Lemma 88.23.5. In the situation above. If $f$ is quasi-compact, then $f_{/T}$ is quasi-compact.

**Proof.**
(Quasi-compact morphisms of formal algebraic spaces are discussed in Formal Spaces, Section 87.17.) We have to show that $(X'_{/T'})_{red} \to (X_{/T})_{red}$ is a quasi-compact morphism of algebraic spaces. By Formal Spaces, Lemma 87.14.5 this is the morphism $Z' \to Z$ where $Z' \subset X'$, resp. $Z \subset X$ is the reduced induced algebraic space structure on $T'$, resp. $T$. It follows that $Z' \to f^{-1}Z = Z \times _ X X'$ is a thickening (a closed immersion defining an isomorphism on underlying topological spaces). Since $Z \times _ X X' \to Z$ is quasi-compact as a base change of $f$ (Morphisms of Spaces, Lemma 67.8.4) we conclude that $Z' \to Z$ is too by More on Morphisms of Spaces, Lemma 76.10.1.
$\square$

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