Lemma 99.9.15. Let $\mathcal X$ be an algebraic stack and $\mathcal X' \subset \mathcal X$ a quasi-compact open substack. Suppose that we have a collection of open substacks $\mathcal{X}_ i \subset \mathcal X$ indexed by $i \in I$ such that $\mathcal{X}' \subset \bigcup _{i \in I} \mathcal{X}_ i$, where we define the union as in Lemma 99.9.14. Then there exists a finite subset $I' \subset I$ such that $\mathcal{X}' \subset \bigcup _{i \in I'} \mathcal{X}_ i$.

Proof. Since $\mathcal X$ is algebraic, there exists a scheme $U$ with a surjective smooth morphism $U \to \mathcal X$. Let $U_ i \subset U$ be the open subscheme representing $\mathcal{X}_ i \times _{\mathcal X} U$ and $U' \subset U$ the open subscheme representing $\mathcal{X}' \times _{\mathcal X} U$. By hypothesis, $U'\subset \bigcup _{i\in I} U_ i$. From the proof of Lemma 99.6.2, there is a quasi-compact open $V \subset U'$ such that $V \to \mathcal{X}'$ is a surjective smooth morphism. Therefore there exists a finite subset $I' \subset I$ such that $V \subset \bigcup _{i \in I'} U_ i$. We claim that $\mathcal{X}' \subset \bigcup _{i \in I'} \mathcal{X}_ i$. Take $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}'_ T)$ for $T \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$. Since $\mathcal{X}' \to \mathcal{X}$ is a monomorphism, we have cartesian squares

$\xymatrix{ V \times _\mathcal {X} T \ar[r] \ar[d] & T \ar[d]^ x \ar@{=}[r] & T \ar[d]^ x \\ V \ar[r] & \mathcal{X}' \ar[r] & \mathcal X }$

By base change, $V \times _{\mathcal X} T \to T$ is surjective. Therefore $\bigcup _{i \in I'} U_ i \times _{\mathcal X} T \to T$ is also surjective. Let $T_ i \subset T$ be the open subscheme representing $\mathcal{X}_ i \times _{\mathcal X} T$. By a formal argument, we have a Cartesian square

$\xymatrix{ U_ i \times _{\mathcal{X}_ i} T_ i \ar[r] \ar[d] & U \times _{\mathcal X} T \ar[d] \\ T_ i \ar[r] & T }$

where the vertical arrows are surjective by base change. Since $U_ i \times _{\mathcal{X}_ i} T_ i \simeq U_ i \times _{\mathcal X} T$, we find that $\bigcup _{i \in I'} T_ i = T$. Hence $x$ is an object of $(\bigcup _{i\in I'} \mathcal{X}_ i)_ T$ by definition of the union. Observe that the inclusion $\mathcal{X}' \subset \bigcup _{i \in I'} \mathcal{X}_ i$ is automatically an open substack. $\square$

There are also:

• 2 comment(s) on Section 99.9: Immersions of algebraic stacks

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05UR. Beware of the difference between the letter 'O' and the digit '0'.