Smooth and topological knot concordance

Choose a $\mathbb{Q}$-basis $v_1,\dots ,v_g$ of $L$. Clearing the finitely many denominators in the coordinates of each $v_i$ gives nonzero integer multiples $x_i\in H\cap L$. The $x_i$ remain $\mathbb{Q}$-linearly independent, so $M=H\cap L$ has rank at least $g$; it has rank at most ${\dim }_{\mathbb{Q}}L=g$. Hence $\mathrm{rank}M=g$.

If $n\mathrm{\ne }0$, $x\in H$, and $nx\in M$, then $nx\in L$ and therefore $x=n^{-1}(nx)\in L$ because $L$ is a $\mathbb{Q}$-subspace. Thus $x\in M$, so $M$ is saturated in $H$.

We prove by induction on $r$ that every saturated subgroup $P\subset {\mathbb{Z}}^r$ is a direct summand. The cases $P=0$ and $P={\mathbb{Z}}^r$ are immediate. Otherwise choose a nonzero $p\in P$, divide its coordinates by their greatest common divisor, and use saturation to see that the resulting primitive vector $e_1$ still lies in $P$. The Euclidean algorithm gives a unimodular change of basis carrying $e_1$ to the first standard basis vector, so write ${\mathbb{Z}}^r=\mathbb{Z}{e}_1\oplus H^{\mathrm{\prime }}$. Every $p^{\mathrm{\prime }}\in P$ has a unique form $a{e}_1+h^{\mathrm{\prime }}$ with $h^{\mathrm{\prime }}\in H^{\mathrm{\prime }}$, and $h^{\mathrm{\prime }}=p^{\mathrm{\prime }}-a{e}_1\in P\cap H^{\mathrm{\prime }}$. Therefore $P=\mathbb{Z}{e}_1\oplus (P\cap H^{\mathrm{\prime }})$. The subgroup $P\cap H^{\mathrm{\prime }}$ is saturated in $H^{\mathrm{\prime }}$; its rank is $\mathrm{rank}P-1$, since the projection of $P$ to $\mathbb{Z}{e}_1$ is onto. Induction makes $P\cap H^{\mathrm{\prime }}$ a direct summand of $H^{\mathrm{\prime }}$, hence $P$ a direct summand of ${\mathbb{Z}}^r$.

Apply step 3.1 to the saturated rank-$g$ subgroup $M\subset H\cong {\mathbb{Z}}^{2g}$ from step 2.1. It is a direct summand of $H$.

Since $M\subset L$ and $V$ vanishes on $L\times L$, its restriction to $M\times M$ vanishes. Together with rank $g$ and the direct-summand conclusion of step 4.1, this proves the claim.

In those collars the image of $\alpha$ is vertical, so write $\alpha (t)=(p,h_0(t))$ near zero and $\alpha (t)=(p,h_1(t))$ near one. Each $h_i$ is an increasing homeomorphism onto its height interval: it is continuous and injective, and the endpoint heights fix its orientation. Choose an increasing homeomorphism $\theta :[0,1]\to [0,1]$ agreeing with the two inverse height functions near the endpoints and interpolate monotonically between them on the remaining middle interval. Then $\beta (s)=\alpha (\theta (s))$ has the same image as $\alpha$ and satisfies $\beta (s)=(p,s)$ on neighborhoods of the source endpoints. In the additional parametrized case take $\theta =\mathrm{id}$. This changes only the parameter used to describe the input image.

Apply [F1] to $\beta$, obtaining an ambient topological isotopy $A_t$ whose final arc $\gamma (s)=A_1(\beta (s))$ is smoothly embedded. Since $A_t$ is fixed near the boundary levels, $\gamma (s)=(p,s)$ near its endpoints. Apply the parametrized clause of [F2] to $\gamma$, obtaining an ambient smooth isotopy $B_t$ fixed near the boundary levels and satisfying $B_1(\gamma (s))=(p,s)$ for every $s$.

