TITLE: Is every saft category cocomplete?
QUESTION [14 upvotes]: 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?

REPLY [5 votes]: 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.