Bridge presentations and unknotting number

Regard $K$ as a compact embedded smooth 1-manifold in $S^3\subset {\mathbb{R}}^4$. By [F1], the exceptional set $E$ of non-Morse directions is null in every smooth atlas of $S^3$. It cannot contain all of $S^3$, since a nonempty coordinate ball has positive Euclidean measure; choose $u\in S^3\setminus E$. Then $x\mapsto u\cdot x$ is Morse on $K$.

Choose $R\in SO(4)$ with $R^T{e}_4=u$. Such a rotation is obtained in the plane spanned by $u$ and $e_4$; if $u=-e_4$, use a half-turn in a plane containing $e_4$. A rotation path $R_t$ from the identity to $R$ is an ambient isotopy of the oriented sphere. On $R(K)$, the fixed height equals $e_4\cdot Rx=u\cdot x$, which is Morse by step 1.1. Thus $K^{\mathrm{\prime }}=R(K)$ is the required representative.

Let $m$ be the unique maximum. The two components of $a\setminus \{m\}$ are strictly monotone in height, so each intermediate regular level meets $a$ in exactly two points. In a Morse chart on $a$ near $m$, the height is $h(m)-t^2$; a short level arc joining the two nearby points and the subarc through $m$ bound a small embedded cap disk. Under [F1], the two points on all lower regular levels form a smooth path of distinct ordered pairs in $S^2$.

Apply [F2] to the two-point isotopy of step 1.1, starting at the cap level. The extended sphere isotopy transports the short cap arc through simple arcs on the intervening levels. Their union is an embedded strip because distinct levels are disjoint; adjoining the cap disk gives an embedded disk with boundary $a\cup {\beta }_0$ for a simple arc ${\beta }_0\subset S_c$.

Let ${\gamma }_0$ be the initial cap arc on the reference $S^2$, and let $f_r$ be the ambient isotopy from [F2] carrying its endpoint pair to the two intersection points on level $S_r$. The arc $f_c^{-1}(\beta )$ is a simple arc with the same endpoints as ${\gamma }_0$. Bigon elimination gives an isotopy $g_t$ of $S^2$, fixed at those endpoints, with $g_1({\gamma }_0)=f_c^{-1}(\beta )$. Precompose the transported cap arc by $g_{\lambda (r)}$ before applying $f_r$, where $\lambda$ is zero near the cap level and increases smoothly to $1$ at $r=c$. Every $g_t$ fixes the reference endpoints, so the joining arcs still have the required endpoint tracks; at $r=c$ the joining arc is exactly $\beta$. The union of this family with the cap disk is therefore an embedded disk bounded by $a\cup \beta$. Reversing height proves the lower-ball statement.

Let $W=\ker (A-B)$. Rank-nullity and the rank hypothesis give $\mathrm{codim}W\le 1$. If $P$ is a positive subspace for $A$ of dimension $n_{+}(A)$, then $\dim (P\cap W)\ge n_{+}(A)-1$. On $P\cap W$ the forms agree, so this intersection is positive for $B$. Thus $n_{+}(B)\ge n_{+}(A)-1$. Interchanging $A$ and $B$ proves $\mathrm{\mid }{n}_{+}(A)-n_{+}(B)\mathrm{\mid }\le 1$.

Apply the same argument to $-A$ and $-B$ to obtain $\mathrm{\mid }{n}_{-}(A)-n_{-}(B)\mathrm{\mid }\le 1$. Therefore $\mathrm{\mid }\mathrm{sign}(A)-\mathrm{sign}(B)\mathrm{\mid }\le \mathrm{\mid }{n}_{+}(A)-n_{+}(B)\mathrm{\mid }+\mathrm{\mid }{n}_{-}(A)-n_{-}(B)\mathrm{\mid }\le 2$.

A smooth embedded circle has empty interior in $S^3$: in every submanifold chart it is a coordinate line, which contains no open three-dimensional ball. Choose $p\in S^3\setminus K$. Choose an orientation-preserving rotation $R\in SO(4)$ carrying $p$ to the fixed point at infinity; a rotation path from the identity to $R$ restricts to an ambient isotopy of $S^3$. Thus $R(K)$ misses infinity, and stereographic coordinates regard it as a compact smooth oriented knot in ${\mathbb{R}}^3$. Apply [F1] to this representative and list the finitely many crossings of the resulting projection. Around each crossing choose a closed disk meeting the image in exactly two transverse arcs, with all such disks pairwise disjoint. Subdivide the remaining compact arcs into finitely many subarcs contained in planar coordinate disks where each arc is a graph over a line interval and where nonincident pieces are separated. Compactness and finiteness give a positive minimum for the separations between the disjoint pieces.

