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Given a presentation $⟨x_1,\dots ,x_n\mid r_1,\dots ,r_m⟩$ of a knot group, apply abelianization to the Fox derivatives $\partial r_j/\partial x_i$. The resulting matrix over $\mathbb{Z}[t^{\pm 1}]$ is an Alexander matrix; suitable maximal minors recover the Alexander polynomial of a knot.

Use generators $a,b,c$ for the three one-handles in the generalized Heegaard diagram. Read a downward crossing of the cut from a left-hand hole to its right-hand partner as the corresponding generator. Traverse the attaching curve around the left-hand $A$ hole, starting at its outgoing downward strand, and then traverse the attaching curve around the left-hand $B$ hole in the same way. Their cut-crossing words areThese are the two relators imposed by the two handle attachments. The remaining boundary is a torus; no third relator should be added. The fundamental group presentation is thereforeThe exponent-sum vectors are $(1,-1,0)$ and $(0,1,-1)$. Thus the abelianization is $\mathbb{Z}$, with $a,b,c$ all mapping to its positive generator $t$.

Apply the Fox derivatives and then this abelianization. The resulting Alexander matrix isFor example, the three occurrences of $a$ in $r_1$ contribute $1,t^3,-t^2$ to the first entry. Every row sums to zero, as the Fox calculus identity predicts.

The cellular boundary in degree one for the infinite cyclic cover is $(t-1,t-1,t-1)$. Its kernel is freely generated by $e_a-e_c,e_b-e_c$. Expressing the two relation boundaries in this basis gives the two-by-two matrix obtained by deleting column $c$ of $F(t)$. Its determinant is the order of the Alexander module of a space. Equivalently, all three maximal matrix minors of $F(t)$ agree up to sign. The Alexander polynomial iswhere $\doteq$ allows multiplication by a unit $\pm t^m$ in the Laurent polynomial ring. Its value at $1$ is $1$.

Both relators have zero exponent sums, so abelianization gives $H_1(Y)={\mathbb{Z}}^3$, with meridian variables $x,y,z$ corresponding to $a,b,c$. The universal abelian cover has deck transformation group ${\mathbb{Z}}^3$ and coefficient group ringThe generalized Heegaard diagram gives a two-dimensional spine with one vertex, three edges and two faces. Its lifted cellular chain complex iswhere chosen lifts of the cells giveThe two columns are the abelianized Fox derivatives of $[a,bc]$ and $[ab,c]$. The Fox calculus identity gives $d_1{d}_2=0$; this can also be checked by multiplying the displayed matrices. There is no three-cell in this spine. One may use the lifted spine because its deformation retraction from $Y$ lifts to the universal abelian cover.

The maximal minors of the Alexander matrix, in row-pair order $(1,2),(1,3),(2,3)$, areTheir greatest common divisor in $R$ is $xyz-1$, since $x-1,y-1,z-1$ have no common nonunit divisor. Accordingly the multivariable Alexander polynomial isThe allowed units are $\pm x^i{y}^j{z}^k$. The single-variable specialization convention can introduce extra $(t-1)$ factors; the answer here is the genuinely multivariable Alexander polynomial.

Keep all four relators and all five generators, in the orders $(r_A,r_B,r_C,r_D)$ and $(d,e,a,b,c)$. Under abelianization, these generators become $(t,t,t^2,t^3,t^2)$. The Fox calculus rules areFor example, $r_B=a{d}^{-1}c{b}^{-1}$ gives the nonzero Fox derivative entries $-a{d}^{-1}$, $1$, $-a{d}^{-1}c{b}^{-1}$, $a{d}^{-1}$ in columns $d,a,b,c$, respectively. After abelianization these are $-t,1,-1,t$. The full Alexander matrix isThe Fox calculus identity supplies a useful check:Deleting the $e$ column, adjacent to the unbounded region across the marked meridian of a knot, givesThus $\det B$ is a representative of the Alexander polynomial of a knot of the figure-eight knot, with its customary ambiguity of multiplication by $\pm t^n$.

