TITLE: Is the filtered colimit of sheaves of abelian groups a sheaf?
QUESTION [7 upvotes]: This might be embarrassingly simple, but I want to be 100% sure I am not missing any subtleties. Let $F_i$, $i\in I$ be a filtered inductive system of sheaves of abelian groups on some site. Take the presheaf colimit $F$. It seems to me that $F$ is a sheaf. Is this true?
It seems to be a consequence of the following: let $\{U_j\longrightarrow U\}_{j
\in J}$ be a covering. Then, for each $i\in I$, $F_i$ being a sheaf, we know that:
$$F_i(U)=lim(\prod_{j\in J}F_i(U_j)\overset{\longrightarrow}{\underset{\longrightarrow}{ }} \prod_{j,j'\in J}F_i(U_j\times_UU_{j'})$$
Now, because filtered colimits commute with finite limits for abelian groups,
it follows that $F$ is a sheaf. Did I miss something?

REPLY [6 votes]: Your proof works if the covering is finite. Hence, the claim is true for noetherian sites. The corresponding statement for sheaf cohomology is also true, see Stacks Project, Tag 0739. But in general filtered colimits (of abelian groups, doesn't really matter) don't commute with infinite products. For example, the canonical map $\mathrm{colim}_n ~\prod_{i \in I} \frac{1}{n} \mathbb{Z} \to \prod_{i \in I} \mathrm{colim}_n \frac{1}{n} \mathbb{Z} = \prod_{i \in I} \mathbb{Q}$ is not surjective when $I$ is infinite, since the denominators of a rational sequence don't have to be bounded. As soon as you have an example for $\mathrm{colim}_n \prod_{i \in I} A_{n,i} \not\cong \prod_{i \in I} \mathrm{colim}_n A_{n,i}$, this is is witnessed by sheaves: Take $X = \sqcup_{i \in I} X_i$ with connected $X_i$ (or just points) and consider the sheaves $\underline{A_{n,i}}$ on $X_i$ which glue to a sheaf $A_n$ on $X$. Then the presheaf colimit $\mathrm{colim}_n A_n$ is not a sheaf.