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 $s_ j = s|_{V_ j}$.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (1)

Comment #8567 by Alejandro González Nevado on

There are also: