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TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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Under dependent choice, a finitary monad on a complete cocomplete locally small category has complete and cocomplete algebras

Statement

Assume dependent choice. If T is a finitary monad on a complete, cocomplete, locally small category C, then its Eilenberg–Moore category CT is complete and cocomplete.

The completeness conclusion itself uses no choice; dependent choice enters only in the construction of coequalizers used for cocompleteness.

Facts & Assumptions

Given: Dependent choice and a finitary monad T on a complete cocomplete locally small category C.

[L1]

The Eilenberg–Moore forgetful functor strictly creates every limit that exists in the base (The Eilenberg–Moore forgetful functor strictly creates every limit that exists in the base).

[L2]

Under dependent choice, algebras for such a finitary monad have coequalizers (Under dependent choice, algebras for a finitary monad on a complete cocomplete locally small category have coequalizers).

[L3]

Over a cocomplete base, a monadic category is cocomplete exactly when it has coequalizers (Over a cocomplete base, a monadic category is cocomplete exactly when it has coequalizers).

[L4]

The Eilenberg–Moore adjunction induces the given monad on the nose (The free–forgetful Eilenberg–Moore adjunction induces the given monad).

Proof

technique · direct
1.1L1

Since C has every small limit, [L1] creates every such limit in CT, including the empty limit. Thus CT is complete without using dependent choice.

1.2L2

Under the stated dependent-choice hypothesis, [L2] gives every coequalizer in CT, including equal parallel maps and the zero-stage case of its construction.

2.1step 1.2L3L4

By [L4], the Eilenberg–Moore adjunction induces T on the nose, so its forgetful functor is monadic. Applying [L3] to the cocomplete base and step 1.2 makes CT cocomplete, including the empty colimit.

3.1step 1.1step 2.1∎

Combining step 1.1 with step 2.1 proves that CT is complete and cocomplete.

Depends on

Used by

Dependency tree · two levels

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Sources