Lemma 100.9.11. Let $[U/R] \to \mathcal{X}$ be a presentation of an algebraic stack. There is a canonical bijection

\[ \text{locally closed substacks }\mathcal{Z}\text{ of }\mathcal{X} \longrightarrow R\text{-invariant locally closed subspaces }Z\text{ of }U \]

which sends $\mathcal{Z}$ to $U \times _\mathcal {X} \mathcal{Z}$. Moreover, a morphism of algebraic stacks $f : \mathcal{Y} \to \mathcal{X}$ factors through $\mathcal{Z}$ if and only if $\mathcal{Y} \times _\mathcal {X} U \to U$ factors through $Z$. Similarly for closed substacks and open substacks.

**Proof.**
By Lemmas 100.9.7 and 100.9.8 we find that the map is a bijection. If $\mathcal{Y} \to \mathcal{X}$ factors through $\mathcal{Z}$ then of course the base change $\mathcal{Y} \times _\mathcal {X} U \to U$ factors through $Z$. Converse, suppose that $\mathcal{Y} \to \mathcal{X}$ is a morphism such that $\mathcal{Y} \times _\mathcal {X} U \to U$ factors through $Z$. We will show that for every scheme $T$ and morphism $T \to \mathcal{Y}$, given by an object $y$ of the fibre category of $\mathcal{Y}$ over $T$, the object $y$ is in fact in the fibre category of $\mathcal{Z}$ over $T$. Namely, the fibre product $T \times _\mathcal {X} U$ is an algebraic space and $T \times _\mathcal {X} U \to T$ is a surjective smooth morphism. Hence there is an fppf covering $\{ T_ i \to T\} $ such that $T_ i \to T$ factors through $T \times _\mathcal {X} U \to T$ for all $i$. Then $T_ i \to \mathcal{X}$ factors through $\mathcal{Y} \times _\mathcal {X} U$ and hence through $Z \subset U$. Thus $y|_{T_ i}$ is an object of $\mathcal{Z}$ (as $Z$ is the fibre product of $U$ with $\mathcal{Z}$ over $\mathcal{X}$). Since $\mathcal{Z}$ is a strictly full substack, we conclude that $y$ is an object of $\mathcal{Z}$ as desired.
$\square$

## Comments (0)

There are also: