Lemma 99.13.7. 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 99.13.7. 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

\[ \begin{matrix} \mathcal{S}\! \mathit{paces}'_{fp, flat, proper, S'}
\\ \downarrow
\\ \mathcal{S}\! \mathit{paces}'_{fp, flat, proper, S} \times _{\mathcal{S}\! \mathit{paces}'_{fp, flat, proper, T}} \mathcal{S}\! \mathit{paces}'_{fp, flat, proper, T'}
\end{matrix} \]

is an equivalence.

**Proof.**
The functor is an equivalence if we drop “proper” from the list of conditions and replace “of finite presentation” by “locally of finite presentation”, see Pushouts of Spaces, Lemma 81.6.7. Thus it suffices to show that given a morphism $X' \to S'$ of an algebraic space to $S'$ which is flat and locally of finite presentation, then $X' \to S'$ is proper if and only if $S \times _{S'} X' \to S$ and $T' \times _{S'} X' \to T'$ are proper. One implication follows from the fact that properness is preserved under base change (Morphisms of Spaces, Lemma 67.40.3) and the other from the fact that properness of $S \times _{S'} X' \to S$ implies properness of $X' \to S'$ by More on Morphisms of Spaces, Lemma 76.10.2.
$\square$

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