Remark 58.6.4. In the situation of Lemma 58.6.3 let us give a more explicit construction of the isomorphism $\text{Gal}(K^{sep}/K) \to \pi _1(X, \overline{x}) = \text{Aut}(F_{\overline{x}})$. Observe that $\text{Gal}(K^{sep}/K) = \text{Aut}(\overline{K}/K)$ as $\overline{K}$ is the perfection of $K^{sep}$. Since $F_{\overline{x}}(Y) = \mathop{\mathrm{Mor}}\nolimits _ X(\mathop{\mathrm{Spec}}(\overline{K}), Y)$ we may consider the map

$\text{Aut}(\overline{K}/K) \times F_{\overline{x}}(Y) \to F_{\overline{x}}(Y), \quad (\sigma , \overline{y}) \mapsto \sigma \cdot \overline{y} = \overline{y} \circ \mathop{\mathrm{Spec}}(\sigma )$

This is an action because

$\sigma \tau \cdot \overline{y} = \overline{y} \circ \mathop{\mathrm{Spec}}(\sigma \tau ) = \overline{y} \circ \mathop{\mathrm{Spec}}(\tau ) \circ \mathop{\mathrm{Spec}}(\sigma ) = \sigma \cdot (\tau \cdot \overline{y})$

The action is functorial in $Y \in \textit{FÉt}_ X$ and we obtain the desired map.

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