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

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In a preadditive category, the coequalizer of a parallel pair is the cokernel of their difference

Statement

Let $f, g : A ⇉ B$ be morphisms in a preadditive category. Then a coequalizer of $f$ and $g$ is exactly a cokernel of $f - g$ , and conversely.

Facts & Assumptions

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

[L1]

In a preadditive category, equalizers are kernels of differences (In a preadditive category, the equalizer of a parallel pair is the kernel of their difference).

[L2]

The opposite of a preadditive category is preadditive (The opposite of a preadditive category is preadditive).

Proof

technique · direct

1.1

In the opposite category, the pair $f, g : A ⇉ B$ becomes a parallel pair $f^{o p}, g^{o p} : B ⇉ A$ . By [L2], that opposite category is again preadditive.

2.1

Applying [L1] there says that an equalizer of $f^{o p}$ and $g^{o p}$ is a kernel of $f^{o p} - g^{o p}$ . Translating back to the original category exchanges equalizers with coequalizers and kernels with cokernels.

L1step 1.1

3.1

Therefore a coequalizer of $f$ and $g$ is exactly a cokernel of $f - g$ .

step 2.1∎

Depends on

Used by

An additive category with all kernels and cokernels has all finite limits and colimits Theorem

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)