If $f:X\longrightarrow Y$ is a universally closed morphism of schemes (that is the map obtained by base extension is closed), then does it imply $f$ is closed? Or, is the assumption of $f$ being closed made in the definition of universally closed.
This question, I ask because, in Hartshorne, in the proof of valuative criterion of properness, in order to prove $f$ is universally closed, hartshorne just proves that, the morphism obtained by base extension is closed, and does not prove $f$ itself is closed. Is it obvious?
We can express $f$ as a trivial base extension of itself (take the identity morphism $Y \to Y$), so universally closed implies closed.