Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-27
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An additive category with kernels is idempotent complete

Statement

Every additive category in which every morphism has a kernel is idempotent complete.

Facts & Assumptions

Given: An additive category C and an idempotent e:AA.

[L1]

In an additive category, hom-sets admit subtraction and there is a zero object (Additive category).

[L2]

On a biproduct, the injection-projection maps satisfy the identity-sum relation (On a biproduct, the injections and projections satisfy the identity-sum relation).

[L3]

The criterion "monic iff kernel zero" holds in a preadditive category with a zero object (In a preadditive category with a zero object, a morphism is monic exactly when its kernel is zero).

[L4]

Idempotent completeness means that every idempotent splits (Idempotent complete category, Idempotent and split idempotent).

Proof

technique · direct
1.1

Let k:KA be the kernel of 1Ae, which exists by hypothesis. Since (1Ae)e=0, the kernel universal property gives a morphism p:AK with kp=e.

L1L4construct
2.1

Because (1Ae)k=0, one has ek=k. Hence kpk=ek=k. The morphisms pk and 1K both satisfy k()=k, so the kernel universal property forces pk=1K. Therefore kp=e and pk=1K, so e splits through K.

L1step 1.1
3.1

Thus every idempotent in C splits. By [L4], the category is idempotent complete. The additive setting is what makes 1Ae meaningful; [L2] and [L3] record the surrounding zero-object calculus on which that subtraction sits.

L2L3L4step 2.1

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