF or under countable choice alone. The local constructions and their original countable-choice strength are preserved.

Let $K,J$ be any disjoint oriented knot components in the links below, and let $D$ be a regular oriented diagram of $K\cup J$. The inclusion $E_K\hookrightarrow S^3\setminus K$ is a homotopy equivalence by radial retraction in the punctured normal tube. By [F5], write $[J]=m[{\mu }_K]$. Let $\mathrm{\ell }(D)$ be the half-sum of mixed crossing signs; [F7] makes it an integer independent of the chosen diagram. At a mixed crossing where $K$ passes over $J$ with sign $ϵ$, switch it. The half-sum changes by $-ϵ$. In the crossing ball, straighten $K$ to the positive $x$-axis: for $ϵ=+1$ the old under-arc of $J$ runs in the positive $y$ direction, and the new arc minus the old arc represents the negative positive-meridian class; for $ϵ=-1$ the class reverses. Thus the switch changes $[J]$ by $-ϵ[{\mu }_K]$, so it changes $m$ by the same integer. After all such switches, $J$ passes over $K$ at every projected intersection. Raise $J$ above $K$ and move its projection into a disjoint disk. This is an isotopy in the complement of $K$; the final pair has a split diagram, so its half-sum is zero, and $[J]=0$ because it lies in a $3$-ball disjoint from $K$. Tracking the equal changes proves $\mathrm{\ell }(D)=m$. In particular, when $J=L_f$ is a framed degree-one parallel of an attaching component $L$, its framing integer is this meridian coefficient. A computed framing value zero then gives the preferred longitude condition by [F6].

A record retains all four cyclic halfports, under/over parity and component names. An RI, RII or RIII record uses, respectively, its actual empty monogon, opposite bigon or triangular face. Its standard local spatial motion is an ambient isotopy. A constant-over or constant-under pickup follows one simple successive overarc or underarc and its specified simple dual-face replacement path: raise the arc into an empty upper slab, or lower it into an empty lower slab, move it relative to its endpoint collars, and return. A finite product-disk sweep realizes this motion. A cyclic, under/over-preserving ribbon isomorphism is an orientation-preserving planar redraw; it is not a mirror change. Carry a sufficiently small normal strip through every motion. In particular an RI changes a blackboard presentation but does not change the transported integer framing. Thus the finite records below specify actual framed attaching isotopies, independently of how an illustration is drawn.

Begin with the literal source panel $F$, with dotted red $R$ and zero-framed green $G$ and blue $B$. Its 36-crossing input has 16 blue self-crossings, 18 blue/red crossings and two green/red crossings. Red is simple and green is its tiny meridian. Choose red and green disks meeting along one radial slit, with the green disk disjoint from the entire blue tube. Put blue transverse to the red disk. At each blue puncture, an untwisted blue-over-green band along a simple disk arc avoiding the other punctures adds a green zero-parallel. Its product arch pushes across an empty bigon, removing that puncture. Such an arc exists because the disk minus the finitely many punctures and its two parallel radial slits is connected. Induction clears the disk; the actual red/green pair cancels outside the retained blue tube. After forgetting that pair, reverse the product-disk pushes to recover the original blue embedding. Its zero framing is retained, since green has zero framing and zero linking with blue. The established source-bound 98-move blue movie and seven-move Figure 1 movie identify this literal blue knot with $C$. Their terminal visit word is $U2,O8,U7,U11,O10,O9,U6,O2,U5,O7,U9,O6,U8,O5,O4,U3,O1,U4,O11,U10,O3,U1$ in the recorded Figure 1 crossing-center order. Hence the $F$ handlebody is $X_0(C)$.

