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