An elementary CW expansion has zero Whitehead torsion

Assume first that $Y$ is connected and put $\pi ={\pi }_1(Y,y)$, $R=\mathbb{Z}[\pi ]$; by [F1] the new cells are $e^{n-1}$ and $e^n$ with $Y=X\cup e^{n-1}\cup e^n$. Choose an oriented lift ${\tilde{e}}^n$ of $e^n$ and an oriented lift ${\tilde{e}}^{n-1}$ of $e^{n-1}$ in the universal cover $\tilde{Y}$. By [F2] the relative cellular chain complex $T_{\bullet }:=C_{\bullet }(\tilde{Y},\tilde{X};R)$ is a bounded finite based free right $R$-complex with exactly two basis vectors, ${\tilde{e}}^{n-1}$ in degree $n-1$ and ${\tilde{e}}^n$ in degree $n$, and all other terms zero.

The boundary of ${\tilde{e}}^n$ in the relative complex is a unit multiple of the free face: by [F1] the characteristic map is a homeomorphism from the interior of $Q^{n-1}$ onto $e^{n-1}$, maps $\mathrm{\partial }{Q}^{n-1}$ into $X$, and maps the complementary boundary into $X$. On the quotient by $X$, this face is one characteristic disk, so its relative incidence degree is $\pm 1$ even when its boundary points are identified in the closed cell. Thus in the cellular chains of $(\tilde{Y},\tilde{X})$ the coefficient of $e^{n-1}$ in $\mathrm{\partial }{e}^n$ is $\pm T_h$ for the deck transformation $T_h$ relating the two chosen lifts, and in right-module coordinates $d^T({\tilde{e}}^n)={\tilde{e}}^{n-1}\cdot (\pm h^{-1})$. Writing $\lambda :=\pm h^{-1}\in R$, the differential of $T_{\bullet }$ has the $1\times 1$ matrix $\lambda$ on right-module coordinate columns (so the coordinate map is left multiplication by $\lambda$).

Define $s({\tilde{e}}^{n-1}):={\tilde{e}}^n{\lambda }^{-1}$ and $s=0$ in all other degrees. Then $ds({\tilde{e}}^{n-1})={\tilde{e}}^{n-1}\lambda {\lambda }^{-1}={\tilde{e}}^{n-1}$ and $sd({\tilde{e}}^n)=s({\tilde{e}}^{n-1}\lambda )={\tilde{e}}^n$, while on the only other degree the complex is zero, so $ds+sd=\mathrm{i}\mathrm{d}$ and $T_{\bullet }$ is contractible with $\mathrm{\#}{B}_{\mathrm{o}\mathrm{d}\mathrm{d}}=\mathrm{\#}{B}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}=1$.

Compute $\tau (T_{\bullet })$. If $n$ is odd, then $T_{\mathrm{o}\mathrm{d}\mathrm{d}}=T_n=R$ and $(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}=d$ with matrix $\lambda$; if $n$ is even, then $T_{\mathrm{o}\mathrm{d}\mathrm{d}}=T_{n-1}=R$ and $(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}=s$ with matrix ${\lambda }^{-1}$. In both cases [F6] gives $\tau (T_{\bullet })=\pm [\lambda ]$ in ${\tilde{K}}_1(R)$, and since $\lambda =\pm h^{-1}$ with $h^{-1}\in \pi$, the definition of $\mathrm{W}\mathrm{h}(\pi )$ in [F8] kills the class: the image of $\tau (T_{\bullet })$ in $\mathrm{W}\mathrm{h}(\pi )$ is $0$.

Write $C_{\bullet }:=C_{\bullet }(\tilde{X};R)$ and $K_{\bullet }:=\mathrm{Cone}({\mathrm{i}\mathrm{d}}_{C_{\bullet }})$, embedded in $\mathrm{Cone}(C_{\ast }(j))$ by the inclusion $(y,x)\mapsto (y,x)$. This inclusion is well defined because in degree $m$ the displayed basis of $C_m(\tilde{Y})$ is the basis of $C_m(\tilde{X})$ together with the lift ${\tilde{e}}^m$ when $m\in \{n-1,n\}$ and together with nothing otherwise, so $C_m(\tilde{X})$ is a direct summand of $C_m(\tilde{Y})$. It is a chain map because $X$ is a subcomplex of $Y$ by [F1], so $d^{\tilde{Y}}$ preserves $C_{\ast }(\tilde{X})$, and because the differential of $\mathrm{Cone}(C_{\ast }(j))$ displayed in [F4] then sends $(y,x)\in C_m(\tilde{X})\oplus C_{m-1}(\tilde{X})$ to $(d^{\tilde{Y}}y+j_{m-1}x,-d^{\tilde{X}}x)\in C_{m-1}(\tilde{X})\oplus C_{m-2}(\tilde{X})$; by [F4] restricted to these submodules it is exactly the differential of $\mathrm{Cone}({\mathrm{i}\mathrm{d}}_{C_{\bullet }})$.

