TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The subobjects of an object in an abelian category form a lattice

Statement

For every object $A$ of an abelian category, the subobjects of $A$ form a bounded lattice. The meet is pullback, the join is the image construction of The join of two subobjects in an abelian category, the bottom element is the zero subobject, and the top element is $1_{A}$ .

Facts & Assumptions

Given: An object $A$ in an abelian category.

[L1]

Any two subobjects of $A$ admit a least upper bound (The join of two subobjects is their least upper bound).

[L2]

Any two subobjects of $A$ admit a greatest lower bound (The meet of two subobjects is their pullback).

[L3]

Subobjects are mutual-factorization classes of monomorphisms into $A$ (Subobject and quotient object as mutual-factorisation classes of monomorphisms and epimorphisms).

[L4]

An abelian category has a zero object, hence zero morphisms (Abelian category).

Proof

technique · direct

1.1

By [L1] and [L2], every pair of subobjects of $A$ has a join and a meet. That gives the binary lattice operations.

L1L2

1.2

By [L4], the unique map $0 \to A$ exists. It is monic because every two maps into $0$ are equal, so by [L3] it represents a subobject of $A$ . Every monomorphism into $A$ factors through $1_{A}$ , and $0 \to A$ factors through every monomorphism into $A$ , so these classes are respectively the top and bottom elements.

L3L4algebra

2.1

Steps 1.1 and 1.2 prove that the subobjects of $A$ form a bounded lattice.

step 1.1step 1.2∎

Depends on

Used by

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

Daniel Murfet, Abelian Categories, Section 4.2 (standard reference, not scraped)