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Homological Gaussian Elimination

1 · Prerequisites

2 · Summary

This page develops Gaussian elimination for complexes over an additive category, in the form used in Khovanov homology computations: an invertible block of a differential is cancelled and the complex is replaced by its Schur-complement reduction, with an explicit strong deformation retract recording the comparison.

The page begins with cochain complexes, cochain maps, homotopies and contractibility over an additive category, and with the dictionary that reindexes them into the published chain convention of Weibel and of Clark–Morrison–Walker. It then fixes the block decomposition $d^{n} = (\begin{array}{cc} a & b \\ c & φ \end{array}) : A \oplus U \to B \oplus V$ with invertible pivot $φ$ , defines the candidate reduction with differential the Schur complement $a - b φ^{- 1} c$ , and proves that the triangular basis changes $L, R$ diagonalize the block, transmit the neighbouring differentials to $(p; 0)$ and $(r 0)$ , and make the reduced arrows square to zero even in the neighbouring degrees.

On that base the central theorem splits the complex as a chain isomorphism onto ${\overset{ˉ}{X}}^{∙} \oplus K$ with $K$ the contractible two-term complex $0 \to U \overset{φ}{\to} V \to 0$ , and the following proposition exhibits the resulting projection, section and contracting homotopy explicitly; the corollary records that the chosen maps are inverse in the homotopy category and, over an abelian category, induce inverse isomorphisms on homology objects. Finite iteration of current pivots composes the retract data as $p = p_{2} p_{1}$ , $ı = ı_{1} ı_{2}$ , $h = h_{1} + ı_{1} h_{2} p_{1}$ , aggregate diagonal pivots may be cancelled in one step with the same result as successive cancellations, and different valid choices give homotopy equivalent reductions, which need not be equal. The page closes with the behaviour under an additive functor, which is exactness-free for the homotopy statement, and with transfer of cochain maps along chosen retracts, which is functorial on homotopy classes but not strictly functorial on cochain maps.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-27Open item page →

Complexes, homotopies and contractibility in an additive category

Definition

Additive setting. Fix an additive category $A$ (Additive category): every hom-set is an abelian group, finite biproducts exist, and every object has an identity morphism. All morphisms below are morphisms of $A$ , sums and negatives are taken in the hom-groups, and $0$ denotes the zero morphism between the indicated objects.

Cochain complexes. A cochain complex $C^{∙} = (C^{n}, d^{n})_{n \in Z}$ consists of objects $C^{n}$ of $A$ and morphisms $d^{n} : C^{n} \to C^{n + 1}$ with $d^{n + 1} d^{n} = 0 for all n \in Z .$ The morphisms $d^{n}$ are the differentials of $C^{∙}$ . No boundedness, finiteness or nonvanishing condition is imposed on the family of objects.

Cochain maps. A cochain map $f : C^{∙} \to D^{∙}$ is a family of morphisms $f^{n} : C^{n} \to D^{n}$ with $f^{n + 1} d_{C}^{n} = d_{D}^{n} f^{n} for all n \in Z .$ Identities and composites of cochain maps are cochain maps, so cochain complexes and cochain maps form a category; this category is written ${Ch}^{∙} (A)$ when the ambient category needs to be recorded.

Homotopies. A homotopy $h : f ≃ g$ between cochain maps $f, g : C^{∙} \to D^{∙}$ is a family of morphisms $h^{n} : C^{n} \to D^{n - 1}$ , one in each degree, such that $f^{n} - g^{n} = d_{D}^{n - 1} h^{n} + h^{n + 1} d_{C}^{n} for all n \in Z .$ Thus $h$ has degree $- 1$ , and the right-hand side is the $n$ -th component of the graded map $d h + h d : = (d_{D}^{n - 1} h^{n} + h^{n + 1} d_{C}^{n})_{n \in Z},$ written with $d$ on both sides. A null homotopy of a cochain map $f : C^{∙} \to D^{∙}$ is a homotopy $f ≃ 0$ to the zero map with the same source and target; $D^{∙}$ need not be the zero complex. A map admitting such a homotopy is null-homotopic.

Homotopy equivalence. A cochain map $f : C^{∙} \to D^{∙}$ is a homotopy equivalence when there is a cochain map $g : D^{∙} \to C^{∙}$ with $g f ≃ 1_{C^{∙}}$ and $f g ≃ 1_{D^{∙}}$ ; the complexes are then homotopy equivalent. The maps $f$ and $g$ are homotopy inverses of one another.

Contractibility. A cochain complex $C^{∙}$ is contractible when its identity is null-homotopic, $1_{C^{∙}} ≃ 0$ : that is, when there is a family of morphisms $h^{n} : C^{n} \to C^{n - 1}$ with $1_{C^{n}} = d^{n - 1} h^{n} + h^{n + 1} d^{n} for all n \in Z .$ A complex is contractible exactly when it is homotopy equivalent to the zero complex, since a homotopy equivalence onto the zero complex is a pair of null-homotopies of the identities.

Degreewise biproducts. If $C^{∙}$ and $D^{∙}$ are cochain complexes, then defining $(C \oplus D)^{n} : = C^{n} \oplus D^{n}, d_{C \oplus D}^{n} : = d_{C}^{n} \oplus d_{D}^{n}$ gives a cochain complex, because $(d_{C}^{n + 1} \oplus d_{D}^{n + 1}) (d_{C}^{n} \oplus d_{D}^{n}) = 0 \oplus 0 = 0$ . The degreewise injections and projections are cochain maps, and the biproduct identities hold in each degree and are therefore identities of cochain maps; hence $C \oplus D$ is a biproduct of $C$ and $D$ . Iterating, finite direct sums of cochain complexes are formed degreewise, and finite direct sums of cochain maps and of homotopies are formed degreewise as well. Under the reindexing below this is the additive structure on complexes over an additive category recorded in The category of complexes in an additive category is additive.

Dictionary with the published chain convention. Reindex by $C_{n} : = C^{- n}$ and $d_{n} : = d^{- n}$ . Then $d_{n} : C_{n} \to C_{n - 1}$ and $d_{n - 1} d_{n} = d^{- (n - 1)} d^{- n} = d^{- n + 1} d^{- n} = 0,$ so $(C_{∙}, d_{∙})$ is an ordinary chain complex. A cochain map $f$ becomes the chain map with components $f_{n} : = f^{- n}$ , and a homotopy $h : f ≃ g$ becomes the chain homotopy $s_{n} : = h^{- n} : C_{n} \to D_{n + 1}$ , because substituting $n \mapsto - n$ into the displayed homotopy equation produces exactly $f_{n} - g_{n} = d_{n + 1}^{D} s_{n} + s_{n - 1} d_{n}^{C} .$ Consequently, when $A$ is abelian, these definitions restrict under this dictionary to the published A chain homotopy and A contractible complex, which are stated for chain complexes in an abelian category. Reindexing reverses the sign of the differential degree ( $+ 1$ for cochains, $- 1$ for chains) and of the homotopy degree ( $- 1$ for cochains, $+ 1$ for chains), and introduces no further sign.

What is not asserted. The definitions use only zero morphisms, addition, negatives, composition and identities. No kernel, cokernel, image, homology object or exactness is assumed or defined, no linear structure on the hom-groups beyond the additive one is used, and no homology object is attached to a complex in an arbitrary additive category; contractibility is the existence of the displayed family $h$ , not the vanishing of homology.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-27Open item page →

An invertible cochain differential block and its candidate reduction

Definition

The block decomposition. Let $X^{∙}$ be a cochain complex in an additive category $A$ (Complexes, homotopies and contractibility in an additive category) and fix an integer $n$ . Suppose that in degrees $n$ and $n + 1$ the objects of $X$ are given as biproducts $X^{n} = A \oplus U, X^{n + 1} = B \oplus V,$ with the injections and projections of these biproducts fixed. With respect to these ordered decompositions write the differential at degree $n$ as the block matrix $d^{n} = (\begin{array}{cc} a & b \\ c & φ \end{array}) : A \oplus U \to B \oplus V,$ where the rows name $B, V$ and the columns name $A, U$ : thus $a : A \to B$ , $b : U \to B$ , $c : A \to V$ and $φ : U \to V$ are the four components, and composition of such block matrices is matrix multiplication (Composition of morphisms between finite biproducts is matrix multiplication).

The pivot. The block $φ : U \to V$ is a pivot when it is an isomorphism, with two-sided inverse $φ^{- 1} : V \to U$ . Only $φ$ is assumed invertible; $a, b, c$ are arbitrary morphisms. The objects $A, U, B, V$ may themselves be biproducts of several objects, in which case $φ$ is an invertible matrix of morphisms and is still required to be a single isomorphism $U \to V$ ; in particular $U$ and $V$ need not be nonzero or indecomposable.

The neighbours. With respect to the same decompositions write the neighbouring differentials as $d^{n - 1} = (\begin{array}{c} p \\ q \end{array}) : X^{n - 1} \to A \oplus U, d^{n + 1} = (\begin{array}{cc} r & s \end{array}) : B \oplus V \to X^{n + 2},$ so that $p : X^{n - 1} \to A$ , $q : X^{n - 1} \to U$ , $r : B \to X^{n + 2}$ and $s : V \to X^{n + 2}$ are the neighbouring components across the two pivot blocks.

The candidate reduction. The candidate reduction ${\overset{ˉ}{X}}^{∙}$ of $X^{∙}$ at the pivot $φ$ is the collection of objects and morphisms ${\overset{ˉ}{X}}^{j} : = X^{j} (j \notin {n, n + 1}), {\overset{ˉ}{X}}^{n} : = A, {\overset{ˉ}{X}}^{n + 1} : = B,$ with differentials ${\overset{ˉ}{d}}^{j} : = d^{j} (j \notin {n - 1, n, n + 1}), {\overset{ˉ}{d}}^{n - 1} : = p, {\overset{ˉ}{d}}^{n} : = a - b φ^{- 1} c, {\overset{ˉ}{d}}^{n + 1} : = r .$ The morphism $a - b φ^{- 1} c : A \to B$ is the Schur complement of the pivot $φ$ in $d^{n}$ . In words: the objects and arrows outside degrees $n, n + 1$ are retained verbatim; the pivot blocks $U, V$ are discarded; and the differential at degree $n$ is replaced by its Schur complement, while the neighbouring differentials lose their components through the discarded blocks and keep $p$ and $r$ .

Status of the construction. The words candidate and reduction are provisional: the definition alone does not assert that ${\overset{ˉ}{d}}^{n} {\overset{ˉ}{d}}^{n - 1} = 0$ or ${\overset{ˉ}{d}}^{n + 1} {\overset{ˉ}{d}}^{n} = 0$ , that is, that ${\overset{ˉ}{X}}^{∙}$ is a cochain complex. The next lemma verifies this, using the identities $c p + φ q = 0$ and $r b + s φ = 0$ that follow from the square-zero composites $d^{n} d^{n - 1} = 0$ and $d^{n + 1} d^{n} = 0$ of the given complex.

