## 6.28 Limits and colimits of presheaves

Let $X$ be a topological space. Let $\mathcal{I} \to \textit{PSh}(X)$, $i \mapsto \mathcal{F}_ i$ be a diagram.

Both $\mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i$ and $\mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i$ exist.

For any open $U \subset X$ we have

\[ (\mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i)(U) = \mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i(U) \]and

\[ (\mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i)(U) = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i(U). \]Let $x \in X$ be a point. In general the stalk of $\mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i$ at $x$ is not equal to the limit of the stalks. But if the index category is finite then it is the case. In other words, the stalk functor is left exact (see Categories, Definition 4.23.1).

Let $x \in X$. We always have

\[ (\mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i)_ x = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_{i, x}. \]

The proofs are all easy.

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