Simple homotopy equivalences have zero torsion

For every finite CW complex $Z$, the composition formula [F3] applied to ${\mathrm{i}\mathrm{d}}_Z\circ {\mathrm{i}\mathrm{d}}_Z$ gives $\tau ({\mathrm{i}\mathrm{d}}_Z)=2\tau ({\mathrm{i}\mathrm{d}}_Z)$, since the identity induces the identity on its Whitehead group. Subtracting gives $\tau ({\mathrm{i}\mathrm{d}}_Z)=0$, componentwise.

An elementary expansion $j:X\to Y$ has $\tau (j)=0$ by [F4].

Let $\phi :X\to Y$ be a cellular isomorphism. By [F5] the class $\tau (\phi )$ is computed from a chosen cellular representative, universal covers, lifts, basepoints, orientations and orders, and by [F2] the class in $\mathrm{W}\mathrm{h}({\pi }_1(Y,y))$ does not depend on these choices. Choose a universal cover $p:\tilde{X}\to X$ and take the cover of $Y$ to be $q:=\phi \circ p:\tilde{X}\to Y$, which is again a universal cover, with lift $\tilde{\phi }:={\mathrm{i}\mathrm{d}}_{\tilde{X}}$, so that $q\circ \tilde{\phi }=\phi \circ p$; with this choice $C_{\ast }(\tilde{\phi })$ is the identity chain map of the based free right $\mathbb{Z}[{\pi }_{1}Y]$-complex $C_{\ast }(\tilde{X})$, by [F7] and the transport of coefficients along ${\phi }_{\ast }$. For any finite based complex $C$, the cone of its identity has contraction $s(y,x)=(0,y)$: $ds+sd=1$. Pair the two cone slots of each vector $e\in C_q$, namely $(e,0)$ in degree $q$ and $(0,e)$ in degree $q+1$. Order these pairs by increasing $q$, with the same within-degree order in both parity bases. The odd-to-even map sends the source slot to its paired target slot with coefficient $1$, plus a term involving $de$, hence in a strictly earlier pair. Its matrix is upper unitriangular in these matched orders. Returning to the prescribed bases only permutes rows and columns, which does not change reduced torsion by [F8]. Thus $\mathrm{Cone}(C_{\ast }(\tilde{\phi }))=\mathrm{Cone}({\mathrm{i}\mathrm{d}}_{C_{\ast }(\tilde{X})})$ has torsion $0$, and $\tau (\phi )=0$ by [F5].

Assume first that $Y$ is connected and write ${\pi }_1{X}_i$ for the fundamental group of the connected complex $X_i$. By [F1] there is a chain $X=X_0\stackrel{f_1}{\to }\cdots \stackrel{f_k}{\to }{X}_k=Y$ with every $f_i$ elementary or a cellular isomorphism and $f\simeq f_k\circ \cdots \circ f_1$; by [F2] homotopic homotopy equivalences have equal torsion, so $\tau (f)=\tau (f_k\circ \cdots \circ f_1)$, and by [F3] applied inductively $\tau (f_k\circ \cdots \circ f_1)=\tau (f_k)+(f_k{)}_{\ast }\tau (f_{k-1}\circ \cdots \circ f_1)={\sum }_{i=1}^k(f_k\circ \cdots \circ f_{i+1}{)}_{\ast }\tau (f_i),$ a sum of transported torsions of the factors. Hence it suffices to prove that each elementary factor and each cellular isomorphism of the sequence has torsion zero in the Whitehead group of its target; for $k=0$ we have $f\simeq {\mathrm{i}\mathrm{d}}_X$ and $\tau (f)=\tau ({\mathrm{i}\mathrm{d}}_X)=0$ by [F2] and step 1.1.

Let $c:Y\to X$ be an elementary collapse, $i:X\hookrightarrow Y$ the corresponding elementary expansion, so that $c\circ i={\mathrm{i}\mathrm{d}}_X$ by [F6]; both $i$ and $c$ are homotopy equivalences by [F1] and [F6]. Applying [F3] to the pair $(i,c)$ gives $\tau (c\circ i)=\tau (c)+c_{\ast }\tau (i)$ in $\mathrm{W}\mathrm{h}({\pi }_{1}X)$, and $\tau (c\circ i)=\tau ({\mathrm{i}\mathrm{d}}_X)=0$ by step 1.1 while $\tau (i)=0$ by step 1.2; hence $\tau (c)=0$.

By steps 1.2, 1.3 and 2.2 every factor of the sequence of step 2.1 has zero torsion, so the transported sum of step 2.1 vanishes and $\tau (f)=0$ in $\mathrm{W}\mathrm{h}({\pi }_1(Y,y))$. This proves the assertion for connected $Y$; the case of disconnected $Y$ follows componentwise, since each $f_i$ restricts to an elementary operation or a cellular isomorphism on the components that it meets and to a homeomorphism of the remaining components, the induced summands are as in [F2] and [F3], and each summand vanishes by the connected argument applied to that component (with the empty sequence handled by step 2.1).