Lemma 7.17.5. Let $\mathcal{C}$ be a site. Let $\mathcal{I} \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$, $i \mapsto \mathcal{F}_ i$ be a filtered diagram of sheaves of sets. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Consider the canonical map

$\Psi : \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i(U) \longrightarrow \left(\mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i\right)(U)$

With the terminology introduced above:

1. If all the transition maps are injective then $\Psi$ is injective for any $U$.

2. If $U$ is quasi-compact, then $\Psi$ is injective.

3. If $U$ is quasi-compact and all the transition maps are injective then $\Psi$ is an isomorphism.

4. If $U$ has a cofinal system of coverings $\{ U_ j \to U\} _{j \in J}$ with $J$ finite and $U_ j \times _ U U_{j'}$ quasi-compact for all $j, j' \in J$, then $\Psi$ is bijective.

Proof. Assume all the transition maps are injective. In this case the presheaf $\mathcal{F}' : V \mapsto \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i(V)$ is separated (see Definition 7.10.9). By Lemma 7.10.13 we have $(\mathcal{F}')^\# = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i$. By Theorem 7.10.10 we see that $\mathcal{F}' \to (\mathcal{F}')^\#$ is injective. This proves (1).

Assume $U$ is quasi-compact. Suppose that $s \in \mathcal{F}_ i(U)$ and $s' \in \mathcal{F}_{i'}(U)$ give rise to elements on the left hand side which have the same image under $\Psi$. This means we can choose a covering $\{ U_ a \to U\} _{a \in A}$ and for each $a \in A$ an index $i_ a \in I$, $i_ a \geq i$, $i_ a \geq i'$ such that $\varphi _{ii_ a}(s) = \varphi _{i'i_ a}(s')$. Because $U$ is quasi-compact we can choose a covering $\{ V_ b \to U\} _{b \in B}$, a map $\alpha : B \to A$ with finite image, and morphisms $V_ b \to U_{\alpha (b)}$ over $U$. Pick $i''\in I$ to be $\geq$ than all of the $i_{\alpha (b)}$ which is possible because the image of $\alpha$ is finite. We conclude that $\varphi _{ii''}(s)$ and $\varphi _{i'i''}(s)$ agree on $V_ b$ for all $b \in B$ and hence that $\varphi _{ii''}(s) = \varphi _{i'i''}(s)$. This proves (2).

Assume $U$ is quasi-compact and all transition maps injective. Let $s$ be an element of the target of $\Psi$. There exists a covering $\{ U_ a \to U\} _{a \in A}$ and for each $a \in A$ an index $i_ a \in I$ and a section $s_ a \in \mathcal{F}_{i_ a}(U_ a)$ such that $s|_{U_ a}$ comes from $s_ a$ for all $a \in A$. Because $U$ is quasi-compact we can choose a covering $\{ V_ b \to U\} _{b \in B}$, a map $\alpha : B \to A$ with finite image, and morphisms $V_ b \to U_{\alpha (b)}$ over $U$. Pick $i \in I$ to be $\geq$ than all of the $i_{\alpha (b)}$ which is possible because the image of $\alpha$ is finite. By (1) the sections $s_ b = \varphi _{i_{\alpha (b)} i}(s_{\alpha (b)})|_{V_ b}$ agree over $V_ b \times _ U V_{b'}$. Hence they glue to a section $s' \in \mathcal{F}_ i(U)$ which maps to $s$ under $\Psi$. This proves (3).

Assume the hypothesis of (4). Let $s$ be an element of the target of $\Psi$. By assumption there exists a finite covering $\{ U_ j \to U\} _{j = 1, \ldots , m} U_ j$, with $U_ j \times _ U U_{j'}$ quasi-compact for all $j, j' \in J$ and for each $j$ an index $i_ j \in I$ and $s_ j \in \mathcal{F}_{i_ j}(U_ j)$ such that $s|_{U_ j}$ is the image of $s_ j$ for all $j$. Since $U_ j \times _ U U_{j'}$ is quasi-compact we can apply (2) and we see that there exists an $i_{jj'} \in I$, $i_{jj'} \geq i_ j$, $i_{jj'} \geq i_{j'}$ such that $\varphi _{i_ ji_{jj'}}(s_ j)$ and $\varphi _{i_{j'}i_{jj'}}(s_{j'})$ agree over $U_ j \times _ U U_{j'}$. Choose an index $i \in I$ wich is bigger or equal than all the $i_{jj'}$. Then we see that the sections $\varphi _{i_ ji}(s_ j)$ of $\mathcal{F}_ i$ glue to a section of $\mathcal{F}_ i$ over $U$. This section is mapped to the element $s$ as desired. $\square$

## 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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0738. Beware of the difference between the letter 'O' and the digit '0'.