Lemma 99.15.7. Let
be a pushout in the category of schemes where T \to T' is a thickening and T \to S is affine, see More on Morphisms, Lemma 37.14.3. Then the functor on fibre categories
is an equivalence.
Lemma 99.15.7. Let
be a pushout in the category of schemes where T \to T' is a thickening and T \to S is affine, see More on Morphisms, Lemma 37.14.3. Then the functor on fibre categories
is an equivalence.
Proof. Using the embedding (99.15.1.1), the description of the image, and the corresponding fact for \mathcal{S}\! \mathit{paces}'_{fp, flat, proper} (Lemma 99.13.7) this reduces to the following statement: given a morphism X' \to S' of an algebraic space to S' which is of finite presentation, flat, proper then X' \to S' has relative dimension \leq 1 if and only if S \times _{S'} X' \to S and T' \times _{S'} X' \to T' have relative dimension \leq 1. One implication follows from the fact that having relative dimension \leq 1 is preserved under base change (Morphisms of Spaces, Lemma 67.34.3). The other follows from the fact that having relative dimension \leq 1 is checked on the fibres and that the fibres of X' \to S' (over points of the scheme S') are the same as the fibres of S \times _{S'} X' \to S since S \to S' is a thickening by More on Morphisms, Lemma 37.14.3. \square
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