The Stacks project

Proof. The construction of the maps $h$, $i$, $\beta _{K, \overline{s}, \overline{t}}$ only depends on the base change of $X$ and $K$ to $\mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$. Thus we may and do assume that $S$ is a strictly henselian scheme with closed point $\overline{s}$. Observe that the local acyclicity of $f$ relative to $K$ is preserved by this base change (for example by Lemma 59.93.4 or just directly by comparing strictly henselian rings in this very special case).

Let $\overline{x}$ be a geometric point of $X_{\overline{s}}$. Or equivalently, let $\overline{x}$ be a geometric point whose image by $f$ is $\overline{s}$. Let us compute the stalk of $i^{-1}\beta _{K, \overline{s}, \overline{t}}$ at $\overline{x}$. First, we have

since pullback preserves stalks, see Lemma 59.36.2. Since we are in the situation $S = \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$ we see that $h : X_{\overline{t}} \to X$ has the property that $X_{\overline{t}} \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{X, \overline{x}}) = F_{\overline{x}, \overline{t}}$. Thus we see that

where the equal sign is Theorem 59.53.1. It follows that the map $(\beta _{K, \overline{s}, \overline{t}})_{\overline{x}}$ is none other than the map $\alpha _{K, \overline{x}, \overline{t}}$ used in Definition 59.93.1. The result follows as we may check whether a map is an isomorphism in stalks by Theorem 59.29.10. $\square$