## 17.19 Constructible sheaves of sets

Let $X$ be a topological space. Given a set $S$ recall that $\underline{S}$ or $\underline{S}_ X$ denotes the constant sheaf with value $S$, see Sheaves, Definition 6.7.4. Let $U \subset X$ be an open of a topological space $X$. We will denote $j_ U$ the inclusion morphism and we will denote $j_{U!} : \mathop{\mathit{Sh}}\nolimits (U) \to \mathop{\mathit{Sh}}\nolimits (X)$ the extension by the empty set described in Sheaves, Section 6.31.

Lemma 17.19.1. Let $X$ be a topological space. Let $\mathcal{B}$ be a basis for the topology on $X$. Let $\mathcal{F}$ be a sheaf of sets on $X$. There exists a set $I$ and for each $i \in I$ an element $U_ i \in \mathcal{B}$ and a finite set $S_ i$ such that there exists a surjection $\coprod _{i \in I} j_{U_ i!}\underline{S_ i} \to \mathcal{F}$.

Proof. Let $S$ be a singleton set. We will prove the result with $S_ i = S$. For every $x \in X$ and element $s \in \mathcal{F}_ x$ we can choose a $U(x, s) \in \mathcal{B}$ and $s(x, s) \in \mathcal{F}(U(x, s))$ which maps to $s$ in $\mathcal{F}_ x$. By Sheaves, Lemma 6.31.4 the section $s(x, s)$ corresponds to a map of sheaves $j_{U(x, s)!}\underline{S} \to \mathcal{F}$. Then

$\coprod \nolimits _{(x, s)} j_{U(x, s)!}\underline{S} \to \mathcal{F}$

is surjective on stalks and hence surjective. $\square$

Lemma 17.19.2. Let $X$ be a topological space. Let $\mathcal{B}$ be a basis for the topology of $X$ and assume that each $U \in \mathcal{B}$ is quasi-compact. Then every sheaf of sets on $X$ is a filtered colimit of sheaves of the form

17.19.2.1
$$\label{modules-equation-towards-constructible-sets} \text{Coequalizer}\left( \xymatrix{ \coprod \nolimits _{b = 1, \ldots , m} j_{V_ b!}\underline{S_ b} \ar@<1ex>[r] \ar@<-1ex>[r] & \coprod \nolimits _{a = 1, \ldots , n} j_{U_ a!}\underline{S_ a} } \right)$$

with $U_ a$ and $V_ b$ in $\mathcal{B}$ and $S_ a$ and $S_ b$ finite sets.

Proof. By Lemma 17.19.1 every sheaf of sets $\mathcal{F}$ is the target of a surjection whose source $\mathcal{F}_0$ is a coproduct of sheaves the form $j_{U!}\underline{S}$ with $U \in \mathcal{B}$ and $S$ finite. Applying this to $\mathcal{F}_0 \times _\mathcal {F} \mathcal{F}_0$ we find that $\mathcal{F}$ is a coequalizer of a pair of maps

$\xymatrix{ \coprod \nolimits _{b \in B} j_{V_ b!}\underline{S_ b} \ar@<1ex>[r] \ar@<-1ex>[r] & \coprod \nolimits _{a \in A} j_{U_ a!}\underline{S_ a} }$

for some index sets $A$, $B$ and $V_ b$ and $U_ a$ in $\mathcal{B}$ and $S_ a$ and $S_ b$ finite. For every finite subset $B' \subset B$ there is a finite subset $A' \subset A$ such that the coproduct over $b \in B'$ maps into the coproduct over $a \in A'$ via both maps. Namely, we can view the right hand side as a filtered colimit with injective transition maps. Hence taking sections over the quasi-compact opens $V_ b$, $b \in B'$ commutes with this coproduct, see Sheaves, Lemma 6.29.1. Thus our sheaf is the colimit of the cokernels of these maps between finite coproducts. $\square$

Lemma 17.19.3. Let $X$ be a spectral topological space. Let $\mathcal{B}$ be the set of quasi-compact open subsets of $X$. Let $\mathcal{F}$ be a sheaf of sets as in Equation (17.19.2.1). Then there exists a continuous spectral map $f : X \to Y$ to a finite sober topological space $Y$ and a sheaf of sets $\mathcal{G}$ on $Y$ with finite stalks such that $f^{-1}\mathcal{G} \cong \mathcal{F}$.

