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Let $X,Y$ Noetherian integral schemes and assume we have an immersion

$$i:X \to Y$$

An immersion is according to https://stacks.math.columbia.edu/tag/01IO a composition of a closed immersion $c: X \to Z$ by an open immersion $o:Z \to Y$. So in our situation there exist an open subscheme $Z$ of $ Y$ such that $X \cong c(X)$ is a closed subscheme of $Z$.

My question is if and why the image $i(X)$ is open in $X' := \overline{i(X)} \subset Y$?

The background of my question is that I read often that in situations as above if one wants to show some topological properties of $X$ one reduces the problem to closed immersion $X' \subset Y$.

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    $\begingroup$ There is not always a scheme structure on the closure compatible with the one on $X$. However, your question is just one of topology: if $X \subseteq Y$ is 'closed inside open', then it equals $U \cap Z$ for $U$ open and $Z$ closed. We may replace $Z$ by $\bar X$, showing that $X$ is open in $\bar X$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Aug 23, 2019 at 13:03
  • $\begingroup$ Oh I forgot to mention that we endow $X'$ with reduced closed subscheme structure in $Y$. $\endgroup$
    – user267839
    Commented Aug 23, 2019 at 13:06
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    $\begingroup$ Ah sorry, I missed the assumption that $X$ is integral. Then reduced induced works, and $X \to \bar X$ is an open immersion. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Aug 23, 2019 at 13:13

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