Ordinary acyclicity over Z does not detect group-ring torsion

Let $C$ be the based free right $R$-complex $0\to C_1=R\stackrel{d_1}{\to }{C}_0=R\to 0$ with differential $d_1(x)=xu$, generated in degree one and zero by one basis vector each. By [F1] $u$ is a unit, and the map $h_0(y)=y{u}^{-1}$, $h_1=0$ satisfies $d_1{h}_0={\mathrm{i}\mathrm{d}}_{C_0}$ and $h_0{d}_1={\mathrm{i}\mathrm{d}}_{C_1}$, so $C$ is contractible with this contraction; by [F2] with $q=1$ its contraction torsion is $\tau (C)=(-1{)}^2[u]=[u]$, which is nonzero in $\mathrm{W}\mathrm{h}(C_5)$ by [F1].

Forgetting the $R$-module structure, the contraction $h$ of step 1.1 is a homomorphism of abelian groups, so it contracts the underlying complex of abelian groups of $C$; hence that underlying complex is contractible, and in particular acyclic, with differential the isomorphism $x\mapsto xu$ and inverse $y\mapsto y{u}^{-1}$. Separately, by [F3] the augmentation $\epsilon :\mathbb{Z}[C_5]\to \mathbb{Z}$ is a ring homomorphism with $\epsilon (1)=1$ and $\epsilon (t^k)=1$ for every $k$, so $\epsilon (u)=1-1-1=-1$, which is a unit of $\mathbb{Z}$; the base change along $\epsilon$ gives the complex $0\to \mathbb{Z}\stackrel{-1}{\to }\mathbb{Z}\to 0$, contracted by $n\mapsto -n$, whose homology also vanishes.

The complex $C$ is bounded and based free over $R=\mathbb{Z}[C_5]$, its contraction torsion class $[u]$ is nonzero in $\mathrm{W}\mathrm{h}(C_5)$ by step 1.1, and its underlying complex of abelian groups is contractible, in particular acyclic, by step 2.1; therefore the false claim fails. This is exactly why the handle chain complex of a cobordism is carried over $\mathbb{Z}[{\pi }_1]$ with its chosen lifts and basis and not over $\mathbb{Z}$, as recorded in [F4]: an abelian acyclicity check cannot see the class $[u]$.