The Stacks project

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

  1. 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 $Y$.

  2. 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. $\square$

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