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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?

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We can express $f$ as a trivial base extension of itself (take the identity morphism $Y \to Y$), so universally closed implies closed.

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  • $\begingroup$ Oh true! That was quite obvious, I am sorry! $\endgroup$ Commented Dec 27, 2013 at 14:50
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    $\begingroup$ I don't think it's obvious when you first learn the definitions since there are a few identifications hidden in there, but arguments like this become more obvious as you get more experience. (In general I think it would be bad terminology to make a definition of something as "universally x" if we didn't have "universally x implies x.") $\endgroup$
    – hunter
    Commented Dec 27, 2013 at 14:53

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