TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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.

The cancellation and epimorphism descriptions of a generator agree

Statement

Let $A$ be a locally small abelian category satisfying AB3, and let $G$ be an object of $A$ . Then the following are equivalent:

$G$ is a generator.
The representable functor $A (G, -)$ is faithful.
For every object $A$ , the canonical morphism $∐_{u \in A (G, A)} G ⟶ A$ is an epimorphism.

Facts & Assumptions

Given: A locally small abelian category $A$ satisfying AB3 and an object $G$ .

[L1]

A generator is exactly a one-object separating set (Generator and cogenerator of a category).

[L2]

AB3 supplies the small coproducts indexed by hom-sets (The axioms AB3 and AB3*).

[L3]

In a locally small category, a separating set is equivalently a jointly faithful family of representables (In a locally small category, separating and coseparating sets are equivalently jointly faithful families of representables).

Proof

technique · direct

1.1L1L3

By [L1] and [L3], condition 1 is equivalent to condition 2: the singleton ${G}$ is separating exactly when the one-member family $A (G, -)$ is faithful.

1.2L2L3construct

Assume condition 2, and fix an object $A$ . By [L2], form the canonical map $e_{A} : ∐_{u \in A (G, A)} G \to A$ whose $u$ -th coproduct injection is sent to $u$ . Let $q : A \to Q$ be its cokernel. If $q \neq 0$ , faithfulness of $A (G, -)$ gives some $u : G \to A$ with $q u \neq 0$ . But $u$ is one of the coproduct components of $e_{A}$ , so $q u = 0$ because $q e_{A} = 0$ , a contradiction. Hence $q = 0$ , and therefore $e_{A}$ is epic. So condition 2 implies condition 3.

2.1L1step 1.2algebra

Assume condition 3. If $f, g : X \to Y$ are distinct, then $h : = f - g \neq 0$ . Apply condition 3 to $X$ : if $h u = 0$ for every $u : G \to X$ , then $h e_{X} = 0$ , and since $e_{X}$ is epic that would force $h = 0$ . So some $u : G \to X$ satisfies $h u \neq 0$ , equivalently $f u \neq g u$ . Thus $G$ separates maps, hence is a generator by [L1]. Therefore condition 3 implies condition 1.

3.1step 1.1step 1.2step 2.1∎

Steps 1.1, 1.2, and 2.1 prove the equivalence of the three descriptions.

Depends on

Used by

Dependency tree · two levels

9 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

Alexandre Grothendieck, Sur quelques points d'algèbre homologique, Barr translation, Proposition 1.9.1 (standard reference, not scraped)
Carlos E. Parra, Manuel Saorin, and Simone Virili, Section 2.1 and Lemma 2.7 (standard reference, not scraped)