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

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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

1.1L1L2

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

1.2L3L4algebra

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.

2.1step 1.1step 1.2∎

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

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)