Lemma 7.10.1. Let $\mathcal{F} : \mathcal{I} \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ be a diagram. Then $\mathop{\mathrm{lim}}\nolimits _\mathcal {I} \mathcal{F}$ exists and is equal to the limit in the category of presheaves.

**Proof.**
Let $\mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i$ be the limit as a presheaf. We will show that this is a sheaf and then it will trivially follow that it is a limit in the category of sheaves. To prove the sheaf property, let $\mathcal{V} = \{ V_ j \to V\} _{j\in J}$ be a covering. Let $(s_ j)_{j\in J}$ be an element of $H^0(\mathcal{V}, \mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i)$. Using the projection maps we get elements $(s_{j, i})_{j\in J}$ in $H^0(\mathcal{V}, \mathcal{F}_ i)$. By the sheaf property for $\mathcal{F}_ i$ we see that there is a unique $s_ i \in \mathcal{F}_ i(V)$ such that $s_{j, i} = s_ i|_{V_ j}$. Let $\phi : i \to i'$ be a morphism of the index category. We would like to show that $\mathcal{F}(\phi ) : \mathcal{F}_ i \to \mathcal{F}_{i'}$ maps $s_ i$ to $s_{i'}$. We know this is true for the sections $s_{i, j}$ and $s_{i', j}$ for all $j$ and hence by the sheaf property for $\mathcal{F}_{i'}$ this is true. At this point we have an element $s = (s_ i)_{i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})}$ of $(\mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i)(V)$. We leave it to the reader to see this element has the required property that $s_ j = s|_{V_ j}$.
$\square$

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