The two early dotted-circle pushes preserve that handlebody. In the outgoing boundary of the dotted unknot, the zero longitude $R_0$ bounds the added solid torus's meridian disk. An untwisted blue-to-$R_0$ band and this disk form a product bigon disjoint from the old-exterior attaching link. Pushing across it isotopes the whole blue tube. For $F$, use the inward red zero-parallel at distance 8, with the recorded cuts at heights 1017 and 1027; the 56-crossing output has the established 255-move marked isotopy to $G$. For $G$, use the outer red zero-parallel at distance 8 and heights 1272 and 1282; the established 69-move marked isotopy gives $H$. The first band adds $+R_0$ with $\mathrm{lk}(B,R)=-1$, and the second adds $-R_0$, so the blue integer is successively $-2$ and $0$. By step 1.2, each of these framing integers is the meridian coefficient of the actual transported pushoff, so the terminal value zero certifies the preferred-longitude framing. Indeed the two parallel band rectangles cancel in the push-off expansion and give $f(B\mathrm{\#}ϵR_0)=f(B)+2ϵ\mathrm{lk}(B,R)$. Green and its entire parallel are carried throughout. Thus $H$ still presents $X_0(C)$, with green and blue both zero-framed.

Use literal $H$ as the dualizable witness. Its blue/red sublink has exactly the two Hopf crossings. Its 16-crossing green/red sublink has the recorded seven-move RI/RII/constant-height movie to the two-crossing Hopf sublink. These are meridional sublinks by their actual attaching isotopies, not by a numerical linking criterion. The green/blue crossing table has seven positive and seven negative crossings, so their mutual linking is zero. Together with step 4.1 this establishes every condition in [F2].

The established 149-move marked movie carries $H$ to $I$. For the $I$–$J$ redraw, remove the six-crossing green finger in each panel and cap its tails by the specified disjoint short cap. Their 29-crossing contexts have the recorded orientation-preserving ribbon isomorphism, including the cap edge $(17,1)$--$(18,1)$. Each finger has two simple parallel under-band sides and a positive blue meridian head; all undercrossings occur in reversed pairs on the same retained edges. Under the context map the heads lie on opposite blue halfedges at $J19$, where blue is under red. Slide the head through that blue product tube below the red overpass. Lower both side strips into an empty lower slab, straighten their simple cores by product-disk pushes relative to the endpoint collars, and reverse this straightening to the target strips. This carries the whole finger and normal strip while fixing the context. In $K$, the three reverse ordinary green-over-blue bands use the cuts on green paths 18, 19, 20 at heights 1500, 1510, 1520 and blue zero-parallels at distances $0.4,0.8,1.2$. Their 128-crossing output has the established 926-move marked equivalence to $J$. Reverse these slides to reach literal $K$. At every ordinary slide, blue retains framing zero and green–blue linking remains zero; the untwisted push-off calculation gives $f(G^{\mathrm{\prime }})=f(G)+f(B)+2ϵ\mathrm{lk}(G,B)=0$ and $\mathrm{lk}(G^{\mathrm{\prime }},B)=\mathrm{lk}(G,B)+ϵf(B)=0$. Consequently the handlebody and the actual zero framings are retained.

Perform the two actual nested green-over-blue bands from K2-nested-1.py. Their green cuts are centered at $(1914,1835.5)$ and $(1892.5,1928)$, of widths $0.3$ and $0.08$, and their blue zero-parallels are the planar normal offsets $0.8$ and $0.4$. Their cores follow the literal orange interruption paths. The second runs through the first deleted green interval, the interior of the first strip, and the first parallel's attachment gap. Put the first strip above old blue/red and below old green; put the second below the entire current green, including the first strip and parallel, and above old blue/red. The first strip is above the future blue parallel and above the second strip at their projected intersections. These strict heights in the disjoint crossing balls, joined along the empty core intervals, give embedded untwisted ribbons with the specified endpoint rectangles. The narrower second attachment lies in the first gap. Thus these are sequential ordinary slides. Their exact marked output is the 100-crossing nested input, and step 5.2's framing formula retains both zero framings and zero mutual linking.

The recorded twelve whole-link moves carry that input to 70 crossings: 56 green/green, 12 blue/green and two blue/red, with no red/green crossing. The blue/red pair has Hopf sign and opposite relative heights. The actual red projection is simple; its green-empty Jordan disk consists of the recorded dual faces 48 and 49, and blue meets it in a single simple crosscut. Flatten this disk and straighten the crosscut to monotone height, carrying the blue tube and fixing the disjoint green strip. The disk then meets blue once. The one/two-handle cancellation is supported in the tube about the outside blue arc and the feet, disjoint from green. It leaves precisely the recorded green56 knot with its transported zero framing. On that knot, the three RIII triples are $\{50,45,51\},\{50,44,52\},\{50,43,53\}$. Literal $L_{64}$ has its five recorded RI/RII/constant-height cleanup moves to $L_{56}$ and then the three RIII triples $\{25,18,23\},\{25,17,21\},\{25,16,20\}$. The resulting complete cyclic/under-over ribbon graphs are identical. Reverse the $L$-side moves to identify the actual cancelled attaching knot with literal $L$, rather than with a knot sharing an invariant.

For $L$–$M$, use the established literal $M59$ movie consisting of 18 RIII moves and the RI removing source crossing 21. On its $M58$ endpoint, perform RIII at halfports $[(17,0),(11,0),(9,1)]$ and pick up the successive five-crossing constant-under arc $[(8,0),(17,0),(9,0),(43,0),(42,0)]$. The pickup removes two crossings. Its full endpoint is the same $L_{56}$ three-RIII endpoint in step 7.1. Reversing the indicated moves therefore identifies literal $L$ with literal $M$, carrying the normal strip.

We now give the actual $M$–$N$ transport. Fix the recorded 26-crossing exterior and its twelve individual port collars. The six proper arcs pair port $i$ with port $11-i$, for $0\le i\le 5$. Start with the source-bound 39-move $M$ prefix: the initial 19 moves of step 8.1, followed by four inverse-RII insertions and sixteen RIII moves transporting the head through the three return pairs and the next body-exit pair. Its endpoint has 66 crossings, with the inserted crossings labeled $100,\dots ,107$. This prefix alone is not a terminal identification. The following explicit suffix supplies the missing coupon and passenger transport while fixing the exterior. All triples below name the actual triangular faces in their immediately preceding cyclic graphs, not arbitrary crossing permutations.

First perform RIII at $\{2,107,18\},\{3,107,38\},\{19,106,18\},\{37,38,107\}$. These correct the over-transported last passenger portions. For the first two passengers, perform in order $\{52,12,101\},\{5,12,103\},\{1,12,105\},\{51,8,101\},\{4,8,103\},\{0,8,105\},\{52,11,100\},\{5,11,102\},\{1,11,104\},\{51,9,100\},\{4,9,102\},\{0,9,104\}$. Each step exchanges the next adjacent out-of-order retained visits on one passenger. Under the retained crossing-role map to $N$, the four complete passenger words are now exactly $[24,25,26,8,10,12]$, $[36,37,38,9,11,13]$, $[43,44,4,40,33,29]$, and $[45,46,3,39,32,28]$, with the actual under/over parities retained. This realizes the passenger block exchange geometrically, instead of postulating it as a substitution.

Transport the coupon by these twelve successive RIII moves: $\{53,103,105\},\{54,101,105\},\{55,101,103\},\{56,103,105\},\{57,101,105\},\{58,101,103\},\{53,102,104\},\{54,100,104\},\{55,100,102\},\{56,102,104\},\{57,100,104\},\{58,100,102\}$. They pass the two three-crossing return patterns through the six-crossing box in its actual three-strand chart. Then use $\{26,101,103\},\{26,100,102\},\{31,101,105\},\{31,100,104\}$. The resulting actual empty bigon removes crossings $106,107$ by RII, and the actual empty monogon removes $103$ by RI. Perform RIII at $\{105,102,104\}$ and remove the resulting empty monogon $104$ by RI. These twenty moves, following step 10.1's sixteen, give a 62-crossing endpoint. Their supports lie in the remaining patch, so every fixed exterior collar and the carried zero-normal strip are retained.

This endpoint is the local graph $N$ encoded by N-pd.json, with its marked port data. The explicit ribbon isomorphism in coupon-suffix-03.json is anchored at source crossing 6 mapping to $N48$ and agrees with the prescribed map of all 26 exterior crossings. It preserves all 248 halfedge adjacencies, even crossing rotations and under/over parities. Following opposite halfedges from each of the twelve collars gives all six complete proper-arc visit words of literal $N$, including both long core arcs, not just their endpoint pairings. The six packet roles are $26\mapsto 1$, $31\mapsto 7$, $100\mapsto 5$, $101\mapsto 6$, $102\mapsto 2$, $105\mapsto 0$. In particular source crossings 26 and 31 survive in this actual movie; the earlier speculative prescription to delete them is unnecessary. The fixed-exterior isomorphism supplies the final relative planar redraw. Thus the preceding transport and endpoint identification prove the framed attaching isotopy to that local $N$ graph; they do not bind it to the published panel $N$.

