The target of a finite cellular mapping cylinder is a simple subcomplex

Take $A=\mathrm{\varnothing }$ in [F2], so that $M_f=W$ is the ordinary mapping cylinder with $i_X=j$, $i_Y=k$ and $p=r$, and $p\circ i_Y={\mathrm{i}\mathrm{d}}_Y$, $p\circ i_X=f$; its cells are the cells of $Y$, the free-end cells $j(e)$ for the cells $e$ of $X$, and the prism cells $e^r\times (0,1)$ of dimension $r+1$ for the $r$-cells $e^r$ of $X$, finitely many in all.

Order the cells of the finite complex $X$ by increasing dimension and, for each $r$-cell $e^r$ of $X$ with characteristic map $\mathrm{\Phi }:D^r\to X$, let $Z(e^r)\supseteq Z_{r-1}$ denote the subcomplex obtained from the previously built subcomplex $Z_{r-1}$ by first attaching the free-end cell $j(e^r)$ and then the prism cell $e^r\times (0,1)$. Its closure $\overline{e^r\times (0,1)}$ is the image of $D^r\times [0,1]$ and its boundary consists of the pieces $D^r\times \{0\}$, $S^{r-1}\times [0,1]$ and $D^r\times \{1\}$; under the identification $(x,0)\sim f(x)$ the first piece maps into $Y\subseteq Z_{r-1}$, under the cellularity of $f$ the middle piece maps into $Y\cup j(X^{(r-1)})\cup \{\text{prisms of cells of }{X}^{(r-1)}\}\subseteq Z_{r-1}$, and the last piece is exactly the closed free-end cell $j(e^r)$ attached in the previous step. The ball pair $(D^r\times [0,1],D^r\times \{1\})$ is homeomorphic to $(D^{r+1},D_{+}^r)$, so the characteristic prism map exhibits an $(r,r+1)$-cell pair with free face $j(e^r)$ corresponding to an upper hemisphere, so $Z_{r-1}\hookrightarrow Z(e^r)$ is an elementary expansion of dimension $r+1$ by [F1].

Performing the steps of step 1.2 for the finitely many cells of $X$ in increasing dimension gives a finite chain of elementary expansions $Y=Z_{-1}\hookrightarrow \cdots \hookrightarrow M_f$ whose composite is $i_Y$, so $i_Y$ is a simple homotopy equivalence by [F3] and $\tau (i_Y)=0$ by [F4].

By step 1.1, $p\circ i_Y={\mathrm{i}\mathrm{d}}_Y$, and by [F2] the strong deformation retraction gives $i_Y\circ p\simeq {\mathrm{i}\mathrm{d}}_{M_f}$, so $p$ is a homotopy inverse of the homotopy equivalence $i_Y$ in the sense of [F7]. Applying [F5] to the composable homotopy equivalences $i_Y$ and $p$ gives $\tau (p\circ i_Y)=\tau (p)+p_{\ast }\tau (i_Y)$ in $\mathrm{W}\mathrm{h}({\pi }_{1}Y)$; the left side is $\tau ({\mathrm{i}\mathrm{d}}_Y)=0$ by [F4] and $\tau (i_Y)=0$ by step 2.1, hence $\tau (p)=0$. Reverse the expansion sequence of step 2.1 and choose the elementary collapse retraction for each pair. Their composite $r:M_f\to Y$ fixes $Y$. If $H$ is the deformation from ${\mathrm{i}\mathrm{d}}_{M_f}$ to $i_{Y}p$ supplied by [F2], then $rH$ is a homotopy from $r$ to $r{i}_{Y}p=p$, relative to $Y$. Thus $p$ is homotopic relative to $Y$ to that collapse composite and has zero torsion; no equality of these retractions is asserted.

Suppose now that $f$ is a homotopy equivalence. By [F7] and step 1.1, $i_Y\circ f=i_Y\circ p\circ i_X\simeq {\mathrm{i}\mathrm{d}}_{M_f}\circ i_X=i_X$, so $i_X$ is homotopic to the composite $i_Y\circ f$; here $f$ is a homotopy equivalence by hypothesis, $i_Y$ is a homotopy equivalence by step 3.1, and a composite of homotopy equivalences is a homotopy equivalence, so $i_X$ is a homotopy equivalence by [F8].

In the situation of step 4.1 the maps $i_X$ and $p$ are homotopy equivalences with $p\circ i_X=f$, so [F5] applies to the pair $(i_X,p)$ and gives $\tau (f)=\tau (p\circ i_X)=\tau (p)+p_{\ast }\tau (i_X)=p_{\ast }\tau (i_X)$ in $\mathrm{W}\mathrm{h}({\pi }_{1}Y)$ by [F6], since $\tau (p)=0$ by step 3.1.

For disconnected $X$ and $Y$ the construction is componentwise: $f$ maps each component of $X$ into a component of $Y$, over a target component $D$ the cylinder is $D$ together with the cylinders on all components of $X$ mapping into $D$, while target components receiving none are unchanged. The expansion sequence of step 1.2 is performed component by component, and the identities of steps 2.1–5.1 hold in the corresponding summands of the Whitehead groups by [F6].