Definition 105.3.1. Thickenings.

1. We say an algebraic stack $\mathcal{X}'$ is a thickening of an algebraic stack $\mathcal{X}$ if $\mathcal{X}$ is a closed substack of $\mathcal{X}'$ and the associated topological spaces are equal.

2. Given two thickenings $\mathcal{X} \subset \mathcal{X}'$ and $\mathcal{Y} \subset \mathcal{Y}'$ a morphism of thickenings is a morphism $f' : \mathcal{X}' \to \mathcal{Y}'$ of algebraic stacks such that $f'|_\mathcal {X}$ factors through the closed substack $\mathcal{Y}$. In this situation we set $f = f'|_\mathcal {X} : \mathcal{X} \to \mathcal{Y}$ and we say that $(f, f') : (\mathcal{X} \subset \mathcal{X}') \to (\mathcal{Y} \subset \mathcal{Y}')$ is a morphism of thickenings.

3. Let $\mathcal{Z}$ be an algebraic stack. We similarly define thickenings over $\mathcal{Z}$ and morphisms of thickenings over $\mathcal{Z}$. This means that the algebraic stacks $\mathcal{X}'$ and $\mathcal{Y}'$ are endowed with a structure morphism to $\mathcal{Z}$ and that $f'$ fits into a suitable $2$-commutative diagram of algebraic stacks.

Comment #2051 by Matthieu Romagny on

In (3), second sentence, change $\mathcal{Y}$ into $\mathcal{Z}$ two times.

Comment #5964 by Dario Weißmann on

typo in (3): are algebraic stack -> are algebraic stacks ( or just delete it and say "are endowed")

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).