Lemma 101.5.6. Let f : \mathcal{X} \to \mathcal{Y} be a monomorphism of algebraic stacks. Then the diagram
is a fibre product square.
Lemma 101.5.6. Let f : \mathcal{X} \to \mathcal{Y} be a monomorphism of algebraic stacks. Then the diagram
is a fibre product square.
Proof. This follows immediately from the fact that f is fully faithful (see Properties of Stacks, Lemma 100.8.4) and the definition of the inertia in Categories, Section 4.34. Namely, an object of \mathcal{I}_\mathcal {X} over a scheme T is the same thing as a pair (x, \alpha ) consisting of an object x of \mathcal{X} over T and a morphism \alpha : x \to x in the fibre category of \mathcal{X} over T. As f is fully faithful we see that \alpha is the same thing as a morphism \beta : f(x) \to f(x) in the fibre category of \mathcal{Y} over T. Hence we can think of objects of \mathcal{I}_\mathcal {X} over T as triples ((y, \beta ), x, \gamma ) where y is an object of \mathcal{Y} over T, \beta : y \to y in \mathcal{Y}_ T and \gamma : y \to f(x) is an isomorphism over T, i.e., an object of \mathcal{I}_\mathcal {Y} \times _\mathcal {Y} \mathcal{X} over T. \square
Comments (0)
There are also: