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I have difficulty to understand the proof of Corollary 5.3.4 in Szamuely, Galois group and fundamental groups. The proof use the following corollary.

Corollary 5.3.3. If $Z \longrightarrow S$ is a connected $S$-scheme and $\phi_1, \phi_2 : Z \longrightarrow X$ are two $S$-morphisms to a finite étale $S$-scheme $X$ with $\phi_1 \circ \overline{z} = \phi_2 \circ \overline{z}$ for some geometric point $\overline{z} : \mathrm{Spec}(\Omega) \longrightarrow Z$, then $\phi_1 = \phi_2$.

Corollary 5.3.4. If $\phi : X \longrightarrow S$ is a connected finite \'etale cover, the nontrivial elements of $\mathrm{Aut}(X\vert S)$ act without fixed points on each geometric fibre.

Proof. Applying Corollary 5.3.3 with $\phi_1 = \phi$, $\phi_2 = \phi \circ \lambda$ for some automorphism $\lambda \in \mathrm{Aut}(X\vert S)$ yields the corollary.

I don't understand this proof. (a) To which geometric point is applied Corollary 5.3.3 ? (b) If $\lambda \in \mathrm{Aut}(X\vert S)$, do we have $\lambda \circ \phi = \phi$ ? (c) If $\overline{s} : \mathrm{Spec}(\Omega) \longrightarrow X$ is a geometric point and if $\xi \in X_{\overline{s}} = X \times_S \mathrm{Spec} (\Omega)$, it is true that $\xi$ is a fixed point under the action of $\lambda \in \mathrm{Aut}(X\vert S)$ if and only if $(\lambda \times_S \mathrm{id})(\xi) = \xi$ ?

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Here's the idea:

(a) Suppose that $\lambda \in \text{Aut}(X/S)$ fixes a geometric point $\bar{z}$ of $X$. Then $\lambda\circ\bar{z} = \bar{z} = \text{id}_X\circ\bar{z}$. Corollary 5.3.3. then implies that $\lambda = \text{id}_X$. So nontrivial automorphisms can't fix any geometric points.

(b) Assuming that I'm interpreting your notation correctly (I don't have the book to hand at the moment!), the group $\text{Aut}(X/S)$ consists of the automorphisms of the $S$-scheme $\phi: X\to S$, so the morphism $\phi$ itself is involved as well. So by definition an element $\lambda\in \text{Aut}(X/S)$ is an automorphism

$$\lambda: (\phi: X\to S)\to(\phi: X\to S)$$

i.e. an automorphism $\lambda: X\to X$ such that $\phi\circ \lambda = \phi$.

(c) With regards to your third question, this is tautologically correct. Indeed, any automorphism $\lambda: X\to X$ acts on the fibre product $X\times_S \text{Spec}(\Omega)$ via the first factor. So saying an element $\xi\in X_{\bar{s}}$ is fixed under this action is by definition saying that $(\lambda \times_S \text{id})(\xi) = \xi$.

I hope this is helpful; let me know if anything is unclear!

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