TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-27

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.

In a preadditive category, the equalizer of a parallel pair is the kernel of their difference

Statement

Let $f, g : A ⇉ B$ be morphisms in a preadditive category. Then an equalizer of $f$ and $g$ is exactly a kernel of the difference $f - g$ , and conversely.

Facts & Assumptions

Given: Parallel morphisms $f, g : A ⇉ B$ in a preadditive category.

[L1]

In a preadditive category each hom-set is an abelian group and composition is bilinear (Preadditive category).

[L2]

An equalizer of $f$ and $g$ is a universal morphism $e : E \to A$ with $f e = g e$ (Equalizers and coequalizers as limits and colimits of a parallel pair).

[L3]

A kernel of a morphism $h$ is an equalizer of $h$ and the zero morphism (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

Proof

technique · direct

1.1

For any morphism $e : E \to A$ , bilinearity in [L1] gives $(f - g) e = f e - g e$ . Therefore $f e = g e$ if and only if $(f - g) e = 0$ .

2.1

By [L2], the universal property of an equalizer of $f$ and $g$ says exactly that every $h : X \to A$ with $f h = g h$ factors uniquely through $e$ . By step 1.1 this is the same as asking that every $h$ with $(f - g) h = 0$ factor uniquely through $e$ .

L2step 1.1

3.1

The condition in step 2.1 is precisely the kernel universal property from [L3]. So equalizers of $f$ and $g$ and kernels of $f - g$ are the same data.

L3step 2.1∎

Depends on

Used by

Dependency tree · two levels

6 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

Saunders Mac Lane, Categories for the Working Mathematician, VIII.2 (standard reference, not scraped)