Grothendieck axioms and sheaf categories

Question

An abelian category, $A$, is said to satisfy axiom AB6 if for every family of filtered diagrams $I_j$ indexed by a set $J$, the canonical map $\mathrm{colim}_{\prod_{I_j}}(\prod_j M_{ij}) \to \prod_J \mathrm{colim}_{I_j} M_{ij}$ is an isomorphism for every collection of functors $M_{ij}\colon I_j \to A$.

Let X be a site. Is it true that $\mathrm{Shv}(X,\mathrm{Ab})$ satisfies AB6? I tried to look for a reference, but I didn't find any, even though I know that AB$4^*$ is not satisfied in general. Moreover, it seems to me that it should be true for a locally compact space, but I am not sure even about that.

Answer 1

The axiom AB6 holds in any category of sheaves on a site that can be constructed using only finite coverings. The reason is the following:

If checking for finite covers suffices, then filtered colimits and limits of sheaves are computed objectwise. (Since the sheaf condition is then a finite limit condition, and finite limits commute with both in Sets/Abelian groups)

Therefore the natural map $$ \text{colim}_{\prod_{J} I_j} ( \prod M_{ij} ) \rightarrow \prod_{j \in J} \text{colim}_{i \in I_j} M_{ij} $$ can be checked valuewise for all $x \in X$ $$ \text{colim}_{\prod_{J} I_j} ( \prod M_{ij}(x) ) \rightarrow \prod_{j \in J} \text{colim}_{i \in I_j} M_{ij}(x), $$ which is an isomorphism since AB6 holds in the category of sets / abelian groups.

(I should remark here that an infinite product of filtered categories is again filtered, which is not hard to show but crucial for this argument)

However, in the absence of this finiteness condition, AB6 might fail, as then the sheaf condition might involve infinite products, and infinite products don't commute with filtered colimits. Therefore a sheafification is required which destroys the argument.