By step 4.1, the handlebody of the local witness $H$ is diffeomorphic to $X_0(C)$. Independently, the published zero-trace equality [F8] identifies $X_0(C)$ with $X_0(K_P)$ for the exact named knots in [F1]. Composing these diffeomorphisms gives a diffeomorphism from the handlebody of $H$ to $X_0(K_P)$. No source-panel identification of the local endpoint in step 12.1 is needed. Reversing an auxiliary knot parameter leaves the zero-framed trace unchanged: conjugating both disk factors of the two-handle extends that reversal with positive four-dimensional orientation.

The local witness $H$ is dualizable by step 5.1, its handlebody is $X_0(C)$ by step 4.1, and step 13.1 identifies that same handlebody with $X_0(K_P)$. The local slides in steps 5.2–6.1 and cancellation in step 7.1 retain their framed geometric conclusions, while steps 8.1–12.1 end at the local $N$ graph. The second named trace description uses [F8], without inferring that this graph is the source panel $N$. Thus $H$ has both stated zero-trace descriptions, with the component and zero-framing conventions preserved.

Transcribe the left knot in [F1], ignoring its red mutation circle. Number its crossings $1,\dots ,11$ at the following approximate pixel centers in the original $5323\times 1843$ source image: $(1540,216),(310,538),(1867,633),(1335,806),(877,840),(175,919),(884,1180),(375,871),(410,1527),(1890,1393),(1443,1230)$. These coordinates identify the crossing neighborhoods, not a metric hypothesis. Start at the underpass of crossing $2$ and follow the strand toward crossing $8$. The complete cyclic under/over visit word is $U2,O8,U7,U11,O10,O9,U6,O2,U5,O7,U9,O6,U8,O5,O4,U3,O1,U4,O11,U10,O3,U1$. Label the outgoing arc after each successive visit by $0,\dots ,21$. Listing each crossing counterclockwise, starting with its incoming under-arc, gives the following PD list: $[(20,15,21,16),(21,6,0,7),(14,19,15,20),(16,13,17,14),(7,12,8,13),(5,11,6,10),(1,9,2,8),(11,1,12,0),(9,5,10,4),(18,4,19,3),(2,18,3,17)]$. Each of the eleven source gaps is used once as an underpass; the matching unbroken source arc is its overpass. The two extra apparent breaks on the horizontal arcs and the bottom break occur at the red mutation circle and are joined without any knot crossing. Thus this is a transcription of the specified diagram, rather than a knot-table identification.

Let $W=(3,-2,3,-2,3,-1,3,-1,2,2,-1)$ be a four-strand braid read left to right, where $+j$ has strand $j$ over strand $j+1$. Close corresponding ends outside the braid without extra crossings. Its PD list, obtained by following its four strands, is $[(12,6,13,5),(6,1,7,2),(2,14,3,13),(14,7,15,8),(8,4,9,3),(15,20,16,21),(4,10,5,9),(21,16,0,17),(17,11,18,10),(11,19,12,18),(19,0,20,1)]$. The arc relabeling $a\mapsto 14-a(\mathrm{m}\mathrm{o}\mathrm{d}22)$ carries the list of step 1.1 to this list: its crossings go respectively to $(6,4,8,11,2,5,1,3,7,9,10)$, with cyclic rotation by two at every crossing. These are equal connected oriented ribbon diagrams with the same overpasses. Their equivalence involves no crossing change: thicken corresponding vertices and edges in the projection sphere, identify these strips by the specified cyclic orders, and extend over the complementary disk faces. One may realize the identification by first matching a thickened spanning tree, then matching the remaining arcs in their disk faces by innermost-bigon isotopies, and finally using the disk Alexander isotopy on the faces. Lifting the crossing neighborhoods with their retained over/under heights gives an ambient isotopy of the knots and their Seifert disk-and-band data. Either direction along the single knot is available; the reversed orientation will be handled as well.