Finite iteration of current invertible-block cancellations

Composition of retract data. Assume $p_1{\mathrm{ı}}_1=1_Y$, ${\mathrm{ı}}_1{p}_1=1_X-(d{h}_1+h_{1}d)$, $p_2{\mathrm{ı}}_2=1_Z$ and ${\mathrm{ı}}_2{p}_2=1_Y-(d{h}_2+h_{2}d)$, and put $p=p_2{p}_1$, $\mathrm{ı}={\mathrm{ı}}_1{\mathrm{ı}}_2$, $h=h_1+{\mathrm{ı}}_1{h}_2{p}_1$. Then $p\mathrm{ı}=p_2{p}_1{\mathrm{ı}}_1{\mathrm{ı}}_2=p_2{1}_Y{\mathrm{ı}}_2=p_2{\mathrm{ı}}_2=1_Z$; moreover $\mathrm{ı}p={\mathrm{ı}}_1{\mathrm{ı}}_2{p}_2{p}_1={\mathrm{ı}}_1(1_Y-d{h}_2-h_{2}d)p_1={\mathrm{ı}}_1{p}_1-{\mathrm{ı}}_{1}d{h}_2{p}_1-{\mathrm{ı}}_1{h}_{2}d{p}_1$, so $1_X-\mathrm{ı}p=(1_X-{\mathrm{ı}}_1{p}_1)+{\mathrm{ı}}_{1}d{h}_2{p}_1+{\mathrm{ı}}_1{h}_{2}d{p}_1=d{h}_1+h_{1}d+{\mathrm{ı}}_{1}d{h}_2{p}_1+{\mathrm{ı}}_1{h}_{2}d{p}_1$. On the other hand $dh+hd=d(h_1+{\mathrm{ı}}_1{h}_2{p}_1)+(h_1+{\mathrm{ı}}_1{h}_2{p}_1)d=d{h}_1+h_{1}d+d{\mathrm{ı}}_1{h}_2{p}_1+{\mathrm{ı}}_1{h}_2{p}_{1}d$, and $d{\mathrm{ı}}_1={\mathrm{ı}}_{1}d$, $p_{1}d=d{p}_1$ because ${\mathrm{ı}}_1,p_1$ are cochain maps, so the two expressions coincide and $1_X-\mathrm{ı}p=dh+hd$.

