Based twisted torsion changes only by determinant and deck units

If a chosen lift of one $q$-cell is translated by $g$, [F2] changes exactly its $d$-vector block in ${\mathbf{c}}_q$ by $\alpha (g)$. The only torsion bracket whose denominator basis changes is the degree-$q$ bracket in [F1], so the torsion is multiplied by $\det (\alpha (g){)}^{(-1{)}^q}$ (or its inverse under the opposite bracket-order convention). This factor belongs to $\mathcal{U}$ by [F3]. Reversing the cell orientation gives the factor $(-1{)}^{d(-1{)}^q}$, also in $\mathcal{U}$. Applying finitely many such changes multiplies by a product of elements of the subgroup $\mathcal{U}$.

In an elementary expansion, order the new cell bases last and clear the old-cell entries of the new boundary column by adding multiples of old basis columns; these are triangular basis changes with determinant $1$. Because the new $q$-cell is a free face, the lifted boundary matrices have the old blocks together with a single invertible diagonal block $\pm \alpha (g)$ between the new $(q+1)$- and $q$-cell blocks; clearing the other entries in that new column by the described triangular basis changes splits off the two-term complex $0\to Q^d\stackrel{\pm \alpha (g)}{\to }{Q}^d\to 0$. The torsion formula [F1] for this two-term summand is $\det (\alpha (g))$ or its inverse according to the degree parity, hence lies in $\mathcal{U}$ by [F3]. The direct-sum torsion formula multiplies the old torsion by that factor, so the class is unchanged. A collapse is the inverse operation. Iterating proves invariance under every finite sequence of these elementary simple moves.