PropositionStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 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.

Each object has a zero member and each member has a negative

Statement

For every object $A$ of an abelian category:

there is a zero member of $A$ ;
every member $x : X \to A$ has a negative member $- x : X \to A$ ;
for every member $x$ , one has $x \equiv 0$ if and only if $x = 0$ as a morphism.

Facts & Assumptions

Given: An abelian category and a member $x : X \to A$ .

[L1]

A zero object supplies zero morphisms between any two objects (A zero object supplies a unique compatible system of zero morphisms).

[L2]

An abelian category is additive, so every hom-set has negatives (Abelian category).

[L3]

Member equivalence is defined by comparison after epimorphisms (Equivalence of members).

Proof

technique · direct

1.1L1L2

By [L1], there is a zero morphism $0_{X, A} : X \to A$ , so $A$ has a zero member. By [L2], the additive inverse $- x : X \to A$ exists, so every member has a negative.

1.2L3assume-hypalgebra

Conversely, if $x \equiv 0$ , choose epimorphisms $u : W \to X$ and $v : W \to Y$ witnessing that equivalence. Then $x u = 0$ , and epicity of $u$ forces $x = 0$ .

2.1L3step 1.1

If $x = 0$ , then $x \equiv 0$ is witnessed by the identity epic $1_{X} : X \to X$ , since $x 1_{X} = 01_{X}$ .

3.1step 1.1step 2.1step 1.2∎

Step 1.1 proves claims 1 and 2, while steps 2.1 and 1.2 prove claim 3.

Depends on

Dependency tree · 2 levels

Sources

Saunders Mac Lane, Categories for the Working Mathematician, VIII.4