Transferring "abelian-ness" across a functor #
If C is an additive category, D is an abelian category,
we have F : C ⥤ D G : D ⥤ C (both preserving zero morphisms),
G is left exact (that is, preserves finite limits),
and further we have adj : G ⊣ F and i : F ⋙ G ≅ 𝟭 C,
then C is also abelian.
See
Notes #
The hypotheses, following the statement from the Stacks project,
may appear surprising: we don't ask that the counit of the adjunction is an isomorphism,
but just that we have some potentially unrelated isomorphism i : F ⋙ G ≅ 𝟭 C.
However Lemma A1.1.1 from [Elephant] shows that in this situation the counit itself
must be an isomorphism, and thus that C is a reflective subcategory of D.
Someone may like to formalize that lemma, and restate this theorem in terms of Reflective.
(That lemma has a nice string diagrammatic proof that holds in any bicategory.)
No point making this an instance, as it requires i.
No point making this an instance, as it requires i and adj.
Auxiliary construction for coimageIsoImage
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Auxiliary construction for coimageIsoImage
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Auxiliary definition: the abelian coimage and abelian image agree. We still need to check that this agrees with the canonical morphism.
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If C is an additive category, D is an abelian category,
we have F : C ⥤ D G : D ⥤ C (both preserving zero morphisms),
G is left exact (that is, preserves finite limits),
and further we have adj : G ⊣ F and i : F ⋙ G ≅ 𝟭 C,
then C is also abelian.
See
Equations
- CategoryTheory.abelianOfAdjunction F G i adj = let_fun this := ⋯; CategoryTheory.Abelian.ofCoimageImageComparisonIsIso
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If C is an additive category equivalent to an abelian category D
via a functor that preserves zero morphisms,
then C is also abelian.
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- One or more equations did not get rendered due to their size.