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I've read about closed immersions being stable under base changes, but I have the following question.

Suppose $f: X \to Y$ is a morphism of $S$-schemes such that $f$ is an immersion and assume furthermore that $f$ is proper. Can we conclude that the base change $\operatorname{Id} \times_S f: X \times_S X \to X \times_S Y$ is a closed immersion?

I already know it should be an immersion since these are stable under base changes, but can we lift "immersion" to "closed immersion" using the fact that $f$ is proper?

Thanks in advance!

soggycornflakes
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  • Sure, but for easier reasons than you may be expecting - a proper map has closed image, so in particular, a proper immersion is an immersion with closed image, which is a closed immersion. So $f$ is a closed immersion and therefore since closed immersions are stable under base change, $Id\times_S f$ is a closed immersion. If this resolves your question, I think it should be a duplicate of that linked post - what do you think? – KReiser Jun 04 '24 at 23:42
  • @KReiser Yes that does help a lot actually! I never thought about a proper morphism having closed image since it only talks about the closedness of projection. But now that I think of it, we have that $Y \times_X X \cong Y$ are isomorphic, right? By the uniqueness up to unique isomorphism of the fibered product? – soggycornflakes Jun 05 '24 at 10:46
  • Yes, exactly - all pullbacks are closed maps implies that the original is a closed map if you pull back along the identity $Y\to Y$. – KReiser Jun 05 '24 at 14:32

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