Definition 106.3.1. Thickenings.

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.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.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.

## Comments (4)

Comment #2051 by Matthieu Romagny on

Comment #2084 by Johan on

Comment #5964 by Dario Weißmann on

Comment #6144 by Johan on