The Stacks project

Lemma 59.73.3. Let $f : X \to Y$ be a surjective morphism of quasi-compact and quasi-separated schemes.

  1. Let $\mathcal{F}$ be a sheaf of sets on $Y_{\acute{e}tale}$. Then $\mathcal{F}$ is constructible if and only if $f^{-1}\mathcal{F}$ is constructible.

  2. Let $\mathcal{F}$ be an abelian sheaf on $Y_{\acute{e}tale}$. Then $\mathcal{F}$ is constructible if and only if $f^{-1}\mathcal{F}$ is constructible.

  3. Let $\Lambda $ be a Noetherian ring. Let $\mathcal{F}$ be sheaf of $\Lambda $-modules on $Y_{\acute{e}tale}$. Then $\mathcal{F}$ is constructible if and only if $f^{-1}\mathcal{F}$ is constructible.

Proof. One implication follows from Lemma 59.71.5. For the converse, assume $f^{-1}\mathcal{F}$ is constructible. Write $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ as a filtered colimit of constructible sheaves (of sets, abelian groups, or modules) using Lemma 59.73.2. Since $f^{-1}$ is a left adjoint it commutes with colimits (Categories, Lemma 4.24.5) and we see that $f^{-1}\mathcal{F} = \mathop{\mathrm{colim}}\nolimits f^{-1}\mathcal{F}_ i$. By Lemma 59.71.8 we see that $f^{-1}\mathcal{F}_ i \to f^{-1}\mathcal{F}$ is surjective for all $i$ large enough. Since $f$ is surjective we conclude (by looking at stalks using Lemma 59.36.2 and Theorem 59.29.10) that $\mathcal{F}_ i \to \mathcal{F}$ is surjective for all $i$ large enough. Thus $\mathcal{F}$ is the quotient of a constructible sheaf $\mathcal{G}$. Applying the argument once more to $\mathcal{G} \times _\mathcal {F} \mathcal{G}$ or the kernel of $\mathcal{G} \to \mathcal{F}$ we conclude using that $f^{-1}$ is exact and that the category of constructible sheaves (of sets, abelian groups, or modules) is preserved under finite (co)limits or (co)kernels inside $\mathop{\mathit{Sh}}\nolimits (Y_{\acute{e}tale})$, $\mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale})$, $\textit{Ab}(Y_{\acute{e}tale})$, $\textit{Ab}(X_{\acute{e}tale})$, $\textit{Mod}(Y_{\acute{e}tale}, \Lambda )$, and $\textit{Mod}(X_{\acute{e}tale}, \Lambda )$, see Lemma 59.71.6. $\square$


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