Homological convention and scope. For a homological, degree-lowering complex the same formulas apply after the reindexing $n \mapsto - n$ of Complexes, homotopies and contractibility in an additive category: in the chain convention the pivot is the corresponding invertible block of $d_{n}$ and the reduced differential is again the Schur complement of that block, with the contracting homotopy of the discarded two-term complex acquiring degree $+ 1$ . No sign is inserted. The construction uses only the additive structure; it assumes no abelian category, no exactness, no projectivity, no boundedness and no homology object.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Triangular basis changes diagonalize an invertible differential block

Statement

Let $X^{∙}$ be a cochain complex in an additive category and let $d^{n} = (\begin{array}{cc} a & b \\ c & φ \end{array}) : A \oplus U \to B \oplus V$ be an invertible-block decomposition as in An invertible cochain differential block and its candidate reduction: the pivot $φ : U \to V$ is an isomorphism, the neighbouring components are $d^{n - 1} = (\begin{array}{c} p \\ q \end{array})$ and $d^{n + 1} = (\begin{array}{cc} r & s \end{array})$ , and $\overset{ˉ}{d} : = a - b φ^{- 1} c$ is the Schur complement. Put $L = (\begin{array}{cc} 1 & - b φ^{- 1} \\ 0 & 1 \end{array}) : B \oplus V \to B \oplus V, R = (\begin{array}{cc} 1 & 0 \\ - φ^{- 1} c & 1 \end{array}) : A \oplus U \to A \oplus U .$ Then:

$L$ and $R$ are isomorphisms, with inverses $L^{- 1} = (\begin{array}{cc} 1 & b φ^{- 1} \\ 0 & 1 \end{array})$ and $R^{- 1} = (\begin{array}{cc} 1 & 0 \\ φ^{- 1} c & 1 \end{array})$ .
$L d^{n} R = (\begin{array}{cc} \overset{ˉ}{d} & 0 \\ 0 & φ \end{array})$ .
$c p + φ q = 0$ and $r b + s φ = 0$ ; consequently $R^{- 1} (\begin{array}{c} p \\ q \end{array}) = (\begin{array}{c} p \\ 0 \end{array})$ and $(\begin{array}{cc} r & s \end{array}) L^{- 1} = (\begin{array}{cc} r & 0 \end{array})$ .
$\overset{ˉ}{d} p = 0$ and $r \overset{ˉ}{d} = 0$ .
The candidate reduction ${\overset{ˉ}{X}}^{∙}$ of An invertible cochain differential block and its candidate reduction is a cochain complex: all composites of consecutive reduced differentials vanish.

Facts & Assumptions

Given: A cochain complex $X^{∙}$ in an additive category, an integer $n$ , a pivot decomposition $X^{n} = A \oplus U$ , $X^{n + 1} = B \oplus V$ with invertible block $φ : U \to V$ , neighbouring components $p, q, r, s$ , the Schur complement $\overset{ˉ}{d} = a - b φ^{- 1} c$ , and the morphisms $L, R, L^{- 1}, R^{- 1}$ displayed above.

[L1]

The components of $d^{n}, d^{n - 1}, d^{n + 1}$ are the blocks $a, b, c, φ$ and $p, q, r, s$ , the pivot satisfies $φ φ^{- 1} = 1_{V}$ and $φ^{- 1} φ = 1_{U}$ , and composition of morphisms between finite biproducts is matrix multiplication (An invertible cochain differential block and its candidate reduction, Composition of morphisms between finite biproducts is matrix multiplication).

[L2]

$X^{∙}$ is a cochain complex, so $d^{n} d^{n - 1} = 0$ and $d^{n + 1} d^{n} = 0$ (Complexes, homotopies and contractibility in an additive category).

Proof

technique · direct

1.1

Multiplying the two matrices with [L1], the first column of $d^{n} R$ is $d^{n} (1_{A}; - φ^{- 1} c) = (a - b φ^{- 1} c; c - φ φ^{- 1} c) = (\overset{ˉ}{d}; 0)$ and the second is $d^{n} (0; 1_{U}) = (b; φ)$ , so $d^{n} R = (\begin{array}{cc} \overset{ˉ}{d} & b \\ 0 & φ \end{array})$ .

L1algebra

1.2

The displayed inverses work: $L L^{- 1} = (\begin{array}{cc} 1 & - b φ^{- 1} \\ 0 & 1 \end{array}) (\begin{array}{cc} 1 & b φ^{- 1} \\ 0 & 1 \end{array}) = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})$ , and symmetrically $L^{- 1} L = 1$ ; likewise $R R^{- 1} = (\begin{array}{cc} 1 & 0 \\ - φ^{- 1} c & 1 \end{array}) (\begin{array}{cc} 1 & 0 \\ φ^{- 1} c & 1 \end{array}) = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})$ and $R^{- 1} R = 1$ , all uses of $φ φ^{- 1} = 1_{V}$ cancelling the middle terms.

L1algebra

1.3

The composite $d^{n} d^{n - 1}$ is the block matrix $(\begin{array}{c} a p + b q \\ c p + φ q \end{array})$ , so $d^{n} d^{n - 1} = 0$ gives $a p + b q = 0$ and $c p + φ q = 0$ ; applying $φ^{- 1}$ on the left to the second equation gives $q = - φ^{- 1} c p$ .

L1L2algebra

1.4

The composite $d^{n + 1} d^{n}$ is the block matrix $(\begin{array}{cc} r a + s c & r b + s φ \end{array})$ , so $d^{n + 1} d^{n} = 0$ gives $r a + s c = 0$ and $r b + s φ = 0$ ; multiplying the second equation on the right by $φ^{- 1}$ gives $r b φ^{- 1} = - s$ .

L1L2algebra

2.1

Applying $L$ to that result, the first column is $L (\overset{ˉ}{d}; 0) = (\overset{ˉ}{d} - b φ^{- 1} 0; 0) = (\overset{ˉ}{d}; 0)$ and the second is $L (b; φ) = (b - b φ^{- 1} φ; φ) = (0; φ)$ , hence $L d^{n} R = (\begin{array}{cc} \overset{ˉ}{d} & 0 \\ 0 & φ \end{array})$ .

step 1.1L1algebra

2.2

The first component of $R^{- 1} (\begin{array}{c} p \\ q \end{array})$ is $1_{A} p + 0 \cdot q = p$ and the second is $φ^{- 1} c p + 1_{U} q = φ^{- 1} c p + q = 0$ by step 1.3, so $R^{- 1} (\begin{array}{c} p \\ q \end{array}) = (\begin{array}{c} p \\ 0 \end{array})$ .

step 1.3L1algebra

2.3

Similarly $(\begin{array}{cc} r & s \end{array}) L^{- 1} = (\begin{array}{cc} r 1_{B} + s \cdot 0 & r b φ^{- 1} + s 1_{V} \end{array}) = (\begin{array}{cc} r & r b φ^{- 1} + s \end{array}) = (\begin{array}{cc} r & 0 \end{array})$ by step 1.4.

step 1.4L1algebra

2.4

$\overset{ˉ}{d} p = (a - b φ^{- 1} c) p = a p - b φ^{- 1} c p = a p + b q = 0$ , substituting $q = - φ^{- 1} c p$ from step 1.3 and then $a p + b q = 0$ .

step 1.3algebra

2.5

$r \overset{ˉ}{d} = r (a - b φ^{- 1} c) = r a - r b φ^{- 1} c = r a + s c = 0$ , substituting $r b φ^{- 1} = - s$ from step 1.4 and then $r a + s c = 0$ .

step 1.4algebra

3.1

Every composite of consecutive differentials of ${\overset{ˉ}{X}}^{∙}$ vanishes. In the two modified degrees these are ${\overset{ˉ}{d}}^{n} d^{n - 1} = \overset{ˉ}{d} p = 0$ by step 2.4 and ${\overset{ˉ}{d}}^{n + 1} {\overset{ˉ}{d}}^{n} = r \overset{ˉ}{d} = 0$ by step 2.5. In the remaining degrees the reduction either keeps the arrows of $X$ or replaces $d^{n - 1}$ by its $A$ -component $p$ and $d^{n + 1}$ by its $B$ -component $r$ : thus $p d^{n - 2} = {pr}_{A} d^{n - 1} d^{n - 2} = 0$ because $d^{n - 1} d^{n - 2} = 0$ , and $d^{n + 2} r = d^{n + 2} d^{n + 1} ı_{B} = 0$ because $d^{n + 2} d^{n + 1} = 0$ , where ${pr}_{A}$ and $ı_{B}$ are the biproduct projection and injection recording the components $p$ and $r$ . All other composites are composites of consecutive differentials of $X$ , hence vanish.

L1L2step 2.4step 2.5algebra

4.1

Step 1.2 proves clause 1; steps 1.1 and 2.1 prove clause 2; steps 1.3, 1.4, 2.2 and 2.3 prove clause 3; steps 2.4 and 2.5 prove clause 4; and step 3.1 proves clause 5. In particular the candidate reduction of the block decomposition is a genuine cochain complex with the neighbouring arrows $p$ and $r$ . ∎

step 1.2step 1.1step 2.1step 1.3step 1.4step 2.2step 2.3step 2.4step 2.5step 3.1

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Gaussian elimination splits a contractible two-term complex

Statement

The cochain map $T : X^{∙} \to {\overset{ˉ}{X}}^{∙} \oplus K$ with components $T^{j} = 1 (j \notin {n, n + 1}), T^{n} = R^{- 1} = (\begin{array}{cc} 1 & 0 \\ φ^{- 1} c & 1 \end{array}), T^{n + 1} = L = (\begin{array}{cc} 1 & - b φ^{- 1} \\ 0 & 1 \end{array})$ is an isomorphism of cochain complexes, with inverse the cochain map $T^{- 1}$ whose components are $1$ in degrees $j \notin {n, n + 1}$ , $R$ in degree $n$ and $L^{- 1}$ in degree $n + 1$ .
$K$ is contractible, with contracting homotopy $k^{n + 1} = φ^{- 1} : V \to U$ and $k^{j} = 0$ for $j \neq n + 1$ .
$X^{∙}$ and ${\overset{ˉ}{X}}^{∙}$ are homotopy equivalent: the projection $\tilde{p} : \overset{ˉ}{X} \oplus K \to \overset{ˉ}{X}$ and the inclusion $\tilde{ı} : \overset{ˉ}{X} \to \overset{ˉ}{X} \oplus K$ satisfy $\tilde{p} \tilde{ı} = 1$ and $1 - \tilde{ı} \tilde{p} = d \tilde{h} + \tilde{h} d$ for the homotopy $\tilde{h}$ vanishing except in degree $n + 1$ , where ${\tilde{h}}^{n + 1} = (\begin{array}{cc} 0 & 0 \\ 0 & φ^{- 1} \end{array})$ , so that $p : = \tilde{p} T$ and $ı : = T^{- 1} \tilde{ı}$ are homotopy inverse cochain maps.
The construction is a chain isomorphism followed by deletion of a contractible summand: it does not identify $X^{∙}$ with ${\overset{ˉ}{X}}^{∙}$ before that summand is split off, and in general $X^{∙}$ and ${\overset{ˉ}{X}}^{∙}$ do not even have the same objects.

