Seifert matrices and the Alexander module

Symmetry of linking. Let $c,d\subset S^3$ be disjoint oriented embedded circles. Their union has empty interior, so choose $q\in S^3\setminus (c\cup d)$ and an orientation-preserving rotation $R\in SO(4)$ of the standard sphere carrying $q$ to $\mathrm{\infty }$, as in [F7]. Then $R(c\cup d)\subset {\mathbb{R}}^3$. The restriction of $R$ to either component complement induces a homology isomorphism by [F6] and carries a positive meridian to a positive meridian, since it preserves the ambient and transported component orientations. By [F3], both ordered linking coefficients are therefore unchanged by $R$. Work with the rotated circles, relabelled $c,d$. Full AC supplies countable choice by [L1]; choose compact oriented embedded surfaces $A,B$ with boundaries $c,d$ by [F8], and now apply [F7] to obtain a regular diagram $D$ of their boundary link in ${\mathbb{R}}^3$. For $X=\mathrm{Int}A$, let $f:X\to S^3\subset {\mathbb{R}}^4$ be the inclusion. A sufficiently small boundary collar misses $d$; choose an open collar region $G$ on which $f$ is transverse to $d$, and protect a smaller collar. Then $K=X\setminus G$ is compact. Choose a relatively compact open $W\subset X$ containing $K$ and avoiding the protected smaller collar, using finitely many coordinate neighbourhoods of $K$. By [F12] choose $\chi :X\to [0,1]$, equal to one near $K$ and with compact support in $W$. Where $\chi =0$, $f$ is already transverse to $d$. For $v$ in a small open ball in ${\mathbb{R}}^4$ put $\mathcal{F}(p,v)=(f(p)+\chi (p)v)\mathrm{/}\mathrm{\parallel }f(p)+\chi (p)v\mathrm{\parallel }$. The denominator is nonzero for $\mathrm{\parallel }v\mathrm{\parallel }<1\mathrm{/}2$. Where $\chi >0$, the derivative in $v$ is $\chi \mathrm{/}\mathrm{\parallel }f+\chi v\mathrm{\parallel }$ times the orthogonal projection onto $T_{\mathcal{F}(p,v)}{S}^3$, hence is surjective. Where $\chi =0$, nonnegativity gives $d\chi =0$, and the old source transversality applies. Thus the total family is transverse to $d$. Parametric transversality and the dense-complement statement [F12] give a transverse slice in every sufficiently small parameter ball. The family equals $f$ off the compact support and converges uniformly in value and first derivative to $f$ there as $v\to 0$, directly by differentiation of the displayed smooth formula on the compact set. It therefore extends by the identity across $\mathrm{\partial }A$ and remains an embedding for sufficiently small $v$: finite compact coordinate subcharts retain a uniform differential lower bound for nearby source points, and compact pairs outside those chart neighbourhoods retain positive image separation. The surface still has oriented boundary $c$ and is transverse to $d$. Apply the identical explicit family to $\mathrm{Int}B$, protecting a collar of $d$ and taking target $c$. No closed double or disjointness of an added half is asserted. Now [F5] applies to these actual transverse spanning surfaces in both orders, with the same boundary diagram $D$. It equates $\mathrm{lk}(c,d)$ and $\mathrm{lk}(d,c)$ with the two diagram half-sums, which agree by [F11]. Linking is therefore symmetric. It is additive in its first cycle by the homological coefficient homomorphism [F3], and symmetry gives additivity in the other cycle.

Independence of the small positive push. In the oriented normal bicollar, any two sufficiently small positive normal sections are joined by linear interpolation through positive sections after shrinking the collar. This homotopy stays disjoint from $F$ and therefore from every representative $y\subset F$. By [F6], the two pushed cycles have the same homology class in the complement of each component of $y$, so their linking sums coincide.

