Choice-free Wall chain finiteness from an explicit finite domination

Put $u=gf$ on $C$. Repeatedly substituting $dh=1-u-hd$ gives, for every integer $s\ge 1$, $d{h}^s=(-1{)}^s{h}^{s}d+{ϵ}_s{h}^{s-1}-{\sum }_{\mathrm{\ell }=0}^{s-1}(-1{)}^{\mathrm{\ell }}{h}^{\mathrm{\ell }}u{h}^{s-1-\mathrm{\ell }},{ϵ}_s=\{\begin{array}{ll}1& s\text{ odd},\\ 0& s\text{ even}\mathrm{.}\end{array}$ For $s=1$ this is the given equation. Multiplying the formula on the right by $h$ and substituting $dh$ in its first term proves the next case: the identity coefficient changes from ${ϵ}_s$ to ${ϵ}_s+(-1{)}^s={ϵ}_{s+1}$ and the last summand acquires index $s$. This is finite induction on a specified identity, with no choices.

Define $D_n={⨁}_{j=0}^n{F}_j$ for $n\ge 0$, and $D_n=0$ for $n<0$. Define maps ${\delta }_n:D_n\to D_{n-1}$ by their finite blocks, with column $j$, row $k$, where $0\le j\le n$ and $0\le k\le n-1$: $({\delta }_n{)}_{kj}=\{\begin{array}{ll}0& j\ge k+2,\\ (-1{)}^{n+k+1}{\mathrm{\partial }}_j& j=k+1,\\ 1_{F_j}-f_j{g}_j& j=k,n+k\text{ odd},\\ f_j{g}_j& j=k,n+k\text{ even},\\ (-1{)}^{n+k}{f}_k{h}^{k-j}{g}_j& j<k\mathrm{.}\end{array}$ All blocks have the displayed source and target, regardless of noncommutativity of $A$. Also define ${\alpha }_n:C_n\to D_n$ to have only its final component $f_n$, define ${\beta }_n:D_n\to C_n$ by the row $(h^n{g}_0,h^{n-1}{g}_1,\dots ,g_n)$, and let $s_n:D_n\to D_{n+1}$ be inclusion into the first $n+1$ summands. These are specified finite formulas in every degree. We next verify that $D$ is a chain complex and that the named maps have all claimed properties; these are not assumed from a matrix citation.

The final column of ${\delta }_n$ has only its final block $+{\mathrm{\partial }}_n$. Therefore ${\delta }_n{\alpha }_n={\alpha }_{n-1}{d}_n$, using $\mathrm{\partial }f=fd$. To check ${\beta }_{n-1}{\delta }_n=d_n{\beta }_n$, consider column $j$. If $j=n$, the assertion is $g_{n-1}{\mathrm{\partial }}_n=d_n{g}_n$. Otherwise put $t=n-j\ge 1$. The superdiagonal contribution is $(-1{)}^t{h}^{t}d{g}_j$ (zero if $j=0$); the diagonal contribution is ${ϵ}_t{h}^{t-1}{g}_j-(-1{)}^{t-1}{h}^{t-1}u{g}_j$. Reindexing the lower contributions by $\mathrm{\ell }=n-1-k$ gives $-{\sum }_{\mathrm{\ell }=0}^{t-2}(-1{)}^{\mathrm{\ell }}{h}^{\mathrm{\ell }}u{h}^{t-1-\mathrm{\ell }}{g}_j$. The total is exactly $d{h}^t{g}_j$ by step 1.1. Thus both intertwining identities hold, without yet assuming ${\delta }^2=0$.

Directly in the blocks, ${\delta }_{n+1}{s}_n+s_{n-1}{\delta }_n=1_{D_n}-{\alpha }_n{\beta }_n\mathrm{.}$ Indeed for rows $k<n$ the off-diagonal blocks cancel when $n$ is replaced by $n+1$, and the two diagonal blocks sum to $1$. In row $n$ only the first term is present: its blocks are $-f_n{h}^{n-j}{g}_j$ for $j<n$ and $1-f_n{g}_n$ for $j=n$. They are exactly the final row of $1-\alpha \beta$. For $n=0$ this is simply ${\delta }_1{s}_0=1-f_0{g}_0$. Also ${\beta }_n{\alpha }_n=g_n{f}_n$.

We prove ${\delta }_{n-1}{\delta }_n=0$ by induction on $n$. It holds at $n=1$ since $D_{-1}=0$. On the first $n$ summands of $D_n$, identified with $s_{n-1}{D}_{n-1}$, step 2.2 and the two intertwining identities give $\begin{array}{rl}{\displaystyle {\delta }_{n-1}{\delta }_n{s}_{n-1}}& {\displaystyle ={\delta }_{n-1}(1-\alpha \beta -s_{n-2}{\delta }_{n-1})}\\ {\displaystyle }& {\displaystyle =(1-\alpha \beta ){\delta }_{n-1}-{\delta }_{n-1}{s}_{n-2}{\delta }_{n-1}}\\ {\displaystyle }& {\displaystyle =s_{n-3}{\delta }_{n-2}{\delta }_{n-1}=0.}\end{array}$ The final equality uses the induction hypothesis, and negative-index terms mean zero. On the remaining summand $F_n$, the two final-column observations give ${\mathrm{\partial }}_{n-1}{\mathrm{\partial }}_n=0$. Thus the composite vanishes on every summand. This proves ${\delta }^2=0$. Now step 2.1 says that $\alpha ,\beta$ are chain maps, and step 2.2 together with $1-\beta \alpha =dh+hd$ gives a chain homotopy equivalence $C\simeq D$.

