Alexander polynomial from the torus-knot group presentation

Abelianizing the presentation in [F1] gives the relation $p[x]=q[y]$. The map $\varphi (x)=q$, $\varphi (y)=p$ is onto because $\gcd (p,q)=1$; its kernel in ${\mathbb{Z}}^2$ is generated by $(p,-q)$, exactly the abelianized relation vector, so the abelianization is $\mathbb{Z}$. The meridian in [F1] maps to $p\alpha +q\beta =1$, so the epimorphism $\varphi$ in the Statement is precisely the positive total-linking homomorphism that classifies the cover in [F4].

For every integer $n$, $(z^n-1)\mathrm{/}(z-1)$ is a finite Laurent sum; for $n<0$ it equals $-(z^{-1}+\cdots +z^n)$. In the lifted attaching word, a positive edge contributes the current deck translate and a negative edge contributes the negative of the preceding translate. The $x^p$ block therefore contributes $(x^p-1)\mathrm{/}(x-1)$ to the $x$-edge coordinate. The following $y^{-q}$ block starts after the $x^p$ translate and contributes $x^p(y^{-q}-1)\mathrm{/}(y-1)$ to the $y$-edge coordinate. These are the complete signed edge contributions of the two-cell in ${\tilde{P}}_{\varphi }$.

The cyclic cover in the Statement has cellular chain groups $C_2=\mathrm{\Lambda }$, $C_1={\mathrm{\Lambda }}^2$, $C_0=\mathrm{\Lambda }$. Its first boundary is $d_1=(t^q-1,t^p-1)$ because lifted $x$- and $y$-edges end $q$ and $p$ deck steps from their starts. Evaluating the two edge contributions in step 1.2 at $x=t^q$, $y=t^p$ gives the second boundary column; the $y$-coordinate is $t^{pq}(t^{-pq}-1)\mathrm{/}(t^p-1)=-(t^{pq}-1)\mathrm{/}(t^p-1)$: $d_2(1)={(\frac{t^{pq}-1}{t^q-1},-\frac{t^{pq}-1}{t^p-1})}^T\mathrm{.}$ This is the cellular boundary of the lifted two-cell.

In $\mathrm{\Lambda }$, $\gcd (t^p-1,t^q-1)\doteq t-1$: for positive exponents the Euclidean algorithm gives this because $\gcd (\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }q\mathrm{\mid })=1$, and negative exponents differ from the positive ones by units. Hence $a=(t^q-1)\mathrm{/}(t-1)$ and $b=(t^p-1)\mathrm{/}(t-1)$ are coprime. The kernel of $d_1$ is the free rank-one submodule generated by $(b,-a{)}^T$: if $au+bv=0$, unique factorization implies $a\mid v$, and then $(u,v)$ is a multiple of $(b,-a)$. The column in step 2.1 is ${\delta }_{p,q}(b,-a{)}^T$. To show ${\delta }_{p,q}\in \mathrm{\Lambda }$, replace signed exponents by $m=\mathrm{\mid }p\mathrm{\mid }$, $n=\mathrm{\mid }q\mathrm{\mid }$, which changes each factor only by a unit. Set $A=(t^m-1)\mathrm{/}(t-1)$ and $B=(t^n-1)\mathrm{/}(t-1)$. They are coprime; since $m,n$ divide $mn$, both divide $(t^{mn}-1)\mathrm{/}(t-1)$. Unique factorization gives $AB\mid (t^{mn}-1)\mathrm{/}(t-1)$, hence ${\delta }_{m,n}=((t^{mn}-1)\mathrm{/}(t-1))\mathrm{/}(AB)\in \mathrm{\Lambda }$, and the signed quotient is a unit multiple. If $\mathrm{\mid }p\mathrm{\mid }=1$ or $\mathrm{\mid }q\mathrm{\mid }=1$, this quotient is a unit, so the same cyclic presentation gives $A_K=0$.

By [F5], $H_1$ of the cyclic cover of $P_{p,q}$ is $\ker d_1\mathrm{/}\mathrm{im}{d}_2\cong \mathrm{\Lambda }\mathrm{/}({\delta }_{p,q})$. The presentation complex and $E_K$ have the same fundamental group and the same homomorphism $\varphi$; the fundamental groups of their cyclic covers are the corresponding kernels, and the deck action on their abelianizations is conjugation by a preimage of $1\in \mathbb{Z}$. Hurewicz therefore identifies their first-homology $\mathrm{\Lambda }$-modules. Full AC [A1] supplies the countable-choice hypothesis of [F3] by [L1]; the deformation retraction then identifies the cyclic-cover homology of $E_K$ with that of $S^3\setminus K$. Thus $A_K\cong \mathrm{\Lambda }\mathrm{/}({\delta }_{p,q})$.

The one-by-one presentation $[{\delta }_{p,q}]$ of $\mathrm{\Lambda }\mathrm{/}({\delta }_{p,q})$ has $E_0$ generated by its sole $1\times 1$ minor, namely ${\delta }_{p,q}$. Isomorphism invariance of the Alexander module and the definition [F6] give $E_0(A_K)=({\delta }_{p,q})$ and hence ${\mathrm{\Delta }}_K\doteq {\delta }_{p,q}$.