The Stacks project

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 6.11.2). By the discussion above we have $(\mathcal{F}')^\# = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i$. By Lemma 6.17.5 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 $. Since $U$ is quasi-compact this means there exists a finite open covering $U = \bigcup _{j = 1, \ldots , m} U_ j$ and for each $j$ an index $i_ j \in I$, $i_ j \geq i$, $i_ j \geq i'$ such that $\varphi _{ii_ j}(s) = \varphi _{i'i_ j}(s')$. Let $i''\in I$ be $\geq $ than all of the $i_ j$. We conclude that $\varphi _{ii''}(s)$ and $\varphi _{i'i''}(s)$ agree on the opens $U_ j$ for all $j$ 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 $. Since $U$ is quasi-compact there exists a finite open covering $U = \bigcup _{j = 1, \ldots , m} U_ j$, for each $j$ an index $i_ j \in I$ and $s_ j \in \mathcal{F}_{i_ j}(U_ j)$ such that $s|_{U_ j}$ comes from $s_ j$ for all $j$. Pick $i \in I$ which is $\geq $ than all of the $i_ j$. By (1) the sections $\varphi _{i_ ji}(s_ j)$ agree over the overlaps $U_ j \cap U_{j'}$. 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). In particular we see that $U$ is quasi-compact and hence by (2) we have injectivity of $\Psi $. Let $s$ be an element of the target of $\Psi $. By assumption there exists a finite open covering $U = \bigcup _{j = 1, \ldots , m} U_ j$, with $U_ j \cap U_{j'}$ 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 $U_ j \cap U_{j'}$ 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 $U_ j \cap U_{j'}$. 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. $\square$