**Proof.**
Let $X \to Y \to Z$ be morphisms. Assume that $X \to Y$ and $Y \to Z$ are separated. The composition

\[ X \to X \times _ Y X \to X \times _ Z X \]

is closed because the first one is by assumption and the second one by Lemma 26.21.9. The same argument works for “quasi-separated” (with the same references).

Let $f : X \to Y$ be a morphism of schemes over a base $S$. Let $S' \to S$ be a morphism of schemes. Let $f' : X_{S'} \to Y_{S'}$ be the base change of $f$. Then the diagonal morphism of $f'$ is a morphism

\[ \Delta _{f'} : X_{S'} = S' \times _ S X \longrightarrow X_{S'} \times _{Y_{S'}} X_{S'} = S' \times _ S (X \times _ Y X) \]

which is easily seen to be the base change of $\Delta _ f$. Thus (3) and (4) follow from the fact that closed immersions and quasi-compact morphisms are preserved under arbitrary base change (Lemmas 26.17.6 and 26.19.3).

If $f : X \to Y$ and $g : U \to V$ are morphisms of schemes over a base $S$, then $f \times g$ is the composition of $X \times _ S U \to X \times _ S V$ (a base change of $g$) and $X \times _ S V \to Y \times _ S V$ (a base change of $f$). Hence (5) and (6) follow from (1) – (4).
$\square$

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