Lemma 41.7.1. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

1. $f$ is unramified and a monomorphism,

2. $f$ is unramified and universally injective,

3. $f$ is locally of finite type and a monomorphism,

4. $f$ is universally injective, locally of finite type, and formally unramified,

5. $f$ is locally of finite type and $X_ s$ is either empty or $X_ s \to s$ is an isomorphism for all $s \in S$.

Proof. We have seen in More on Morphisms, Lemma 37.6.8 that being formally unramified and locally of finite type is the same thing as being unramified. Hence (4) is equivalent to (2). A monomorphism is certainly universally injective and formally unramified hence (3) implies (4). It is clear that (1) implies (3). Finally, if (2) holds, then $\Delta : X \to X \times _ S X$ is both an open immersion (Morphisms, Lemma 29.35.13) and surjective (Morphisms, Lemma 29.10.2) hence an isomorphism, i.e., $f$ is a monomorphism. In this way we see that (2) implies (1).

Condition (3) implies (5) because monomorphisms are preserved under base change (Schemes, Lemma 26.23.5) and because of the description of monomorphisms towards the spectra of fields in Schemes, Lemma 26.23.11. Condition (5) implies (4) by Morphisms, Lemmas 29.10.2 and 29.35.12. $\square$

Comment #4977 by Tim Holzschuh on

I guess $y$ and $Y$ should be $s$ and $S$ in $(5)$.

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