The cut-open cover presentation of the Alexander module

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.