In each crossing disk, choose a nearby rational point $q$ and move the crossing there by a small planar isotopy supported in that disk. Choose two linearly independent rational direction vectors close to the two tangent directions there. Small rational multiples in both signs give four rational endpoints about $q$, in the same cyclic order as the branch germs. Replace the branches near $q$ by the two straight diameters through these endpoints, joining back inside the corresponding four disjoint sectors; this is a small isotopy of the crossing. Outside these disks, approximate each graph arc from step 1.1 by a sufficiently fine polygonal graph and perturb its finitely many vertices to rational coordinates. Choose all perturbations smaller than one third of the positive separation bound, so distinct nonincident pieces remain disjoint and every adjacent segment stays in its graph chart.

Orient the polygonal edges by the original circle parameter, pair the four crossing half-edges as on the original branches, and retain each crossing's over/under label. In each graph chart interpolate linearly between the original graph and the rational polygonal graph; the graphs remain embedded, their boundary endpoints agree, and their disjoint neighborhoods prevent new intersections. In each crossing disk the two interpolated branches keep the same four sectors and nonzero crossing determinant at the fixed rational point. Smooth each bend inside a smaller disk disjoint from all crossings. This gives an isotopy through diagrams with exactly the original crossing data. Lift the over/under heights continuously; strict height inequalities at each crossing persist, so this is an isotopy of embedded circles, which extends to an ambient isotopy by [F4]. The resulting rational polygonal data satisfy all clauses of [F2].

The polygonal drawing has finitely many bends. If the basepoint is a bend in the geometric interior of its edge, choose a short forward segment within that edge containing no crossing or other bend and move the basepoint into its interior. The adjusted point satisfies the algorithm's input condition. Moving the point within the same oriented edge crosses no crossing vertex, so it changes neither the cyclic order of crossing encounters nor the first branch reached at any crossing. Thus the adjustment leaves the crossing-change list and updated code independent of its choice. The code has finitely many edges and crossing records by [F1]. Following its coherent half-edge pairings from the basepoint returns to the starting edge after finitely many incidences, so the traversal terminates. It visits each crossing record once at first encounter, giving exactly $c$ processing operations.

Each operation either leaves the over/under label unchanged or exchanges the two labels once. It does not change vertices, edges, cyclic orders, branch pairings, orientations, or the rational drawing certificate; hence all validity conditions in [F1] remain satisfied. The resulting list has at most $c$ entries, and every recorded first underpass has become an overpass while first overpasses remain over.

Cut the based circle at a small crossing-free subarc around the basepoint. Orient the cut subarc from the endpoint where the remaining long arc ends to the endpoint where it begins. Identify the plane with the equatorial sphere $S_0\subset S^3$ minus one point, and write the projected long arc as $\gamma (t)\in S_0$, $0\le t\le 1$, and the cut subarc as $\delta (s)\in S_0$, $0\le s\le 1$. Thus $\gamma (1)=\delta (0)$ and $\delta (1)=\gamma (0)$. Choose $a=1\mathrm{/}2$ and lift these paths by $\widehat{\gamma }(t)=(\sqrt{1-a^2{\cos }^2(\pi t)}\gamma (t),a\cos (\pi t)),\widehat{\delta }(s)=(\sqrt{1-a^2{\cos }^2(\pi s)}\delta (s),-a\cos (\pi s))\mathrm{.}$ The first height strictly decreases from $a$ to $-a$ and the second strictly increases from $-a$ to $a$; their first derivatives vanish at the joins and their second derivatives match. Reparametrize the projected paths near their endpoints to match the smooth based-diagram jets. The resulting closed knot is smooth and has exactly one maximum and one minimum. The projected cut subarc is disjoint from the rest of the diagram except at its endpoints.

At every double point of $\gamma$, the first encounter occurs at smaller $t$ and therefore has strictly larger fourth coordinate; hence the lifted knot has exactly the prescribed overpassing branch at every crossing. The long lifted arc is embedded because its fourth coordinate is strictly monotone, and the base subarc is disjoint from its projection away from the endpoints, so their union is an embedding. To identify this lift with the knot represented by the diagram, let $z_0$ be the smooth height function of any realizing lift over the same smoothed projected circle, and let $z_1$ be the height function specified in step 1.1. For $0\le u\le 1$, set $z_u=(1-u)z_0+u{z}_1$. This is a jointly smooth family with values in $(-1,1)$, since the compact image of $z_0$ lies in $(-1,1)$ and $\mathrm{\mid }{z}_1\mathrm{\mid }\le 1\mathrm{/}2$. At every crossing the first-encounter inequality holds for both $z_0$ and $z_1$, so it holds strictly for every $z_u$; distinct projected points remain distinct, and the projected derivative is nonzero. Thus the corresponding lifts form a smooth isotopy of embeddings of the compact circle. By [F2] it extends to an ambient isotopy, proving that the constructed lift represents the same knot as the descending diagram.

Choose the regular level $S_0$ between the unique minimum and maximum. It cuts the lifted knot into an upper proper arc with one maximum and a lower proper arc with one minimum. Choose one simple joining arc on $S_0$. By [F3], each proper arc bounds an embedded disk with that same joining arc; push their interiors into the opposite open balls, so the disks intersect only in the joining arc. Their union, after smoothing its corner, is an embedded genus-zero spanning surface for the knot.

Apply [F4] to the spanning disk of step 3.1. The lifted knot, and therefore the represented descending diagram by step 2.1, is ambient isotopic to the standard unknot.

Switching over and under does not change the oriented smoothing at the distinguished crossing. Thus [F1] gives the same Seifert circles, disk records, and bands at every other crossing. Only the distinguished band changes: its two possible half-twist directions differ by one full twist. By [F2], both disk-and-band constructions give connected embedded oriented spanning surfaces, and each surface retracts to the finite graph with one vertex per disk and one edge per band.

Let $e$ be the distinguished edge of that graph. If $e$ is a bridge, every graph cycle avoids $e$, so choose the corresponding homology basis outside the changed band; all Seifert-pairing entries are then identical. If $e$ is not a bridge, the graph with $e$ removed is connected. Choose a spanning tree of that graph; its fundamental cycles, together with the fundamental cycle using $e$, form an integral homology basis. All basis curves except the last can be represented disjoint from the distinguished band.

The crossing change is supported in a ball disjoint from every other band and disk. Hence the basis curves avoiding $e$, and their positive push-offs, have identical linking numbers before and after the change. Their pairings with the last basis curve also remain the same, because the other curve and its push-off miss that ball. The last curve passes once through the distinguished band; changing the band by a full twist adds one signed meridional turn to its positive push-off about its core, so its self-linking in [F3] changes by exactly $+1$ or $-1$. Thus only one diagonal entry changes. The matrix difference is zero in the bridge case and is $\pm v{v}^T$ in the non-bridge case, so its rank is at most one.

Apply [F1] to choose an admissible finite code for a regular diagram of a representative of $K$, and choose a basepoint in the interior of one edge away from every crossing and geometric bend. There is such a point: a one-component code has at least one nondegenerate polygonal edge segment, including in the crossing-free case, and only finitely many crossings and bends are forbidden. Choose an interior point of such a segment outside that finite set.

Apply [F2] to that finite code. The algorithm terminates after the finite crossing records are scanned and produces a finite crossing-change list; after those changes, each crossing is first encountered as an overpass.

By [F3], the code after the listed changes represents the standard unknot. Thus the chosen code and its finite crossing-change list witness the asserted existence for $K$.

The annulus $A$ is incompressible: a compression would have boundary essential in $A$ and nullhomotopic in $E$, contrary to the assumed injection. The stated prohibition of disks is precisely boundary-incompressibility. A knot is a nonsplit link, and [F2] supplies irreducibility for the disk-exchange isotopies in the imported result. Hence the $m=2$ case of [F1] applies to this specified $P$ and the fixed bridge sphere $S$. Its conclusion and final puncture redistribution give the one-circle form with one puncture in each bridge ball, rather than merely the existence of an unrelated decomposing sphere.

To realize that finite disk-move sequence smoothly, keep the initial smooth knot and bridge sphere fixed and round the finitely many arcs and disk seams at each stage. The moves in the source proof are disk exchange, pushing a surface band across a spanning subdisk, and replacing a spanning subdisk by a parallel surface disk. Each has a disk-product neighborhood: on a band push, move its interval coordinate using a bump function stationary near the retained boundary intervals; on disk replacement, take a small normal parallel copy and round the joining seam. A puncture slide uses the same product at a bridge endpoint and has velocity tangent to the fixed knot. Disk exchange across a sphere uses the knot-free ball from [F2], with its collars retained by [F4]. Choose these neighborhoods sufficiently small to avoid the other same-side disks and all nonincident pieces; there are only finitely many compact pieces at each stage. These operations are smooth versions of the prescribed moves, with the same incidences. Extend their embedding isotopies by [F3]. Thus the PL input yields smooth spheres meeting the fixed knot twice and smooth disjoint spanning-disk systems for the fixed bridge pair.

In the final configuration of [F1], every spanning-disk intersection arc has exactly one puncture endpoint and one endpoint on $P\cap S$. Each puncture lies on exactly one spanning disk in its bridge ball, so there are exactly two such arcs. There are no circle components or other arcs by the imported normal form. Small foot perturbations make their circle endpoints distinct, preserving every incidence. All changes of $P$ were isotopies preserving the knot setwise, so [F5] preserves the specified oriented factors. This proves the stated normal form, including opposite-side punctures.

Suppose $b(K)=1$. By [A1] and [F1], the nonempty set of maximum counts has least value $1$, so choose a Morse representative $K^{\mathrm{\prime }}$ with exactly one local maximum. On the circle, nondegenerate critical points alternate between maxima and minima; compactness gives finitely many, and a function on a circle has at least one minimum. Hence $h\mathrm{\mid }{K}^{\mathrm{\prime }}$ has exactly one minimum. Choose a regular value between these two critical values. Its level sphere meets $K^{\mathrm{\prime }}$ in exactly two points, and the upper and lower balls contain one proper arc each, with respectively one maximum and one minimum.

Fix a simple arc $\beta$ on the level sphere joining the two intersection points. Apply [F2] to the upper arc and the lower arc, using this same $\beta$. The resulting bridge disks lie in opposite balls and meet only along $\beta$; their union, with its corner smoothed, is an embedded disk whose boundary is $K^{\mathrm{\prime }}$. Thus $K^{\mathrm{\prime }}$ has genus zero, and [F3] makes it ambient isotopic to the standard unknot.

Conversely, choose an orientation-preserving rotation in $SO(4)$ carrying the fixed coordinate 2-plane of the standard unknot $U$ to the plane spanned by $e_1$ and $e_4$. A path of rotations from the identity gives an ambient isotopy, and on the rotated circle the fixed height is a sine or cosine with exactly one maximum. Hence $b(U)\le 1$. Every Morse function on a compact circle has at least one maximum, so $b(U)\ge 1$ and $b(U)=1$. Ambient-isotopic knots have the same set of representative maximum counts, so every unknot has bridge number one.

Make the basepoint adjustment in [F1] if necessary, then apply the procedure to $D$. Since $D$ has $c(D)$ crossing records, the algorithm terminates after that finite scan and returns at most $c(D)$ crossing changes.

After those changes every crossing is first encountered as an overpass, so [F2] says the updated code represents the standard unknot. It follows from [F3] that the returned list witnesses $u(K)\le c(D)$. In particular the set defining $u(K)$ is nonempty and its minimum is finite.

We first justify using bridge spheres for the height-based minimum in [F1]. In the product of regular height levels $S^2\times (-1,1)$, a maximum can be raised without changing the number or kinds of critical points: take a small cap containing only that maximum, perturb its horizontal location so that its central vertical ray misses the rest of the knot, narrow the cap inside its original small chart to fit a tube about that ray, and lift it along the ray, lengthening its two descending ends monotonically. The horizontal location is available arbitrarily close to the initial cap: the projection of the remaining finite collection of compact smooth arcs has empty interior, as follows by covering each arc by finitely many intervals of derivative bound $C$, whose projected images lie in disks of radius $C\delta$ and have total area tending to zero as $\delta \to 0$. Compact disjointness then gives a narrow tube about the ray avoiding the rest of the knot; its two thin parallel sides are the lengthened ends. A horizontal cap perturbation and the subsequent vertical move have embedding tracks with no new extrema; the moving cap has the translated local quadratic maximum and the ends have strictly positive or negative height derivative. Extend with [F7]. Do this finitely many times, choosing distinct rays avoiding earlier cap tracks, until all maxima lie above a regular level above all minima. The upper and lower pieces are trivial tangles with simultaneous disjoint spanning disks: sweep their endpoints between successive critical heights, extending the finite point motions on each level sphere by disjoint small coordinate vector fields. Transport disjoint joining arcs by these level isotopies. At a cap choose its joining arc in its small cap chart, away from the finitely many earlier arcs (move those arcs off that chart by a level isotopy before the birth). Their swept traces, capped at the extrema, give pairwise disjoint disks. Conversely, for any smooth bridge sphere and displayed disks, [F6] first carries the sphere to a regular level. Push each bridge through its own disk to a thin collar of its foot; in disk parameters use a positive rounded arch whose height is $\epsilon \sin (\pi t)$ above, and $-\epsilon \sin (\pi t)$ below, with compatible endpoint collars. The disjoint same-side disks allow these isotopies simultaneously; [F7] extends them, and rounding at endpoints gives exactly one maximum per upper arc and one minimum per lower arc. Thus the minimum number of upper arcs over bridge spheres is exactly the height-defined $b$. In particular a minimum-height representative supplies a bridge sphere with $b$ upper arcs, and any factor bridge sphere with $n$ upper arcs gives $b(\text{factor})\le n$.

Choose bridge representatives realizing $b(K)$ and $b(J)$, which is possible by [F1] and step 1.1. Each comes with its bridge sphere and disjoint upper and lower bridge-spanning disks.

If either factor is the unknot, [F8] gives equality immediately. Suppose therefore both are nontrivial. Choose a bridge sphere $S$ of $L=K\mathrm{\#}J$ with $b=b(L)$ upper arcs, by step 1.1. Transport the connected-sum construction's two-puncture sphere through the representative isotopy; this gives a smooth sphere $P$ cutting $L$ into the specified factors. Choose a sufficiently small product tube adapted to its two transverse punctures and form $A=P\setminus \mathrm{int}N(L)$. The annulus core is a positive or negative meridian in the tube: both its boundary components bound the normal disks $P\cap N(L)$ and its core is homotopic within $A$ to either boundary. By [F4], composing ${\pi }_1(A)=\mathbb{Z}\to {\pi }_1(E_L)$ with abelianization sends its generator to $\pm 1\in \mathbb{Z}$. The inclusion is therefore injective.

Choose one point of each knot on its bridge sphere where an upper bridge arc and a lower bridge arc meet. Take product-coordinate balls around the two points so each ball meets the sphere in a disk, the knot in one short arc crossing that disk once, and each of the two incident bridge-spanning disks in a half-disk. Remove the ball interiors and glue the punctured spheres by the connected-sum map from [F2]. The two selected upper bridge arcs join into one proper upper arc, and the two selected lower bridge arcs join into one proper lower arc. The new upper spanning disk is the union of the two punctured upper disks and the gluing rectangle; two disks joined along one boundary interval form a disk. The same construction gives the lower spanning disk. Every unselected bridge arc and disk remains unchanged and disjoint from these disks. Thus the connected-sum position has $b(K)+b(J)-1$ upper arcs, so $b(K\mathrm{\#}J)\le b(K)+b(J)-1$.

There is no boundary compression of $A$. If there were, [F5] would give a parallelism to one of the two annuli $A_0$ on the boundary torus between its meridional boundaries. This implies that one factor is the unknot, as can be checked with the markings: write the tube over the corresponding subinterval of the core knot as $D^2\times I$, with side $A_0=\mathrm{\partial }{D}^2\times I$ and end disks the two disks $P\cap N(L)$. Attach the annulus parallelism to this side, retaining its rim collars. It merely enlarges the radial disk coordinate in each $I$-fiber, so the result is a ball with boundary the original $P$ after rounding. Its knot arc is the core $\{0\}\times I$, and it cobounds with an arc on that ball's boundary a disk given by a fixed radial interval times $I$, rounded at its ends. Capping this arc on $P$ produces an unknot. The same construction works for either complementary annulus of the torus; it uses its actual product parameter, not just a homological slope. By [F8] the resulting unknot is one of the given factors, contradicting nontriviality. Thus all hypotheses of [F3] hold for this particular $P$, and we may apply its isotopy and disk changes without changing $(L,S)$ or the specified factors.

In the resulting normal form $P\cap S=C$ is one circle disjoint from $L$. The two disks $P_{\pm }=P\cap B_{\pm }$ each contain one puncture. This is the opposite-side puncture conclusion in the Statement of [F3], rather than an inference from the existence of two arcs alone. Each $P_{\pm }$ is a proper disk cutting its bridge ball into two balls: cap it by either of the two disks on $\mathrm{\partial }{B}_{\pm }$ bounded by its rim, round, and apply [F6]; the side not containing the exterior of $B_{\pm }$ is the required ball. Their collars give compatible markings. Let $S_1,S_2$ be the disks on $S$ bounded by $C$, labelled by the two sides of $P$, and put $r_i=\mathrm{\mid }L\cap S_i\mathrm{\mid }$. Then $r_1+r_2=2b$. In each original bridge ball exactly one bridge arc crosses $P_{\pm }$: it is the arc carrying its unique puncture. Its two endpoints on $S$ lie on opposite sides of $C$, because the bridge crosses that separating disk once. Every other bridge arc is wholly on one side. Consequently $r_i$ is odd.

Cap each of the two cut ball-arcs of $L$ by a boundary arc on $P$ running from the upper puncture to the lower puncture and crossing $C$ once, chosen as the two type-I traces joined by an arc of $C$ between their distinct foot endpoints, then rounded slightly in the collar of $C$ so that it crosses $C$ transversely just once. Push the two caps into their respective boundary collars. On each side this cap is a trivial ball-arc meeting the capped bridge sphere $S_i\cup$(cap disk) once. Its upper and lower portions extend the severed bridge arcs. Every uncut bridge disk remains in its original region. Each of the two cut bridge disks is cut along its one trace into two disks; on each side join its disk piece to the narrow cap strip. This is a disk attached along one boundary interval, so gives a spanning disk for the extended upper or lower arc. Choose the cap strips in disjoint small quadrant collars of the traces; other disks miss them, and on a given side all disk pieces were originally disjoint. Thus these are simultaneous disjoint spanning-disk systems. The upper and lower ambient pieces are balls: attach the cap half-balls along the displayed windows on $P_{\pm }$ and use [F6]. We have therefore constructed actual bridge spheres for the two specified oriented factors, not merely diagrams with a proposed number of bridges. Each has $r_i+1$ knot endpoints on its sphere, the extra endpoint being the single crossing of the cap. It follows that their upper arc counts are $n_i=(r_i+1)\mathrm{/}2$, and $n_1+n_2=b+1$. Cap orientation is chosen to close the original oriented arc, so [F8] identifies the factors with $K,J$, up to their harmless interchange.

Step 1.1 applied to the two factor bridge spheres gives $b(K)\le n_1$ and $b(J)\le n_2$. Hence $b(K)+b(J)-1\le n_1+n_2-1=b=b(K\mathrm{\#}J)$. Together with step 3.1 and the unknot case in step 2.2, this proves the formula for every pair of smooth oriented knots in the Statement. The constructions use finitely many extrema, disks, cuts and collars; their smooth extension and Schoenflies uses are precisely the stated countable-choice suppliers.

At $\omega =-1$ the form is nonsingular for every knot. Indeed $V+V^T\equiv V-V^T(\mathrm{m}\mathrm{o}\mathrm{d}2)$, so its determinant is odd by [F5] and in particular is not zero. Thus $H_{-1}=2(V+V^T)$ is nonsingular. By [F4], $-1$ is not an Alexander root and the actual matrix signature is precisely ${\sigma }_{-1}(K)$; its surface and basis independence also follows from [F4]. This removes any need to use a root-average convention at $-1$.

For two successive knots in a crossing-change sequence, use the two corresponding bases from [F2]. If their Seifert matrix difference is $\pm v{v}^T$, their $H_{-1}$ difference is $\pm 4v{v}^T$, of rank at most one; if the difference is zero, the same bound holds. By [F3], the absolute signature change is at most two. Step 1.1 identifies the signatures of these particular matrices with the knot invariants, regardless of which diagram surfaces occur at other stages.

By [F1], take a sequence of length $u(K)$ ending at the standard unknot. Its disk Seifert surface has an empty matrix and signature zero. Apply the estimate of step 2.1 to each consecutive pair and add the finitely many inequalities. The triangle inequality yields $\mathrm{\mid }{\sigma }_{-1}(K)\mathrm{\mid }\le 2u(K)$, including a length-zero sequence. Together with step 1.1 this proves both claims.

By [A1], the hypotheses of [F1] hold, so it supplies both implications of the asserted bridge-number characterization, with the specified oriented standard unknot.

Apply [F2] to the same knot. It gives the asserted signature inequality with the same unknotting-number convention. Combining it with step 1.1 proves the conjunction without changing either interface.