Lemma 38.5.3. Let $S$, $X$, $\mathcal{F}$, $s$ be as in Definition 38.5.1. Let $(S', s') \to (S, s)$ be any morphism of pointed schemes. Let $(Z_ k, Y_ k, i_ k, \pi _ k, \mathcal{G}_ k, \alpha _ k)_{k = 1, \ldots , n}$ be a complete dévissage of $\mathcal{F}/X/S$ over $s$. Given this data let $X', Z'_ k, Y'_ k, i'_ k, \pi '_ k$ be the base changes of $X, Z_ k, Y_ k, i_ k, \pi _ k$ via $S' \to S$. Let $\mathcal{F}'$ be the pullback of $\mathcal{F}$ to $X'$ and let $\mathcal{G}'_ k$ be the pullback of $\mathcal{G}_ k$ to $Z'_ k$. Let $\alpha '_ k$ be the pullback of $\alpha _ k$ to $Y'_ k$. If $S'$ is affine, then $(Z'_ k, Y'_ k, i'_ k, \pi '_ k, \mathcal{G}'_ k, \alpha '_ k)_{k = 1, \ldots , n}$ is a complete dévissage of $\mathcal{F}'/X'/S'$ over $s'$.

**Proof.**
By Lemma 38.4.4 we know that the base change of a one step dévissage is a one step dévissage. Hence it suffices to prove that formation of $\mathop{\mathrm{Coker}}(\alpha _ k)$ commutes with base change and that condition (2) of Definition 38.5.1 is preserved by base change. The first is true as $\pi '_{k, *}\mathcal{G}'_ k$ is the pullback of $\pi _{k, *}\mathcal{G}_ k$ (by Cohomology of Schemes, Lemma 30.5.1) and because $\otimes $ is right exact. The second because by the same token we have

with obvious notation. $\square$

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