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Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeNov 4th 2013

at total category I have added after the definition and after the first remark these two further remarks:

+– {: .num_remark}

###### Remark

Since the Yoneda embedding is a full and faithful functor, a total category $C$ induces an idempotent monad $Y \circ L$ on its category of presheaves, hence a modality. One says that $C$ is a totally distributive category if this modality is itself the right adjoint of an adjoint modality.

=–

+– {: .num_remark}

###### Remark

The $(L \dashv Y)$-adjunction of a total category is closely related to the $(\mathcal{O} \dashv Spec)$-adjunction discussed at Isbell duality and at function algebras on ∞-stacks. In that context the $L Y$-modality deserves to be called the affine modality.

=–

• CommentRowNumber2.
• CommentAuthorTim Campion
• CommentTimeDec 21st 2015

The article currently says something to the effect of “cototal categories are more rare than total categories”. But it occurs to me that $Top$ is cototal by Day’s criterion (it’s complete, mono-complete, and has a cogenerator given by the indiscrete space on two elements). In fact, since being a topological functor is self-dual, and since $Set$ is cototal, any category which is topological over $Set$ is cototal – I’ll add this to the article as a class of examples. I don’t know of a reason to expect categories of a more “algebraic” nature to be cototal, but at least this suggests that many categories of “spaces” might be cototal.

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeDec 21st 2015
• (edited Dec 21st 2015)

Yes, good observation.

It’s known for example that $Grp$ is not cototal, and neither is say the category of commutative rings $CRing$. An easy way to see this is to produce continuous functors $C \to Set$ that are not representable, e.g., for $C = Grp$, the classical example is the class-indexed product of representables $hom(G,-)$ where $G$ ranges over all simple groups. (For any group $H$, $Hom(G, H)$ will be trivial once the simple group $G$ has cardinality greater than $H$, so the product of $Hom(G, H)$ over all simple $G$ will still be a set.) A similar example can be cooked up for commutative rings; see e.g. this MO answer. I guess algebraic categories with a plentiful supply of simple objects would be amenable to similar constructions.

• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeJun 11th 2017

I added some more examples to total category. (One is that Ab is cototal as well as total.)

• CommentRowNumber5.
• CommentAuthorMarc
• CommentTimeJun 27th 2018

corrected year of Ross Street’s publication and inserted a link to the article at the AMS journal website