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Suppose that the map $f$ in the following diagram is a separated morphism (i.e. $\Delta_{X/S}:X\rightarrow X\times_{S}X$ is a closed immersion). I want to prove that $p_{2}$ is also a separated morphism. $$\require{AMScd}$$ \begin{CD} X\times_{S}Y @>{p_{1}}>> X\\ @VV{p_{2}}V @VV{f}V\\ Y @>{h}>> S \end{CD} To prove that $p_{2}$ is also separated we have to show that the diagonal morphism $\Delta_{X\times_{S}Y/Y}: X\times_{S}Y\rightarrow (X\times_{S}Y)\times_{Y}(X\times_{S}Y)$ is a closed immersion.

My strategy was to construct a cartesian diagram containing the $\Delta_{X/S}$ and $\Delta_{X\times_{S}Y/X}$ and use the fact that closed immersions are stable under base change, i.e. if we have a cartesian diagram $$\require{AMScd}$$ \begin{CD} Z @>>> Y\\ @VVV @VVV\\ X @>>> S \end{CD} such that $X\rightarrow S$ is a closed immersion, then also $Z\rightarrow Y$ is a closed immersion.

Unfortunately I couldn't find such a cartesian diagram.

Jozef
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I guess that your right horizontal $g$ in your first diagram is your $f: X \to S$ and your $p_2: X \times_S Y \to Y$ is the pullback of $f$ along horizontal $Y \to S$ (you called it also $f$ but in your first sentence you reserved $f$ for $X \to S$. Let call the horizontal arrow $Y \to S$ $h$. Clearly the diagram below is a pullback because $X \times_{ X \times_S X} (X\times_{S}Y \times_Y X\times_{S}Y) = X \times_{ X \times_S X} X \times_S X \times_S Y=X\times_{S}Y$ (use universal property of fiber product).

\begin{CD} X\times_{S}Y @>{p_{1}}>> X\\ @VV\Delta_{}V @VV\Delta_{X/S}V\\ X\times_{S}Y \times_Y X\times_{S}Y @>{pr \times_h pr}>> X \times_S X \end{CD}

Now closed immersions are preserved under base change and you assumed $\Delta_{X/S}:X\rightarrow X\times_{S}X$ be closed immersion.

  • I see that I messed up a lot of things in the notation, I am really sorry for this. I will make an edit such that everything will be right. – Jozef Jun 07 '20 at 15:39
  • I still have some struggles with where the map $pr\times_{h} pr$ comes from, and therefore why it makes this diagram commutative. – Jozef Jun 07 '20 at 16:18
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    the $pr: X \times_S Y \to X$ is your horizontal $p_1$ in the diagram and $h: Y \to S$. It's a general fact for categories with fiber products that maps $a: A \to A', b: B \to B', c: C \to C'$ induce $a \times_c b: A \times_C B \to A' \times_{C'} B'$. The diagram in my answer is commutative by the argument in line 5. Recall general canceling and associative rules of fiber products $A \times_B B =B$ and $A \times_B (C \times_B D) = (A \times_B C) \times_B D$. That's all follows generally from universal properties of fiber products. –  Jun 07 '20 at 16:24