Well-definedness of the Seifert pairing

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.