Fox differential calculus and the Alexander invariant of a link

I am teaching a course in knot theory, and I would like to describe the presentation of the Alexander module of a link obtained via Fox differential calculus. In doing this, I should prove the following fact.

Let $L$ be an $n$-component link in $S^3$ with complement $M$, and let $x_0\in M$ be a basepoint. Let $\widetilde M$ be the maximal abelian covering of $M$, and let $\widetilde{M}_0$ be the preimage of $x_0$ in $\widetilde{M}$. Fox differential calculus shows how to obtain a presentation matrix for $H_1(\widetilde{M},\widetilde{M}_0;\mathbb{Z})$ (with coefficients in $\mathbb{Z}[t_1^{\pm 1},\ldots, t_n^{\pm 1}]$) from a presentation of $\pi_1(M)$.

The usual proof of this fact uses cellular homology, and proceeds more or less as follows: 1. Once a finite presentation of the group $G$ is given, if $X$ is the $2$-dimensional cellular complex associated to the presentation (with only one $0$-cell, one $1$-cell for every generator and one $2$-cell for every relation), then the matrix associated to the presentation of $G$ provides a presentation of $H_1(\widetilde{X},\widetilde{X}_0;\mathbb{Z})$; 2. Using Tietze's Theorem, one shows that the module presented by the Fox matrix of a presentation of a group does not depend on the chosen presentation; 3. One shows that the link complement $M$ admits a realization as a cellular complex with only one $0$-cell.

In fact, point (3) allows to provide a presentation of $\pi_1 (M)$ whose Fox derivatives compute $H_1(\widetilde{M},\widetilde{M}_0)$, by point (1). Then (2) allows us to use any other presentation to get the same result.

My question is: is it possible to prove that $H_1(\widetilde{M},\widetilde{M}_0)$ is presented by the Fox matrix of any presentation of $\pi_1 (M)$ without relying on the fact that a link complement admits a cellular structure with only one $0$-cell (and maybe without using cellular homology too much)? I feel that there should be a proof that does not use cellular homology, and only relies on Hurewicz Theorem.

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Let $p:(\tilde{X},y)\rightarrow (X,x)$ be the universal cover of a CW-complex $X$. If $x$ is the only $0$_cell, then every $1$-cell has a unique lift to the universal cover having $y$ as its initial point, and these lifts form a basis for the $1$-chains over the group ring of $\pi_1(X,x)$. The Fox derivative is an axiomatization of the computation of the boundary of a two cell lifted to $\tilde{X}$ so that its initial point in a corresponding relator is $y$.
If you have more $0$-cells then the $1$-skeleton gives you information about the fundamental groupoid of $X$ restricted to the points which are the $0$-cells. As you want data about fundamental group, you could choose a maximal tree in the $1$-skeleton of $X$. The generators of $\pi_1(X)$ correspond to the edges of the $1$-skeleton that aren't in the tree. You can associate a relator to each two cell by keeping track of what order, and in what direction its boundary crosses those edges. The Fox Calculus should now work, because you could alternatively collapse the tree to a point, and not change the homotopy type of the $CW$-complex.
To compute the first homology of any cover as a module over $\mathbb{Z}$[\pi_1(X)]$you just need to interpret the answer in the universal cover in the appropriate cover. To effect your general proof of invariance, you just need to produce change of basis matrices corresponding to changing trees, and doing elementary moves on the two complex. To avoid the cell complex use group cohomology. If$M$is a left$R[G]$-module a crossed homomorphism is a map$\phi:G\rightarrow M$satisfying$\phi(xy)=\phi(x)+x.\phi(y)$for any$x,y\in G$. The crossed homomorphisms are an$R$-module. A principle crossed homomorphism is one of the form$\phi(g)=v-g.v$for some$v\in M$. The quotient of the crossed homomorphisms by the principle homomorphsisms is$H^1(G,M)$. You are interested in this when$G=\pi_1(X)$and$M=H_1(X)$. Given a presentation of your group, with generators$x_i$and relators$r_j$, there is a presentation of the crossed homomorphisms, as those crossed homomorphisms on the free group generated by the$x_i$that annihilate the relators. It can be shown that the space of crossed homomorphisms on the free group are generated as an$R[F(x_i)]$module by the Fox partial derivatives, and you are off and running, as you are really just interested in calculuating group cohomology. - Dear prof. Frohman, your post completely answers my question. However, I was wondering if it is possible to prove that Fox Calculus provides a presentation for the module$H_1(\widetilde{M},\widetilde{M}_0)$without referring to cellular homology. In fact, I believe that this is true even without the hypothesis that$M$is a CW complex, under the only assumption that$M$is path connected (and has a finitely presented fundamental group). (See below the second part of my comment). – Roberto Frigerio Nov 14 '11 at 21:34 For example, if we were interested in the absolute module$H_1(\widetilde{M})$(where$\widetilde{M}$is the maximal abelian covering of$M$), then this module is clearly isomorphic to$G'/G''$(where$G=\pi_1 (M)$,$G'=[G,G]$and$G''=[G',G']$), on which$G/G'$acts by conjugation. (See below the last part of my comment). – Roberto Frigerio Nov 14 '11 at 21:35 With some effort, one may start from this remark and deduce a presentation of$H_1(\widetilde{M})$from a presentation of$G$, just by using Hurewicz Theorem (I am not sure, but this is probably done in the book by Burde and Zieschang). I was wondering if something similar could work for proving that the usual recipe involving Fox Calculus provides a presentation of$H_1(\widetilde{M},\widetilde{M}_0)$, even in the case when$M$is not a CW complex. – Roberto Frigerio Nov 14 '11 at 21:35 As the fundamental group of the abelian cover is just the commutator subgroup of the fundamental group, and the first homology of that is just it's abelianization, then build a CW complex from your presentation of the fundamental group, take the abelian cover of that and compute its first homology, using Fox calculus, and that is necessarily computing the first homology of$\tilde{M}$? – Charlie Frohman Nov 15 '11 at 0:48 I agree that$H_1(\widetilde{M})$is canonically isomorphic (as a$G/G'$-module) to$G'/G''$, where$G/G'$acts on$G'/G''$by conjugation. However, the Alexander module of a link is usually defined as the relative homology module$H_1(\widetilde{M};\widetilde{M}_0)\$, and Fox calculus provides a presentation of this module. I was wondering if there is a proof of this fact which avoids the use of cellular homology. – Roberto Frigerio Nov 15 '11 at 18:13