A Seifert matrix presents the one-variable Alexander module

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 $z=0,r\ge 0$.

Let $E$ be the compact exterior of this adapted tube and $S=F\cap E$. Then $S$ is properly embedded and its boundary on $\mathrm{\partial }E$ projects once positively to $K$, because it is the outer edge of the inward boundary collar. It bounds $S$ in $E$, so its homology class maps to zero in $H_1(E)$. By [F2] it is the unique degree-one class in that kernel, hence the preferred longitude. Thus the hypotheses of [F3] hold.

The cut-open cover supplied by [F3] has first homology the cokernel of $V-t{V}^T$ over $\mathrm{\Lambda }$, with $t$ moving in the positive-meridian direction.

Let $p:{\tilde{X}}_K\to X_K$ be the total-linking cover from [F4], and let $H:X_K\times I\to X_K$ be the strong deformation retraction onto $E$ from [F1]. By [F6], lift $H(p(-),s)$ starting at the identity of ${\tilde{X}}_K$. The lifted homotopy fixes $p^{-1}(E)$ pointwise, since the corresponding base paths are constant, and its final map lands in $p^{-1}(E)$. It is therefore a strong deformation retraction onto $p^{-1}(E)$, which is connected. For every deck transformation $d$, the homotopies $\tilde{H}(d(-),s)$ and $d\tilde{H}(-,s)$ lift the same base homotopy and have the same initial map $d$; uniqueness in [F6] makes them equal. Thus the induced isomorphism on $H_1$ is deck-equivariant. The restricted covering $p^{-1}(E)\to E$ has subgroup the kernel of total linking restricted to ${\pi }_1(E)$. By [F2] and [F4], this homomorphism sends the positive meridian to $+1$, so it is the homomorphism defining the cut-open cover in step 3.1. Classification [F5], now over the common base $E$, identifies these connected covers and their positive deck generators. The resulting isomorphism on $H_1$ is consequently $\mathrm{\Lambda }$-linear and identifies the cokernel in step 3.1 with $A_K$.