Facts & Assumptions

Given: A cochain complex $X^{∙}$ in an additive category with a pivot decomposition at degree $n$ , its candidate reduction ${\overset{ˉ}{X}}^{∙}$ , the two-term complex $K$ , and the maps $T, T^{- 1}, L, R, L^{- 1}, R^{- 1}, \tilde{p}, \tilde{ı}, \tilde{h}$ displayed above.

[L1]

The candidate reduction is a cochain complex; $L d^{n} R = (\begin{array}{cc} \overset{ˉ}{d} & 0 \\ 0 & φ \end{array})$ ; $R^{- 1} (p; q) = (p; 0)$ , $(r s) L^{- 1} = (r 0)$ and $r b φ^{- 1} = - s$ (Triangular basis changes diagonalize an invertible differential block).

[L2]

The decomposition of $X$ at degrees $n, n + 1$ , the candidate reduction ${\overset{ˉ}{X}}^{∙}$ with objects $A, B$ in those degrees and the two-term complex $K$ with differential $φ$ are as in the block definition; in particular ${\overset{ˉ}{d}}^{n - 1} = p$ , ${\overset{ˉ}{d}}^{n} = \overset{ˉ}{d}$ , ${\overset{ˉ}{d}}^{n + 1} = r$ and the differential of $\overset{ˉ}{X} \oplus K$ in degree $n$ is $(\begin{array}{cc} \overset{ˉ}{d} & 0 \\ 0 & φ \end{array})$ (An invertible cochain differential block and its candidate reduction).

[L3]

Cochain maps, homotopies, homotopy equivalence, contractibility and degreewise biproducts are defined by componentwise equations, and the identity of a zero object is the zero morphism (Complexes, homotopies and contractibility in an additive category).

Proof

technique · direct

1.1

Away from degrees $n - 1, n, n + 1$ the components of $T$ are identities and ${\overset{ˉ}{X}}^{∙} \oplus K$ agrees with $X^{∙}$ in the two adjacent degrees of each such case, so $T$ commutes with the differentials there; at degree $n - 1$ one has $T^{n} d^{n - 1} = R^{- 1} (p; q) = (p; 0) = {\overset{ˉ}{d}}^{n - 1} T^{n - 1}$ by [L1] and [L2].

L1L2algebra

1.2

At degree $n$ , $T^{n + 1} d^{n} = L d^{n} = (\begin{array}{cc} \overset{ˉ}{d} & 0 \\ 0 & φ \end{array}) R^{- 1} = d_{\overset{ˉ}{X} \oplus K}^{n} T^{n}$ , using $L d^{n} R = diag (\overset{ˉ}{d}, φ)$ and $R^{- 1} R = 1$ from [L1].

L1L2algebra

1.3

$K$ is a cochain complex: its only composite of consecutive differentials is $d_{K}^{n + 1} d_{K}^{n} = 0 \cdot φ = 0$ , the differentials into and out of the zero objects $K^{j}$ being zero morphisms.

L2L3algebra

1.4

$K$ is contractible with the displayed $k$ : in degree $n$ one has $d_{K}^{n - 1} k^{n} + k^{n + 1} d_{K}^{n} = 0 + φ^{- 1} φ = 1_{U}$ , in degree $n + 1$ one has $d_{K}^{n} k^{n + 1} + k^{n + 2} d_{K}^{n + 1} = φ φ^{- 1} + 0 = 1_{V}$ , and in every other degree both terms are zero morphisms on a zero object.

L3algebra

2.1

At degree $n + 1$ , $T^{n + 2} d^{n + 1} = (\begin{array}{cc} r & s \end{array})$ and $d_{\overset{ˉ}{X} \oplus K}^{n + 1} T^{n + 1} = (\begin{array}{cc} r & 0 \end{array}) L = (\begin{array}{cc} r & - r b φ^{- 1} \end{array}) = (\begin{array}{cc} r & s \end{array})$ , using $r b φ^{- 1} = - s$ from [L1]. Hence $T$ is a cochain map.

L1L2step 1.2algebra

2.2

The family $T^{'}$ with components $1$ in degrees $j \notin {n, n + 1}$ , $R$ in degree $n$ and $L^{- 1}$ in degree $n + 1$ is a two-sided inverse of $T$ componentwise: $T^{' n} T^{n} = R R^{- 1} = 1$ , $T^{n} T^{' n} = R^{- 1} R = 1$ , $T^{' n + 1} T^{n + 1} = L^{- 1} L = 1$ , $T^{n + 1} T^{' n + 1} = L L^{- 1} = 1$ , and the remaining components are identities.

L1step 1.2algebra

2.3

In the biproduct ${\overset{ˉ}{X}}^{∙} \oplus K$ the projection $\tilde{p}$ onto $\overset{ˉ}{X}$ and the inclusion $\tilde{ı}$ of $\overset{ˉ}{X}$ satisfy $\tilde{p} \tilde{ı} = 1_{\overset{ˉ}{X}}$ . For the homotopy $\tilde{h}$ that vanishes in all degrees except $n + 1$ , where ${\tilde{h}}^{n + 1} = (\begin{array}{cc} 0 & 0 \\ 0 & φ^{- 1} \end{array})$ in the coordinates $B \oplus V \to A \oplus U$ , one computes degreewise: in degree $n$ both $1 - \tilde{ı} \tilde{p} = (\begin{array}{cc} 0 & 0 \\ 0 & 1_{U} \end{array})$ and $d \tilde{h} + \tilde{h} d = {\tilde{h}}^{n + 1} (\begin{array}{cc} \overset{ˉ}{d} & 0 \\ 0 & φ \end{array}) = (\begin{array}{cc} 0 & 0 \\ 0 & 1_{U} \end{array})$ ; in degree $n + 1$ both $1 - \tilde{ı} \tilde{p} = (\begin{array}{cc} 0 & 0 \\ 0 & 1_{V} \end{array})$ and $d \tilde{h} + \tilde{h} d = (\begin{array}{cc} \overset{ˉ}{d} & 0 \\ 0 & φ \end{array}) {\tilde{h}}^{n + 1} = (\begin{array}{cc} 0 & 0 \\ 0 & 1_{V} \end{array})$ ; in all other degrees $\tilde{h} = 0$ and $\tilde{ı} \tilde{p} = 1$ .

L2L3step 1.3algebra

3.1

The family $T^{'}$ is a cochain map: for every $j$ , using the equation $T^{j + 1} d^{j} = d_{\overset{ˉ}{X} \oplus K}^{j} T^{j}$ for the cochain map $T$ and the componentwise inverse identities of step 2.2, one has $T^{' j + 1} d_{\overset{ˉ}{X} \oplus K}^{j} = T^{' j + 1} d_{\overset{ˉ}{X} \oplus K}^{j} T^{j} T^{' j} = T^{' j + 1} T^{j + 1} d^{j} T^{' j} = d^{j} T^{' j}$ . Hence $T^{'} = T^{- 1}$ is the displayed inverse cochain map.

step 2.2step 1.1step 1.2step 2.1algebra

4.1

Define $p : = \tilde{p} T : X^{∙} \to {\overset{ˉ}{X}}^{∙}$ , $ı : = T^{- 1} \tilde{ı} : {\overset{ˉ}{X}}^{∙} \to X^{∙}$ and $h : = T^{- 1} \tilde{h} T$ . Then $p ı = \tilde{p} T T^{- 1} \tilde{ı} = \tilde{p} \tilde{ı} = 1$ and $1 - ı p = T^{- 1} (1 - \tilde{ı} \tilde{p}) T = T^{- 1} (d \tilde{h} + \tilde{h} d) T = d (T^{- 1} \tilde{h} T) + (T^{- 1} \tilde{h} T) d = d h + h d$ , using $T d = d T$ and $T^{- 1} d = d T^{- 1}$ for the chain isomorphisms $T, T^{- 1}$ of steps 1.1 to 2.2. Hence $p$ and $ı$ are cochain maps that are homotopy inverse, so $X^{∙}$ and ${\overset{ˉ}{X}}^{∙}$ are homotopy equivalent.

step 2.3step 3.1algebra

5.1

Steps 1.1, 1.2 and 2.1 show that $T$ is a cochain map and steps 2.2 and 3.1 show that $T^{'}$ is a two-sided inverse cochain map, so $T$ is an isomorphism of complexes with inverse $T^{- 1} = T^{'}$ ; steps 1.3 and 1.4 show that $K$ is a complex contractible via $φ^{- 1}$ ; and step 4.1 transports the direct-sum deformation retract along $T$ to the homotopy equivalence of $X^{∙}$ with ${\overset{ˉ}{X}}^{∙}$ . Because the construction replaces the objects $A \oplus U$ , $B \oplus V$ by $A$ , $B$ and modifies the neighbouring differentials, it never asserts an equality of complexes between $X^{∙}$ and ${\overset{ˉ}{X}}^{∙}$ : the deletion of $K$ is a homotopy equivalence only after the chain isomorphism $T$ . ∎

step 1.1step 1.2step 2.1step 2.2step 3.1step 1.4step 4.1

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Explicit strong deformation retract from Gaussian cancellation

Statement

Let $X^{∙}$ be a cochain complex in an additive category with an invertible-block decomposition $d^{n} = (\begin{array}{cc} a & b \\ c & φ \end{array}) : A \oplus U \to B \oplus V$ and Schur complement $\overset{ˉ}{d} = a - b φ^{- 1} c$ , let ${\overset{ˉ}{X}}^{∙}$ be the candidate reduction, and let $X^{∙} ≃ {\overset{ˉ}{X}}^{∙}$ be the homotopy equivalence of Gaussian elimination splits a contractible two-term complex. Define graded maps $p : X^{∙} \to {\overset{ˉ}{X}}^{∙}$ , $ı : {\overset{ˉ}{X}}^{∙} \to X^{∙}$ and $h : X^{∙} \to X^{∙}$ by the components $p^{n} = (\begin{array}{cc} 1 & 0 \end{array}), p^{n + 1} = (\begin{array}{cc} 1 & - b φ^{- 1} \end{array}), ı^{n} = (\begin{array}{c} 1 \\ - φ^{- 1} c \end{array}), ı^{n + 1} = (\begin{array}{c} 1 \\ 0 \end{array}),$ with all other components of $p$ and $ı$ identities, and by $h^{n + 1} = (\begin{array}{cc} 0 & 0 \\ 0 & φ^{- 1} \end{array}), h^{j} = 0 (j \neq n + 1) .$ Then $p$ and $ı$ are cochain maps, and with the conventions of Complexes, homotopies and contractibility in an additive category for the components of composites, $p ı = 1_{{\overset{ˉ}{X}}^{∙}}, 1_{X^{∙}} - ı p = d h + h d, p h = 0, h ı = 0, h^{2} = 0.$ Here $(p h)^{j} = p^{j - 1} h^{j}$ , $(h ı)^{j} = h^{j} ı^{j}$ and $(h^{2})^{j} = h^{j - 1} h^{j}$ , so each side condition is a statement about the indicated composite in the degree written. In particular $p$ and $ı$ are the homotopy inverses of the homotopy equivalence of the previous theorem, exhibited by the explicit homotopy $h$ .

Facts & Assumptions

Given: A cochain complex $X^{∙}$ in an additive category with the pivot decomposition at degree $n$ , its reduction ${\overset{ˉ}{X}}^{∙}$ , and the graded maps $p, ı, h$ displayed above.

[L1]

The chain isomorphism $T$ of Gaussian elimination splits a contractible two-term complex has components $T^{n} = R^{- 1} = (\begin{array}{cc} 1 & 0 \\ φ^{- 1} c & 1 \end{array})$ and $T^{n + 1} = L = (\begin{array}{cc} 1 & - b φ^{- 1} \\ 0 & 1 \end{array})$ , identities elsewhere, and the transported homotopy equivalence has the form $p = \tilde{p} T$ , $ı = T^{- 1} \tilde{ı}$ , $h = T^{- 1} \tilde{h} T$ with ${\tilde{h}}^{n + 1} = (\begin{array}{cc} 0 & 0 \\ 0 & φ^{- 1} \end{array})$ and $\tilde{p}, \tilde{ı}$ the projection onto and inclusion of the reduction summand.

[L2]

The blocks satisfy $L d^{n} R = diag (\overset{ˉ}{d}, φ)$ , $R^{- 1} (p; q) = (p; 0)$ , $(r s) L^{- 1} = (r 0)$ , $q = - φ^{- 1} c p$ , $r b φ^{- 1} = - s$ , $\overset{ˉ}{d} p = 0$ and $r \overset{ˉ}{d} = 0$ (Triangular basis changes diagonalize an invertible differential block, Gaussian elimination splits a contractible two-term complex).

[L3]

Composites of the graded maps above are formed degreewise — $(p h)^{j} = p^{j - 1} h^{j}$ , $(h ı)^{j} = h^{j} ı^{j}$ and $(h^{2})^{j} = h^{j - 1} h^{j}$ — and the components of $p$ and $ı$ in degrees $j \notin {n, n + 1}$ are identities, so $ı^{j} p^{j} = 1_{X^{j}}$ there, sums and negatives being those of the additive ambient category (Complexes, homotopies and contractibility in an additive category).

Proof

technique · direct

1.1

The components of $p$ are those of $\tilde{p} T$ : in degree $n$ , $p^{n} = (\begin{array}{cc} 1 & 0 \end{array}) R^{- 1} = (\begin{array}{cc} 1 & 0 \end{array})$ ; in degree $n + 1$ , $p^{n + 1} = (\begin{array}{cc} 1 & 0 \end{array}) L = (\begin{array}{cc} 1 & - b φ^{- 1} \end{array})$ ; and in the remaining degrees both factors are identities. Since $p$ is a composite of cochain maps, it is a cochain map.

L1algebra

1.2

The components of $ı$ are those of $T^{- 1} \tilde{ı}$ : in degree $n$ , $ı^{n} = R (\begin{array}{c} 1 \\ 0 \end{array}) = (\begin{array}{c} 1 \\ - φ^{- 1} c \end{array})$ ; in degree $n + 1$ , $ı^{n + 1} = L^{- 1} (\begin{array}{c} 1 \\ 0 \end{array}) = (\begin{array}{c} 1 \\ 0 \end{array})$ ; and in the remaining degrees both factors are identities. So $ı$ is a cochain map.

L1algebra

1.3

The components of $h$ are those of $\tilde{h}$ transported by the identities in degrees other than $n + 1$ : $(T^{- 1} \tilde{h} T)^{n + 1} = R {\tilde{h}}^{n + 1} L = (\begin{array}{cc} 0 & 0 \\ 0 & φ^{- 1} \end{array})$ , because $L$ preserves the $V$ -coordinate, ${\tilde{h}}^{n + 1}$ records only it with $φ^{- 1}$ , and $R$ leaves the element $(0; φ^{- 1} v)$ unchanged; in degree $j \neq n + 1$ one has ${\tilde{h}}^{j} = 0$ and hence $h^{j} = 0$ .

L1algebra

2.1

$p ı = 1_{{\overset{ˉ}{X}}^{∙}}$ : in degree $n$ one has $p^{n} ı^{n} = (\begin{array}{cc} 1 & 0 \end{array}) (\begin{array}{c} 1 \\ - φ^{- 1} c \end{array}) = 1_{A}$ ; in degree $n + 1$ one has $p^{n + 1} ı^{n + 1} = (\begin{array}{cc} 1 & - b φ^{- 1} \end{array}) (\begin{array}{c} 1 \\ 0 \end{array}) = 1_{B}$ ; in every other degree $p^{j} ı^{j} = 1 \cdot 1 = 1$ .

step 1.1step 1.2algebra

2.2

$(1 - ı p)^{n} = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}) - (\begin{array}{c} 1 \\ - φ^{- 1} c \end{array}) (\begin{array}{cc} 1 & 0 \end{array}) = (\begin{array}{cc} 0 & 0 \\ φ^{- 1} c & 1 \end{array})$ and $(d h + h d)^{n} = d^{n - 1} h^{n} + h^{n + 1} d^{n} = (\begin{array}{cc} 0 & 0 \\ 0 & φ^{- 1} \end{array}) (\begin{array}{cc} a & b \\ c & φ \end{array}) = (\begin{array}{cc} 0 & 0 \\ φ^{- 1} c & 1 \end{array})$ , using $h^{n} = 0$ and the block form of $d^{n}$ .

step 1.3L2algebra

2.3

$(1 - ı p)^{n + 1} = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}) - (\begin{array}{c} 1 \\ 0 \end{array}) (\begin{array}{cc} 1 & - b φ^{- 1} \end{array}) = (\begin{array}{cc} 0 & b φ^{- 1} \\ 0 & 1 \end{array})$ and $(d h + h d)^{n + 1} = d^{n} h^{n + 1} + h^{n + 2} d^{n + 1} = (\begin{array}{cc} a & b \\ c & φ \end{array}) (\begin{array}{cc} 0 & 0 \\ 0 & φ^{- 1} \end{array}) = (\begin{array}{cc} 0 & b φ^{- 1} \\ 0 & 1 \end{array})$ , using $h^{n + 2} = 0$ .

step 1.3L2algebra

2.4

$(p h)^{j} = p^{j - 1} h^{j} = 0$ for all $j$ : for $j = n + 1$ this is $p^{n} h^{n + 1} = (\begin{array}{cc} 1 & 0 \end{array}) (\begin{array}{cc} 0 & 0 \\ 0 & φ^{- 1} \end{array}) = (\begin{array}{cc} 0 & 0 \end{array})$ , and for $j \neq n + 1$ the factor $h^{j}$ is zero.

step 1.1step 1.3L3algebra

2.5

$(h ı)^{j} = h^{j} ı^{j} = 0$ for all $j$ : for $j = n + 1$ this is $h^{n + 1} ı^{n + 1} = (\begin{array}{cc} 0 & 0 \\ 0 & φ^{- 1} \end{array}) (\begin{array}{c} 1 \\ 0 \end{array}) = (\begin{array}{c} 0 \\ 0 \end{array})$ , and for $j \neq n + 1$ the factor $h^{j}$ is zero.

step 1.2step 1.3L3algebra

2.6

$(h^{2})^{j} = h^{j - 1} h^{j} = 0$ for all $j$ : if $j \neq n + 1$ then $h^{j} = 0$ , and if $j = n + 1$ then $h^{j - 1} = h^{n} = 0$ .

step 1.3L3algebra

3.1

In every degree $j \notin {n, n + 1}$ one has $h^{j} = 0$ and $ı^{j} p^{j} = 1 \cdot 1 = 1$ , hence $(1 - ı p)^{j} = 0 = (d h + h d)^{j}$ . Together with steps 2.2 and 2.3 this gives $1 - ı p = d h + h d$ .

step 1.3step 2.2step 2.3L3algebra

4.1

Steps 2.1, 2.2, 2.3 and 3.1 show $p ı = 1$ and $1 - ı p = d h + h d$ for the explicit cochain maps $p, ı$ and homotopy $h$ , and steps 2.4 to 2.6 verify the three side conditions $p h = 0$ , $h ı = 0$ and $h^{2} = 0$ in the degree conventions stated. Hence the displayed data is an explicit strong deformation retract of $X^{∙}$ onto ${\overset{ˉ}{X}}^{∙}$ : a chosen retraction, a chosen section and a chosen contracting homotopy with the side conditions above. ∎

step 2.1step 2.2step 2.3step 3.1step 2.4step 2.5step 2.6

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Gaussian cancellation preserves homotopy type and abelian-category homology

Statement

Let $X^{∙}$ be a cochain complex in an additive category $A$ with a pivot decomposition at degree $n$ , let ${\overset{ˉ}{X}}^{∙}$ be the candidate reduction at that pivot, and let $p : X^{∙} \to {\overset{ˉ}{X}}^{∙}, ı : {\overset{ˉ}{X}}^{∙} \to X^{∙}, h$ be the explicit cochain maps and contracting homotopy of Explicit strong deformation retract from Gaussian cancellation, so that $p ı = 1_{{\overset{ˉ}{X}}^{∙}}$ , $1_{X^{∙}} - ı p = d h + h d$ and $p h = 0 = h ı = h^{2}$ .

Homotopy type. Under the reindexing dictionary $C_{n} : = C^{- n}$ of Complexes, homotopies and contractibility in an additive category, the complexes $X^{∙}$ and ${\overset{ˉ}{X}}^{∙}$ become chain complexes $X_{∙}$ and ${\overset{ˉ}{X}}_{∙}$ and the maps $p, ı$ become chain maps whose homotopy classes are mutually inverse isomorphisms in the homotopy category $K (A)$ of The homotopy category of chain complexes. Thus $X^{∙}$ and ${\overset{ˉ}{X}}^{∙}$ are isomorphic in the homotopy category, over every additive category $A$ .
Homology. If $A$ is abelian, then for every integer $n$ the induced maps on the homology objects of the reindexed chain complexes are inverse isomorphisms, $H_{n} (ı) H_{n} (p) = 1_{H_{n} (X)}, H_{n} (p) H_{n} (ı) = 1_{H_{n} (\overset{ˉ}{X})},$ where the homology objects are those of Homology object of a chain complex and, in the cochain indexing, are the homology objects of $X^{∙}$ and ${\overset{ˉ}{X}}^{∙}$ in degree $- n$ .
What is not claimed. No identification of $X^{∙}$ with ${\overset{ˉ}{X}}^{∙}$ as complexes is asserted before the contractible summand is split off: by Gaussian elimination splits a contractible two-term complex the isomorphism only exists after passing to the biproduct ${\overset{ˉ}{X}}^{∙} \oplus K$ with the contractible two-term complex $K$ , and the objects of $X^{∙}$ and ${\overset{ˉ}{X}}^{∙}$ in degrees $n$ and $n + 1$ are in general different.