Set $T={⨁}_{j=0}^m{F}_j$. For all $n\ge m$, $D_n$ is canonically $T$, because all subsequent summands are zero. For $n\ge m+1$, the square matrix ${\delta }_n:T\to T$ depends only on the parity of $n$. Write $p$ for its value when $n$ is even. The formulas show that its value when $n$ is odd is $1-p$: the off-diagonal blocks change sign and the diagonal blocks are complementary. Taking two consecutive such degrees in ${\delta }^2=0$ gives $p(1-p)=0=(1-p)p$, hence $p^2=p$. No assertion that the homotopy idempotent $fg$ itself is strictly idempotent has been made.

Choose the explicit even integer $N=2(m+1)$, and let $q=p:T\to T$. Define $E_n=D_n$ for $0\le n<N$, $E_N=qT\subseteq T=D_N$, and $E_n=0$ otherwise. Restrict ${\delta }_N$ to $qT$ at the top, and keep ${\delta }_n$ below it. The square-zero identities remain valid by step 3.1. The maps $J:E\to D$ are the evident inclusions. Define $I:D\to E$ to be identity below $N$, $q$ in degree $N$ and zero above $N$. At degree $N$, ${\delta }_{N}q={\delta }_N$ because $1-q={\delta }_{N+1}$ and ${\delta }_N{\delta }_{N+1}=0$. At degree $N+1$, $q{\delta }_{N+1}=q(1-q)=0$. These are the only boundary degrees in checking that $I$ is a chain map; $J$ is a chain map by restriction, and $IJ=1_E$.

Define $t_n:D_n\to D_{n+1}$ to be zero for $n<N$ and the canonical identity $T\to T$ for $n\ge N$. Then $1_D-JI=\delta t+t\delta \mathrm{.}$ Below $N$ both sides are zero. In degree $N$ the right side is ${\delta }_{N+1}=1-q$, which is the left side. Above $N$ it is the sum of the two complementary matrices $p$ and $1-p$, hence identity, as is the left side. Thus $I,J$ give a specified chain homotopy equivalence between $D$ and the bounded complex $E$.

The modules $E_n$ below $N$ are finite direct sums of the supplied finite free modules, so their concatenated bases are finite. The module $qT$ is generated by the finitely many vectors $q(e_i)$, where $(e_i)$ is that concatenated basis of $T$, and $T=qT\oplus (1-q)T$: write $x=qx+(1-q)x$, and apply $q$ to an element in the intersection to show it is zero. For completeness its projectivity is choice-free. Given a surjection $v:U\to V$ and a right-linear $w:qT\to V$, take, for each of the finitely many $e_i$, a lift in $U$ of $w(q{e}_i)$. Their simultaneous existence is [F3]. Extend this finite assignment right-linearly to a map $\tilde{w}:T\to U$. Then $v\tilde{w}=wq$, and restriction to $qT$ is the requested lift. The same proof with $q=1$ gives projectivity of the lower finite free modules. This verifies the actual lifting property, not the general claim that arbitrary free modules are projective in ZF.

Set $a=I\alpha$, $b=\beta J$. These are chain maps and $1_C-ba=d(h+\beta t\alpha )+(h+\beta t\alpha )d,$ since $1_C-\beta JI\alpha =(1_C-\beta \alpha )+\beta (1_D-JI)\alpha$. Similarly, using $IJ=1_E$ and step 2.2, $1_E-ab={\delta }_E(IsJ)+(IsJ){\delta }_E\mathrm{.}$ Indeed $I(1_D-\alpha \beta )J=1_E-I\alpha \beta J$. These are the explicit required homotopies $H=h+\beta t\alpha$ and $L=IsJ$.

Every degree of $D$, every matrix product used to define its maps, and each component of the homotopies uses a finite sum and finitely many compositions of supplied maps. The single family indexed by all nonnegative degrees is defined by these formulas; it is not obtained by independently selecting a lift or basis in each degree. The only lifts selected in the entire projectivity verification are a finite list for one supplied lifting problem. If the two input complexes vanish below a common integer $b$, reindex $C_n^{\mathrm{\prime }}=C_{n+b}$, $F_n^{\mathrm{\prime }}=F_{n+b}$ with the same differentials and domination equations, apply the construction, and reindex the result back. No infinite recursion of choices and no unbounded-below input is involved. The proof therefore establishes exactly the stated ZF conclusion for the required bounded-below setting.