Aggregate pivots. Let $X^n=A\oplus U_1\oplus U^{\mathrm{\prime }}$ and $X^{n+1}=B\oplus V_1\oplus V^{\mathrm{\prime }}$ with pivot blocks $U=U_1\oplus U^{\mathrm{\prime }}$, $V=V_1\oplus V^{\mathrm{\prime }}$ and $\mathrm{\Phi }=\mathrm{diag}({\phi }_1,{\mathrm{\Phi }}^{\mathrm{\prime }})$, where ${\phi }_1:U_1\to V_1$ and ${\mathrm{\Phi }}^{\mathrm{\prime }}:U^{\mathrm{\prime }}\to V^{\mathrm{\prime }}$ are isomorphisms; write $d^n$ as the block matrix with rows $B,V_1,V^{\mathrm{\prime }}$ and columns $A,U_1,U^{\mathrm{\prime }}$ as $\left(\begin{array}{ccc}a& b_1& b^{\mathrm{\prime }}\\ c_1& {\phi }_1& 0\\ c^{\mathrm{\prime }}& 0& {\mathrm{\Phi }}^{\mathrm{\prime }}\end{array}\right)$, and write $d^{n-1}=(p;q_1;q^{\mathrm{\prime }})$ and $d^{n+1}=(r{s}_1{s}^{\mathrm{\prime }})$ accordingly. Cancelling $\mathrm{\Phi }$ at once, ${\mathrm{\Phi }}^{-1}=\mathrm{diag}({\phi }_1^{-1},{\mathrm{\Phi }}^{\mathrm{\prime }-1})$ multiplies out to $b{\mathrm{\Phi }}^{-1}c=b_1{\phi }_1^{-1}{c}_1+b^{\mathrm{\prime }}{\mathrm{\Phi }}^{\mathrm{\prime }-1}{c}^{\mathrm{\prime }}$, so by [L2] the reduced differential is $a-b_1{\phi }_1^{-1}{c}_1-b^{\mathrm{\prime }}{\mathrm{\Phi }}^{\mathrm{\prime }-1}{c}^{\mathrm{\prime }}$ and the neighbouring arrows are $p$ and $r$. Cancelling first ${\phi }_1$ in the reassociated decomposition $(A\oplus U^{\mathrm{\prime }})\oplus U_1$, $(B\oplus V^{\mathrm{\prime }})\oplus V_1$ of [L4], the pivot matrix is $\left(\begin{array}{cc}\tilde{a}& \tilde{b}\\ \tilde{c}& {\phi }_1\end{array}\right)$ with $\tilde{a}=\left(\begin{array}{cc}a& b^{\mathrm{\prime }}\\ c^{\mathrm{\prime }}& {\mathrm{\Phi }}^{\mathrm{\prime }}\end{array}\right)$, $\tilde{b}=(b_1;0)$, $\tilde{c}=(c_{1}0)$, so the new Schur complement is $\tilde{a}-\tilde{b}{\phi }_1^{-1}\tilde{c}=\left(\begin{array}{cc}a-b_1{\phi }_1^{-1}{c}_1& b^{\mathrm{\prime }}\\ c^{\mathrm{\prime }}& {\mathrm{\Phi }}^{\mathrm{\prime }}\end{array}\right)$, a complex by [L3], whose $(U^{\mathrm{\prime }},V^{\mathrm{\prime }})$-pivot is ${\mathrm{\Phi }}^{\mathrm{\prime }}$ and whose neighbouring arrows are $(p;q^{\mathrm{\prime }})$ and $(r{s}^{\mathrm{\prime }})$; cancelling ${\mathrm{\Phi }}^{\mathrm{\prime }}$ there gives reduced differential $a-b_1{\phi }_1^{-1}{c}_1-b^{\mathrm{\prime }}{\mathrm{\Phi }}^{\mathrm{\prime }-1}{c}^{\mathrm{\prime }}$, incoming arrow $p$ and outgoing arrow $r$. The two orders therefore produce the same objects and the same three reduced arrows; iterating the two-block comparison cancels $\mathrm{diag}({\phi }_1,\dots ,{\phi }_k)$ in one step with the same result as the successive cancellations.

Side conditions of the composite. With the data of step 1.1, $ph=p_2{p}_1{h}_1+p_2{p}_1{\mathrm{ı}}_1{h}_2{p}_1=p_2\cdot 0+p_2\cdot 1_Y\cdot h_2{p}_1=p_2{h}_2{p}_1=0$, using $p_1{h}_1=0$ and $p_2{h}_2=0$; likewise $h\mathrm{ı}=h_1{\mathrm{ı}}_1{\mathrm{ı}}_2+{\mathrm{ı}}_1{h}_2{p}_1{\mathrm{ı}}_1{\mathrm{ı}}_2=0+{\mathrm{ı}}_1{h}_2\cdot 1_Y\cdot {\mathrm{ı}}_2={\mathrm{ı}}_1{h}_2{\mathrm{ı}}_2=0$, using $h_1{\mathrm{ı}}_1=0$ and $h_2{\mathrm{ı}}_2=0$; and $h^2=h_1^2+h_1{\mathrm{ı}}_1{h}_2{p}_1+{\mathrm{ı}}_1{h}_2{p}_1{h}_1+{\mathrm{ı}}_1{h}_2{p}_1{\mathrm{ı}}_1{h}_2{p}_1=0+0+{\mathrm{ı}}_1{h}_2(p_1{h}_1)+{\mathrm{ı}}_1{h}_2\cdot 1_Y\cdot h_2{p}_1={\mathrm{ı}}_1{h}_2^2{p}_1=0$, using $h_1^2=0$ and $h_2^2=0$. Hence the composite data satisfies all three side conditions of [L1].