Facts & Assumptions

Given: A cochain complex $X^{∙}$ in an additive category $A$ with the pivot decomposition at degree $n$ , its candidate reduction ${\overset{ˉ}{X}}^{∙}$ , the explicit cochain maps $p, ı$ and contracting homotopy $h$ of Explicit strong deformation retract from Gaussian cancellation, the chain isomorphism $T$ of Gaussian elimination splits a contractible two-term complex, and — for clause 2 — the additional assumption that $A$ is abelian.

[L1]

$p$ and $ı$ are cochain maps satisfying $p ı = 1_{{\overset{ˉ}{X}}^{∙}}$ and $1_{X^{∙}} - ı p = d h + h d$ , with $h$ of degree $- 1$ and $p h = 0$ , $h ı = 0$ , $h^{2} = 0$ (Explicit strong deformation retract from Gaussian cancellation).

[L2]

Reindexing $C_{n} : = C^{- n}$ , $d_{n} : = d^{- n}$ turns a cochain complex over an additive category into a chain complex over the same category, a cochain map into a chain map and a degree- $(- 1)$ cochain homotopy $h$ into the chain homotopy $s_{n} : = h^{- n}$ of degree $+ 1$ with $f_{n} - g_{n} = d_{n + 1}^{D} s_{n} + s_{n - 1} d_{n}^{C}$ ; no sign is inserted (Complexes, homotopies and contractibility in an additive category).

[L3]

For an additive category, $K (A)$ has the chain complexes as objects and the homotopy classes $[f]$ modulo null-homotopic chain maps as morphisms, with composition induced from representatives; $[f] = [g]$ exactly when $f - g$ is null-homotopic (The homotopy category of chain complexes, Homotopy classes of chain maps).

[L4]

In an abelian category, a chain complex has cycle subobjects $Z_{n} (C) = \ker (d_{n})$ with inclusions $k_{n}$ , boundary subobjects $B_{n} (C) = im (d_{n + 1})$ , the factorization $d_{n + 1} = k_{n} β_{n} e_{n}$ through the boundary-to-cycle map $β_{n} : B_{n} (C) \to Z_{n} (C)$ , and homology objects $H_{n} (C) = coker (β_{n})$ with quotient $q_{n} : Z_{n} (C) \to H_{n} (C)$ ; a chain map $u$ has a unique induced $H_{n} (u)$ characterized by $H_{n} (u) q_{n}^{C} = q_{n}^{D} Z_{n} (u)$ , where $Z_{n} (u)$ is the cycle map carried by $u$ ; and $H_{n}$ is additive, so $H_{n} (1) = 1$ and $H_{n} (u + v) = H_{n} (u) + H_{n} (v)$ (Chain complex in an abelian category, Cycle and boundary subobjects of a complex, The boundary subobject factors through the cycle subobject, Homology object of a chain complex, A chain map carries cycles to cycles and boundaries to boundaries, A chain map induces a well-defined map on homology, Homology is an additive functor).

[L5]

The cycle inclusion $k_{n}$ is a kernel and hence a monomorphism, and the homology quotient $q_{n}$ is a cokernel and hence an epimorphism (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers, Every equalizer is a monomorphism, and every coequalizer is an epimorphism).

[L6]

The isomorphism $T : X^{∙} \to {\overset{ˉ}{X}}^{∙} \oplus K$ has components $T^{n} = R^{- 1}$ and $T^{n + 1} = L$ in degrees $n, n + 1$ and identities elsewhere, so it is not an isomorphism of $X^{∙}$ with ${\overset{ˉ}{X}}^{∙}$ itself, and the objects in degrees $n, n + 1$ are $A \oplus U, B \oplus V$ on the source and $A, B$ on the reduction (Gaussian elimination splits a contractible two-term complex, Explicit strong deformation retract from Gaussian cancellation).

Proof

technique · direct

1.1

Clause 1. By [L1], $p ı = 1_{{\overset{ˉ}{X}}^{∙}}$ and $ı p = 1_{X^{∙}} - (d h + h d)$ . Under the reindexing [L2] these are chain maps $p_{∙}, ı_{∙}$ between $X_{∙}$ and ${\overset{ˉ}{X}}_{∙}$ with $p_{∙} ı_{∙} = 1$ and $ı_{∙} p_{∙} = 1_{X_{∙}} - w$ , where $w_{n} : = d_{n + 1} s_{n} + s_{n - 1} d_{n}$ and $s_{n} : = h^{- n}$ . The family $s$ exhibits $w ≃ 0$ , so $w$ is null-homotopic and $1_{X_{∙}} - ı_{∙} p_{∙} = w$ is null-homotopic; by [L3] therefore $[ı_{∙}] [p_{∙}] = [ı_{∙} p_{∙}] = [1_{X_{∙}}]$ and $[p_{∙}] [ı_{∙}] = [p_{∙} ı_{∙}] = [1_{{\overset{ˉ}{X}}_{∙}}]$ , so the two classes are mutually inverse isomorphisms in $K (A)$ .

L1L2L3algebra

1.2

Clause 2, first composite. Assume $A$ abelian. By [L2] the reindexed complexes are chain complexes in $A$ , and $p_{∙} ı_{∙} = 1_{{\overset{ˉ}{X}}_{∙}}$ strictly by [L1]. Functoriality and additivity of $H_{n}$ in [L4] give $H_{n} (p) H_{n} (ı) = H_{n} (p_{∙} ı_{∙}) = H_{n} (1_{{\overset{ˉ}{X}}_{∙}}) = 1_{H_{n} (\overset{ˉ}{X})}$ .

L1L2L4algebra

1.3

Cycle-level computation. Let $k_{n} : Z_{n} (X) \to X_{n}$ be the cycle inclusion, so $d_{n} k_{n} = 0$ , and let $d_{n + 1} = k_{n} β_{n} e_{n}$ be the factorization of [L4]. Then $w_{n} k_{n} = d_{n + 1} s_{n} k_{n} + s_{n - 1} d_{n} k_{n} = k_{n} β_{n} (e_{n} s_{n} k_{n})$ , because the second summand vanishes and the first is $d_{n + 1} (s_{n} k_{n})$ . As a difference of the chain maps $1_{X_{∙}}$ and $ı_{∙} p_{∙}$ , the map $w$ is a chain map, so [L4] gives a cycle map $Z_{n} (w)$ with $k_{n} Z_{n} (w) = w_{n} k_{n} = k_{n} β_{n} (e_{n} s_{n} k_{n})$ ; by [L5] the inclusion $k_{n}$ is monic, so $Z_{n} (w) = β_{n} (e_{n} s_{n} k_{n})$ .

L4L5algebra

2.1

The homotopy term induces zero. Applying the homology quotient $q_{n} : Z_{n} (X) \to H_{n} (X)$ to step 1.3 gives $q_{n} Z_{n} (w) = q_{n} β_{n} (e_{n} s_{n} k_{n}) = 0$ , since $q_{n}$ is the cokernel of $β_{n}$ by [L4]. The characterizing property $H_{n} (w) q_{n} = q_{n} Z_{n} (w)$ of [L4] therefore gives $H_{n} (w) q_{n} = 0$ , and $q_{n}$ is epic by [L5], so $H_{n} (w) = 0$ for every $n$ .

L4L5step 1.3algebra

3.1

Clause 2, second composite. By [L1] and [L2], $ı_{∙} p_{∙} = 1_{X_{∙}} - w$ with $w$ as in step 1.1, so functoriality and additivity of $H_{n}$ in [L4] give $H_{n} (ı) H_{n} (p) = H_{n} (1_{X_{∙}}) + H_{n} (- w) = 1_{H_{n} (X)} - H_{n} (w)$ , which equals $1_{H_{n} (X)}$ by step 2.1. Together with step 1.2 the two induced maps are inverse isomorphisms.

L1L2L4step 2.1algebra

4.1

Conclusion. Step 1.1 proves clause 1: the classes are inverse in $K (A)$ over an arbitrary additive category. Steps 1.2 and 2.1, 3.1 prove clause 2: over an abelian category $H_{n} (p)$ and $H_{n} (ı)$ are mutually inverse isomorphisms on every homology object. Clause 3 is the qualification carried by [L6]: the displayed identities are those of the deformation retract of $X^{∙}$ onto ${\overset{ˉ}{X}}^{∙}$ and of the chain isomorphism onto ${\overset{ˉ}{X}}^{∙} \oplus K$ , so no equality or canonical identification of the complexes $X^{∙}$ and ${\overset{ˉ}{X}}^{∙}$ is being asserted. ∎

step 1.1step 1.2step 2.1step 3.1L6

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Finite iteration of current invertible-block cancellations

Statement

Let $X^{∙}$ be a cochain complex in an additive category.

Iteration. Suppose a finite sequence of Gaussian cancellations is performed on $X^{∙}$ , each step cancelling an invertible pivot block in the current complex, so that each step is a block decomposition as in An invertible cochain differential block and its candidate reduction and the current complex is replaced by its candidate reduction. Then the composite of the steps is a strong deformation retract of $X^{∙}$ onto the final reduction, with the explicit data described in clause 2.
Composition of retract data. If $X^{∙} ⇄ Y^{∙}$ has data $(p_{1}, ı_{1}, h_{1})$ and $Y^{∙} ⇄ Z^{∙}$ has data $(p_{2}, ı_{2}, h_{2})$ in the sense of Explicit strong deformation retract from Gaussian cancellation, then $p = p_{2} p_{1}, ı = ı_{1} ı_{2}, h = h_{1} + ı_{1} h_{2} p_{1}$ satisfy $p ı = 1_{Z^{∙}}$ , $1_{X^{∙}} - ı p = d h + h d$ , $p h = 0$ , $h ı = 0$ and $h^{2} = 0$ , so they are strong deformation retract data of $X^{∙}$ onto $Z^{∙}$ .
Aggregate pivots. If a decomposition of $X^{n}, X^{n + 1}$ is presented with pivot blocks that are finite biproducts $U = U_{1} \oplus \dots \oplus U_{k}$ , $V = V_{1} \oplus \dots \oplus V_{k}$ and a block-diagonal isomorphism $Φ = diag (φ_{1}, \dots, φ_{k}) : U \to V$ , then a single cancellation with pivot $Φ$ is available, and the resulting reduction is the complex obtained by cancelling $φ_{1}, \dots, φ_{k}$ successively in the current Schur-complement complexes.
Choices and limits. Different valid finite choices of cancellations yield reductions that are homotopy equivalent but not generally equal complexes: there is no canonical reduced complex, no guarantee that a reduction is smaller, and no assertion about infinite sequences of cancellations.

Facts & Assumptions

Given: A cochain complex $X^{∙}$ in an additive category, its invertible-block decompositions at the chosen degrees, the explicit strong deformation retracts attached to single cancellations, and the composites described in the statement.

[L1]

A single cancellation with pivot $φ : U \to V$ in the current complex gives cochain maps $p, ı$ and a homotopy $h$ of degree $- 1$ with $p ı = 1$ and $1 - ı p = d h + h d$ , $p h = 0$ , $h ı = 0$ , $h^{2} = 0$ (Explicit strong deformation retract from Gaussian cancellation).

[L2]

The candidate reduction at the pivot replaces the objects $A \oplus U, B \oplus V$ in degrees $n, n + 1$ by $A, B$ , keeps all other objects and arrows, keeps the neighbouring components $p$ of $d^{n - 1}$ and $r$ of $d^{n + 1}$ , and replaces $d^{n}$ by the Schur complement $a - b φ^{- 1} c$ ; the pivot blocks $U, V$ may themselves be biproducts and $φ$ may be any isomorphism between them (An invertible cochain differential block and its candidate reduction).

[L3]

The candidate reduction is a cochain complex, and the identities $L d^{n} R = diag (\overset{ˉ}{d}, φ)$ , $R^{- 1} (p; q) = (p; 0)$ , $(r s) L^{- 1} = (r 0)$ hold for every pivot (Triangular basis changes diagonalize an invertible differential block).

[L4]

Composition of morphisms between finite biproducts is matrix multiplication, and finite biproducts may be reassociated: splitting $A \oplus U_{1} \oplus U^{'}$ as $(A \oplus U^{'}) \oplus U_{1}$ and $B \oplus V_{1} \oplus V^{'}$ as $(B \oplus V^{'}) \oplus V_{1}$ is a biproduct decomposition again (Composition of morphisms between finite biproducts is matrix multiplication, An invertible cochain differential block and its candidate reduction).

[L5]

Cochain maps are closed under composition, the equations $p ı = 1$ , $1 - ı p = d h + h d$ , $p h = 0$ , $h ı = 0$ , $h^{2} = 0$ are degreewise identities of morphisms, and a cochain map $u$ satisfies $d u = u d$ in the graded sense (Complexes, homotopies and contractibility in an additive category).

Proof

technique · direct

1.1

Composition of retract data. Assume $p_{1} ı_{1} = 1_{Y}$ , $ı_{1} p_{1} = 1_{X} - (d h_{1} + h_{1} d)$ , $p_{2} ı_{2} = 1_{Z}$ and $ı_{2} p_{2} = 1_{Y} - (d h_{2} + h_{2} d)$ , and put $p = p_{2} p_{1}$ , $ı = ı_{1} ı_{2}$ , $h = h_{1} + ı_{1} h_{2} p_{1}$ . Then $p ı = p_{2} p_{1} ı_{1} ı_{2} = p_{2} 1_{Y} ı_{2} = p_{2} ı_{2} = 1_{Z}$ ; moreover $ı p = ı_{1} ı_{2} p_{2} p_{1} = ı_{1} (1_{Y} - d h_{2} - h_{2} d) p_{1} = ı_{1} p_{1} - ı_{1} d h_{2} p_{1} - ı_{1} h_{2} d p_{1}$ , so $1_{X} - ı p = (1_{X} - ı_{1} p_{1}) + ı_{1} d h_{2} p_{1} + ı_{1} h_{2} d p_{1} = d h_{1} + h_{1} d + ı_{1} d h_{2} p_{1} + ı_{1} h_{2} d p_{1}$ . On the other hand $d h + h d = d (h_{1} + ı_{1} h_{2} p_{1}) + (h_{1} + ı_{1} h_{2} p_{1}) d = d h_{1} + h_{1} d + d ı_{1} h_{2} p_{1} + ı_{1} h_{2} p_{1} d$ , and $d ı_{1} = ı_{1} d$ , $p_{1} d = d p_{1}$ because $ı_{1}, p_{1}$ are cochain maps, so the two expressions coincide and $1_{X} - ı p = d h + h d$ .

L1L5algebra

1.2

Aggregate pivots. Let $X^{n} = A \oplus U_{1} \oplus U^{'}$ and $X^{n + 1} = B \oplus V_{1} \oplus V^{'}$ with pivot blocks $U = U_{1} \oplus U^{'}$ , $V = V_{1} \oplus V^{'}$ and $Φ = diag (φ_{1}, Φ^{'})$ , where $φ_{1} : U_{1} \to V_{1}$ and $Φ^{'} : U^{'} \to V^{'}$ are isomorphisms; write $d^{n}$ as the block matrix with rows $B, V_{1}, V^{'}$ and columns $A, U_{1}, U^{'}$ as $(\begin{array}{ccc} a & b_{1} & b^{'} \\ c_{1} & φ_{1} & 0 \\ c^{'} & 0 & Φ^{'} \end{array})$ , and write $d^{n - 1} = (p; q_{1}; q^{'})$ and $d^{n + 1} = (r s_{1} s^{'})$ accordingly. Cancelling $Φ$ at once, $Φ^{- 1} = diag (φ_{1}^{- 1}, Φ^{' - 1})$ multiplies out to $b Φ^{- 1} c = b_{1} φ_{1}^{- 1} c_{1} + b^{'} Φ^{' - 1} c^{'}$ , so by [L2] the reduced differential is $a - b_{1} φ_{1}^{- 1} c_{1} - b^{'} Φ^{' - 1} c^{'}$ and the neighbouring arrows are $p$ and $r$ . Cancelling first $φ_{1}$ in the reassociated decomposition $(A \oplus U^{'}) \oplus U_{1}$ , $(B \oplus V^{'}) \oplus V_{1}$ of [L4], the pivot matrix is $(\begin{array}{cc} \tilde{a} & \tilde{b} \\ \tilde{c} & φ_{1} \end{array})$ with $\tilde{a} = (\begin{array}{cc} a & b^{'} \\ c^{'} & Φ^{'} \end{array})$ , $\tilde{b} = (b_{1}; 0)$ , $\tilde{c} = (c_{1} 0)$ , so the new Schur complement is $\tilde{a} - \tilde{b} φ_{1}^{- 1} \tilde{c} = (\begin{array}{cc} a - b_{1} φ_{1}^{- 1} c_{1} & b^{'} \\ c^{'} & Φ^{'} \end{array})$ , a complex by [L3], whose $(U^{'}, V^{'})$ -pivot is $Φ^{'}$ and whose neighbouring arrows are $(p; q^{'})$ and $(r s^{'})$ ; cancelling $Φ^{'}$ there gives reduced differential $a - b_{1} φ_{1}^{- 1} c_{1} - b^{'} Φ^{' - 1} c^{'}$ , 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 $diag (φ_{1}, \dots, φ_{k})$ in one step with the same result as the successive cancellations.

L2L3L4algebra

2.1

Side conditions of the composite. With the data of step 1.1, $p h = p_{2} p_{1} h_{1} + p_{2} p_{1} ı_{1} h_{2} p_{1} = p_{2} \cdot 0 + p_{2} \cdot 1_{Y} \cdot h_{2} p_{1} = p_{2} h_{2} p_{1} = 0$ , using $p_{1} h_{1} = 0$ and $p_{2} h_{2} = 0$ ; likewise $h ı = h_{1} ı_{1} ı_{2} + ı_{1} h_{2} p_{1} ı_{1} ı_{2} = 0 + ı_{1} h_{2} \cdot 1_{Y} \cdot ı_{2} = ı_{1} h_{2} ı_{2} = 0$ , using $h_{1} ı_{1} = 0$ and $h_{2} ı_{2} = 0$ ; and $h^{2} = h_{1}^{2} + h_{1} ı_{1} h_{2} p_{1} + ı_{1} h_{2} p_{1} h_{1} + ı_{1} h_{2} p_{1} ı_{1} h_{2} p_{1} = 0 + 0 + ı_{1} h_{2} (p_{1} h_{1}) + ı_{1} h_{2} \cdot 1_{Y} \cdot h_{2} p_{1} = ı_{1} h_{2}^{2} p_{1} = 0$ , using $h_{1}^{2} = 0$ and $h_{2}^{2} = 0$ . Hence the composite data satisfies all three side conditions of [L1].

L1L5step 1.1algebra

3.1

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

L1step 1.1step 2.1algebra

4.1

Different choices. Suppose two finite sequences of cancellations lead from $X^{∙}$ to reductions ${\overset{ˉ}{X}}^{∙}$ and ${\overset{ˉ}{X}}^{' ∙}$ ; by step 3.1 there are strong deformation retract data $(p, ı, h)$ of $X^{∙}$ onto ${\overset{ˉ}{X}}^{∙}$ and $(p^{'}, ı^{'}, h^{'})$ of $X^{∙}$ onto ${\overset{ˉ}{X}}^{' ∙}$ . Define $u : = p^{'} ı : {\overset{ˉ}{X}}^{∙} \to {\overset{ˉ}{X}}^{' ∙}$ and $v : = p ı^{'} : {\overset{ˉ}{X}}^{' ∙} \to {\overset{ˉ}{X}}^{∙}$ ; then $u v = p^{'} ı p ı^{'} = p^{'} (1_{X^{∙}} - (d h + h d)) ı^{'} = 1_{{\overset{ˉ}{X}}^{'}} - d (p^{'} h ı^{'}) - (p^{'} h ı^{'}) d$ and similarly $v u = 1_{\overset{ˉ}{X}} - d (p h^{'} ı) - (p h^{'} ı) d$ , using that $p^{'}, ı, p, ı^{'}$ are cochain maps and $p^{'} ı^{'} = 1$ , $p ı = 1$ . Hence the two reductions are homotopy equivalent, with explicit comparison maps. They need not be equal: in the complex $k \overset{1}{\to} k \overset{0}{\to} k \overset{1}{\to} k$ 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.

L1L2step 3.1algebra

5.1

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. ∎

step 1.1step 2.1step 1.2step 3.1step 4.1L2

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Additive functors preserve chosen Gaussian cancellations

Statement

Let $F : A \to B$ be an additive functor between additive categories, let $X^{∙}$ be a cochain complex in $A$ with a pivot decomposition at degree $n$ , Schur complement $\overset{ˉ}{d} = a - b φ^{- 1} c$ , candidate reduction ${\overset{ˉ}{X}}^{∙}$ and two-term complex $K$ as in An invertible cochain differential block and its candidate reduction and Gaussian elimination splits a contractible two-term complex, and let $(p, ı, h)$ be the strong deformation retract data of Explicit strong deformation retract from Gaussian cancellation.

