The Stacks project

Lemma 7.17.8. 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. Consider the canonical map

\[ \Psi : \mathop{\mathrm{colim}}\nolimits _ i \Gamma (\mathcal{C}, \mathcal{F}_ i) \longrightarrow \Gamma (\mathcal{C}, \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i) \]

We have the following:

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

  2. If $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ is quasi-compact, then $\Psi $ is injective.

  3. If $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ is quasi-compact and all the transition maps are injective then $\Psi $ is an isomorphism.

  4. Assume there exists a set $S \subset \mathop{\mathrm{Ob}}\nolimits (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}))$ with the following properties:

    1. for every surjection $\mathcal{F} \to *$ there exists a $\mathcal{K} \in S$ and a map $\mathcal{K} \to \mathcal{F}$ such that $\mathcal{K} \to *$ is surjective,

    2. for $\mathcal{K} \in S$ the product $\mathcal{K} \times \mathcal{K}$ is quasi-compact.

    Then $\Psi $ is bijective.

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 similary for $b$. For $i' \geq i$ denote

\[ E_{i'} = \text{Equalizer}(a_{i'}, b_{i'}) \subset * \quad \text{and}\quad E = \text{Equalizer}(a, b) \subset * \]

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

\[ \coprod \nolimits _{i \in I} * \times _{s, \mathcal{F}} \mathcal{F}_ i \to * \]

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

\[ \xymatrix{ \mathcal{K} \times \mathcal{K} \ar@<1ex>[r] \ar@<-1ex>[r] & \mathcal{K} \ar[d] \ar[r] & {*} \ar[d]^ s \\ & \mathcal{F}_ i \ar[r] & \mathcal{F} \ar@{=}[r] & \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i } \]

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 similary for $b$. We have $a = b$ by the commutativity of the diagram above. For $i' \geq i$ denote

\[ E_{i'} = \text{Equalizer}(a_{i'}, b_{i'}) \subset \mathcal{K} \times \mathcal{K} \quad \text{and}\quad E = \text{Equalizer}(a, b) \subset \mathcal{K} \times \mathcal{K} \]

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 $i' \geq i$. $\square$


Comments (2)

Comment #7341 by Alejandro González Nevado on

TeX typo in subindex of first coproduct in the proof (3rd paragraph).

Comment #7342 by Alejandro González Nevado on

Same TeX typo in the last coproduct of the proof (last paragraph) and in the sentence "this is done shown".


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