Smooth disjoint representatives. Remove a thin open boundary collar from $F$, obtaining a compact core $N\subset \mathrm{Int}F$ homeomorphic to $F$. Apply [F2] in the boundaryless surface $\mathrm{Int}F$ to $N$, and reattach the collar. The disk-and-band construction in [F2] gives a homeomorphism from a thickening of a finite connected graph to $N$, hence to $F$. Give the model its standard smooth structure by rounding the disk-and-band corners. Both surfaces are smooth and Hausdorff with finite chart covers, so [F13] replaces this homeomorphism by a diffeomorphism. Choose a spanning tree of the graph. Each edge outside the tree closes along the unique tree path to a simple graph cycle; these cycles freely generate its fundamental group. In the thickening they have simple smooth representatives: use parallel tracks in edge bands and join them by arcs in vertex disks. The finitely many tracks and joining arcs can be drawn polygonally in these model coordinates with only transverse double crossings, no triple points, and then rounded away from crossings. The graph retraction and [F2] make their classes an integral homology basis. Transport these smooth circles by the diffeomorphism. For any integral class take finitely many parallel copies of these circles, orienting copies negatively for negative coefficients. In the disk-and-band coordinates the copies can be chosen to have only transverse double crossings, with no triple points. Choose mutually disjoint small disks about those finitely many crossings. At each crossing replace the two crossing arcs by the disjoint oriented smoothing, joining incoming ends to outgoing ends and agreeing with the old arcs near the disk boundary. Such arcs occupy the two sectors pairing adjacent compatible ends and can be rounded smoothly inside the disk. The old arcs minus the new arcs form a singular one-cycle supported in that disk, since their endpoint boundaries cancel. It bounds there explicitly by coning each oriented edge to the disk center in convex disk coordinates; the radial endpoint edges cancel. Thus each replacement preserves integral homology. After all replacements the result is a compact smooth embedded oriented one-manifold without boundary, with finitely many circle components. The zero class may use the empty cycle, including when $g=0$. Each $a_i$ in the given basis can consequently be represented in this way, without requiring that different basis representatives be mutually disjoint.

Independence of cycle representatives. Write a class $x$ as ${\sum }_i{m}_i[a_i]$ and a class $y$ as ${\sum }_j{n}_j[a_j]$ using the given basis, represented by the multicurves of step 1.3. If $x^{\mathrm{\prime }}$ is another disjoint-union representative of $x$, then $x^{\mathrm{\prime }}-{\sum }_i{m}_i{a}_i=\mathrm{\partial }C$ for a 2-chain $C$ on $F$. Pushing $C$ to a sufficiently small positive normal level gives a chain disjoint from $y$ whose boundary is the difference between the positive push-offs of the two representatives. Hence they determine the same class in the homology of the complement of each component of $y$, so their linking sums agree by [F3]. If $y^{\mathrm{\prime }}$ represents $y$, symmetry from step 1.1 changes the comparison to linking with $x^{+}$; the chain on $F$ bounding $y-y^{\mathrm{\prime }}$ is disjoint from $x^{+}$ and gives the same conclusion. Therefore the value is independent of both representatives.

Bilinearity and the matrix formula. Additivity of linking in step 1.1 and representative independence in step 2.1 show that the value is additive in each homology variable. For $x={\sum }_i{m}_i[a_i]$ and $y={\sum }_j{n}_j[a_j]$, additivity gives $\mathrm{lk}(x^{+},y)={\sum }_{i,j}{m}_i{n}_j\mathrm{lk}(a_i^{+},a_j)={\sum }_{i,j}{m}_i{n}_j{V}_{ij}$. This is exactly the pairing defined by the matrix, proving the claim.

