# Locally Closed Immersion

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?

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.

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 $$X$$.