The Stacks project

Proof. Since the base change of a smooth morphism is smooth, it suffices to show that smooth morphisms are locally acyclic. Let $\overline{x}$ be a geometric point of $X$ with image $\overline{s} = f(\overline{x})$. Let $\overline{t}$ be a geometric point of $\mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, f(\overline{x})})$. Since we are trying to prove a property of the ring map $\mathcal{O}^{sh}_{S, \overline{s}} \to \mathcal{O}^{sh}_{X, \overline{x}}$ (see discussion following Definition 59.93.1) we may and do replace $f : X \to S$ by the base change $X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}}) \to \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$. Thus we may and do assume that $S$ is the spectrum of a strictly henselian local ring and that $\overline{s}$ lies over the closed point of $S$.

We will apply Lemma 59.86.5 to the diagram

and the sheaf $\mathcal{F} = \underline{M}$ where $M = \mathbf{Z}/n\mathbf{Z}$ for some integer $n$ prime to the characteristic of the residue field of $\overline{x}$. We know that the map $f^{-1}R^ qg_*\mathcal{F} \to R^ qh_*e^{-1}\mathcal{F}$ is an isomorphism by smooth base change, see Theorem 59.89.2 (the assumption on torsion holds by our choice of $n$). Thus Lemma 59.86.5 gives us the middle equality in

For the outer two equalities we use that $S = \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$. Since $\overline{t}$ is the spectrum of a separably closed field we conclude that

which is what we had to show (see discussion following Definition 59.93.1). $\square$