Is every saft category cocomplete? - MathOverflow

Is every saft category cocomplete?

2010-12-12T21:17:53Z

Here is a word that I think should be adopted by the category theorists. (If there is another synonymous word already in existence, please let me know.)

Definition: A category $C$ is saft if every cocontinuous functor $C \to D$ has a right adjoint.

The word "saft" is an abbreviation for "special adjoint functor theorem (SAFT)", of which there are many, because SAFTs always take the form "If a category $C$ is XYZ, then it is saft." For example: it suffices for a category to be some cocompletion of some small subcategory (F. Ulmer, The adjoint functor theorem and the Yoneda embedding, Illinois Journal of Mathematics, 1971 vol. 15 (3) pp. 355-361).

It is an easy exercise that a category $C$ is saft if and only if every continuous functor $C^{\rm op} \to \text{Set}$ is representable. In particular any saft category is complete, because any putative limit corresponds to a continuous functor, and hence necessarily representable.

On the other hand, I don't see any particular reason why a saft category is necessarily cocomplete, except that every example I know is, and every SAFT uses cocompleteness as one of the conditions.

Question: Is a saft category necessarily cocomplete?

Answer by Buschi Sergio for Is every saft category cocomplete?

2010-12-12T23:34:12Z

Let $I$ a small category: then $\Delta: C \to C^I$ is a cocontinuous funtor (a colimit of costant funcror is a punctual colimit), then a right adjoint is the limit functor, then $C$ has the limits.

If this condition were true also for general categories $I$ then $C$ has large limits, and then is a (large) complete and cocomplete preorder.

If $P: I\to C$, $I$ small, and the category $C(P)$ of cocones from $P$ to some objet $X\in C$ is little then the limit of the natural (vertex proiection) $C(P)\to C$ is the colimit of $P$.

The condition: $P: C^{op}\to Set$ is representable, is equivalent to the condition: $P$ has a left adjoint .

If exixst a $P: C^{op}\to Set$ preserving limits but not the $\lambda$-directed (filtrant) colimits for any regular cardinal $\lambda$ then $C$ cannot be locally presentable (seeT.1.66 p.52 in Adámek and Rosicky, Locally Presentable and Accessible Categories Cambridge University Press, Cambridge, (1994))

Answer by Todd Trimble for Is every saft category cocomplete?

2010-12-13T01:14:58Z

Theo, your question is in the neighborhood of what is called "total cocompleteness" or "totality". A category $C$ is total if the Yoneda embedding $y: C \to Set^{C^{op}}$ has a left adjoint. There is a similar notion of totality in the enriched case.

Total categories have this "saft" property you are discussing: every cocontinuous functor from a total category to a locally small category has a right adjoint.

Total categories were studied by Max Kelly in this paper. Kelly discusses this "saft" condition, but calls it "compact": a category $C$ is by definition compact if $C^{op} \to Set$ is representable precisely when it preserves those limits which exist in $C^{op}$.

In theorem 5.6 (page 15 of 25), Kelly gives a string of implications, including this "total implies compact". However, on the following page he refers to an example to the effect that compact categories need not have coequalizers, which would answer your question in the negative. I unfortunately do not have access to the paper of Adamek where the example is treated in full (and I do not know the example), but the compact category in question is the category of algebras of some monad on the category of graphs.

Hope this is somewhat helpful. If Mike Shulman or Steve Lack see this, they may have more to say.

Edit: I may as well add a few comments on total categories which may come in handy (and which may help explain why every example Theo thought of was actually cocomplete).

Every category that is monadic over $Set$ is total (no rank condition needed!).
Presheaf categories are total.
Every reflective (full) subcategory of a total category is total.
As a corollary of the last two results, every locally presentable category (the category of models of a small limit sketch) is total.
Every category which is topological over $Set$ is total.

The last result is given in a paper by Tholen, who remarks that the notions of compact/saft, cocompact, total, and cototal all coincide if the given category has a set of generators and a set of cogenerators.

See also the nLab article on total category.

Answer by Mike Shulman for Is every saft category cocomplete?

2010-12-13T20:25:13Z

This is just an extension of Todd's answer to summarize Adamek's example, which is a bit convoluted. By a graph we mean a set equipped with a binary relation, call it ∼. If A is a graph, let $A^{(3)}$ be the set of triples (x,y,z) such that x∼y∼z. And let F(A) be the power set $P(A^{(3)})$, equipped with the relation defined by ∅∼X for all nonempty X. Adamek's category is the category of algebras for the endofunctor F of the category of graphs, i.e. of graphs A equipped with a graph-morphism F(A)→A. He proves that the forgetful functor from this category to Graphs has a left adjoint, namely $A\mapsto A \sqcup F(A)$, and is monadic. But he gives the following example of a pair of parallel morphisms of F-algebras that have no coequalizer.

Let A be the set {p,q} with the empty relation ∼, and let B be the set {s,t} with s∼t only. Then F(A) and F(B) are both the graph {∅} with the empty relation, and we make A and B into F-algebras by sending ∅ to p and s, respectively. Now let f:A→B send p to s and q to t, while g:A→B sends p and q both to s. Adamek goes on to prove that if f and g had a coequalizer in F-algebras, then one could construct from this a weakly initial P-algebra, where P is the powerset endofunctor on Set; from this one could then construct an initial P-algebra, hence a fixed point of P, which contradicts Cantor's diagonal argument.

I don't have time to summarize the construction of the weakly initial P-algebra from a coequalizer of f and g, but I'm making this CW, so anyone else who wants to add it, feel free.