Lemma 17.17.3. 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 (17.17.2.1). 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 5.23.13. 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. $\square$

Comment #3647 by Brian Conrad on

Typo on line 4 of the proof: "maps map"

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