Hopf link and two-component unlink Alexander modules

Every point of $S^3\setminus H$ has $z\mathrm{\ne }0$ and $w\mathrm{\ne }0$. The map $(z,w)\mapsto (z\mathrm{/}\mathrm{\mid }z\mathrm{\mid },w\mathrm{/}\mathrm{\mid }w\mathrm{\mid },\mathrm{\mid }z\mathrm{\mid })$ has inverse $(u,v,s)\mapsto (su,\sqrt{1-s^2}v)$, so the open complement is $T^2\times (0,1)$ and a compact exterior is $T^2\times I$. The two phase loops are positive meridians of the two coordinate-circle components, hence form the basis of ${\pi }_1(E_H)=H_1(E_H;\mathbb{Z})={\mathbb{Z}}^2$. By [F3], the maximal abelian cover is the universal cover ${\mathbb{R}}^2\times I$, whose first homology vanishes. The compact-exterior/open-complement identification [F7] therefore gives $A_H=0$.

For the standard unknot $K=\{w=0\}$ in $S^3\subset {\mathbb{C}}^2$, the exterior of the tube $\mathrm{\mid }w\mathrm{\mid }\le ϵ$ is homeomorphic to $D_{\sqrt{1-{ϵ}^2}}^2\times S^1$ by $(z,v)\mapsto (z,\sqrt{1-\mathrm{\mid }z{\mathrm{\mid }}^2}v)$; hence it is a solid torus. For the split unlink use its given separating sphere, with one component and its tube in each ball. The exterior is a collared union along that sphere of two ball-minus-tube pieces. Each piece is a solid-torus unknot exterior with the interior of one disjoint standard ball removed. Reattaching the ball does not change fundamental group, by van Kampen with the simply connected sphere overlap. Thus each piece has group $\mathbb{Z}$ generated by its positive meridian. Van Kampen on their separating-sphere union gives ${\pi }_1(E_U)\cong \mathbb{Z}\ast \mathbb{Z}$ with these meridian generators. This does not identify $E_U$ with a handlebody or assert a graph deformation retraction.

By [F1], the two positive meridians form the coordinate basis; the coordinate-sum definition [F2] therefore sends both to $1$, so its kernel in ${\mathbb{Z}}^2$ is generated by $(1,-1)$. Quotienting the universal cover ${\mathbb{R}}^2\times I$ by this rank-one translation subgroup gives $S^1\times \mathbb{R}\times I$. Its first homology is $\mathbb{Z}$ generated by the circle. Every deck transformation is represented by a translation, and the path of translations from the identity to that translation is a homotopy, so $t$ acts as $1$. Thus the cyclic module is ${\mathrm{\Lambda }}_1\mathrm{/}(t-1)\mathrm{\ne }0$, whereas $A_H{\otimes }_{{\mathrm{\Lambda }}_2,\eta }{\mathrm{\Lambda }}_1=0$ because $A_H=0$.

Let $G=S^1\vee S^1$ map its two based oriented loops to these meridians in $E_U$. It induces the computed fundamental-group isomorphism and intertwines abelianization to ${\mathbb{Z}}^2$. Lift it to the connected maximal abelian covers after fixing the base lift. The induced map on fundamental groups identifies the kernels of abelianization, hence is an isomorphism. Degree-one Hurewicz then makes its map on first homology an isomorphism. This is Laurent-linear: conjugation by each meridian implements its deck action on the kernel abelianization, and the map intertwines those conjugations. The same argument for total linking to $\mathbb{Z}$ compares the cyclic-cover $H_1$. Only first homology is compared; the separating sphere is not removed by a false homotopy equivalence. The lifted graph cellular boundary is $C_1={\mathrm{\Lambda }}_2^2\stackrel{(t_1-1,t_2-1)}{\to }{C}_0={\mathrm{\Lambda }}_2$, so this graph complex computes the desired first homology.

The quotient ${\mathrm{\Lambda }}_2\mathrm{/}(t_1-1)$ is $\mathbb{Z}[t_2^{\pm 1}]$, a domain by [F5], so $t_1-1$ is prime. It does not divide $t_2-1$, whose image under $t_1\mapsto 1$ is the nonzero element $t_2-1$. If $(a,b)$ lies in the kernel, $(t_1-1)a=-(t_2-1)b$, so primality forces $t_1-1$ to divide $b$. Write $b=(t_1-1)c$; cancellation in the domain gives $a=-(t_2-1)c$. Thus the kernel is generated by $(-(t_2-1),t_1-1)$; it is free because its second coordinate can vanish only when $c=0$. Hence $A_U\cong {\mathrm{\Lambda }}_2$.

Under total linking, both loop edges map to $t$, so [F6] gives the cyclic-cover boundary $(t-1,t-1):{\mathrm{\Lambda }}_1^2\to {\mathrm{\Lambda }}_1$. Since ${\mathrm{\Lambda }}_1$ is a domain, its kernel consists of $(a,-a)$ and is free of rank one. This confirms the stated cyclic-cover module and completes the example.