Proof. We can write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as a directed limit of finite sober spaces, see Topology, Lemma 5.23.14. Of course the transition maps $X_{i'} \to X_ i$ are spectral and hence by Topology, Lemma 5.24.5 the maps $p_ i : X \to X_ i$ are spectral. For some $i$ we can find opens $U_{a, i}$ and $V_{b, i}$ of $X_ i$ whose inverse images are $U_ a$ and $V_ b$, see Topology, Lemma 5.24.6. The two maps

$\beta , \gamma : \coprod \nolimits _{b \in B} j_{V_ b!}\underline{S_ b} \longrightarrow \coprod \nolimits _{a \in A} j_{U_ a!}\underline{S_ a}$

whose coequalizer is $\mathcal{F}$ correspond by adjunction to two families

$\beta _ b, \gamma _ b : S_ b \longrightarrow \Gamma (V_ b, \coprod \nolimits _{a \in A} j_{U_ a!}\underline{S_ a}), \quad b \in B$

of maps of sets. Observe that $p_ i^{-1}(j_{U_{a, i}!}\underline{S_ a}) = j_{U_ a!}\underline{S_ a}$ and $(X_{i'} \to X_ i)^{-1}(j_{U_{a, i}!}\underline{S_ a}) = j_{U_{a, i'}!}\underline{S_ a}$. It follows from Sheaves, Lemma 6.29.3 (and using that $S_ b$ and $B$ are finite sets) that after increasing $i$ we find maps

$\beta _{b, i}, \gamma _{b, i} : S_ b \longrightarrow \Gamma (V_{b, i}, \coprod \nolimits _{a \in A} j_{U_{a, i}!}\underline{S_ a}) , \quad b \in B$

which give rise to the maps $\beta _ b$ and $\gamma _ b$ after pulling back by $p_ i$. These maps correspond in turn to maps of sheaves

$\beta _ i, \gamma _ i : \coprod \nolimits _{b \in B} j_{V_{b, i}!}\underline{S_ b} \longrightarrow \coprod \nolimits _{a \in A} j_{U_{a, i}!}\underline{S_ a}$

on $X_ i$. Then we can take $Y = X_ i$ and

$\mathcal{G} = \text{Coequalizer}\left( \xymatrix{ \coprod \nolimits _{b = 1, \ldots , m} j_{V_{b, i}!}\underline{S_ b} \ar@<1ex>[r] \ar@<-1ex>[r] & \coprod \nolimits _{a = 1, \ldots , n} j_{U_{a, i}!}\underline{S_ a} } \right)$

We omit some details. $\square$

Lemma 17.19.4. Let $X$ be a spectral topological space. Let $\mathcal{B}$ be the set of quasi-compact open subsets of $X$. Let $\mathcal{F}$ be a sheaf of sets as in Equation (17.19.2.1). Then there exist finitely many constructible closed subsets $Z_1, \ldots , Z_ n \subset X$ and finite sets $S_ i$ such that $\mathcal{F}$ is isomorphic to a subsheaf of $\prod (Z_ i \to X)_*\underline{S_ i}$.

Proof. By Lemma 17.19.3 we reduce to the case of a finite sober topological space and a sheaf with finite stalks. In this case $\mathcal{F} \subset \prod _{x \in X} i_{x, *}\mathcal{F}_ x$ where $i_ x : \{ x\} \to X$ is the embedding. We omit the proof that $i_{x, *}\mathcal{F}_ x$ is a constant sheaf on $\overline{\{ x\} }$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).