Ordinary acyclicity forgets nonzero group-ring torsion

In $R=\mathbb{Z}[t]\mathrm{/}(t^5-1)$, direct multiplication gives $(1-t^2-t^3)(1-t-t^4)=1$: the coefficient vector in the basis $(1,t,t^2,t^3,t^4)$ is $(1,0,0,0,0)$. Thus $u$ is a unit with the stated inverse. Its coefficient vector is $(1,0,-1,-1,0)$, unlike every $\pm t^k$, so it is not a trivial unit.

Since $C_5$ is abelian, $R$ is commutative. Determinant sends $G{L}_m(R)$ to $R^{\times }$, is multiplicative, and sends elementary matrices to $1$; hence it descends to $K_1(R)$ and then gives a homomorphism $\mathrm{Wh}(C_5)\to R^{\times }\mathrm{/}⟨\pm t^k⟩$. The determinant of the one-by-one matrix $(u)$ is $u$, whose coset is nontrivial by step 1.1. Therefore $[u]\mathrm{\ne }0$ in $\mathrm{Wh}(C_5)$.

The differential $u:R\to R$ is invertible, so the displayed two-term complex has contraction $u^{-1}$ and is acyclic as an $R$-complex and after forgetting to abelian groups. Its degree-one based torsion is $[u]\mathrm{\ne }0$ by [F1] and step 2.1. This exhibits why ordinary homology alone does not retain the chosen group-ring bases and their torsion.

Take the finite presentation complex $X$ of $C_5=⟨t\mid t^5⟩$: one vertex, one loop and one two-cell. Apply [F3] to $X$ and the nonzero class $[u]$ to obtain a finite CW homotopy equivalence $i:X\hookrightarrow Y$ with $\tau (i)=[u]$. By [F4] and step 2.1 it is not simple. Its relative group-ring complex may be chosen as the two-term invertible-matrix complex constructed in [F3], so its ordinary relative homology also vanishes. ∎