Construct the smooth handle basis and its intersection matrix. Cap the single boundary component by a disk and apply [F1] to obtain the oriented polygon normal form ${\prod }_{h=1}^g[a_h,b_h]$; remove the capping disk to recover the one-boundary surface. In this model the $h$th handle has two standard core curves, one meridional and one longitudinal, which meet once. Choose their orientations so the ordered tangent pair is positive at that crossing; curves from distinct handles are disjoint. The finite chart and disk-band construction in [F1] realizes this model on the given smooth surface: draw the curves as finite piecewise-smooth arcs in its bands and smooth their finitely many corners in disjoint coordinate disks, leaving their crossings fixed. In the model the surface retracts to the graph of these $2g$ core generators, so their homology classes form an integral basis; the same is true on $F$. Their intersection matrix is therefore the block sum $J_0={⨁}_{h=1}^g\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, with zero entries between distinct handles.

Compare positive and negative pushes in the constructed handle basis. Use the curves $b_i,b_j$ constructed in step 1.1 and let $C_i=b_i\times [-ϵ,ϵ]$ be the product strip, oriented by $(n,u)$ so that $\mathrm{\partial }{C}_i=b_i^{+}-b_i^{-}$. Write $V^0$ for the Seifert matrix in this basis. By [F5] choose a compact oriented spanning surface ${\mathrm{\Sigma }}_j$ for $b_j$, and choose regular diagrams for the resulting link pairs. For $i=j$ the desired skew-entry identity is $0=0$, and the diagonal surface-intersection entry is zero by the handle model in step 1.1; hence take $i\mathrm{\ne }j$ in the intersection construction. The boundary curves $b_i^{+},b_i^{-},b_j$ are pairwise disjoint. First let $X=\mathrm{Int}{\mathrm{\Sigma }}_j$ with inclusion $f:X\to S^3\subset {\mathbb{R}}^4$ and closed target $\mathrm{\partial }{C}_i$. Choose an open collar $G$ of $b_j$ disjoint from this target and protect a smaller collar. Its complement $K=X\setminus G$ is compact. By [F7] choose a smooth $\chi :X\to [0,1]$, equal to one near $K$, compactly supported in a relatively compact open neighbourhood of $K$ avoiding the smaller collar. Then $\chi =0$ only where $f$ is already transverse. For $\mathrm{\parallel }v\mathrm{\parallel }<1\mathrm{/}2$ define $\mathcal{F}(p,v)=(f(p)+\chi (p)v)\mathrm{/}\mathrm{\parallel }f(p)+\chi (p)v\mathrm{\parallel }$. Where $\chi >0$ its parameter derivative is $\chi \mathrm{/}\mathrm{\parallel }f+\chi v\mathrm{\parallel }$ times the tangent projection onto $T{S}^3$, hence surjective; where $\chi =0$, $d\chi =0$ and old transversality applies. Thus the family is transverse. By [F7] every small parameter ball has a transverse slice. Compact support and direct differentiation give uniform $C^1$ convergence to the inclusion. A sufficiently small slice therefore extends by the unchanged boundary to an embedded spanning surface: finite compact coordinate subcharts keep a differential lower bound near the diagonal, and the remaining compact pairs keep positive image separation. Denote this surface again by ${\mathrm{\Sigma }}_j$; it is transverse to $\mathrm{\partial }{C}_i$. Let $P=C_i\cap b_j$. These points are finite, and the tangent of $b_j$ is not in $T{C}_i$ there because the handle curves cross transversely in $F$. Consequently $C_i$ is already transverse to ${\mathrm{\Sigma }}_j$ near $P$. It is also transverse near $\mathrm{\partial }{C}_i\cap {\mathrm{\Sigma }}_j$, since ${\mathrm{\Sigma }}_j$ was made transverse to $\mathrm{\partial }{C}_i$. Put $U=S^3\setminus b_j$ and $X=\mathrm{Int}{C}_i\setminus P$. The target $Z={\mathrm{\Sigma }}_j\setminus b_j$ is a closed embedded boundaryless submanifold of $U$: ${\mathrm{\Sigma }}_j$ is compact and all its boundary was removed. The inclusion $X\to U$ is a map between boundaryless manifolds. Protect a closed inner collar of $\mathrm{\partial }{C}_i$ and, at every point of $P$, a sufficiently small punctured closed disk neighbourhood. These are closed subsets of $X$, and the inclusion is already transverse to $Z$ on their neighbourhoods. Choose these neighbourhoods small enough to retain all the original endpoint data. For the inclusion $f:X\to U\subset S^3\subset {\mathbb{R}}^4$, choose an open protected region $G$ containing all boundary ends and punctured