Finite iteration. A sequence of length one is a single cancellation, which is [L1]. For a sequence of length $m\ge 2$, apply the inductive hypothesis to the first $m-1$ cancellations, obtaining a strong deformation retract of $X^{\bullet }$ onto the intermediate complex $Y^{\bullet }$ given by data $(p_1,{\mathrm{ı}}_1,h_1)$, and let $(p_2,{\mathrm{ı}}_2,h_2)$ be the data of the last cancellation, performed in the current complex $Y^{\bullet }$ with an invertible pivot, so that it is a strong deformation retract of $Y^{\bullet }$ onto the final reduction $Z^{\bullet }$; such data is supplied by [L1] for that pivot. Steps 1.1 and 2.1 then show that $p=p_2{p}_1$, $\mathrm{ı}={\mathrm{ı}}_1{\mathrm{ı}}_2$, $h=h_1+{\mathrm{ı}}_1{h}_2{p}_1$ are strong deformation retract data of $X^{\bullet }$ onto $Z^{\bullet }$. By induction on the length, every finite sequence of cancellations with invertible current pivots yields such a composite retract.

Different choices. Suppose two finite sequences of cancellations lead from $X^{\bullet }$ to reductions ${\bar{X}}^{\bullet }$ and ${\bar{X}}^{\mathrm{\prime }\bullet }$; by step 3.1 there are strong deformation retract data $(p,\mathrm{ı},h)$ of $X^{\bullet }$ onto ${\bar{X}}^{\bullet }$ and $(p^{\mathrm{\prime }},{\mathrm{ı}}^{\mathrm{\prime }},h^{\mathrm{\prime }})$ of $X^{\bullet }$ onto ${\bar{X}}^{\mathrm{\prime }\bullet }$. Define $u:=p^{\mathrm{\prime }}\mathrm{ı}:{\bar{X}}^{\bullet }\to {\bar{X}}^{\mathrm{\prime }\bullet }$ and $v:=p{\mathrm{ı}}^{\mathrm{\prime }}:{\bar{X}}^{\mathrm{\prime }\bullet }\to {\bar{X}}^{\bullet }$; then $uv=p^{\mathrm{\prime }}\mathrm{ı}p{\mathrm{ı}}^{\mathrm{\prime }}=p^{\mathrm{\prime }}(1_{X^{\bullet }}-(dh+hd)){\mathrm{ı}}^{\mathrm{\prime }}=1_{{\bar{X}}^{\mathrm{\prime }}}-d(p^{\mathrm{\prime }}h{\mathrm{ı}}^{\mathrm{\prime }})-(p^{\mathrm{\prime }}h{\mathrm{ı}}^{\mathrm{\prime }})d$ and similarly $vu=1_{\bar{X}}-d(p{h}^{\mathrm{\prime }}\mathrm{ı})-(p{h}^{\mathrm{\prime }}\mathrm{ı})d$, using that $p^{\mathrm{\prime }},\mathrm{ı},p,{\mathrm{ı}}^{\mathrm{\prime }}$ are cochain maps and $p^{\mathrm{\prime }}{\mathrm{ı}}^{\mathrm{\prime }}=1$, $p\mathrm{ı}=1$. Hence the two reductions are homotopy equivalent, with explicit comparison maps. They need not be equal: in the complex $k\stackrel{1}{\to }k\stackrel{0}{\to }k\stackrel{1}{\to }k$ over a field, cancelling at degree $0$ leaves the two-term complex with $k$ in degrees $2,3$ and cancelling at degree $2$ leaves the two-term complex with $k$ in degrees $0,1$; these complexes are both contractible, hence homotopy equivalent, but their degree-$0$ objects are $0$ and $k$, so they are not equal.

Conclusion. Step 1.1 and step 2.1 give the composition formulas of clause 2 together with all five identities; step 3.1 gives clause 1 by induction; step 1.2 verifies clause 3, including that a block-diagonal aggregate pivot may be cancelled in one step with the same outcome as the successive cancellations; and step 4.1 gives clause 4, producing explicit homotopy inverse comparison maps between reductions obtained from different choices and an example where the reductions are not equal. The statement asserts nothing about infinite sequences of cancellations, about termination of any automatic procedure, or about the size of the reduction when no invertible pivot is available at a chosen degree. ∎