Concatenate $A_{2t}$ for $0\le t\le 1\mathrm{/}2$ with $B_{2t-1}{A}_1$ for $1\mathrm{/}2\le t\le 1$. They agree at the joining time. This is a continuous ambient family of homeomorphisms starting at the identity, fixed on a common neighborhood of the two boundary levels. Its terminal map sends the image of $\beta$, and hence the image of $\alpha$, to $\{p\}\times [0,1]$. In the additional parametrized case $\theta =\mathrm{id}$, so it sends $\alpha (s)$ to $(p,s)$ for every $s$. This proves both conclusions without an added choice hypothesis.

Cut a short trivial arc from $K$ inside a small ball. Identify its complementary ball with the unit ball $B^3\subset {\mathbb{R}}^3$ so that the remaining arc is a smooth embedding $a:[0,1]\to B^3$, its endpoints lie on $S^2$, and its interior lies in the open ball. Straighten its endpoint collars to be radial and transverse to $S^2$. A small perturbation supported away from the endpoints makes $a$ avoid the origin; compactness and the fact that a one-dimensional embedded arc has empty interior allow the perturbation to be chosen arbitrarily small in $C^1$, so it remains embedded and fixed on the endpoint collars. Put $f(s)=\parallel a(s){\parallel }^2$. Then $f>0$ on $[0,1]$, $f(0)=f(1)=1$, $f<1$ on $(0,1)$, and transversality gives $f^{\mathrm{\prime }}(0)<0<f^{\mathrm{\prime }}(1)$.

The zeros of $f^{\mathrm{\prime }}$ lie in a compact subinterval of $(0,1)$ because $f^{\mathrm{\prime }}$ is nonzero on the endpoint collars. Cover the zero set by finitely many intervals carrying smooth functions ${\varphi }_i$ supported in $(0,1)$ whose derivatives equal $1$ on smaller intervals still covering the zero set. For $u=(u_i)$ define $f_u=f+{\sum }_i{u}_i{\varphi }_i$. On a neighborhood of every zero of $f^{\mathrm{\prime }}$ the total family $\mathcal{F}(s,u)=f_u^{\mathrm{\prime }}(s)$ is a submersion in the parameter direction, since some $\mathrm{\partial }\mathcal{F}\mathrm{/}\mathrm{\partial }{u}_i={\varphi }_i^{\mathrm{\prime }}(s)=1$. On the compact complement of that neighborhood, $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }$ has a positive lower bound, so for all sufficiently small $u$ no new zeros occur there. Thus the total map is transverse to $0$ for parameters in a sufficiently small ball. By [F3], choose a parameter in that ball for which $0$ is a regular value of $f_u^{\mathrm{\prime }}$. The critical points of $f_u$ are then nondegenerate and finite. Shrink the parameter ball so $0<f_u<1$ on the perturbation support. The radial arcs $a_{\lambda }(s)=\sqrt{f_{\lambda u}(s)\mathrm{/}f(s)}a(s)$, $0\le \lambda \le 1$, form a smooth isotopy of embeddings fixed near the endpoints: the formula is smooth since $f>0$, and sufficiently small $C^1$ perturbations preserve immersion and injectivity on the compact source. By [F4], extend this isotopy in an ambient isotopy supported away from the removed trivial arc, so replacing $a$ by $a_u$ preserves the knot type of its closure. Rename $a_u$ as $a$ and $f_u$ as $f$.

