Ind-completion

Remarks

So an ind-object is a filtered colimit of representable presheaves. In terms of the diagrams $X \colon \mathcal{I} \to \mathcal{C}$ and $Y \colon \mathcal{J} \to \mathcal{C}$, the morphisms are $$\mathrm{Ind}(\mathcal{C})(X, Y) = \lim_{i} \operatorname{colim}_{j} \mathcal{C}(X_i, Y_j).$$ The functor $y$ gives a fully faithful functor $\mathcal{C} \to \mathrm{Ind}(\mathcal{C})$, and $\mathrm{Ind}(\mathcal{C})$ has all small filtered colimits. It is universal: every functor from $\mathcal{C}$ to a category with filtered colimits extends, uniquely up to isomorphism, to a filtered-colimit-preserving functor on $\mathrm{Ind}(\mathcal{C})$.

Example: let $\mathcal{C}$ be a skeleton of the category of finite sets. Every set is the filtered colimit of its finite subsets, and this gives an equivalence between $\mathrm{Ind}(\mathcal{C})$ and the category of sets.

Non-example: let $\mathcal{C} = \mathbf{1}$, the category with one object and one morphism. A presheaf on $\mathbf{1}$ is a set, and the only representable is a one-point set. A filtered category is nonempty and connected, so every filtered colimit of one-point sets along identities is a one-point set. So a presheaf with two elements, or with none, is not an ind-object.