Free cocompletion of thin categories

Question

Let $C$ be a thin category. Is there a cocomplete thin category $C\to C'$ such that for every functor $C\to D$ into a cocomplete thin category there is a unique, up to isomorphism, cocontinous functor $C'\to D$ making the diagram commute? I think a natural candidate would be the power set $P(ob(C))$, but then, I think, for every subset $A\subseteq ob(C)$ the equality $A=\bigcup_{c\in C}\{d\in C\mid \exists d\to c\}$ should hold. This is wrong for simple examples.

Answer 1

I will use the language of order theory.

Given a preordered set $X$, let $\hat{X}$ be the set of downward-closed subsets of $X$, i.e. $A \subseteq X$ such that $x_0 \le x_1$ and $x_1 \in A$ implies $x_0 \in A$. Let ${\downarrow} \{ x \} = \{ x' \in X : x' \le x \}$.

Theorem. For every preordered set $Y$ with infinitary joins and every monotone map $f : X \to Y$, there is a monotone map $\bar{f} : \hat{X} \to Y$ with the following property:

• For every monotone map $g : \hat{X} \to Y$, if $f (x) \le g ({\downarrow} \{ x \})$ for every $x \in X$, then $\bar{f} (A) \le g (A)$ for every $A \in D (X)$.

Moreover, every such $\bar{f} : \hat{X} \to Y$ preserves infinitary joins.

Indeed, simply define $\bar{f} (A) = \bigvee_{x \in A} f (x)$ and check that this works.