In the unit ball $B^4=\{(x,t)\in {\mathbb{R}}^3\times \mathbb{R}:\parallel x{\parallel }^2+t^2\le 1\}$ define $\mathrm{\Omega }=\{(s,t):0\le s\le 1,t^2\le 1-f(s)\}$ and $D=\{(a(s),t):(s,t)\in \mathrm{\Omega }\}$. The map $(s,t)\mapsto (a(s),t)$ is an injective immersion, hence an embedding because $\mathrm{\Omega }$ is compact. The function $q(s)=(1-f(s))\mathrm{/}(s(1-s))$ extends smoothly and positively to $s=0,1$, with $q(0)=-f^{\mathrm{\prime }}(0)>0$ and $q(1)=f^{\mathrm{\prime }}(1)>0$. Consequently $(s,v)\mapsto (s,v\sqrt{q(s)})$ is a smooth diffeomorphism from the round disk $v^2\le s(1-s)$ onto $\mathrm{\Omega }$. The boundary of $\mathrm{\Omega }$ is regular: at an interior boundary point $t\mathrm{\ne }0$, and at its two endpoints $f^{\mathrm{\prime }}(s)\mathrm{\ne }0$, so the differential of $f(s)+t^2$ is nonzero. Thus $D$ is a smooth disk, its boundary lies in $S^3$, and its interior lies in the open four-ball. By [F6], its boundary is exactly the boundary knot specified by properness.

The two boundary arcs of $D$ are the lifts $(a(s),\pm \sqrt{1-f(s)})$ and meet the equatorial $S^2=\{t=0\}\cap S^3$ only at their endpoints. Each hemisphere of $S^3$ is a standard three-ball. Away from a small collar of its equator, projection $(x,t)\mapsto x$ is a smooth diffeomorphism onto the corresponding truncated ball and identifies the lifted arc with $a$. At an endpoint, use the equator collar coordinate $t\ge 0$ (or $-t\ge 0$ in the lower hemisphere). Since $f^{\mathrm{\prime }}(0)\mathrm{\ne }0$ and $f^{\mathrm{\prime }}(1)\mathrm{\ne }0$, the implicit function theorem writes the lifted arc near each endpoint as $s=\sigma (t^2)$, with $\sigma$ smooth; in a chart $y$ on the equator its tangential coordinates are $y(t)=y_0+O(t^2)$, while $t$ is the inward collar coordinate. Choose a cutoff $\chi (y,t)$ supported in this chart and equal to $1$ near the arc, and use $(y,t)\mapsto (y-\chi (y,t)(y(t)-y_0),t)$. Its derivative is the identity plus $O(t)$, so after shrinking the collar it is a diffeomorphism and straightens the arc to $\{y_0\}\times [0,ϵ)$. This is the same standard ball-arc collar as the radial endpoint collar of $a$. Thus the lifted upper arc is the original ball-arc tangle and the lower one is its image under $(x,t)\mapsto (x,-t)$, up to collar-supported isotopy. The boundary orientation traverses one lifted arc in the direction of $a$ and the other in the reverse direction. By [F2] and [F5], closing these two arcs with the fixed connected-sum gluing gives precisely the oriented type $K\mathrm{\#}{K}^{†}$. Therefore $\mathrm{\partial }D$ has the required oriented knot type.

On $D$, the squared radial function is $\rho (s,t)=f(s)+t^2$. Its interior critical points are exactly $(s,0)$ where $f^{\mathrm{\prime }}(s)=0$. At each such point the Hessian is $\mathrm{diag}(f^{\mathrm{\prime }\mathrm{\prime }}(s),2)$, which is nondegenerate and has index $0$ when $f^{\mathrm{\prime }\mathrm{\prime }}(s)>0$ and index $1$ when $f^{\mathrm{\prime }\mathrm{\prime }}(s)<0$ by [F7]. Thus there are no index-$2$ critical points, and step 4.1 identifies the boundary knot as $K\mathrm{\#}{K}^{†}$. This proves the statement.