The quotient of $\mathrm{Cone}(C_{\ast }(j))$ by $K_{\bullet }$ is $T_{\bullet }$: the quotient in degree $m$ has basis the images of the complementary basis vectors, namely the lift ${\tilde{e}}^m$ for $m\in \{n-1,n\}$ and none otherwise, matching the basis of $T_m$ of step 1.1, and the induced differential sends the class of $({\tilde{e}}^n,0)$ to the class of $(d^{\tilde{Y}}{\tilde{e}}^n,0)$, whose $C_{n-1}(\tilde{X})$-part dies in the quotient and whose remaining part is the relative boundary computed in step 1.2; equivalently $(y,x)\mapsto q(y)$ for the relative quotient map $q$ is a chain map with kernel $K_{\bullet }$. Hence $0\to K_{\bullet }\to \mathrm{Cone}(C_{\ast }(j))\to T_{\bullet }\to 0$ is degreewise based exact after reordering the displayed basis of $\mathrm{Cone}(C_{\ast }(j){)}_m$ in each degree as the basis of $K_m$ followed by the image of the basis of $T_m$; a reordering of a displayed basis changes torsion in ${\tilde{K}}_1(R)$ by a sum of classes of permutation matrices, each of which is $0$ or $[-1]=0$ there ([F7], clause 2).

All three complexes of step 2.1 are contractible, bounded and based free: $T_{\bullet }$ by step 1.3, $K_{\bullet }=\mathrm{Cone}({\mathrm{i}\mathrm{d}}_{C_{\bullet }})$ by the explicit contraction $s(y,x)=(0,y)$, since $d(0,y)=(y,-dy)$ and $s(dy+x,-dx)=(0,dy+x)$ give $ds+sd=\mathrm{i}\mathrm{d}$, and $\mathrm{Cone}(C_{\ast }(j))$ by [F5] applied to the homotopy equivalence $j$ of [F10]. Hence [F7] gives $\tau (\mathrm{Cone}(C_{\ast }(j)))=\tau (K_{\bullet })+\tau (T_{\bullet })$ in ${\tilde{K}}_1(R)$.

Claim: $\tau (\mathrm{Cone}({\mathrm{i}\mathrm{d}}_{C_{\bullet }}))=0$ for every bounded finite based free right $R$-complex $C_{\bullet }$. The contraction $s$ of step 3.1 is available, so by [F6] the torsion is the class of the matrix of $(d+s{)}_{\mathrm{o}\mathrm{d}\mathrm{d}}$ in the displayed degree-ordered bases of $K_{\mathrm{o}\mathrm{d}\mathrm{d}}$ and $K_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$. Each basis vector $e\in C_q$ occupies exactly two slots of $K_{\bullet }$, namely $(e,0)\in K_q$ and $(0,e)\in K_{q+1}$, and the formula of step 3.1 matches them: the source slot maps to its matched slot with coefficient $1$ plus one correction term, namely $(de,0)$ when $q$ is odd and $(0,-de)$ when $q$ is even. Order both bases by increasing degree and, within a degree, with the second summand before the first; this makes the matching order-preserving, and the correction term of a source slot always lies in a strictly earlier target slot, because for odd $q$ it lies in degree $q-1$ and for even $q$ it lies in the second summand of the degree-$q$ target block whose matched slot is in the first summand of that block. Hence in these matched orders the matrix is upper unitriangular, since each correction is in an earlier row than its matched diagonal entry, and has class $0$ by [F9]. Returning to the prescribed degree-ordered bases permutes rows and columns; these permutations contribute only classes of $-1$, killed in ${\tilde{K}}_1(R)$ by [F7]. Thus the torsion in the prescribed bases is $0$ in ${\tilde{K}}_1(R)$.

Combining steps 3.1 and 4.1 with step 1.4: $\tau (\mathrm{Cone}(C_{\ast }(j)))=\tau (T_{\bullet })=(-1{)}^{n+1}[\lambda ]$ in ${\tilde{K}}_1(R)$. Its image is $0$ in $\mathrm{W}\mathrm{h}(\pi )$ by step 1.4, and therefore $\tau (j)=0$ in $\mathrm{W}\mathrm{h}(\pi )$, because $\tau (j)$ is by [F3] the image of $\tau (\mathrm{Cone}(C_{\ast }(j)))$ under the quotient map $K_1(R)\to \mathrm{W}\mathrm{h}(\pi )$. This proves the first assertion for connected $Y$; it also proves that the identity map of a based cellular complex of any finite CW complex has zero torsion.

Componentwise: for arbitrary finite CW complexes $X,Y$, the two new cells of the elementary expansion lie in a single component $D$ of $Y$, and ${\pi }_0(j):{\pi }_0(X)\to {\pi }_0(Y)$ is a bijection since $Y=X\cup e^{n-1}\cup e^n$ with the free face attached inside $X$ by [F1]. By the componentwise definition of $\tau$ in [F3] and additivity over direct sums in [F7], $\tau (j)$ is the tuple whose $D$-entry is the torsion of the restriction $j{\mathrm{\mid }}_C:C\to D$ of the component with the new cells and whose other entries are the torsions of the identity inclusions of the remaining components, each of which vanishes by step 5.1; the restriction $j{\mathrm{\mid }}_C$ falls under steps 1.1 through 5.1, so $\tau (j)=0$ in ${⨁}_{E\in {\pi }_0(Y)}\mathrm{W}\mathrm{h}({\pi }_{1}E)$. The relative complex of $(Y,X)$ is concentrated in degrees $n-1$ and $n$ with the single entry $\pm g$ of step 1.2, $g=h^{-1}$, and $[\pm g]=0$ in $\mathrm{W}\mathrm{h}({\pi }_{1}Y)$ by [F8].