The Stacks project

Proof. The normal scheme $X$ is geometrically unibranch (Properties, Lemma 28.15.2). Hence Lemma 58.10.7 applies to $X$. Thus $\pi _1(\eta , \overline{\eta }) \to \pi _1(X, \overline{\eta })$ is surjective and top horizontal arrow of the commutative diagram

is fully faithful. The left vertical arrow is the equivalence of Theorem 58.6.2 and the right vertical arrow is the equivalence of Lemma 58.6.3. The lower horizontal arrow is induced by the map of the proposition. By Lemmas 58.11.1 and 58.11.2 we see that the essential image of $c$ consists of $\text{Gal}(K^{sep}/K)\textit{-Sets}$ isomorphic to sets of the form

with $L_ i/K$ finite separable such that $X$ is unramified in $L_ i$. Thus if $M \subset K^{sep}$ is as in the statement of the lemma, then $\text{Gal}(K^{sep}/M)$ is exactly the subgroup of $\text{Gal}(K^{sep}/K)$ acting trivially on every object in the essential image of $c$. On the other hand, the essential image of $c$ is exactly the category of $S$ such that the $\text{Gal}(K^{sep}/K)$-action factors through the surjection $\text{Gal}(K^{sep}/K) \to \pi _1(X, \overline{\eta })$. We conclude that $\text{Gal}(K^{sep}/M)$ is the kernel. Hence $\text{Gal}(K^{sep}/M)$ is a normal subgroup, $M/K$ is Galois, and we have a short exact sequence

by Galois theory (Fields, Theorem 9.22.4 and Lemma 9.22.5). The proof is done. $\square$