TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30

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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.
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The category of complexes in an abelian category is abelian

Statement

If $A$ is an abelian category, then $Ch (A)$ is an abelian category.

Facts & Assumptions

Given: An abelian category $A$ .

[L1]

An abelian category is an additive category in which every morphism has a kernel and a cokernel, and every coimage-to-image comparison is an isomorphism (Abelian category).

[L2]

$Ch (A)$ is additive (The category of complexes in an additive category is additive).

[L3]

Kernels and cokernels of chain maps are computed degreewise (The kernel of a chain map is computed degreewise, The cokernel of a chain map is computed degreewise).

[L4]

Images, coimages, and the coimage-to-image comparison of a chain map are computed degreewise (Images and coimages of chain maps are computed degreewise).

Proof

technique · direct

1.1

By [L2] and [L3], $Ch (A)$ is additive and every chain map has a kernel and a cokernel.

L2L3

1.2

Let $f : C \to D$ be a chain map. By [L4], the coimage-to-image comparison of $f$ is the family of the coimage-to-image comparisons of the component maps $f_{n}$ . Each of those is an isomorphism by [L1], so the family is an isomorphism of complexes. Therefore the defining clauses of [L1] hold in $Ch (A)$ .

L1L4algebra

2.1

Steps 1.1 and 1.2 prove that $Ch (A)$ is abelian.

L1step 1.1step 1.2∎

Depends on

Used by

A sequence of chain maps is exact exactly when it is exact degreewise Corollary
A subcomplex is the kernel of its quotient map Proposition

Dependency tree · two levels

11 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)
The Stacks Project, Lemma 12.13.3 (standard reference, not scraped)