Lemma 38.4.5. Let $S$, $X$, $\mathcal{F}$, $x$, $s$ be as in Definition 38.4.2. Let $(Z, Y, i, \pi , \mathcal{G}, z, y)$ be a one step dévissage of $\mathcal{F}/X/S$ at $x$. Let $(S', s') \to (S, s)$ be a morphism of pointed schemes which induces an isomorphism $\kappa (s) = \kappa (s')$. Let $(Z', Y', i', \pi ', \mathcal{G}')$ be as constructed in Lemma 38.4.4 and let $x' \in X'$ (resp. $z' \in Z'$, $y' \in Y'$) be the unique point mapping to both $x \in X$ (resp. $z \in Z$, $y \in Y$) and $s' \in S'$. If $S'$ is affine, then $(Z', Y', i', \pi ', \mathcal{G}', z', y')$ is a one step dévissage of $\mathcal{F}'/X'/S'$ at $x'$.

**Proof.**
By Lemma 38.4.4 $(Z', Y', i', \pi ', \mathcal{G}')$ is a one step dévissage of $\mathcal{F}'/X'/S'$ over $s'$. Properties (1) – (4) of Definition 38.4.2 hold for $(Z', Y', i', \pi ', \mathcal{G}', z', y')$ as the assumption that $\kappa (s) = \kappa (s')$ insures that the fibres $X'_{s'}$, $Z'_{s'}$, and $Y'_{s'}$ are isomorphic to $X_ s$, $Z_ s$, and $Y_ s$.
$\square$

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