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TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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The meet of two subobjects is their pullback

Statement

Let B and C be subobjects of an object A in an abelian category, represented by monomorphisms b:BA and c:CA. Then the meet BC in the subobject order is represented by the pullback of b and c.

Facts & Assumptions

Given: Monomorphisms b:BA and c:CA.

[L1]

Pullbacks are defined by their commutative square and universal property (Pullbacks and pushouts as limits and colimits of cospans and spans).

[L3]

Subobject inequalities are factorization relations between representatives (Subobjects and quotient objects form oppositely oriented partially ordered collections).

Proof

technique · direct
1.1

Form a pullback square tikzcd P \arrow[r, "q"] \arrow[d, "p"'] & C \arrow[d, "c"] \\ B \arrow[r, "b"'] & A. By [L2], the pullback leg p is monic, so the composite m:=bp=cq:PA is monic as well. Because m factors through both b and c, it is a lower bound of the two subobjects.

L1L2L3
1.2

Let n:NA be any lower bound. Then n=bu=cv for some u:NB and v:NC. By the pullback universal property [L1], there is a unique w:NP with pw=u and qw=v. Therefore mw=bu=n, so n factors through m. Hence nm.

L1L3construct
2.1

Steps 1.1 and 1.2 show that m is a lower bound above every other lower bound. By [L3], the pullback subobject is exactly BC.

L3step 1.1step 1.2

Depends on

Used by

Dependency tree · two levels

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Sources