The functor $C\to Ind(C)$ and filtered colimits

Question

Suppose that I have a category $C$ which has all finite limits and colimits. I have read that in general the canonical functor $C \to Ind(C)$ does not behave nicely with respect to filtered colimits.

Suppose that $I$ is a filtered category and a $F$ functor from $I$ to $C$. I would like to know that if the colimit (when computed in $C$) exists, it agrees with the colimit (when computed in $Ind(C)$).

Now I understand this to be false in general (this is the not behaving nicely part). However, perhaps under some extra reasonable assumptions on $C$ it is known to be true? For instance, I am willing to assume that $C$ is quasi-abelian and has enough projectives and injectives.

Answer 1

Since $\mathrm{Ind}(C)$ is the free cocompletion of $C$ under filtered colimits, this is only true if the colimit of $F$ is absolute, i.e. preserved by every functor whatsoever.

In more detail, suppose $C\to \mathrm{Ind}(C)$ preserves the colimit of $F$. Then any functor $G:C\to D$ to a category $D$ with filtered colimits factors through a functor $\mathrm{Ind}(C)\to D$ preserving filtered colimits. Thus, $G$ is the composite $C\to \mathrm{Ind}(C)\to D$ of two functors which preserve the colimit of $F$, hence also preserves the colimit of $F$. This can be extended to the case when $D$ doesn't have all filtered colimits by embedding it into its co-presheaf category $[D,\mathrm{Set}]^{\mathrm{op}}$.

In particular, there are no reasonable hypotheses on $C$ which will make this true for all $F$.