Lemma 69.6.5. Notation and assumptions as in Situation 69.6.1. If

1. $f$ is universally injective,

2. $f_0$ is locally of finite type,

then $f_ i$ is universally injective for some $i \geq 0$.

Proof. Recall that a morphism $X \to Y$ is universally injective if and only if the diagonal $X \to X \times _ Y X$ is surjective (Morphisms of Spaces, Definition 66.19.3 and Lemma 66.19.2). Observe that $X_0 \to X_0 \times _{Y_0} X_0$ is of locally of finite presentation (Morphisms of Spaces, Lemma 66.28.10). Hence the lemma follows from Lemma 69.6.4 by considering the morphism $X_0 \to X_0 \times _{Y_0} X_0$. $\square$

Comment #791 by Kestutis Cesnavicius on

One can add a similar lemma for 'open immersion' by using that etale universally injective morphisms are open immersions, i.e., by combining this lemma, 66.51.2, and 69.6.2.

Comment #802 by on

OK, you could, but actually this follows from the slightly stronger Lemma 69.5.2. Right?

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