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

\[ \xymatrix{ T \ar[r] \ar[d] & T' \ar[d] \\ S \ar[r] & S' } \]

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

\[ \mathcal{C}\! \mathit{urves}_{S'} \longrightarrow \mathcal{C}\! \mathit{urves}_ S \times _{\mathcal{C}\! \mathit{urves}_ T} \mathcal{C}\! \mathit{urves}_{T'} \]

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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