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

1.1

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 $g + g^{'}$ .

L1L3givenalgebra

2.1

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 $u g v$ .

L1L2step 1.1algebra∎

Depends on

Used by

Composition of homotopy classes is well defined Lemma
Homotopy equivalence is an equivalence relation on complexes Proposition
Null-homotopic maps form a two-sided additive ideal Proposition

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra (standard reference, not scraped)