disks at $P$, on which $f$ is transverse to $Z$; protect smaller such regions. Then $K=X\setminus G$ is compact. Choose a relatively compact open $W\subset X$ containing $K$ and missing the smaller protected regions, and use [F7] to choose $\chi \in [0,1]$ equal to one near $K$, with compact support in $W$. Where $\chi =0$ old transversality holds. Use the explicit family $\mathcal{F}(p,v)=(f(p)+\chi (p)v)\mathrm{/}\mathrm{\parallel }f(p)+\chi (p)v\mathrm{\parallel }$. As above, where $\chi >0$ the parameter derivative surjects onto $T{S}^3$ and where $\chi =0$ the old derivative is unchanged, so the total family is transverse to $Z$. The compact set $f(\mathrm{supp}\chi )$ has positive distance from $b_j$; choosing $v$ small enough that the uniform displacement is less than half that distance keeps the entire family in $U$. Parametric transversality and the dense-complement statement [F7] now give an arbitrarily small transverse parameter. Direct differentiation on the compact support gives uniform $C^1$ convergence. Finite compact coordinate subcharts and off-diagonal separation, as in the preceding phase, ensure embeddedness. The family is identically the original inclusion on smaller punctured endpoint disks and boundary collars, so extends by the identity at $P$ and across $\mathrm{\partial }{C}_i$, fixes all endpoint intersection germs, and creates no new intersection with $b_j$. Thus the original compact strip and spanning surface have been made transverse, preserving exactly the required endpoint sets. To apply the boundary-preimage theorem within its hypotheses, remove pairwise disjoint small open disks $D_p^{\circ }\subset \mathrm{Int}{C}_i$ about the points $p\in P$, small enough that their punctured parts lie in the protected neighborhoods, and put $C_i^0=C_i\setminus {\bigcup }_{p\in P}{D}_p^{\circ }$. The inclusion $C_i^0\to U$ is defined on a compact surface with boundary; it is transverse to $Z$, and its restriction to each boundary component is transverse to $Z$ by the protected endpoint germs on $\mathrm{\partial }{C}_i$ and the local half-arc at each $p$. The boundary-preimage theorem gives a compact oriented one-manifold $C_i^0\cap Z$. Adjoining the protected half-arcs from the inner circles to their centers gives a compact oriented one-manifold $C_i\cap {\mathrm{\Sigma }}_j$ with boundary exactly $(\mathrm{\partial }{C}_i\cap {\mathrm{\Sigma }}_j)\cup (C_i\cap b_j)$. No added double half or disjointness assertion about one is used. Orient its tangent line by requiring $(n_{C_i},n_{{\mathrm{\Sigma }}_j},\mathrm{\ell })$ to be positive in $S^3$, where the first two vectors are the positive normals of the oriented surfaces. The local determinant check at an endpoint on $\mathrm{\partial }{C}_i$ gives boundary sign minus the ordered boundary-curve/surface sign; at an endpoint on $\mathrm{\partial }{\mathrm{\Sigma }}_j$ it gives boundary sign plus the ordered surface/boundary-curve sign. Explicitly, the first local model is $C_i=\{z=0,x\ge 0\}$ with basis $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$ and boundary tangent $-{\mathrm{\partial }}_y$, and ${\mathrm{\Sigma }}_j=\{y=0\}$ with basis $({\mathrm{\partial }}_x,{\mathrm{\partial }}_z)$; the intersection tangent is inward $+{\mathrm{\partial }}_x$. The second model is ${\mathrm{\Sigma }}_j=\{z=0,x\ge 0\}$ with basis $(-{\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$ and boundary tangent $+{\mathrm{\partial }}_y$, and $C_i=\{y=0\}$ with basis $({\mathrm{\partial }}_x,{\mathrm{\partial }}_z)$; the intersection tangent is again inward $+{\mathrm{\partial }}_x$. Reversing either surface orientation reverses both signs in its endpoint identity. Thus the zero signed boundary sum is $0=-(\mathrm{\partial }{C}_i\cdot {\mathrm{\Sigma }}_j)+(C_i\cdot b_j),$ and therefore $b_i^{+}\cdot {\mathrm{\Sigma }}_j-b_i^{-}\cdot {\mathrm{\Sigma }}_j=C_i\cdot b_j\mathrm{.}$ If $u,v$ are the positive tangents to $b_i,b_j$ at one of their crossings and $n$ is the positive normal to $F$, then the strip tangent basis is $(n,u)$ and adjoining $v$ gives $(n,u,v)$. By [F3] this has the ambient sign of $(u,v)$ in $F$, and [F4] fixes the ordered intersection convention; hence $C_i\cdot b_j=b_i{\cdot }_F{b}_j$. Applying [F5] to the transverse link pairs yields $V_{ij}^0-\mathrm{lk}(b_i^{-},b_j)=b_i{\cdot }_F{b}_j\mathrm{.}$

The pair $(b_i^{-},b_j)$ can be translated through the normal collar by $+ϵ$ while keeping the two curves disjoint; it becomes $(b_i,b_j^{+})$. Hence $\mathrm{lk}(b_i^{-},b_j)=\mathrm{lk}(b_i,b_j^{+})$ by [F2]. Symmetry of linking in [F2] changes the latter to $\mathrm{lk}(b_j^{+},b_i)=V_{ji}^0$. Therefore step 2.1 gives $V_{ij}^0-V_{ji}^0=b_i{\cdot }_F{b}_j$.

Identify the intersection matrix in every basis. Step 3.1 proves $V^0-(V^0{)}^T=J_0$. For any other integral basis $a_j={\sum }_i{b}_i{P}_{ij}$, the change matrix $P$ lies in ${\mathrm{G}\mathrm{L}}_{2g}(\mathbb{Z})$. Bilinearity of the Seifert pairing [F2] and of surface intersection gives $V=P^T{V}^{0}P$ and $J_F=P^T{J}_{0}P$. Hence $V-V^T=J_F$, $J_F^T=-J_F$, and $\det J_F=(\det P{)}^2\det J_0=1$. If $g=0$, the matrices are empty and the determinant is the empty product $1$.

The boundary class. The boundary circle $\mathrm{\partial }S$ follows the inward conormal collar of $F$ once around $K$, so its projection to the core has degree $+1$. It bounds the oriented surface $S$ in $E$, hence maps to zero in $H_1(E;\mathbb{Z})$. By [F2] it is the unique degree-one class in the boundary kernel, namely the preferred longitude. The collar removed from $F$ is an annulus, so [F3] also identifies $H_1(S;\mathbb{Z})$ with $H_1(F;\mathbb{Z})$.

Homology of the cut complement. In an interior product chart $F\times (-\delta ,\delta )$, the complement has the two half-collars $r<0$ and $r>0$; on either half replace $r$ during the homotopy by $(1-s)r+s\mathrm{sign}(r)\delta$ for $0\le s\le 1$. At the boundary collar, the adapted tube gives the corresponding punctured half-disk fibers, and radial projection to their outer boundary arcs is the same deformation. These maps agree with the identity off the collar and exhibit the cut-open complement $Y$ as homotopy equivalent to $S^3\setminus F$. Since $H_2(F)=0$ by [F3], Alexander duality in [F4] gives ${\tilde{H}}_0(S^3\setminus F)=0$, hence $Y$ is connected. In degree one, [F4] and the universal-coefficient sequence identify $H_1(Y;\mathbb{Z})$ with $\mathrm{Hom}(H_1(F;\mathbb{Z}),\mathbb{Z})$. The natural duality map sends $[\gamma ]$ to $([a]\mapsto \mathrm{lk}(\gamma ,a))$; the meridian normalization in [F5] fixes its sign.

The stacked space is the positive-linking cover. Product neighborhoods of the cut faces show that the quotient map ${\tilde{E}}_F\to E$ is a covering: away from $S$ it is the identity on a sheet, and at $S$ the two product half-collars of adjacent copies give the two half-neighborhoods of a point in $E$. The stack is connected because each $Y_n$ is connected by step 1.2 and consecutive copies meet along the connected surface $S$. The shift $t$ is a deck transformation with quotient group $\mathbb{Z}$. To fix the sign, let $k$ be the positive tangent to $K$ and $u$ the inward conormal of $F$ at its boundary. The outward-normal-first boundary convention makes $(k,u)$ positive in $TF$, so $(n,k,u)$ is positive in $T{S}^3$ for the declared positive normal $n$. The positive meridian disk then has oriented basis $(u,n)$; at its intersection with $F$ its positive boundary tangent is $n$, so it crosses from the negative to the positive side. Under the specified gluing this moves from $Y_n$ to $Y_{n+1}$ and has monodromy $+1$. Every loop's sheet change is its signed intersection with $S$, which equals its total linking with $K$ because $\mathrm{\partial }S$ is parallel to $K$ and the omitted boundary collar misses loops in $E$. Therefore the cover subgroup is the kernel of the one-variable total-linking homomorphism, and [F7] identifies this as the cyclic cover with the stated positive deck generator.

Mayer–Vietoris presentation. Cover the stack by open thickenings of the copies $Y_n$; consecutive thickenings overlap in one copy of $S$ and nonconsecutive ones are disjoint. A singular chain has compact image and therefore meets only finitely many sheets, so the finite-chain Mayer–Vietoris sequences pass to the direct limit over finite consecutive stacks. Identifying translates with powers of $t$ gives the exact sequence $H_1(S;\mathbb{Z})\otimes \mathrm{\Lambda }\stackrel{i_{-}-t{i}_{+}}{\to }{H}_1(Y;\mathbb{Z})\otimes \mathrm{\Lambda }⟶H_1({\tilde{E}}_F;\mathbb{Z})⟶H_0(S;\mathbb{Z})\otimes \mathrm{\Lambda }\stackrel{1-t}{\to }{H}_0(Y;\mathbb{Z})\otimes \mathrm{\Lambda }\mathrm{.}$ Both $S$ and $Y$ are connected, so the last map is multiplication by $1-t$. If $q\in \mathrm{\Lambda }$ and $(1-t)q=0$, a nonzero $q$ has a largest exponent $m$ with nonzero coefficient $q_m$, while the coefficient of $t^{m+1}$ in $(1-t)q$ is $-q_m\mathrm{\ne }0$; hence $q=0$. The last map is therefore injective. Exactness makes the preceding map onto $H_1({\tilde{E}}_F)$ with kernel exactly the image of $i_{-}-t{i}_{+}$; using $H_1(S)\cong H_1(F)$ from step 1.1 gives the asserted cokernel presentation.

Put $h=h_1$. Because $h$ preserves the ambient orientation and the orientations of $K,F$, [F3] sends each positive normal push-off of a curve $a_i$ to a positive normal push-off of $h(a_i)$, up to a homotopy within the positive collar. By [F2] the induced map on the complement preserves the positive-meridian coefficient, so $\mathrm{lk}(a_i^{+},a_j)=\mathrm{lk}(h(a_i{)}^{+},h(a_j))\mathrm{.}$ Thus the transported basis $b_i=h_{\ast }[a_i]$ of $H_1(F^{\mathrm{\prime }};\mathbb{Z})$ has exactly the same matrix $V$.

