I know this is true with etaleness added (SGA I, 5.1; in that case the morphism is an open immersion) or if the morphism is in addition proper (since by ZMT the map is then finite and the fibers are all either isomorphic to the residue field or empty, so the morphism is a closed immersion). Is this true in general? I don't know if holds, for instance, that an unramified morphism is an open morphism onto its scheme-theoretic image (which would imply the result by the same argument as in SGA). I am interested in whether this can yield a "functorial" characterization of immersions (since the above remarks yield functorial characterizations of open and closed immersions: etale (resp. unramified) is equivalent to a nilpotent lifting property under finite type hypotheses, radicial is a condition on functors of points, and properness can be checked via the valuative criterion).



Does unramified + radicial imply that a morphism is an immersion?
No, this is not true, things can go wrong with the topology. For example, consider a smooth curve $X$ with a point $p$, and the natural morphism $(X \smallsetminus \{p\}) \sqcup \{p\} \to X$.