Suppose first that a smooth proper disk $D\subset B^4$ has boundary $K$. Choose an interior point $p\in D$ and a sufficiently small round four-ball $E$ about $p$ contained in $\mathrm{int}{B}^4$. In orthonormal coordinates at $p$, the local graph theorem writes $D$ as $z=g(u)$ over a disk in its tangent plane, with $g(0)=0$ and $Dg(0)=0$. For small radius $R$, the function $r\mapsto r^2+g(r\theta {)}^2$ has derivative $2r+O(r^3)>0$ for every unit direction $\theta$ and $0<r\le R$. Hence $D\cap \mathrm{\partial }E$ is one radial graph over the tangent circle. Replacing $g$ by $\lambda g$, $0\le \lambda \le 1$, and solving the same strictly increasing radial equation gives an isotopy through smooth embeddings of $S^1$ in $\mathrm{\partial }E\cong S^3$ to the equatorial circle. By [F4], with target $\mathrm{\partial }E\cong S^3$, this isotopy extends to an ambient isotopy of $S^3$; hence $D\cap \mathrm{\partial }E$ is ambient isotopic to the standard unknot. The family of round balls moving the center of $E$ to $0$ along a path in $\mathrm{int}{B}^4$ is an isotopy of compact embeddings supported in $\mathrm{int}{B}^4$; apply [F4] with target ${\mathbb{R}}^4$ to extend it. Hence $B^4\setminus \mathrm{int}E$ is diffeomorphic, with boundary collars, to the standard shell $S^3\times I$. Removing $\mathrm{int}(D\cap E)$ from $D$ gives an annulus in that shell with boundary $K$ and a standard unknot.

Conversely, suppose $K$ is smoothly concordant to the standard unknot. Reverse the interval coordinate if needed so the unknot is the lower boundary. Glue a standard four-ball along the lower $S^3$ end of $S^3\times I$. The union is a standard four-ball with an added boundary collar: explicitly, the collar extends the radial coordinate of the unit ball from radius $1$ to radius $2$, and radial rescaling identifies it with the unit ball. Cap the lower unknot boundary of the annulus by the standard disk from [F3]. The annulus and cap meet in matching product collars, so their union is a smooth properly embedded disk with boundary $K$.

The product collars at the two boundary circles identify the annulus from step 1.1 with a smooth concordance, after orienting it so the lower boundary is the negative of its induced knot orientation and the upper boundary is positive. If necessary, reverse the interval coordinate to place the unknot at the lower end; a half-turn of the standard round circle about a diameter is an orientation-preserving ambient rotation that reverses its circle orientation, so the endpoint still has the standard oriented unknot type. Reversing the annulus orientation when required gives the boundary convention in [F1]. Therefore $K$ is smoothly concordant to the standard unknot.

Steps 1.1 and 2.1 prove that a smooth slice disk gives smooth concordance to the unknot, and step 1.2 proves the converse.

Reflexivity is witnessed by the product annulus $K_0\times I\subset S^3\times I$, oriented so its boundary is $-K_0\times \{0\}\bigsqcup K_0\times \{1\}$. If a representative is changed by an ambient isotopy $H_t$, reparameterize time so $H_t$ is constant near $t=0$ and $t=1$. Its trace $(x,t)\mapsto (H_t(x),t)$ then has product endpoint collars and is a concordance annulus; it is smooth in the smooth case and locally flat in the topological case because it is itself smooth. For an existing concordance, glue the trace to the matching product collar: use the reversed trace at the lower end and the forward trace at the upper end. The glued surface is an annulus in the same category, so replacing either endpoint by an ambient-isotopic representative preserves concordance. Thus the relation is well-defined on ambient-isotopy classes.

If an annulus $A$ gives a concordance from $K_0$ to $K_1$, apply $(x,t)\mapsto (x,1-t)$ to its image. Reorient the resulting annulus so the boundary at the new lower end is $-K_1$ and the boundary at the new upper end is $K_0$. The map and reorientation preserve smoothness in the smooth category and local flatness and product collars in the topological category. This gives symmetry.

