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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 and an idempotent .
In an additive category, hom-sets admit subtraction and there is a zero object (Additive category).
On a biproduct, the injection-projection maps satisfy the identity-sum relation (On a biproduct, the injections and projections satisfy the identity-sum relation).
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).
Idempotent completeness means that every idempotent splits (Idempotent complete category, Idempotent and split idempotent).
Proof
Let be the kernel of , which exists by hypothesis. Since , the kernel universal property gives a morphism with .
Because , one has . Hence . The morphisms and both satisfy , so the kernel universal property forces . Therefore and , so splits through .
Thus every idempotent in splits. By [L4], the category is idempotent complete. The additive setting is what makes meaningful; [L2] and [L3] record the surrounding zero-object calculus on which that subtraction sits.
Depends on
Used by
- FALSE: every idempotent splits False statement
Dependency tree · two levels
16 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
- The Stacks Project, Section 12.3, Lemma 12.3.16 (standard reference, not scraped)
- Dixy Msapato, The Karoubi envelope and weak idempotent completion of an extriangulated category, Proposition 2.2 (standard reference, not scraped)