Construction of an immersion of schemes using the fibered product

Question

I'm trying to understand the proof of Lemma 26.21.9. in the stack project.

1. How does one obtain the following commutative diagram mentioned in the proof lemma?

So far I've tried to construct the following diagram

2. The 3 lemmas in the proof mention well known properties stable under base change, it's not clear for me how to exploit them in the commutative diagram.

Answer 1

It suffices to show that it satisfies the universal property of a fiber product diagram: that is, given an object $Q$ with morphisms $q_1:Q\to T$ and $q_2:Q\to X\times_S Y$ so that $\Delta_{T/S}\circ q_1=(f\times_S g)\circ q_2$ as morphisms $Q\to T\times_S T$, then there's a unique morphism $\alpha:Q\to X\times_T Y$ so that $q_1=p_1\circ \alpha$ and $q_2=p_2\circ\alpha$.

As $\Delta_{T/S}\circ q_1 =(q_1\times_S q_1)$ and $(f\times_S g)\circ q_2 = ((f\circ q_2)\times_S (g\circ q_2))$, the condition that these morphisms agree is the condition that $f\circ q_2 = g\circ q_2$ as morphisms $Q\to X\to T$ and $Q\to Y\to T$. This gives a unique map $Q\to X\times_T Y$ by the definition of the fiber product, and thus $X\times_T Y$ is the fiber product of $T\stackrel{\Delta_{T/S}}{\to} T\times_S T$ and $X\times_S Y\to T\times_S T$.

As to how one might think about this, it's instructive to give the diagram a gander in the category of sets. Then you can actually get your hands on some elements and compute what the fiber product should be: an element of $T\times_{T\times_S T} (X\times_S Y)$ is an element of $T\times X\times Y$ so that it's image under the two ways around the diagram are the same - but this is exactly the set of elements of $X\times Y$ which give the same image under the maps $X\to T$ and $Y\to T$, or $X\times_T Y$.

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As to why this suffices to prove the claims:

• $T\to S$ separated is the fact $\Delta_{T/S}$ a closed immersion by definition, and as closed immersions are stable under base change, this implies that $X\times_T Y\to X\times_S Y$ is a closed immersion;
• $T\to S$ quasi-separated is the fact that $\Delta_{T/S}$ is quasi-compact by definition, and as quasi-compact morphisms are stable under base change, this implies that $X\times_T Y\to X\times_S Y$ is quasi-compact.

If you're struggling with what stable under base change means, here's a reminder: a property $\mathcal{P}$ of a morphism being stable under base change means that if $f:X\to Y$ has property $\mathcal{P}$ and $g:Y'\to Y$ is any morphism, then letting $X'=X\times_Y Y'$ and $f'=X'\to Y'$, then $f'$ has property $\mathcal{P}$ too.