Let $A$ be a concordance from $K_0$ to $K_1$ and $B$ one from $K_1$ to $K_2$ in the same category. Glue the upper product collar of $A$ to the lower product collar of $B$ along their common oriented knot $K_1$. The two collar coordinates combine to a single collar across the gluing hypersurface, so the annuli join without a corner in the smooth category and their local-flat pair charts join in the topological category. Gluing two annuli along one boundary circle gives an annulus; the intermediate boundary orientations $K_1$ and $-K_1$ cancel, leaving boundary $-K_0\bigsqcup K_2$. This gives transitivity.

Steps 1.1–1.3 prove reflexivity, symmetry, transitivity, and representative independence for both relations.

First consider smooth concordances. Pick one point on each endpoint knot. By [F4], apply separate endpoint ambient isotopies carrying the marked points to a common point $p\in S^3$; extend them over disjoint product collars of each concordance and interpolate by the identity between the collars. This changes the representatives only within their ambient-isotopy classes. Choose in each annulus a smooth spanning arc with image equal to $\{p\}\times I$ near its endpoints. Apply [F8] to each arc. Its ambient smooth isotopy is fixed near both end levels and carries the spanning arc to the vertical product arc; applying its terminal diffeomorphism to the annulus preserves smoothness, orientation, properness and the endpoint knots and their product collars.

For the TOP implication, first code the actual input as compact metric data. Parametrize the two annuli by the standard compact annulus, and choose finitely many local pair charts covering each, together with their inverses and the given end-collar maps. Such a finite selection is possible by compactness and finite choice; using countably many relatively compact chart restrictions instead would require only ${\mathrm{A}\mathrm{C}}_{\omega }$. Include the endpoint knots and the chosen connected-sum representatives from [F2]. For these smooth knots and the finite smooth coordinate maps used in their sums, also code the derivatives of every finite order, their continuity moduli and rational difference-quotient remainder bounds; thus the inner model sees smooth knots, not merely continuous embeddings. Continuous maps on these standard metric domains are coded by their values on fixed countable rational dense sets and uniform continuity moduli on compact exhaustions; include inverse moduli for embeddings and chart inverses. All this countable numerical data is one real $r$.

For each straightened annulus $A$, height restricts to a submersion along the vertical core, since the tangent of that core has height derivative $1$. Choose a sufficiently small strip in $A$ parametrized by $(u,t)\mapsto (a(u,t),t)$ with $a(0,t)=p$, using the product collars at its ends. Complete its tangent line in each three-dimensional level to an oriented frame and take small normal disks within that level. The resulting smooth family of embeddings ${\mathrm{\Phi }}_t:D^3\to S^3$ satisfies $A\cap ({\mathrm{\Phi }}_t(D^3)\times \{t\})={\mathrm{\Phi }}_t(D^1)\times \{t\}$, and is independent of $t$ near each end. Uniform radii exist by compactness; separation from the compact remainder of $A$ excludes other sheets. The initial small coordinate ball is standard: in a fixed coordinate chart, rescaling a local embedding toward its derivative gives an isotopy to a linear ball after shrinking, and [F3] extends that ball isotopy in $S^3$. Its complement is consequently a three-ball. Apply [F3] again to the compact ball isotopy ${\mathrm{\Phi }}_t$ in the boundaryless target $S^3$, obtaining diffeomorphisms $G_t$ with $G_t{\mathrm{\Phi }}_0={\mathrm{\Phi }}_t$. Thus $(x,t)\mapsto (G_t(x),t)$ identifies the entire tube and its complement with the products of the initial ball and its complement. In these product coordinates, remove the two tube interiors from two copies of $S^3\times I$ and glue their boundary cylinders by the fixed orientation-reversing map, independent of $t$. The ambient result is the product of two three-balls glued along their boundary spheres with $I$, hence $S^3\times I$. Removing the open strip from each annulus leaves a rectangle; glue the rectangles along their two side intervals in the connecting cylinder, compatibly with their orientations. Their lower and upper boundary knots are $-(K_0\mathrm{\#}{J}_0)$ and $K_1\mathrm{\#}{J}_1$ by [F2]. The resulting connected orientable surface has two boundary circles and Euler characteristic $1+1-2=0$, so it is an annulus. Round the product corners in the fixed local coordinates; the endpoint product collars are retained. This gives the smooth concordance.

