Lemma 7.17.7. 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). By Lemma 7.17.2 the object $U$ is quasi-compact, hence $\Psi$ is injective by (2). To prove surjectivity, let $s$ be an element of the target of $\Psi$. By assumption there exists a finite covering $\{ U_ j \to U\} _{j = 1, \ldots , m}$, with $U_ j \times _ U U_{j'}$ quasi-compact for all $1 \leq j, j' \leq m$ 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. $\square$

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).