LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Chain homotopy is compatible with addition and composition

Statement

Let $B_{∙}, C_{∙}, D_{∙}, E_{∙}$ be chain complexes in an abelian category $A$ , and let $f, g : C_{∙} \to D_{∙}$ be chain maps with $f ≃ g$ .

If $f^{'}, g^{'} : C_{∙} \to D_{∙}$ are chain maps with $f^{'} ≃ g^{'}$ , then $f + f^{'} ≃ g + g^{'}$ .
If $v : B_{∙} \to C_{∙}$ and $u : D_{∙} \to E_{∙}$ are chain maps, then $u f v ≃ u g v$ .

Facts & Assumptions

Given: An abelian category $A$ , a chain homotopy $s : f ≃ g$ , a chain homotopy $t : f^{'} ≃ g^{'}$ , and composable chain maps $v : B_{∙} \to C_{∙}$ , $u : D_{∙} \to E_{∙}$ between complexes in $A$ .

[L1]

A chain homotopy is a degree- $1$ family satisfying $f - g = d s + s d$ (A chain homotopy).

[L2]

Identities and composites of chain maps are chain maps (Identities and composites of chain maps are chain maps).

[L3]

Because an abelian category is additive, its category of chain complexes is additive, so sums of parallel chain maps are defined degreewise (The category of complexes in an additive category is additive).

Proof

technique · direct

1.1L1L3givenalgebra

By [L1], we have $f - g = d s + s d$ and $f^{'} - g^{'} = d t + t d$ . Using [L3], add these equalities to obtain $(f + f^{'}) - (g + g^{'}) = d (s + t) + (s + t) d,$ so $s + t$ is a homotopy from $f + f^{'}$ to $g + g^{'}$ .

2.1L1L2step 1.1algebra∎

Since $u$ and $v$ are chain maps by [L2], their differentials commute in the usual way. Therefore $u f v - u g v = u (f - g) v = u (d s + s d) v = d (u s v) + (u s v) d,$ so $u s v$ is a chain homotopy from $u f v$ to $u g v$ .

Depends on

Dependency tree · 2 levels

Sources

Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra