Lemma 76.15.15 (06BF)—The Stacks project

Processing math: 100%

Table of contents
Part 4: Algebraic Spaces
Chapter 76: More on Morphisms of Spaces
Section 76.15: Universal first order thickenings
Lemma 76.15.15 (cite)

previous lemma

numberstags

Lemma 76.15.15. Let $S$ be a scheme. Let $Z \to Y \to X$ be formally unramified morphisms of algebraic spaces over $S$ .

If $Z \subset Z'$ is the universal first order thickening of $Z$ over $X$ and $Y \subset Y'$ is the universal first order thickening of $Y$ over $X$ , then there is a morphism $Z' \to Y'$ and $Y \times _{Y'} Z'$ is the universal first order thickening of $Z$ over $Y$ .
There is a canonical exact sequence

$i^*\mathcal{C}_{Y/X} \to \mathcal{C}_{Z/X} \to \mathcal{C}_{Z/Y} \to 0$

where the maps come from Lemma 76.15.8 and $i : Z \to Y$ is the first morphism.

Proof. The map $h : Z' \to Y'$ in (1) comes from Lemma 76.15.8. The assertion that $Y \times _{Y'} Z'$ is the universal first order thickening of $Z$ over $Y$ is clear from the universal properties of $Z'$ and $Y'$ . By Lemma 76.5.6 we have an exact sequence

$(i')^*\mathcal{C}_{Y \times _{Y'} Z'/Z'} \to \mathcal{C}_{Z/Z'} \to \mathcal{C}_{Z/Y \times _{Y'} Z'} \to 0$

where $i' : Z \to Y \times _{Y'} Z'$ is the given morphism. By Lemma 76.5.5 there exists a surjection $h^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{Y \times _{Y'} Z'/Z'}$ . Combined with the equalities $\mathcal{C}_{Y/Y'} = \mathcal{C}_{Y/X}$ , $\mathcal{C}_{Z/Z'} = \mathcal{C}_{Z/X}$ , and $\mathcal{C}_{Z/Y \times _{Y'} Z'} = \mathcal{C}_{Z/Y}$ this proves the lemma. $\square$

Comments (0)

There are also:

2 comment(s) on Section 76.15: Universal first order thickenings

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$ ). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Comments (0)

Post a comment