The Stacks project

Proof. Since the functor $f_!$ is exact, it suffices to show that $f_!\mathcal{F}$ is constructible for any constructible sheaf $\mathcal{F}$ of $\Lambda $-modules on $X_{\acute{e}tale}$. The question is local on $Y$ and hence we may and do assume $Y$ is affine. Then $X$ is quasi-compact and quasi-separated, see Morphisms, Definition 29.21.1. Say $X = \bigcup _{i = 1, \ldots , n} X_ i$ is a finite affine open covering. By Lemma 63.4.7 we see that it suffices to show that $f_{i, !}\mathcal{F}|_{X_ i}$ and $f_{ii', !}\mathcal{F}|_{X_ i \cap X_{i'}}$ are constructible where $f_ i : X_ i \to Y$ and $f_{ii'} : X_ i \cap X_{i'} \to Y$ are the restrictions of $f$. Since $X_ i$ and $X_ i \cap X_{i'}$ are quasi-compact and separated this means we may assume $f$ is separated. By Zariski's main theorem (in the form of More on Morphisms, Lemma 37.43.4) we can choose a factorization $f = g \circ j$ where $j : X \to X'$ is an open immersion and $g : X' \to Y$ is finite and of finite presentation. Then $f_! = g_! \circ j_!$ by Lemma 63.3.13. By Étale Cohomology, Lemma 59.73.1 we see that $j_!\mathcal{F}$ is constructible on $X'$. The morphism $g$ is finite hence $g_! = g_*$ by Lemma 63.3.4. Thus $f_!\mathcal{F} = g_!j_!\mathcal{F} = g_*j_!\mathcal{F}$ is constructible by Étale Cohomology, Lemma 59.73.9. $\square$