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
(\pi _{k, *}\mathcal{G}_ k)_{\xi _ k} \otimes _{\mathcal{O}_{Y_ k, \xi _ k}} \kappa (\xi _ k) \otimes _{\kappa (\xi _ k)} \kappa (\xi '_ k) \cong (\pi '_{k, *}\mathcal{G}'_ k)_{\xi '_ k} \otimes _{\mathcal{O}_{Y'_ k, \xi '_ k}} \kappa (\xi '_ k)
with obvious notation. \square
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