[Math] Is the stalk of the (co)limit of sheaves equal to the (co)limit of the stalks

ag.algebraic-geometrysheaf-theory

More precisely, if $\mathcal F_i$ is a system of sheaves, is it the case that
$$
(\lim \mathcal F_i)_p = \lim ((\mathcal F_i)_p)
$$
and similarly for colimits? I can see how to get a map
$$
(\lim \mathcal F_i)_p \rightarrow \lim ((\mathcal F_i)_p)
$$
by taking the stalks in the diagram for $\lim \mathcal F_i$ and then using the universal property of $(\lim \mathcal F_i)_p$, but I can't see how I should use the sheaf condition to go the other way.

I learning algebraic geometry and taking a class with Hartshorne's book, and we showed that a map of sheaves is injective/surjective iff it is injective/surjective on stalks. Since one can check injectivity/surjectivity by computing the kernal/cokernal, I am thinking that what I wrote above would be a generalization of this fact.

Looking at sheafication as a adjoint and how it interacts with limits and colimits really made me feel more comfortable with sheafication, so I'm hoping that looking the interaction between sheaves and their stalks in terms of limits and colimits will give me some intuition.

Best Answer

Let $F$ be a sheaf on $X$ and $p \in X$. Then $F_p$ is just the pullback $i^{-1} F$, where $i : \{p\} \to X$ is the inclusion of a point. Now $i^{-1}$ is left adjoint to $i_*$, thus cocontinuous, i.e. preserves all colimits. This shows that the canonical morphism $\mathrm{colim}_i(F_p) \to \mathrm{colim}_i(F)_p$ is an isomorphism.

Now for limits, we get a canonical morphism $(\mathrm{lim}_i F)_p \to \mathrm{lim}_i(F_p)$. This is almost never an isomorphism (neither injective nor surjective). Consider infinite products and see what happens. However, the (left) exactness of $F \mapsto F_p$ means that this functor preserves finite limits.

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