Complexes and pivots. $F (X^{∙})$ , with differentials $F (d^{n})$ , is a cochain complex in $B$ ; the pivot $F (φ)$ is invertible with inverse $F (φ^{- 1})$ ; and with respect to the biproduct decompositions $F (X^{n}) = F (A) \oplus F (U)$ , $F (X^{n + 1}) = F (B) \oplus F (V)$ whose structure maps are the $F$ -images of those of $X^{∙}$ , the differential $F (d^{n})$ has the entrywise image matrix $(\begin{array}{cc} F (a) & F (b) \\ F (c) & F (φ) \end{array})$ .
The corresponding cancellation. The reduction of $F (X^{∙})$ at the pivot $F (φ)$ is $F ({\overset{ˉ}{X}}^{∙})$ : its objects and neighbouring arrows are the $F$ -images of those of ${\overset{ˉ}{X}}^{∙}$ , and its differential in degree $n$ is the Schur complement $F (a) - F (b) F (φ)^{- 1} F (c) = F (\overset{ˉ}{d})$ .
Retract data. The images $F (p), F (ı), F (h)$ satisfy $F (p) F (ı) = 1$ , $1 - F (ı) F (p) = F (d) F (h) + F (h) F (d)$ , $F (p) F (h) = 0$ , $F (h) F (ı) = 0$ and $F (h)^{2} = 0$ , so they are strong deformation retract data of $F (X^{∙})$ onto $F ({\overset{ˉ}{X}}^{∙})$ . Moreover $F (K)$ is the two-term complex $F (U) \overset{F (φ)}{\to} F (V)$ with vanishing neighbouring terms, contractible via $F (φ^{- 1})$ , and $F (T), F (T^{- 1})$ remain mutually inverse cochain isomorphisms between $F (X^{∙})$ and $F ({\overset{ˉ}{X}}^{∙} \oplus K)$ .
Scope. Clauses 1 to 3 use only additivity: no exactness of $F$ is assumed or needed. If $B$ is abelian, the image retract maps induce inverse isomorphisms on the homology of $F (X^{∙})$ and $F ({\overset{ˉ}{X}}^{∙})$ , by Gaussian cancellation preserves homotopy type and abelian-category homology. No comparison of $F (H_{n} (X))$ with $H_{n} (F (X))$ is asserted; these expressions both make sense when $A$ and $B$ are abelian, but comparing them is a separate question about commuting $F$ with homology.

Facts & Assumptions

Given: An additive functor $F : A \to B$ between additive categories, a cochain complex $X^{∙}$ in $A$ with the pivot decomposition at degree $n$ , its reduction ${\overset{ˉ}{X}}^{∙}$ , the two-term complex $K$ , the chain isomorphism $T$ , and the explicit cochain maps $p, ı$ and homotopy $h$ of the strong deformation retract.

[L1]

$p, ı$ are cochain maps and $h$ has degree $- 1$ , with $p ı = 1_{{\overset{ˉ}{X}}^{∙}}$ , $1_{X^{∙}} - ı p = d h + h d$ , $p h = 0$ , $h ı = 0$ and $h^{2} = 0$ (Explicit strong deformation retract from Gaussian cancellation).

[L2]

The decomposition $X^{n} = A \oplus U$ , $X^{n + 1} = B \oplus V$ has $φ : U \to V$ invertible, $d^{n} = (\begin{array}{cc} a & b \\ c & φ \end{array})$ , $d^{n - 1} = (p; q)$ , $d^{n + 1} = (r s)$ , and the candidate reduction replaces degrees $n, n + 1$ by $A, B$ with $d^{n}$ replaced by $\overset{ˉ}{d} = a - b φ^{- 1} c$ and neighbouring arrows $p$ and $r$ (An invertible cochain differential block and its candidate reduction).

[L3]

$T : X^{∙} \to {\overset{ˉ}{X}}^{∙} \oplus K$ is an isomorphism of cochain complexes with inverse $T^{- 1}$ , where $K$ has $K^{n} = U$ , $K^{n + 1} = V$ , vanishing terms elsewhere and differential $φ$ , and $K$ is contractible with contracting homotopy $φ^{- 1}$ in degree $n + 1$ (Gaussian elimination splits a contractible two-term complex).

[L4]

An additive functor preserves composition and identities, and its induced maps on hom-groups are homomorphisms: $F (f + g) = F (f) + F (g)$ and hence $F (- f) = - F (f)$ (Additive functor).

[L5]

An additive functor preserves finite biproducts, so the $F$ -images of the injections and projections of a finite biproduct exhibit $F (A \oplus U)$ as a biproduct $F (A) \oplus F (U)$ with the same identity-sum relations; it also preserves zero morphisms; and composition of morphisms between finite biproducts is matrix multiplication (An additive functor preserves finite biproducts, An additive functor preserves zero morphisms, Composition of morphisms between finite biproducts is matrix multiplication, Complexes, homotopies and contractibility in an additive category).

[L6]

In an abelian category, the maps of a Gaussian strong deformation retract induce mutually inverse maps on every homology object of the reindexed chain complexes (Gaussian cancellation preserves homotopy type and abelian-category homology).

Proof

technique · direct

1.1

Complex and pivot. Since $d^{n + 1} d^{n} = 0$ in $X^{∙}$ , [L4] gives $F (d^{n + 1}) F (d^{n}) = F (d^{n + 1} d^{n}) = F (0_{X^{n}, X^{n + 2}})$ , which is the zero morphism by [L5]; thus $F (X^{∙})$ is a cochain complex. Likewise $F (φ) F (φ^{- 1}) = F (φ φ^{- 1}) = F (1_{V}) = 1_{F (V)}$ and $F (φ^{- 1}) F (φ) = 1_{F (U)}$ , so $F (φ)$ is invertible with the displayed inverse.

L4L5algebra

1.2

Image matrices. By [L5] the $F$ -images of the injections and projections of $X^{n} = A \oplus U$ and $X^{n + 1} = B \oplus V$ exhibit $F (X^{n})$ as $F (A) \oplus F (U)$ and $F (X^{n + 1})$ as $F (B) \oplus F (V)$ . Writing $d^{n} = i_{B} a p_{A} + i_{B} b p_{U} + i_{V} c p_{A} + i_{V} φ p_{U}$ with the biproduct structure maps [L2], additivity of $F$ on hom-groups, preservation of composition and the biproduct relations give $F (d^{n}) = F (i_{B}) F (a) F (p_{A}) + F (i_{B}) F (b) F (p_{U}) + F (i_{V}) F (c) F (p_{A}) + F (i_{V}) F (φ) F (p_{U})$ , whose matrix with respect to the image decompositions is $(\begin{array}{cc} F (a) & F (b) \\ F (c) & F (φ) \end{array})$ by the matrix convention of [L5]. The same computation applies to $d^{n - 1}$ and $d^{n + 1}$ , giving the image neighbouring components $F (p), F (q), F (r), F (s)$ .

L4L5algebra

1.3

Retract identities are preserved. Applying [L4] to the identities of [L1] and using [L5] for the zero morphisms: $F (p) F (ı) = F (p ı) = F (1_{{\overset{ˉ}{X}}^{∙}}) = 1_{F ({\overset{ˉ}{X}}^{∙})}$ ; $F (d) F (h) + F (h) F (d) = F (d h + h d) = F (1_{X^{∙}} - ı p) = 1_{F (X^{∙})} - F (ı) F (p)$ ; $F (p) F (h) = F (p h) = F (0) = 0$ ; $F (h) F (ı) = F (h ı) = 0$ ; and $F (h)^{2} = F (h^{2}) = F (0) = 0$ . Since $p, ı$ are cochain maps, $F (p), F (ı)$ are cochain maps by [L4].

L1L4L5algebra

2.1

The contractible summand and the isomorphism. By [L3] and [L4], $F (T) F (T^{- 1}) = F (T T^{- 1}) = F (1) = 1$ and $F (T^{- 1}) F (T) = 1$ , so $F (T)$ is an isomorphism of complexes with inverse $F (T^{- 1})$ ; and $F (K)$ has objects $F (U), F (V)$ in degrees $n, n + 1$ , vanishing terms elsewhere with zero differentials, and differential $F (φ)$ , with $F (φ) F (φ^{- 1}) = 1_{F (V)}$ and $F (φ^{- 1}) F (φ) = 1_{F (U)}$ from step 1.1, so $F (φ^{- 1})$ is a contracting homotopy for $F (K)$ .

L3L4L5step 1.1algebra

2.2

The reduction of $F (X^{∙})$ is $F ({\overset{ˉ}{X}}^{∙})$ . By step 1.2 the reduction problem for $F (X^{∙})$ in degrees $n, n + 1$ is the image matrix $(\begin{array}{cc} F (a) & F (b) \\ F (c) & F (φ) \end{array})$ with pivot $F (φ)$ , and by step 1.1 that pivot is invertible; [L4] gives $F (a) - F (b) F (φ^{- 1}) F (c) = F (a - b φ^{- 1} c) = F (\overset{ˉ}{d})$ , and applying $F$ to the remaining data of [L2] gives objects $F (A), F (B)$ in degrees $n, n + 1$ , neighbouring arrows $F (p), F (r)$ and the unchanged images of the outside objects and arrows. Hence the candidate reduction of $F (X^{∙})$ at this pivot is exactly $F ({\overset{ˉ}{X}}^{∙})$ , its differential in degree $n$ being the image $F (\overset{ˉ}{d})$ of the Schur complement.

L2L4step 1.1step 1.2algebra

3.1

Conclusion. Step 1.1 shows that $F (X^{∙})$ is a complex with invertible pivot $F (φ)$ , step 1.2 computes the image matrices, and step 2.2 identifies the reduction of $F (X^{∙})$ with $F ({\overset{ˉ}{X}}^{∙})$ , which is clause 2 and the matrix assertion of clause 1. Step 1.3 verifies all five strong deformation retract identities for $F (p), F (ı), F (h)$ , and step 2.1 shows that $F (K)$ is contractible via $F (φ^{- 1})$ and that $F (T)$ is an isomorphism, which is clause 3. Only additivity, preservation of finite biproducts and preservation of zero morphisms are used, so no exactness hypothesis enters; if $B$ is abelian, [L6] applied to the image cancellation identified in step 2.2 gives inverse homology maps. This compares the homology of the two image complexes, not the image under $F$ of a homology object in $A$ . ∎

L6step 2.1step 2.2step 1.3step 1.2step 1.1algebra

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Transferred maps are functorial up to homotopy, with strict naturality limits

Statement

Let $A$ be an additive category and let chosen strong deformation retract data be given as in Explicit strong deformation retract from Gaussian cancellation: $(p_{X}, ı_{X}, h_{X}) for X^{∙} onto {\overset{ˉ}{X}}^{∙}, (p_{Y}, ı_{Y}, h_{Y}) for Y^{∙} onto {\overset{ˉ}{Y}}^{∙}, (p_{Z}, ı_{Z}, h_{Z}) for Z^{∙} onto {\overset{ˉ}{Z}}^{∙} .$ For every cochain map $f : X^{∙} \to Y^{∙}$ define the transfer $\overset{ˉ}{f} : = p_{Y} f ı_{X} : {\overset{ˉ}{X}}^{∙} \to {\overset{ˉ}{Y}}^{∙},$ and for every homotopy $s : f ≃ g$ define $\overset{ˉ}{s} : = p_{Y} s ı_{X}$ .