Forthe Fox calculus product rule and $\partial d^{-1}/\partial d=-d^{-1}$ give, in generator order $(a,b,c,d)$,Multiplication of every row by the unit $t$ gives the equivalent Alexander matrixFor example, deleting the $a$-column gives determinant $-(t^2-t+1)$, the Alexander polynomial of a knot of the trefoil up to a unit.

The free part of the abelianization is generated by $[y]$, so the fibration class sends $y\mapsto 1$ and $x\mapsto 0$. Applying Fox calculus to the relator and then substituting $x=1$, $y=t$ givesThis has breadth two. Since $K$ is assumed fibered, the breadth is twice the genus, so its fiber is a genus-one surface with one boundary component:

The monodromy can also be read directly from the relator. Put $x_i=y^{i}x{y}^{-i}$. Reidemeister rewriting givesThus the kernel is freely generated by $(x_{-1},x_0)$, and conjugation by $y$ sendsOn first homology, in the ordered basis $([x_{-1}],[x_0])$, a representative isIts characteristic polynomial is $t^2-4t+1$, agreeing up to the unit $-t$ with $4-t-t^{-1}$.

Orient the diagram and assign its regions an Alexander numbering with $R_0$ numbered zero. The abelianization $\varphi :{\pi }_1(E_K)\to ⟨t⟩$ sends $a_i$ to $t^{m_i}$, where $m_i$ is the number of $R_i$. Since $R_1$ is adjacent to $R_0$, $m_1=\pm 1$.

Form the square matrixfrom the Fox derivatives with respect to $a_i$ for $i>1$. This is the Alexander matrix with the $a_1$ column deleted. The Fox identity implies that its maximal minors differ by the factors $\varphi (a_i)-1$, and the standard presentation of the Alexander module therefore givesbecause $m_1=\pm 1$. Thus ${\Delta }_K(t)$ is $\det A_1$ up to a unit $\pm t^r$.

For the free presentation $G=F/R$, conjugation in $F$ gives the relation modulewhere $f\in F$ is any lift of $g\in G$. A different lift differs by an element of $R$, whose inner conjugation acts trivially on the abelianization, so this is a well-defined $\mathbb{Z}G$-module action.

Choose free generators $x_1,\dots ,x_n$ of $F$. The presentation relation sequence becomeswhere $d_1(e_i)={\bar{x}}_i-1$. In Fox calculus, the first map isThe Fox identity $r-1={\sum }_i(\partial r/\partial x_i)(x_i-1)$ gives $d_{1}j=0$, and the standard lifting argument in the free group proves exactness.

Split the sequence at the augmentation ideal $I_G$. Applying ${\mathrm{Hom}}_{\mathbb{Z}G}(-,M)$ toand using projectivity of $(\mathbb{Z}G{)}^n$ givesThe other short exact sequence $0\to I_G\to \mathbb{Z}G\to \mathbb{Z}\to 0$ identifies the last group with $H^2(G,M)$. A one-cocycle on $F$ is a derivation and is determined freely by its values on $x_1,\dots ,x_n$, so $H^1(F,M)$ is $M^n$ modulo principal derivations. Principal derivations vanish on $R$, and restriction sends a derivation $d$ to the $G$-map $r[R,R]\mapsto d(r)$. The preceding cokernel sequence therefore descends to the Mac Lane exact sequence for a free presentation

The left map need not be injective. Take $F=⟨t⟩\cong \mathbb{Z}$, $R=⟨t^m⟩$, $G\cong C_m$, and the trivial module $M=\mathbb{Z}/m\mathbb{Z}$, where $m>1$. Then $H^1(F,M)\cong M$, while restriction sends the derivation determined by $d(t)=a$ toThus the left map is zero although its domain is nonzero, and $R$ is nontrivial.