The Stacks project

Proof. It is clear that (1) implies (2).

Assume (2). For every $x \in X$ we can find an $i$ and an affine open neighbourhood $V_ x \subset U_ i$ of $x$. Hence we can find a finite affine open covering $X = \bigcup V_ j$ such that for each $j$ there exists a finite decomposition $V_ j = \coprod V_{j, k}$ by locally closed constructible subsets such that $\mathcal{F}|_{V_{j, k}}$ is finite locally constant. By Topology, Lemma 5.15.5 each $V_{j, k}$ is constructible as a subset of $X$. By Topology, Lemma 5.28.7 we can find a finite stratification $X = \coprod X_ l$ with constructible locally closed strata such that each $V_{j, k}$ is a union of $X_ l$. Thus (3) holds.

Assume (3) holds. Let $U \subset X$ be an affine open. Then $U \cap X_ i$ is a constructible locally closed subset of $U$ (for example by Properties, Lemma 28.2.1) and $U = \coprod U \cap X_ i$ is a partition of $U$ as in Definition 59.71.1. Thus (1) holds. $\square$

Proof. First, we can assume the covering is finite as $X$ is quasi-compact in the spectral topology (Topology, Lemma 5.23.2 and Properties, Lemma 28.2.4). Observe that each $T_ i$ is a quasi-compact and quasi-separated scheme in its own right (because it is constructible in $X$; details omitted). Thus we can find a finite partition $T_ i = \coprod T_{i, j}$ into locally closed constructible parts of $T_ i$ such that $\mathcal{F}|_{T_{i, j}}$ is finite locally constant (Lemma 59.71.2). By Topology, Lemma 5.15.12 we see that $T_{i, j}$ is a constructible locally closed subscheme of $X$. Then we can apply Topology, Lemma 5.28.7 to $X = \bigcup T_{i, j}$ to find the desired partition of $X$. $\square$

Proof. The statement means if $X = \bigcup U_ i$ is an open covering such that $\mathcal{F}|_{U_ i}$ is constructible, then $\mathcal{F}$ is constructible. If $U \subset X$ is affine open, then $U = \bigcup U \cap U_ i$ and $\mathcal{F}|_{U \cap U_ i}$ is constructible (it is trivial that the restriction of a constructible sheaf to an open is constructible). It follows from Lemma 59.71.2 that $\mathcal{F}|_ U$ is constructible, i.e., a suitable partition of $U$ exists. $\square$

Proof. By Lemma 59.71.4 this reduces to the case where $X$ and $Y$ are affine. By Lemma 59.71.2 it suffices to find a finite partition of $X$ by constructible locally closed subschemes such that $f^{-1}\mathcal{F}$ is finite locally constant on each of them. To find it we just pull back the partition of $Y$ adapted to $\mathcal{F}$ and use Lemma 59.64.2. $\square$

Proof. We prove (3). We will use the criterion of Homology, Lemma 12.10.3. Suppose that $\varphi : \mathcal{F} \to \mathcal{G}$ is a map of constructible sheaves of $\Lambda $-modules. We have to show that $\mathcal{K} = \mathop{\mathrm{Ker}}(\varphi )$ and $\mathcal{Q} = \mathop{\mathrm{Coker}}(\varphi )$ are constructible. Similarly, suppose that $0 \to \mathcal{F} \to \mathcal{E} \to \mathcal{G} \to 0$ is a short exact sequence of sheaves of $\Lambda $-modules with $\mathcal{F}$, $\mathcal{G}$ constructible. We have to show that $\mathcal{E}$ is constructible. In both cases we can replace $X$ with the members of an affine open covering. Hence we may assume $X$ is affine. Then we may further replace $X$ by the members of a finite partition of $X$ by constructible locally closed subschemes on which $\mathcal{F}$ and $\mathcal{G}$ are of finite type and locally constant. Thus we may apply Lemma 59.64.6 to conclude.

The proofs of (1) and (2) are very similar and are omitted. $\square$

Proof. Proof of (1). Let $X = \coprod X_ i$ be a partition of $X$ by locally closed constructible subschemes such that both $\mathcal{F}$ and $\mathcal{G}$ are finite locally constant over the parts (use Lemma 59.71.2 for both $\mathcal{F}$ and $\mathcal{G}$ and choose a common refinement). Then apply Lemma 59.64.5 to the restriction of the map to each part.

The proof of (2) and (3) is omitted. $\square$

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. $\square$

Proof. The question immediately reduces to the case where $X$ is affine. Since any two partitions of $X$ with constructible locally closed strata have a common refinement of the same type and since pullbacks commute with tensor product we reduce to Lemma 59.64.7. $\square$

Proof. Omitted. Hint: affine locally you can use the same stratification. $\square$