The Stacks project

Proof. Proof of (1). 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).

Proof of (2). Assume $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ is quasi-compact. Recall that $\Gamma (\mathcal{C}, \mathcal{F}) = \mathop{\mathrm{Mor}}\nolimits (*, \mathcal{F})$ for all $\mathcal{F}$ in $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$. Let $a_ i, b_ i : * \to \mathcal{F}_ i$ and for $i' \geq i$ denote $a_{i'}, b_{i'} : * \to \mathcal{F}_{i'}$ the composition with the transition maps of the system. Set $a = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} a_{i'}$ and similarly for $b$. For $i' \geq i$ denote

By Categories, Lemma 4.19.2 we have $E = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} E_{i'}$. It follows that $\coprod _{i' \geq i} E_{i'} \to E$ is a surjective map of sheaves. Hence, if $E = *$, i.e., if $a = b$, then because $*$ is quasi-compact, we see that $E_{i'} = *$ for some $i' \geq i$, and we conclude $a_{i'} = b_{i'}$ for some $i' \geq i$. This proves (2).

Proof of (3). Assume $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ is quasi-compact and all transition maps are injective. Let $a : * \to \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ be a map. Then $E_ i = a^{-1}(\mathcal{F}_ i) \subset *$ is a subsheaf and we have $\mathop{\mathrm{colim}}\nolimits E_ i = *$ (by the reference above). Hence for some $i$ we have $E_ i = *$ and we see that the image of $a$ is contained in $\mathcal{F}_ i$ as desired.

Proof of (4). Let $S \subset \mathop{\mathrm{Ob}}\nolimits (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}))$ satisfy (4)(a), (b). Applying (4)(a) to $\text{id} : * \to *$ we find there exists a $\mathcal{K} \in S$ such that $\mathcal{K} \to *$ is surjective. The maps $\mathcal{K} \times \mathcal{K} \to \mathcal{K} \to *$ are surjective. By (4)(b) and Lemma 7.17.5 we conclude that $\mathcal{K}$ and $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ are quasi-compact. Thus $\Psi$ is injective by (2). Set $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$. Let $s : * \to \mathcal{F}$ be a global section of the colimit. Since $\coprod \mathcal{F}_ i \to \mathcal{F}$ is surjective, we see that the projection

is surjective. By (4)(a) we obtain $\mathcal{K} \in S$ and a map $\mathcal{K} \to \coprod _{i \in I} * \times _{s, \mathcal{F}} \mathcal{F}_ i$ with $\mathcal{K} \to *$ surjective. Since $\mathcal{K}$ is quasi-compact we obtain a factorization $\mathcal{K} \to \coprod _{i' \in I'} * \times _{s, \mathcal{F}} \mathcal{F}_{i'}$ for some finite subset $I' \subset I$. Let $i \in I$ be an upper bound for the finite subset $I'$. The transition maps define a map $\coprod _{i' \in I'} \mathcal{F}_{i'} \to \mathcal{F}_ i$. This in turn produces a map $\mathcal{K} \to * \times _{s, \mathcal{F}} \mathcal{F}_ i$. In other words, we obtain $\mathcal{K} \in S$ with $\mathcal{K} \to *$ surjective and a commutative diagram

Observe that the top row of this diagram is a coequalizer. Hence it suffices to show that after increasing $i$ the two induced maps $a_ i, b_ i : \mathcal{K} \times \mathcal{K} \to \mathcal{F}_ i$ are equal. This is done shown in the next paragraph using the exact same argument as in the proof of (2) and we urge the reader to skip the rest of the proof.

For $i' \geq i$ denote $a_{i'}, b_{i'} : \mathcal{K} \times \mathcal{K} \to \mathcal{F}_{i'}$ the composition of $a_ i, b_ i$ with the transition maps of the system. Set $a = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} a_{i'} : \mathcal{K} \times \mathcal{K} \to \mathcal{F}$ and similarly for $b$. We have $a = b$ by the commutativity of the diagram above. For $i' \geq i$ denote

By Categories, Lemma 4.19.2 we have $E = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} E_{i'}$. It follows that $\coprod _{i' \geq i} E_{i'} \to E$ is a surjective map of sheaves. Since $a = b$ we have $E = \mathcal{K} \times \mathcal{K}$. As $\mathcal{K} \times \mathcal{K}$ is quasi-compact by (4)(b), we see that $E_{i'} = \mathcal{K} \times \mathcal{K}$ for some $i' \geq i$, and we conclude $a_{i'} = b_{i'}$ for some $i' \geq i$. $\square$