The braid-axis torsion product gives the twisted Burau determinant

Let $\alpha (g)=\rho (g)\psi (g)$ denote the coefficient matrix. Under [A1], the supplied filling certificate and [F7] identify ${\pi }_1(X)$ as the quotient of ${\pi }_1(Y)$ that kills the axis meridian $x$. Pulling the coefficient system back from $X$ therefore gives $\alpha (x)=I_d$ and $\psi (x)=1$; the core of $V$ is represented on $T$ by $g_n$, with matrix $M_n$. The algorithm forms the full positive-meridian basis on $X$ before applying $\chi$, while the compatible filling certificate specifies the resulting pulled-back or extended coefficient systems on $Y,T,V$ and makes them agree on overlaps. The map ${\pi }_1(Y)\to {\pi }_1(X)$ is surjective, so the pulled-back deck map on $Y$ is still onto and the coefficient-unit images for the two exteriors agree. Since $D\mathrm{\ne }0$, $M_n-I_d$ is invertible.

First compute the core-circle spine $V_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$ with one $0$-cell and one $1$-cell for $g_n$. Its twisted boundary is $M_n-I_d$, so this core complex is acyclic and its torsion is $D^{-1}$. This core is not used as a CW pair containing $T$; a separate relative model and its torsion comparison are established before applying the gluing sequence. Give $T$ the standard one-vertex CW structure with $1$-cells $x,g_n$ and one commutator $2$-cell $x{g}_n{x}^{-1}{g}_n^{-1}$. The Fox boundary blocks are $d_1=(0,M_n-I_d{)}^{\mathsf{T}}$ and $d_2=(I_d-M_n,0)$: hence $d_1$ is onto, $d_2$ is injective onto $\ker d_1$, and $T$ is acyclic. In the torsion determinant product, the factors $\det (I_d-M_n)$ and $\det ((M_n-I_d{)}^{-1})$ cancel up to $(-1{)}^d$, so $\tau (T)=(-1{)}^d$.

For $X$, Fox differentiation of $r_i=g_i^{-1}{\beta }_{\ast }(g_i)$ gives block row $M_i^{-1}({\tilde{B}}_{i,\ast }-e_i)$, where $e_i$ is the $i$th identity block row and $M_i=\rho (g_i)\psi (g_i)$. Thus, after deleting the $g_n$ column, the Fox block is $\mathrm{diag}(M_1^{-1},\dots ,M_{n-1}^{-1})({\bar{B}}_{\rho }(\beta )-I)$, whose determinant is $({\prod }_{i=1}^{n-1}\det (M_i^{-1}))N$. The prefactor is a nonzero coefficient unit. The $d_1$ block for $g_n$ is $M_n-I_d$, which is invertible, so $d_1$ is onto and projection from $\ker d_1$ to the first $n-1$ coordinate blocks is an isomorphism. By [F5], the projected $d_2$ is the displayed deleted Fox block. Therefore $N\mathrm{\ne }0$ implies $d_2$ injective and, by dimensions and $d_1{d}_2=0$, the complex is acyclic; if the complex is acyclic, the same projection makes $d_2$ injective and forces $N\mathrm{\ne }0$. Hence nonacyclicity is equivalent to $N=0$.

Give $D^2$ its boundary vertex, boundary edge $x$, and meridian disk cell $a$, and give $S^1$ its vertex and core edge $g_n$. Their product is a finite CW model $V_{\mathrm{r}\mathrm{e}\mathrm{l}}$ of the actual solid torus containing exactly the specified torus CW structure as its boundary subcomplex. Besides the cells of $T$, it has the $2$-cell $a\times \{\ast \}$, still denoted $a$, and the $3$-cell $b=a\times g_n$. Put $A=M_n-I_d$ and let $q$ denote the torus $2$-cell, with the orientation in step 2.1. For a coefficient row $v$, the product boundaries are $\mathrm{\partial }a(v)=x(v),\mathrm{\partial }q(v)=x(-vA),\mathrm{\partial }b(v)=q(v)+a(vA)\mathrm{.}$ The first formula is the meridian disk boundary, the second is the torus Fox boundary from step 2.1, and the third is the oriented boundary of disk times core edge; its two terms have cancelling images under $\mathrm{\partial }$. Replace the degree-two block $q(v)$ by $q^{\mathrm{\prime }}(v)=q(v)+a(vA)$. This block-triangular basis change has determinant one, $\mathrm{\partial }{q}^{\mathrm{\prime }}=0$, and $\mathrm{\partial }b(v)=q^{\mathrm{\prime }}(v)$. Split off the unit two-term pair $(b,q^{\mathrm{\prime }})$, then the unit pair $(a,x)$. The remaining based complex is exactly the core complex with $\mathrm{\partial }{g}_n(v)=vA$. Each split pair has identity incidence and torsion one up to the sign unit, and the triangular change contributes determinant one. Hence $V_{\mathrm{r}\mathrm{e}\mathrm{l}}$ has the same homology and torsion class as $V_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$, so it is acyclic and its torsion is $D^{-1}$ up to the declared units. This calculation supplies the missing based comparison while retaining $T$ in the actual gluing model.

