Lemma 69.6.11. Notation and assumptions as in Situation 69.6.1. If

1. $f$ is a monomorphism,

2. $f_0$ is locally of finite type,

then $f_ i$ is a monomorphism for some $i \geq 0$.

Proof. Recall that a morphism is a monomorphism if and only if the diagonal is an isomorphism. The morphism $X_0 \to X_0 \times _{Y_0} X_0$ is locally of finite presentation by Morphisms of Spaces, Lemma 66.28.10. Since $X_0 \times _{Y_0} X_0$ is quasi-compact and quasi-separated we conclude from Lemma 69.6.10 that $\Delta _ i : X_ i \to X_ i \times _{Y_ i} X_ i$ is an isomorphism for some $i \geq 0$. For this $i$ the morphism $f_ i$ is a monomorphism. $\square$

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