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

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A subcomplex is the kernel of its quotient map

Statement

If $S_{∙} ↪ C_{∙}$ is a subcomplex, then the quotient map $C_{∙} \to C_{∙} / S_{∙}$ has kernel $S_{∙}$ . Conversely, every monomorphism of complexes identifies its source, up to unique isomorphism, with a subcomplex of its target.

Facts & Assumptions

Given: A subcomplex $S_{∙} ↪ C_{∙}$ and a monomorphism $m : A_{∙} \to C_{∙}$ .

[L1]

The quotient complex $C / S$ is defined degreewise from the quotients $C_{n} / S_{n}$ (Quotient complex).

[L2]

Kernels of chain maps are computed degreewise (The kernel of a chain map is computed degreewise).

[L3]

In $Ch (A)$ , monomorphisms are morphisms in an abelian category (The category of complexes in an abelian category is abelian).

[L4]

In an abelian category, a morphism is monic exactly when its kernel is zero (In an abelian category, monic means zero kernel and epic means zero cokernel).

Proof

technique · direct

1.1

Let $q : C \to C / S$ be the quotient map. By [L1], each component $q_{n} : C_{n} \to C_{n} / S_{n}$ has kernel represented by $S_{n} ↪ C_{n}$ . Then [L2] says the kernel complex of $q$ is exactly $S_{∙}$ .

L1L2

2.1

Now let $m : A \to C$ be monic. Its kernel in $Ch (A)$ is therefore the zero complex. By [L2], that kernel is assembled degreewise from the kernels of the component maps $m_{n}$ , so each $\ker (m_{n})$ is zero. Then [L4] makes every $m_{n}$ monic in $A$ . Since $m$ is a chain map, the commuting squares $m_{n - 1} d_{n}^{A} = d_{n}^{C} m_{n}$ show that the family of monomorphisms $m_{n} : A_{n} ↪ C_{n}$ is stable under the differentials. Hence these degreewise monomorphisms exhibit $A_{∙}$ as a subcomplex of $C_{∙}$ , unique up to the usual unique isomorphism of represented subobjects.

L2L3L4givenalgebra∎

Depends on

Used by

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Dependency tree · two levels

16 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

The Stacks Project, Lemma 12.13.3 (standard reference, not scraped)
Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra (standard reference, not scraped)