We use certificates, not an unqualified assertion that homeomorphism existence is absolute. For every open chart domain record its rational-ball code (relative rational half-balls at a boundary chart), compact exhaustion, and positive-distance margins putting the images of exhaustion pieces in the corresponding open image. Record both map/inverse identities on dense sets. For the pair condition record the parametrization of the linear plane section and its identities with the annulus parametrization; on smaller charts also record positive separation from the compact remainder of the annulus outside that patch. A finite parameter-patch cover and a Lebesgue-number certificate cover the entire annulus. Properness is certified on the compact subannuli $S^1\times [1\mathrm{/}j,1-1\mathrm{/}j]$ by positive separation from the ambient end levels. Equalities with the oriented endpoint parametrizations and the product collars are dense-set equalities with their continuity moduli. The smooth derivative identities are certified by rational difference quotients and their coded remainder bounds at each finite order. These certificates express input validity using only rational inequalities, integer indices and quantification over natural numbers. They therefore have the same truth in $L[r]$ and in the ambient universe; in particular the coded input is valid in $L[r]$.

The preceding construction proves the smooth implication.

Apply [F6] inside $L[r]$, which satisfies full Choice by [F7]. It gives a locally flat annulus for the two connected sums. In that model choose its finite pair-chart cover, inverse maps, compact-exhaustion moduli, open-image margins and end-collar certificates of the same form as in step 2.2; collect them into a real $s\in L[r]$. Decode $s$ in the ambient universe by completion from the rational dense domains. The coded forward and inverse moduli give continuous inverse maps on their coded domains; the dense-set identities extend to all points, and the open-image and separation certificates make these actual local pair charts with no additional surface branch. The finite cover certificate, properness margins, and endpoint/collar identities likewise persist. Thus the decoded object is an actual oriented proper locally flat annulus in the ambient $S^3\times I$, with the required endpoint connected sums and product collars. This transfer uses arithmetic certificates for the particular compact objects, not preservation of arbitrary sets or homeomorphisms between inner models.

Steps 1.2, 2.2, 3.2 earn the required compact specialization of the external TOP input under the stated ${\mathrm{A}\mathrm{C}}_{\omega }$, without assuming ambient full AC. Together with the retained smooth construction in steps 1.1–3.1, this proves both implications in the unchanged Statement. No annulus-adapted transverse tube or ambient complement is inferred from the bare-arc theorem [F5]; that former local proof gap is bypassed by the explicitly external input [F6].

By [F1], choose a connected compact smooth oriented Seifert surface $\mathrm{\Sigma }\subset S^3$ with $\mathrm{\partial }\mathrm{\Sigma }=K$. By [F3], choose $0<\epsilon <1$ and identify a collar of $S^3=\mathrm{\partial }{B}^4$ with $S^3\times [0,\epsilon )$, where $(x,u)$ maps to $(1-u)x$. Apply the collar theorem in [F3] also to $\mathrm{\Sigma }$ and write $r\ge 0$ for its inward boundary-collar coordinate. Choose a smooth cutoff $\chi (r)$ equal to $1$ near $r=0$ and $0$ before the collar ends, and a constant $c\in (0,\epsilon )$; set $h=\chi (r)r+(1-\chi (r))c$ in the collar and $h=c$ beyond it. Since $0\le r,c<\epsilon <1$, $h$ takes values in $[0,1)$; it is smooth, vanishes exactly on $\mathrm{\partial }\mathrm{\Sigma }$, and is positive on the interior.

The graph $F=\{((1-h(x))x):x\in \mathrm{\Sigma }\}\subset B^4$ is embedded because the radial collar projection to $S^3$ recovers $x$. It is compact and connected, its boundary is exactly $K$, and all its nonboundary points lie in the interior of $B^4$. Its smooth charts give the local-flat pair charts in [F4]. The graph is homeomorphic to $\mathrm{\Sigma }$, so it has genus $g(\mathrm{\Sigma })$ from [F2]. Thus this one surface contributes an element to each of ${\mathcal{S}}_{\mathrm{s}\mathrm{m}}(K)$ and ${\mathcal{S}}_{\mathrm{t}\mathrm{o}\mathrm{p}}(K)$.

Each set is therefore a nonempty subset $A\subseteq {\mathbb{N}}_0$. Choose one $n\in A$. The finite set $A\cap \{0,1,\dots ,n\}$ is nonempty, so it has a least element $m$ by the ordinary finite ordering. If $a\in A$ and $a<m$, then $a\le n$ and hence $a\in A\cap \{0,1,\dots ,n\}$, contradicting the choice of $m$. Thus $m$ is least in $A$. Applying this argument to each of the two sets gives both minima; no well-ordering principle or additional choice axiom is used.

In the smooth category, [F1] makes the smooth concordance classes a quotient set. [F2] ensures that replacing either connected-sum input by a concordant knot leaves the resulting class unchanged, so connected sum defines a binary operation on that quotient. [F3] supplies associativity, commutativity, and the two-sided identity on oriented knot types; these equalities pass to the quotient because the operation is well-defined.

For every $K$, [F4] gives a smooth disk bounded by $K\mathrm{\#}{K}^{†}$. By [F5], this connected sum is smoothly concordant to the standard unknot, so $[K]+[K^{†}]=[U]$ in the smooth quotient. Together with the commutative monoid structure from step 1.1, every element has an inverse; thus the smooth quotient is an abelian group with the stated inverse.

In the locally flat topological category, [F1] again gives a quotient set, and the TOP implication of [F2] makes connected sum independent of both concordance representatives. Associativity, commutativity and the unknot identity descend from [F3] exactly as in step 1.1. For each smooth knot representative $K$, the smooth disk from [F4] and its smooth concordance to $U$ from [F5] are also locally flat by the surface/collar convention. Therefore $[K]+[K^{†}]=[U]$ in the TOP quotient. Every element has the stated inverse, so this quotient too is an abelian group. The Countable Choice strength here is that of the bounded compatibility specialization [F2], not an unstated full-AC hypothesis.

Finite-chain model and independence. Write each rational chain as a finite rational sum of parametrized singular simplices. Within each labeled intersection factor, take the finite family of its simplices and iterated faces, identifying equal parametrized faces so that cancellations remain exact. Never identify simplices across the two factors merely because their maps agree: their perturbations must be independent. Preserve the given smooth product end collars and truncate at a smaller positive collar level. The remaining target is the boundaryless interior $B^{4\circ }$. Apply [F7] inductively over the finite face family, extending compatible face homotopies and fixing prescribed smooth faces. This gives a smooth chain replacement respecting the boundary operator, with the same rational coefficients; the homotopy prisms give relative homology. Applying [F7] with any prescribed open neighborhood as target keeps the replacement there. For general position, keep the two labeled families separate and make each pair of their open face strata transverse to the diagonal in $B^{4\circ }\times B^{4\circ }$. Choose faces in increasing dimension, fixing their previously treated subfaces. In Euclidean coordinates a cutoff-supported translation of the unfixed map is submersive in its parameter where the cutoff is positive; near the fixed part the pair is already transverse. The relative and parametric clauses of [F6] therefore give the required perturbation. Extend a chosen face homotopy consistently to all incident simplices in that factor, but not to the other factor. Finitely many face-pair conditions can be imposed simultaneously. Subdivision into finitely many compact coordinate pieces and sufficiently small parameters keep all changes inside prescribed open neighborhoods and away from the end collars. For complementary two-chains the diagonal preimages are finite signed sets; lower-dimensional face pairs are empty. Define the intersection by summing these signs with the rational product coefficients. This agrees with the given transverse interior count. Now fix $Q$ and two choices $P,P^{\mathrm{\prime }}$ bounding $\alpha$, using facewise-transverse representatives with their identical end collars. Their difference $C=P-P^{\mathrm{\prime }}$ is a smooth rational two-cycle compactly supported in $B^{4\circ }$. By [F3], choose a finite continuous rational three-chain $Z$ there with $\mathrm{\partial }Z=C$. Smooth $Z$ relative to its prescribed boundary $C$ by [F7], identifying faces within the $Z$ family only. Keeping $Q$ independent and fixed, perturb the remaining $Z$ faces relative to $C$, whose intersections with $Q$ are already transverse. Each simplex-pair diagonal preimage is a compact oriented one-manifold; codimension-one faces give its endpoints, while codimension-two faces have negative expected dimension. The local half-space model and [F4] give the boundary signs. Sum with the rational product coefficients: internal face endpoints cancel because the same face map occurs with opposite boundary coefficients. What remains is $(\mathrm{\partial }Z)\cdot Q=C\cdot Q$, together with the term from $Z$ and $\mathrm{\partial }Q$. That latter term is empty since $Z$ is compactly supported in the interior and $\mathrm{\partial }Q=\beta \subset S^3$. By [F2] the total signed boundary count is zero, so $P\cdot Q=P^{\mathrm{\prime }}\cdot Q$. Apply the same finite construction to homotopy prisms to obtain independence of the replacements. Interchanging the two two-dimensional factors introduces no sign and proves independence of the second chain. All operations are rational-linear, giving the same assertions for finite rational combinations. No embedded-manifold realization of $Z$ is used.

By [F5], choose an oriented Seifert surface $S_{\beta }\subset S^3$ for each knot component of $\beta$, and similarly choose a cap for each component of $\alpha$. It suffices by bilinearity to consider one knot component of each. Choose a collar depth $0<ϵ$ in $B^4$. Make $Q$ a graph of $S_{\beta }$ in the boundary collar with depth zero on $\beta$ and strictly between $0$ and $ϵ\mathrm{/}2$ on its interior. To give this graph a literal product end collar, choose $0<\delta <ϵ\mathrm{/}2$, attach $\beta \times [0,\delta ]$, and join its inner edge to the graph inside a tubular neighborhood of $\beta$ disjoint from $\alpha$, leaving the graph unchanged outside that neighborhood; choose the join below depth $ϵ\mathrm{/}2$. This adds no intersections with $P$ and changes none of the signs computed below. Make $P$ consist of the product collar $\alpha \times [0,ϵ]$ followed by a cap at depth $ϵ$. Then $P$ and $Q$ intersect exactly above $\alpha \cap S_{\beta }$. At each such point, the oriented tangent basis of the collar of $P$ is $(n_{\mathrm{o}\mathrm{u}\mathrm{t}},a)$, where $a$ is the positive tangent to $\alpha$ and $n_{\mathrm{o}\mathrm{u}\mathrm{t}}$ is the outward normal of $B^4$; the tangent basis of $Q$ is a positive basis $(s_1,s_2)$ of $S_{\beta }$, up to radial components which do not change the determinant. By [F4], $(n_{\mathrm{o}\mathrm{u}\mathrm{t}},a,s_1,s_2)$ is positive exactly when $(a,s_1,s_2)$ is positive in the boundary orientation. Thus the $4$-dimensional intersection sign is exactly the $3$-dimensional sign of $\alpha \cap S_{\beta }$. Hence $P\cdot Q=\mathrm{lk}(\alpha ,\beta )$ by [F1].