Let $c_1,\dots ,c_n$ be any chosen integral basis on $F^{\mathrm{\prime }}$. There is a unique integer matrix $P$ with $[c_j]={\sum }_i{P}_{ij}{b}_i$; since both lists are integral bases, $P\in {\mathrm{G}\mathrm{L}}_n(\mathbb{Z})$. Bilinearity [F1] gives the matrix in the $c$ basis as $V^{\mathrm{\prime }}=P^{T}VP$. By [F4], this unimodular congruence is one of the defining S-equivalence moves, so the two matrices are S-equivalent.

Matrices of the two inclusions. Use the dual coordinate of $\gamma \in H_1(Y;\mathbb{Z})$ at $a_i$ given by $\mathrm{lk}(\gamma ,a_i)$ from [F3]. The $j$th source basis vector under $i_{-}$ is the negative push $a_j^{-}$. Its $i$th dual coordinate is $\mathrm{lk}(a_j^{-},a_i)$. By symmetry [F2], this equals $\mathrm{lk}(a_i,a_j^{-})$. Translating both curves by a small positive normal amount preserves their disjointness and changes the pair to $(a_i^{+},a_j)$, so this coordinate is $V_{ij}$ by [F1]. Thus the $j$th column of $i_{-}$ is the $j$th column of $V$. For $i_{+}$ the $i$th dual coordinate of the image of $a_j$ is $\mathrm{lk}(a_j^{+},a_i)=V_{ji}$ by [F1], so the matrix of $i_{+}$ is $V^T$.

The deck variable multiplies the $i_{+}$ term by $t$ under the seam convention in [F3]. Therefore the coordinate matrix of $i_{-}-t{i}_{+}$ is $V-t{V}^T$ by step 1.1. The exact Mayer–Vietoris sequence in [F3] identifies its cokernel with $H_1({\tilde{E}}_F;\mathbb{Z})$, proving both claims.

Choose a tube adapted to $F$. The smooth boundary collar of $F$ gives $c:K\times [0,\delta )\to F$ with $c(s,0)=s$; in local ambient charts extend it smoothly a small distance to negative collar parameter, shrinking it so its image stays in the domain of a tubular chart ${\mathrm{\Phi }}_0$ from [F1]. Let $\pi$ be the chart's projection to the zero section. The map $\mathrm{\Psi }:K\times (-\delta ,\delta )\to K\times \mathbb{R},\mathrm{\Psi }(s,r)=(\pi (c(s,r)),r)$ is two-dimensional. At $r=0$, its derivative in the $(s,r)$ variables has block form $\left(\begin{array}{cc}1& \ast \\ 0& 1\end{array}\right)$, so it is invertible. The inverse function theorem makes $\mathrm{\Psi }$ a local diffeomorphism near $K\times \{0\}$. After shrinking $\delta$, each circle map $s\mapsto \pi (c(s,r))$ is $C^1$-close to the identity and hence a degree-one local diffeomorphism of the circle; it is therefore a diffeomorphism. Thus $\mathrm{\Psi }$ is a diffeomorphism on the smaller collar, and reparameterizing by it writes the extended collar as a graph $q(s,r)$ in the normal disks, with $q(s,0)=0$ and ${\mathrm{\partial }}_{r}q(s,0)=u(s)$, the inward conormal. Use the oriented framing in [F1] with $u(s)$ as first vector and a transverse vector $v(s)$ as second. The map $(s,r,z)\mapsto {\mathrm{\Phi }}_0(s,q(s,r)+zv(s))$ has invertible derivative along the zero section and agrees with $c(s,r)$ for $z=0,r\ge 0$. The inverse function theorem gives local coordinates; compactness gives a common radius, and after shrinking, distinct fibers do not collide. Reparameterizing each fiber disk by these coordinates yields an embedded tubular map for which the surface is exactly the radial half-collar $z=0,r\ge 0$.