Composition and based-pair sum formulas for Whitehead torsion

Choose cellular representatives of $f,g$ and compatible lifts $\tilde{f},\tilde{g}$ of $gf$; by [F3] the lifted chain maps are right-linear chain homotopy equivalences after transporting coefficients, and by [F3] again the composite $C_{\ast }(\tilde{g})C_{\ast }(\tilde{f})$ differs from the lift of $gf$ by a deck twist, which does not change classes in $\mathrm{W}\mathrm{h}$. Hence the algebraic composition formula of [F2] applies to the transported based complexes and gives the composition formula after applying the group isomorphism $g_{\ast }$ to the coefficient ring of the middle complex; this is the displayed formula, since the transport of $\tau (f)$ from $\mathrm{W}\mathrm{h}({\pi }_1(Y))$ to $\mathrm{W}\mathrm{h}({\pi }_1(Z))$ is exactly $g_{\ast }\tau (f)$ by [F5].

For clause 2 write $A^{\mathrm{\prime }}=p_X^{-1}A\subset \tilde{X}$ and $B^{\mathrm{\prime }}=p_Y^{-1}B\subset \tilde{Y}$, and transport all source coefficients through $f_{X\ast }$ to $R=\mathbb{Z}[{\pi }_1(Y,y)]$. In each degree the lifted cells of $X$ split into those over $A$ and those outside $A$, and similarly for $(Y,B)$. Hence [F4] gives two degreewise based exact rows $0\to C_{\ast }(A^{\mathrm{\prime }})\to C_{\ast }(\tilde{X})\to C_{\ast }(\tilde{X},A^{\mathrm{\prime }})\to 0$ and $0\to C_{\ast }(B^{\mathrm{\prime }})\to C_{\ast }(\tilde{Y})\to C_{\ast }(\tilde{Y},B^{\mathrm{\prime }})\to 0$, joined by the three chain maps induced by $f_A,f_X,f_{\mathrm{r}\mathrm{e}\mathrm{l}}$. A lift of the cellular pair map preserves the subcomplexes, so the diagram commutes.

Taking algebraic mapping cones of the three vertical maps in step 1.2 gives the degreewise based exact sequence $0\to \mathrm{Cone}(C_{\ast }(A^{\mathrm{\prime }})\to C_{\ast }(B^{\mathrm{\prime }}))\to \mathrm{Cone}(C_{\ast }(\tilde{X})\to C_{\ast }(\tilde{Y}))\to \mathrm{Cone}(f_{\mathrm{r}\mathrm{e}\mathrm{l}})\to 0.$ The first and middle cones are contractible: for the first, decompose $A,B$ componentwise and use the induced chain equivalences of [F4]; for the middle use [F3]. The last cone is contractible as well. Explicitly, a graded basis splitting of the exact sequence gives a graded section $s$ of the quotient and defect $\delta =ds-sd$ with values in the first cone; if $h$ contracts the first cone, $s^{\mathrm{\prime }}=s-h\delta$ is a chain section because $d\delta +\delta d=0$ and $dh+hd=1$. The last cone is then a chain retract of the contractible middle cone, so it inherits a contraction. By the cone criterion of [F3], $f_{\mathrm{r}\mathrm{e}\mathrm{l}}$ is a chain homotopy equivalence.

Apply the based exact sequence formula of [F2] to step 2.1. The cone bases in each degree are concatenations of the bases of the lifted subcomplex and quotient cells, up to cell permutations whose classes vanish in $\mathrm{W}\mathrm{h}$ by [F3]. Hence the middle cone torsion is the sum of the first and last cone torsions. The first is the image $j_{\ast }\tau (f_A)$ under componentwise coefficient extension: the induced complex over $R$ is obtained by extending the component coefficient rings and their chosen bases, so its contraction matrix is the scalar extension of the contraction matrix for $f_A$. The last is $\tau (f_{\mathrm{r}\mathrm{e}\mathrm{l}})$ by definition of algebraic cone torsion. Passing to $\mathrm{W}\mathrm{h}({\pi }_{1}Y)$ proves $\tau (f_X)=j_{\ast }\tau (f_A)+\tau (f_{\mathrm{r}\mathrm{e}\mathrm{l}})$.

Clause 3 is the algebraic statement of [F2] as proved from Basis-change, direct-sum and based exact-sequence formulas; clause 1 is step 1.1 and clause 2 is steps 1.2, 2.1 and 3.1. No step used a choice principle beyond finitely many cell and lift choices, and no step used a smooth, handle or cobordism statement.