Implication Details
Assumptions: Cauchy complete, essentially countable, locally finite
Conclusions: ℵ₁-accessible
Proof: The proof is similar to Thm. 2.2.2 in Makkai-Pare (which only establishes -accessibility) and to MO/509853 (which concerns -accessibility in the finite case).
Let be a category that is locally finite, essentially countable, and Cauchy complete. By replacing with an equivalent category if necessary, we may assume that it is countable and that all Hom-sets are finite sets. Let be an -filtered category and let be a diagram. We will prove that the colimit of exists and is a retract of some , in fact for arbitrarily large . (Note that one cannot in general prove that it is isomorphic to some ; this is false.) It follows that -filtered colimits exist and that every object of is -presentable.
Let be a fixed object. Then is a diagram of finite sets. Regard it as a diagram in and let be its colimit. Thus, for every we have maps that are jointly surjective, and two elements in and become equal in iff there exists a cospan such that these elements map to the same element in .
We claim that is finite. Otherwise, there would exist infinitely many pairwise distinct elements in . Choose objects such that has a preimage in . Since is -filtered, there exists a cocone in . We then obtain preimages in , which must be pairwise distinct because are pairwise distinct. This contradicts the finiteness of . Therefore, is finite.
For every element of the finite set , choose a preimage in some . Since is filtered, there exists that works for all elements simultaneously. Equivalently, the canonical map is surjective.
Now consider the free cocompletion and the colimit of the diagram . Since colimits in are computed objectwise, We have shown above that is finite and that there exists some such that the canonical map is surjective. Since is countable and is -filtered, there exists a cocone in . It follows that is surjective for all . Equivalently, is an epimorphism in . For every morphism , the map is likewise an epimorphism. We will now show that some is in fact a split epimorphism in .
For every , since is a surjective map of finite sets, there exists a map such that . The issue is that need not be natural in . To analyze this, let be a morphism in . Then we have a diagram of finite sets: The outer rectangle commutes, and the square on the right commutes. Hence, the square on the left commutes after composition with . Since is finite, it follows that there exists some morphism such that the left square commutes after composition with A priori, the index depends on the morphism . However, since is -filtered and has only countably many morphisms, we may choose independently of . Thus, by construction, is natural in . Moreover, Therefore, is a split epimorphism in . Since is Cauchy complete, it follows that is representable, represented by a retract of . Since is fully faithful, this object is a colimit of .