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TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-27
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An additive functor is left exact exactly when it preserves kernels

Statement

Let F:C→D be an additive functor between additive categories. Then F is left exact if and only if it preserves kernels.

Facts & Assumptions

Given: An additive functor F:C→D.

[L1]

Additive categories are preadditive with finite biproducts (Additive category).

[L2]

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).

[L3]

An additive functor preserves finite biproducts, hence finite products (An additive functor preserves finite biproducts).

[L4]

Proof

technique · direct
1.1L4

If F is left exact, then it preserves all finite limits by definition, so in particular it preserves kernels because a kernel is a finite limit.

1.2L1L2L3

Conversely, assume F preserves kernels. By [L1] the source and target are preadditive, and by [L3] the functor preserves finite products. If e:E→A equalizes f,g:A⇉B, then [L2] identifies e as a kernel of f−g. Since F is additive, F(f−g)=Ff−Fg, so the image of e is a kernel of Ff−Fg, hence again an equalizer of Ff and Fg by [L2]. Therefore F preserves equalizers.

2.1L4step 1.2

Now [L4] applied to step 1.2 shows that F preserves all finite limits. So F is left exact.

3.1step 1.1step 2.1∎

Thus left exactness and kernel preservation are equivalent for additive functors.

Depends on

Used by

Dependency tree · two levels

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Sources