Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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A square with monic legs is a pullback exactly when it identifies the source with the intersection subobject

Statement

Consider a commutative square in an abelian category

XBCDmn

whose right and bottom legs are monomorphisms. Let I→D be the intersection of the subobjects represented by m and n. Then the square is a pullback if and only if the induced morphism X→I is an isomorphism.

Facts & Assumptions

Given: The displayed commutative square with monic right and bottom legs.

[L1]

In an abelian category, pullbacks of cospans exist and are computed by the construction of A pullback is the kernel of the difference of the two legs, and dually for pushouts.

Proof

technique · direct
1.1L1L2

By [L1], the pullback P→D of the two monics m and n exists. Because the square defining P is a common lower bound of [m] and [n], and every other common lower bound factors uniquely through that pullback, [L2] says that P→D represents their intersection subobject.

2.1L2step 1.1

If the displayed square is a pullback, then its source X→D is another representative of the same greatest lower bound from step 1.1. Therefore the induced morphism X→I is an isomorphism.

3.1step 1.1step 2.1∎

Conversely, if the induced map X→I is an isomorphism, then composing the pullback square representing I from step 1.1 with that isomorphism yields the displayed square. Pullbackness is invariant under replacing the corner object by an isomorphic one, so the displayed square is a pullback.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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