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Is the direct limit of sheaves a sheaf? Is the inverse limit of sheaves a sheaf? I guess another way of saying it is whether sheafification commutes with direct limit or inverse limit. While we are at it, does sheafification commutes with direct sum or product?

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Since no one has mentioned it, I'll just quickly say that sheafification is a left exact left adjoint, so it preserves all colimits and all finite limits. In particular it is an exact functor. – Zhen Lin Sep 11 '12 at 1:12
up vote 3 down vote accepted

Keenan Kidwell has answered your main question but I think I can add a little exposition earned from doing all the $\textrm{Ch II}\S1$ exercises in Hartshorne.

A finite (presheaf) product or coproduct of sheaves is again a sheaf, since $\mathscr{F} \oplus_{\mathsf{PSh}} \mathscr{G} \cong \mathscr{F} \times_{\mathsf{PSh}} \mathscr{G}$. Note that in general even finite colimits of sheaves have to be taken in $\mathsf{PSh}$ and then sheafified, e.g. cokernels.

However even a countably infinite family of sheaves can have the property that the direct sum(as presheaves) is not a sheaf! Take our space $\mathbb{N}_{\geq 0}$ with the discrete topology, and let $\mathscr{F}_i$ for $i \geq 0$ be the skyscraper sheaf with stalk $\mathscr{F}_{i,i}=\mathbb{Z}$ and zero elsewhere. Now we see that $\oplus_{i,\mathsf{PSh}} \mathscr{F}_i$ isn't a sheaf! Take the open sets $\{j\}_{j \geq 0}$ and the compatible family of sections $i_j(j)$ for $j \in \mathscr{F}_j(\{j\}) \cong \mathbb{Z}$ and where $i_j : \mathscr{F}_j \hookrightarrow \oplus_{i,\mathsf{PSh}} \mathscr{F}_{i}$.

EDIT: It occurs to me here that it's easier to show in general that if $\{ U_i \}$ is a family of disjoint open sets then $\mathscr{F}(\cup U_i) = \prod_i \mathscr{F}(U_i)$ if $\mathscr{F}$ is a sheaf. This immediately shows that this direct sum is not a sheaf.

It's a little dense, but if you do your part to break up that paragraph you'll see this family has no gluing, precisely because direct sums only have finitely many nonzero terms.

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It doesn't really make sense to ask "is the direct limit of sheaves a sheaf?" I assume you mean to ask whether or not the natural choice for a presheaf direct limit of a directed system of sheaves a sheaf. The answer is no in general. You have to sheafify. That is, given a directed system of sheaves $(\mathscr{F}_i,\varphi_{ij})$, the presheaf defined by $U\mapsto\varinjlim_i\mathscr{F}_i(U)$, taken with respect to the maps $\varphi_{ij}(U)$, is usually only a presheaf. The associated sheaf is what we call $\varinjlim_i\mathscr{F}$, and you can verify that it has the properties to be the (i.e. categorical) direct limit of the $\mathscr{F}_i$. I believe one case where the direct limit presheaf is already a sheaf is when $X$ is a Noetherian topological space (i.e. every subset is quasi-compact).

For an inverse system of sheaves $(\mathscr{F}_i,\varphi_{ij})$, the presheaf $U\mapsto\varprojlim_i\mathscr{F}_i(U)$ actually is a sheaf, $\varprojlim_i\mathscr{F}_i$, and it is the categorical inverse limit of the $\mathscr{F}_i$.

In general, for colimits of diagrams of sheaves, one takes the natural presheaf (take colimits open set by open set), and then has to sheafify, while for limits, the natural presheaf is already a sheaf.

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It's worth mentioning that the reason the presheaf limit of a diagram of sheaves is a sheaf while the same need not be true for colimits is because the forgetful functor from sheaves to presheaves is a right adjoint and so preserves limits (its left adjoint is the sheafification functor.) – SL2 Sep 11 '12 at 0:01

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