Locally Closed Immersion

Question

My question refers to a step in the proof of Prop. 7.4.1 (pages 312-313) in Bosch's "Algebraic Geometry and Commutative Algebra". Here the excerpt:

Let $X$ be relative $S$ scheme. The goal is to show that the diagonal morphism $\Delta_X: X \to X \times_S X$ is a locally closed immersion.

Remark: A morphism of schemes $f: X \to Y$ is called a locally closed immersion if $f$ factorize as $X \xrightarrow{h}Z \xrightarrow{g}Y$ where $h$ is a closed immersion and $g$ is an open immersion.

The author reduces the problem to verification that for affine subset $U \subset X$ the diagonal map $\Delta_U:U \to U \times_{S'} U$ in following diagram is a closed immersion:

$$ \require{AMScd} \begin{CD} U @>{\Delta_U} >> U \times_{S'} U \\ @VVcanV @VVcanV \\ X @>{\Delta_X}>> X \times_{S} X \end{CD} $$

My question is why does this already imply that $\Delta_X$ is a locally closed immersion?

Is it a base change argument? I know that (open/closed) immersions are stable under base change but does the base change argument work "in another direction"? namely that if $P$ is a property stable under base change and $k:X\to Y$ has $P$ and $l:X' \to X$ is some morphism then $Y \times_X X' \to X'$ has also $P$.

But this work in exactly the opposite direction then the problem we have here.

Does anybody see how the auther here conclude that $\Delta_X$ is locally closed immersion?

Answer 1

As stated in your excerpt: $i_U : U \times_{S'} U \to X \times_S X$ is an open embedding.

Choosing such $U_x$ for varying $x \in X$ results in an open cover $X = \bigcup_x U_x$.

As each $U \times_{S'} U$ is open in $X\times_S X$, so is $V := \bigcup_{x}U_x\times_{S'_x}U_x$. Note however, that it is not clear (and in general not true!), that $X\times_S X = \bigcup_x U_x \times_{S'_x} U_x$. Hence we got an open embedding $i:V \to X$.

Now as $\Delta_X |_U = i_U \circ \Delta_U$ factorizes over $i_U$, it actually takes values in $U \times_{S'} U$. Hence $\Delta_X$ factorizes over the open embedding $i: V \to X$, i.e. $\Delta_X = i \circ \Delta_X|^V$, where $\Delta_X|^V : X \to V$ is the "corestriction" to $V$.

It therefore suffices to show that $\Delta_X|^V$ is a closed embedding.

Note that being a closed embedding is a property that is local on the target: $f : X \to Y$ is a closed embedding iff. $f| : f^{-1} V \to V$ is a closed embedding for any open $V$ in an open cover $Y = \bigcup_V V$.

In our case we have the open cover $V = \bigcup_x U_x \times_{S'_x} U_x$ and we know that $\Delta_X|^V$ restricts exactly to $\Delta_U : U \to U\times_{S'} U$, which is shown to be a closed embedding.

Answer 2

We know that $U \times_S U \to X \times_S X$ is on open immersion for every open $U \subset X$. Also, we know that $\Delta_U: U \to U \times_S U$ is a closed immersion for every affine open $U$. But the map $\Delta_U$ is just $\Delta_X: X \to X \times_S X$, restricted to $U$. Also, $U$ is exactly the preimage of $U \times_S U$ under $\Delta_X$.

Take the open set $V = \bigcup_U U \times_S U \subset X \times_S X$. We know that $\Delta_X$ factors over $V$, i.e. it is the composition $X \to V \to X \times_S X$. I claim that $X \to V$ is closed immersion. Indeed, we have a covering of $V$ by sets $U \times U$, such that $U \to U \times U$ is a closed immersion. Hence the image of $X$ in $V$ is also closed (because it is closed in each of the $U \times U$).

General notes about closedness: A subset $A \subset X$ of some topological space $X$, that is closed inside some $U$ need not be closed. But if we cover $X$ by open sets $U_i$, and each $U_i \cap A$ is closed in $U_i$, then $A$ is closed in $X$.