Lemma 59.71.8. Let $X$ be a quasi-compact and quasi-separated scheme. Let $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathcal{F}_ i$ be a filtered colimit of sheaves of sets, abelian sheaves, or sheaves of modules.

If $\mathcal{F}$ and $\mathcal{F}_ i$ are constructible sheaves of sets, then the ind-object $\mathcal{F}_ i$ is essentially constant with value $\mathcal{F}$.

If $\mathcal{F}$ and $\mathcal{F}_ i$ are constructible sheaves of abelian groups, then the ind-object $\mathcal{F}_ i$ is essentially constant with value $\mathcal{F}$.

Let $\Lambda $ be a Noetherian ring. If $\mathcal{F}$ and $\mathcal{F}_ i$ are constructible sheaves of $\Lambda $-modules, then the ind-object $\mathcal{F}_ i$ is essentially constant with value $\mathcal{F}$.

**Proof.**
Proof of (1). We will use without further mention that finite limits and colimits of constructible sheaves are constructible (Lemma 59.64.6). For each $i$ let $T_ i \subset X$ be the set of points $x \in X$ where $\mathcal{F}_{i, \overline{x}} \to \mathcal{F}_{\overline{x}}$ is not surjective. Because $\mathcal{F}_ i$ and $\mathcal{F}$ are constructible $T_ i$ is a constructible subset of $X$ (Lemma 59.71.7). Since the stalks of $\mathcal{F}$ are finite and since $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathcal{F}_ i$ we see that for all $x \in X$ we have $x \not\in T_ i$ for $i$ large enough. Since $X$ is a spectral space by Properties, Lemma 28.2.4 the constructible topology on $X$ is quasi-compact by Topology, Lemma 5.23.2. Thus $T_ i = \emptyset $ for $i$ large enough. Thus $\mathcal{F}_ i \to \mathcal{F}$ is surjective for $i$ large enough. Assume now that $\mathcal{F}_ i \to \mathcal{F}$ is surjective for all $i$. Choose $i \in I$. For $i' \geq i$ denote $S_{i'} \subset X$ the set of points $x$ such that the number of elements in $\mathop{\mathrm{Im}}(\mathcal{F}_{i, \overline{x}} \to \mathcal{F}_{\overline{x}})$ is equal to the number of elements in $\mathop{\mathrm{Im}}(\mathcal{F}_{i, \overline{x}} \to \mathcal{F}_{i', \overline{x}})$. Because $\mathcal{F}_ i$, $\mathcal{F}_{i'}$ and $\mathcal{F}$ are constructible $S_{i'}$ is a constructible subset of $X$ (details omitted; hint: use Lemma 59.71.7). Since the stalks of $\mathcal{F}_ i$ and $\mathcal{F}$ are finite and since $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} \mathcal{F}_{i'}$ we see that for all $x \in X$ we have $x \not\in S_{i'}$ for $i'$ large enough. By the same argument as above we can find a large $i'$ such that $S_{i'} = \emptyset $. Thus $\mathcal{F}_ i \to \mathcal{F}_{i'}$ factors through $\mathcal{F}$ as desired.

Proof of (2). Observe that a constructible abelian sheaf is a constructible sheaf of sets. Thus case (2) follows from (1).

Proof of (3). We will use without further mention that the category of constructible sheaves of $\Lambda $-modules is abelian (Lemma 59.64.6). For each $i$ let $\mathcal{Q}_ i$ be the cokernel of the map $\mathcal{F}_ i \to \mathcal{F}$. The support $T_ i$ of $\mathcal{Q}_ i$ is a constructible subset of $X$ as $\mathcal{Q}_ i$ is constructible (Lemma 59.71.7). Since the stalks of $\mathcal{F}$ are finite $\Lambda $-modules and since $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathcal{F}_ i$ we see that for all $x \in X$ we have $x \not\in T_ i$ for $i$ large enough. Since $X$ is a spectral space by Properties, Lemma 28.2.4 the constructible topology on $X$ is quasi-compact by Topology, Lemma 5.23.2. Thus $T_ i = \emptyset $ for $i$ large enough. This proves the first assertion. For the second, assume now that $\mathcal{F}_ i \to \mathcal{F}$ is surjective for all $i$. Choose $i \in I$. For $i' \geq i$ denote $\mathcal{K}_{i'}$ the image of $\mathop{\mathrm{Ker}}(\mathcal{F}_ i \to \mathcal{F})$ in $\mathcal{F}_{i'}$. The support $S_{i'}$ of $\mathcal{K}_{i'}$ is a constructible subset of $X$ as $\mathcal{K}_{i'}$ is constructible. Since the stalks of $\mathop{\mathrm{Ker}}(\mathcal{F}_ i \to \mathcal{F})$ are finite $\Lambda $-modules and since $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} \mathcal{F}_{i'}$ we see that for all $x \in X$ we have $x \not\in S_{i'}$ for $i'$ large enough. By the same argument as above we can find a large $i'$ such that $S_{i'} = \emptyset $. Thus $\mathcal{F}_ i \to \mathcal{F}_{i'}$ factors through $\mathcal{F}$ as desired.
$\square$

## Comments (1)

Comment #9506 by nkym on

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