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Let $f:X \to Y$ be a proper morphism between schemes, therefore separated, of finite type, and universally closed.

My question is how to deduce that $f(X)$ is closed in $Y$?

By definition I can only expect the closeness of the induced morphism $X \times _Y Y' \xrightarrow{pr_{Y'}} Y'$ of base change by $Y \to Y'$.

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    $\begingroup$ The base change of $f$ along $\mathrm{id}_Y:Y\to Y$ is just $f$. So $f$ is closed. $\endgroup$ – Keenan Kidwell Nov 29 '17 at 0:41
  • $\begingroup$ @Keenan Kidwell: So do you mean that it follows by uniqueness of pullback? For $id_Y$ we would have obviously $X \cong X \times _Y Y$, so $X$ would be the pull back for $f$ and $id_Y$ and by pullback property $id_Y \circ pr_Y = f$? $\endgroup$ – KarlPeter Nov 29 '17 at 0:53
  • $\begingroup$ We would be living in a sad world if "universally closed" would not imply "closed". $\endgroup$ – MooS Nov 29 '17 at 7:50

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