Maps and homotopies. Each transfer $\overset{ˉ}{f}$ is a cochain map, each $\overset{ˉ}{s}$ is a homotopy $\overset{ˉ}{f} ≃ \overset{ˉ}{g}$ of degree $- 1$ , and consequently transfer is well defined on homotopy classes of cochain maps.
Functoriality up to homotopy. The identity transfers strictly, $\overline{1_{X^{∙}}} = 1_{{\overset{ˉ}{X}}^{∙}}$ , and for composable cochain maps $f : X^{∙} \to Y^{∙}$ , $g : Y^{∙} \to Z^{∙}$ , $\overline{(g f)} - \overset{ˉ}{g} \overset{ˉ}{f} = d k + k d, k : = p_{Z} g h_{Y} f ı_{X},$ so $\overline{(g f)} ≃ \overset{ˉ}{g} \overset{ˉ}{f}$ : transfer preserves identities and composition on homotopy classes, but it is not asserted to be a strict functor on cochain maps.
Strictness fails. Transfer need not preserve composition strictly: for over a field $k$ , take $Y^{∙} = K = (k \overset{1}{\to} k)$ in degrees $0, 1$ and $X^{∙} = Y^{∙} \oplus K$ . The split-off retract onto ${\overset{ˉ}{X}}^{∙} = Y^{∙}$ admits cochain maps $f, g$ with $\overset{ˉ}{f} = 0 = \overset{ˉ}{g}$ and $\overline{(g f)} = 1_{{\overset{ˉ}{X}}^{∙}}$ , so $\overline{(g f)} - \overset{ˉ}{g} \overset{ˉ}{f} \neq 0$ .
Strict naturality, and what is not claimed. If cochain maps $f : X^{∙} \to Y^{∙}$ and $g : Y^{∙} \to Z^{∙}$ commute with the chosen retract data, that is $f ı_{X} = ı_{Y} \overset{ˉ}{f}$ , $p_{Y} f = \overset{ˉ}{f} p_{X}$ , $g ı_{Y} = ı_{Z} \overset{ˉ}{g}$ and $p_{Z} g = \overset{ˉ}{g} p_{Y}$ , then $\overline{(g f)} = \overset{ˉ}{g} \overset{ˉ}{f}$ strictly. Transfer depends on the chosen retracts and homotopies; no choice-free, canonical or confluent transfer, and no independence of the chosen data, is claimed.

Facts & Assumptions

Given: An additive category $A$ with three chosen strong deformation retract data $(p_{X}, ı_{X}, h_{X})$ , $(p_{Y}, ı_{Y}, h_{Y})$ , $(p_{Z}, ı_{Z}, h_{Z})$ as in the statement, composable cochain maps $f : X^{∙} \to Y^{∙}$ and $g : Y^{∙} \to Z^{∙}$ , and, separately, parallel cochain maps $u, v : X^{∙} \to Y^{∙}$ with a homotopy $s : u ≃ v$ .

[L1]

For each of the three pairs, $p$ and $ı$ are cochain maps, $h$ has degree $- 1$ , and $p ı = 1$ , $1 - ı p = d h + h d$ , $p h = 0$ , $h ı = 0$ , $h^{2} = 0$ (Explicit strong deformation retract from Gaussian cancellation).

[L2]

The split-off retract of the theorem: if $X^{∙} = {\overset{ˉ}{X}}^{∙} \oplus K$ is a biproduct in which $K$ is the contractible two-term complex with differential the identity in degrees $n, n + 1$ , then the projection $p$ , the inclusion $ı$ and the homotopy $h$ with $h^{n + 1} = (\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array})$ and $h^{j} = 0$ for $j \neq n + 1$ are strong deformation retract data of $X^{∙}$ onto ${\overset{ˉ}{X}}^{∙}$ (Gaussian elimination splits a contractible two-term complex, Explicit strong deformation retract from Gaussian cancellation).

[L3]

A homotopy $s$ of degree $- 1$ between cochain maps satisfies $f - g = d s + s d$ ; composites and sums of cochain maps are cochain maps, and a cochain map $u$ satisfies $d u = u d$ in the graded sense; homotopy is an equivalence relation compatible with composition, so the homotopy classes of cochain maps are the morphisms of the homotopy category under the reindexing dictionary of Complexes, homotopies and contractibility in an additive category (Homotopy classes of chain maps, The homotopy category of chain complexes).

Proof

technique · direct

1.1

Transfer of maps and homotopies. The composite $\overset{ˉ}{f} = p_{Y} f ı_{X}$ of cochain maps is a cochain map, so $d \overset{ˉ}{f} = \overset{ˉ}{f} d$ by [L3]. If $f - g = d s + s d$ , then $\overset{ˉ}{f} - \overset{ˉ}{g} = p_{Y} (f - g) ı_{X} = p_{Y} d s ı_{X} + p_{Y} s d ı_{X} = d (p_{Y} s ı_{X}) + (p_{Y} s ı_{X}) d$ , using $p_{Y} d = d p_{Y}$ and $d ı_{X} = ı_{X} d$ ; hence $\overset{ˉ}{s} = p_{Y} s ı_{X}$ is a degree- $(- 1)$ homotopy $\overset{ˉ}{f} ≃ \overset{ˉ}{g}$ . Therefore homotopic maps have homotopic transfers, and transfer is well defined on homotopy classes.

L1L3algebra

1.2

Functoriality up to homotopy. The identity transfers strictly: $\overline{1_{X^{∙}}} = p_{X} 1_{X^{∙}} ı_{X} = p_{X} ı_{X} = 1_{{\overset{ˉ}{X}}^{∙}}$ . For composable $f, g$ set $k : = p_{Z} g h_{Y} f ı_{X}$ ; then $\overset{ˉ}{g} \overset{ˉ}{f} = p_{Z} g ı_{Y} p_{Y} f ı_{X} = p_{Z} g (1_{Y^{∙}} - (d h_{Y} + h_{Y} d)) f ı_{X} = \overline{(g f)} - (p_{Z} g d h_{Y} f ı_{X} + p_{Z} g h_{Y} d f ı_{X})$ , and since $p_{Z}, g, f, ı_{X}$ are cochain maps the two correction terms are $d (p_{Z} g h_{Y} f ı_{X})$ and $(p_{Z} g h_{Y} f ı_{X}) d$ , that is $d k$ and $k d$ . Hence $\overset{ˉ}{g} \overset{ˉ}{f} = \overline{(g f)} - (d k + k d)$ , equivalently $\overline{(g f)} - \overset{ˉ}{g} \overset{ˉ}{f} = d k + k d$ , so $\overline{(g f)} ≃ \overset{ˉ}{g} \overset{ˉ}{f}$ with this sign convention.

L1L3algebra

1.3

Strictness fails. Take $A$ the category of vector spaces over a field, let $K$ be the two-term complex $k \overset{1}{\to} k$ concentrated in degrees $0, 1$ with zero neighbouring terms, let ${\overset{ˉ}{X}}^{∙}$ be a second copy of it and $X^{∙} = {\overset{ˉ}{X}}^{∙} \oplus K$ , and use the split-off retract of [L2] with $ı$ the inclusion of the first summand, $p$ the projection onto it and $h^{1} = (\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array})$ , $h^{0} = h^{2} = 0$ . In each degree let $f = (\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array})$ and $g = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ in the coordinates ${\overset{ˉ}{X}}^{0} \oplus K^{0}$ respectively ${\overset{ˉ}{X}}^{1} \oplus K^{1}$ ; the components in degrees $0$ and $1$ agree, so both maps commute with the only nonzero differential $d^{0} = 1$ , so $f$ and $g$ are cochain maps. Then $g f = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}) = ı p$ , and the transfers are $\overset{ˉ}{f} = p f ı = 0$ and $\overset{ˉ}{g} = p g ı = 0$ because $f ı$ and $g ı$ land in the complementary summand killed by $p$ , while $\overline{(g f)} = p (g f) ı = p ı = 1_{{\overset{ˉ}{X}}^{∙}}$ . Hence $\overline{(g f)} - \overset{ˉ}{g} \overset{ˉ}{f} = 1_{{\overset{ˉ}{X}}^{∙}} \neq 0$ , so transfer is not a strict functor on cochain maps.

L2L3algebra

1.4

Strict naturality for commuting morphisms. Let $f : X^{∙} \to Y^{∙}$ and $g : Y^{∙} \to Z^{∙}$ satisfy $f ı_{X} = ı_{Y} \overset{ˉ}{f}$ , $p_{Y} f = \overset{ˉ}{f} p_{X}$ , $g ı_{Y} = ı_{Z} \overset{ˉ}{g}$ and $p_{Z} g = \overset{ˉ}{g} p_{Y}$ . Then $\overline{(g f)} = p_{Z} g f ı_{X} = p_{Z} g ı_{Y} \overset{ˉ}{f} = \overset{ˉ}{g} p_{Y} ı_{Y} \overset{ˉ}{f} = \overset{ˉ}{g} \overset{ˉ}{f}$ , the last step by [L1]; under these hypotheses the transfer $p_{Y} f ı_{X} = \overset{ˉ}{f} p_{X} ı_{X} = \overset{ˉ}{f}$ is the given $\overset{ˉ}{f}$ , so the computation compares the transfer of the composite with the composite of the transfers. In this situation identity and composition are preserved strictly, not merely up to homotopy.

L1L3algebra

2.1

Conclusion. Step 1.1 shows that transfer sends cochain maps to cochain maps and homotopic maps to homotopic maps, so it is well defined on homotopy classes of cochain maps; step 1.2 shows that it preserves identities strictly and composition up to the explicit homotopy $p_{Z} g h_{Y} f ı_{X}$ , so it is functorial on homotopy classes while not being a strict functor on cochain maps; step 1.3 exhibits cochain maps with $\overset{ˉ}{f} = \overset{ˉ}{g} = 0$ and $\overline{(g f)} = 1$ , which establishes that failure; and step 1.4 gives strict functoriality on the subcategory of morphisms commuting with the chosen retract data. Since the transfer uses the chosen projections and inclusions, while the displayed comparison homotopy also uses the chosen homotopies, clause 4 records that no choice-free, canonical or confluent transfer and no independence of the chosen data is being claimed. ∎

step 1.1step 1.2step 1.3step 1.4L3

5 · Examples, counterexamples and false statements

None yet.

Homological Gaussian Elimination

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Complexes, homotopies and contractibility in an additive category

Definition

An invertible cochain differential block and its candidate reduction

Definition

Triangular basis changes diagonalize an invertible differential block

Statement

Facts & Assumptions

Proof

Gaussian elimination splits a contractible two-term complex

Statement

Facts & Assumptions

Proof

Explicit strong deformation retract from Gaussian cancellation

Statement

Facts & Assumptions

Proof

Gaussian cancellation preserves homotopy type and abelian-category homology

Statement

Facts & Assumptions

Proof

Finite iteration of current invertible-block cancellations

Statement

Facts & Assumptions

Proof

Additive functors preserve chosen Gaussian cancellations

Statement

Facts & Assumptions

Proof

Transferred maps are functorial up to homotopy, with strict naturality limits

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources