TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29

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.

Epimorphy is detected by members

Statement

For a morphism $g : B \to C$ in an abelian category, the following are equivalent:

$g$ is epic.
For every member $z : Z \to C$ , there exists a member $y : Y \to B$ with $g y \equiv z .$

Facts & Assumptions

Given: A morphism $g : B \to C$ .

[L1]

Members modulo equivalence correspond exactly to subobjects (Members modulo equivalence correspond to subobjects).

[L3]

The image is the least subobject through which a morphism factors (The image is the least subobject through which a morphism factors).

[L4]

Pullbacks exist, and pullbacks of epimorphisms are epimorphisms (Pullbacks and pushouts as limits and colimits of cospans and spans, The pullback of an epimorphism is an epimorphism).

Proof

technique · direct

1.1L4assume-hypconstruct

Assume $g$ is epic. For a member $z : Z \to C$ , form the pullback of $z$ along $g$ with projections $y : P \to B$ and $α : P \to Z$ . By [L4], $α$ is epic, and the pullback equation $g y = z α$ shows $g y \equiv z$ .

1.2L1L3assume-hypalgebra

Assume condition 2, and write $g$ as an epi-mono factorization $B \overset{e}{\to} I \overset{m}{\to} C$ . Apply condition 2 to the identity member $1_{C} : C \to C$ . Then there is $y : Y \to B$ with $g y \equiv 1_{C}$ , so [L1] says that $g y$ and $1_{C}$ determine the same whole subobject of $C$ . Since $g y$ factors through $g$ , [L3] gives $[im (g y)] \leq [im (g)] = [⁣ m ⁣]$ , hence $[1_{C}] \leq [⁣ m ⁣]$ , which forces $[⁣ m ⁣] = [1_{C}]$ . Thus $g$ is epic.

2.1step 1.1step 1.2∎

Therefore conditions 1 and 2 are equivalent.

Depends on

Used by

Four lemma in an abelian category Theorem
Nine lemma in an abelian category Theorem
Nine lemma variants by which rows are assumed exact Theorem
Sharp nine lemma Theorem
Weak four lemma with the exactness hypotheses named Theorem

Dependency tree · two levels

19 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

Saunders Mac Lane, Categories for the Working Mathematician, Theorem VIII.4.3(iii) (standard reference, not scraped)