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CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27
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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 Xˉ∙ be the candidate reduction at that pivot, and let p:X∙→Xˉ∙,ı:Xˉ∙→X∙,h be the explicit cochain maps and contracting homotopy of Explicit strong deformation retract from Gaussian cancellation, so that pı=1Xˉ∙, 1X∙−ıp=dh+hd and ph=0=hı=h2.

  1. Homotopy type. Under the reindexing dictionary Cn:=C−n of Complexes, homotopies and contractibility in an additive category, the complexes X∙ and Xˉ∙ become chain complexes X∙ and Xˉ∙ 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 Xˉ∙ are isomorphic in the homotopy category, over every additive category A.
  2. 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, Hn(ı)Hn(p)=1Hn(X),Hn(p)Hn(ı)=1Hn(Xˉ), 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 Xˉ∙ in degree −n.
  3. What is not claimed. No identification of X∙ with Xˉ∙ 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 Xˉ∙⊕K with the contractible two-term complex K, and the objects of X∙ and Xˉ∙ 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 Xˉ∙, 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 pı=1Xˉ∙ and 1X∙−ıp=dh+hd, with h of degree −1 and ph=0, hı=0, h2=0 (Explicit strong deformation retract from Gaussian cancellation).

[L2]

Reindexing Cn:=C−n, dn:=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 sn:=h−n of degree +1 with fn−gn=dn+1Dsn+sn−1dnC; 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 Zn(C)=ker⁡(dn) with inclusions kn, boundary subobjects Bn(C)=im⁡(dn+1), the factorization dn+1=knβnen through the boundary-to-cycle map βn:Bn(C)→Zn(C), and homology objects Hn(C)=coker⁡(βn) with quotient qn:Zn(C)→Hn(C); a chain map u has a unique induced Hn(u) characterized by Hn(u)qnC=qnDZn(u), where Zn(u) is the cycle map carried by u; and Hn is additive, so Hn(1)=1 and Hn(u+v)=Hn(u)+Hn(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 kn is a kernel and hence a monomorphism, and the homology quotient qn 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∙→Xˉ∙⊕K has components Tn=R−1 and Tn+1=L in degrees n,n+1 and identities elsewhere, so it is not an isomorphism of X∙ with Xˉ∙ itself, and the objects in degrees n,n+1 are A⊕U,B⊕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], pı=1Xˉ∙ and ıp=1X∙−(dh+hd). Under the reindexing [L2] these are chain maps p∙,ı∙ between X∙ and Xˉ∙ with p∙ı∙=1 and ı∙p∙=1X∙−w, where wn:=dn+1sn+sn−1dn and sn:=h−n. The family s exhibits w≃0, so w is null-homotopic and 1X∙−ı∙p∙=w is null-homotopic; by [L3] therefore [ı∙][p∙]=[ı∙p∙]=[1X∙] and [p∙][ı∙]=[p∙ı∙]=[1Xˉ∙], 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∙=1Xˉ∙ strictly by [L1]. Functoriality and additivity of Hn in [L4] give Hn(p)Hn(ı)=Hn(p∙ı∙)=Hn(1Xˉ∙)=1Hn(Xˉ).

L1L2L4algebra
1.3

Cycle-level computation. Let kn:Zn(X)→Xn be the cycle inclusion, so dnkn=0, and let dn+1=knβnen be the factorization of [L4]. Then wnkn=dn+1snkn+sn−1dnkn=knβn(ensnkn), because the second summand vanishes and the first is dn+1(snkn). As a difference of the chain maps 1X∙ and ı∙p∙, the map w is a chain map, so [L4] gives a cycle map Zn(w) with knZn(w)=wnkn=knβn(ensnkn); by [L5] the inclusion kn is monic, so Zn(w)=βn(ensnkn).

L4L5algebra
2.1

The homotopy term induces zero. Applying the homology quotient qn:Zn(X)→Hn(X) to step 1.3 gives qnZn(w)=qnβn(ensnkn)=0, since qn is the cokernel of βn by [L4]. The characterizing property Hn(w)qn=qnZn(w) of [L4] therefore gives Hn(w)qn=0, and qn is epic by [L5], so Hn(w)=0 for every n.

L4L5step 1.3algebra
3.1

Clause 2, second composite. By [L1] and [L2], ı∙p∙=1X∙−w with w as in step 1.1, so functoriality and additivity of Hn in [L4] give Hn(ı)Hn(p)=Hn(1X∙)+Hn(−w)=1Hn(X)−Hn(w), which equals 1Hn(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 Hn(p) and Hn(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 Xˉ∙ and of the chain isomorphism onto Xˉ∙⊕K, so no equality or canonical identification of the complexes X∙ and Xˉ∙ is being asserted. ∎

step 1.1step 1.2step 2.1step 3.1L6

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