Use the relative-cellular gluing operation [F9] for the actual decomposition in the supplied certificate. For $V$ use $V_{\mathrm{r}\mathrm{e}\mathrm{l}}$ from step 3.1, not its core spine. For the exterior pieces the operation uses the compatible relative CW representatives and the supplied based simple-spine comparisons. Equivalently, a boundary map into a spine is represented by its finite cellular mapping cylinder: each added prism over a torus cell pairs with its source face, with lifted incidence a unit; these pairs are removed in decreasing cell dimension. Their torsion changes lie in the declared unit subgroup. Thus the based torsion classes used below are those of the supplied spines, without identifying the boundary torus with a subcomplex of a core circle or of an abstract presentation. In the common relative models each cell of $T$ occurs in both pieces, and each other cell occurs in exactly one. Inclusion into the first piece and its negative into the second, followed by the difference quotient to the union, therefore give the degreewise split short exact sequence $0⟶C_{\ast }(T)⟶C_{\ast }(Y_{\mathrm{r}\mathrm{e}\mathrm{l}})\oplus C_{\ast }(V_{\mathrm{r}\mathrm{e}\mathrm{l}})⟶C_{\ast }(X_{\mathrm{r}\mathrm{e}\mathrm{l}})⟶0.$ Choose the shared lifts compatibly; any permitted adjustment changes only the declared determinant/deck units. Since $T$ and $V_{\mathrm{r}\mathrm{e}\mathrm{l}}$ are acyclic by steps 2.1 and 3.1, the long exact homology sequence identifies $H_{\ast }(Y;Q^d)$ with $H_{\ast }(X;Q^d)$. Thus $Y$ is acyclic exactly when $X$ is. Subsequent references to the gluing sequence mean these relative models and their certified based comparisons.

In the acyclic case, choose boundary bases in the short exact sequence compatibly: lift a basis of each quotient boundary space, append a basis of the subcomplex boundary space, and lift the resulting boundary bases to chain groups. The change-of-basis matrices are block triangular with diagonal blocks the matrices for the subcomplex and quotient, so their determinants multiply degree by degree; the alternating torsion product therefore gives $\tau (Y)\tau (V)=\pm \tau (T)\tau (X)$. As $\tau (T)=\pm 1$ and both signs lie in the declared unit subgroup, step 2.1 yields $\tau (Y)\doteq \tau (X)D$.

Suppose $X$ and hence $Y$ are acyclic over $Q$. The finite based chain modules over $\mathrm{\Lambda }=\mathbb{F}[H]$ are free. If $z\in C_1(Y;{\mathrm{\Lambda }}^d)$ is a $1$-cycle, acyclicity over $Q$ gives $z\mathrm{/}1=d_2(y)\mathrm{/}s$ for some $y\in C_2(Y;{\mathrm{\Lambda }}^d)$ and nonzero $s\in \mathrm{\Lambda }$. By [F8], equality in the fraction field means that some nonzero $u\in \mathrm{\Lambda }$ annihilates $sz-d_2(y)$; since $C_1(Y;{\mathrm{\Lambda }}^d)$ is free over the domain $\mathrm{\Lambda }$, multiplication by $u$ is injective, so $sz=d_2(y)$. Every degree-one homology class is therefore killed by a nonzero element of $\mathrm{\Lambda }$.

Apply [F2] to the certified spine of $Y$ and delete the $g_n$ column, whose denominator is $D\mathrm{\ne }0$. For $s_i={\beta }_{\ast }(g_i)(x^{-1}{g}_{i}x{)}^{-1}$, the $g_1,\dots ,g_n$ Fox blocks evaluate to $\tilde{B}-I$, where $\tilde{B}$ is the full cumulative-meridian Fox matrix, and the $x$-column blocks are $I_d-M_i$ with $M_i=\rho (g_i)\psi (g_i)$. Because ${\beta }_{\ast }(g_n)=g_n$, the last row of $\tilde{B}-I$ is zero; after deleting its $g_n$ column, the remaining matrix has upper-left block ${\bar{B}}_{\rho }(\beta )-I$, arbitrary upper-right block, zero lower-left block, and lower-right block $I_d-M_n$. It is block upper triangular, so its determinant is $(-1{)}^{d}ND$. Dividing by $D$ gives $(-1{)}^{d}N$, which represents $N$ modulo the sign unit. Consequently $\tau (Y)\doteq N$.

If the complex is acyclic, combine steps 5.1 and 6.1 to obtain $\tau (X)D\doteq N$, hence $\tau (X)\doteq N\mathrm{/}D$. If it is nonacyclic, step 2.2 gives $N=0$, and the declared convention sets $\tau (X)=0=N\mathrm{/}D$. This proves the statement in both branches.