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TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-27
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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→A with fe=ge (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
1.1L1

For any morphism e:E→A, bilinearity in [L1] gives (f−g)e=fe−ge. Therefore fe=ge if and only if (f−g)e=0.

2.1L2step 1.1

By [L2], the universal property of an equalizer of f and g says exactly that every h:X→A with fh=gh 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.

3.1L3step 2.1∎

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.

Depends on

Used by

Dependency tree · two levels

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Sources