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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.

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.