Lemma 100.9.14. Let $\mathcal X$ be an algebraic stack and $\mathcal{X}_ i \subset \mathcal X$ a collection of open substacks indexed by $i \in I$. Then there exists an open substack, which we denote $\bigcup _{i\in I} \mathcal{X}_ i \subset \mathcal X$, such that the $\mathcal{X}_ i$ are open substacks covering it.

Proof. We define a fibred subcategory $\mathcal{X}' = \bigcup _{i \in I} \mathcal{X}_ i$ by letting the fiber category over an object $T$ of $(\mathit{Sch}/S)_{fppf}$ be the full subcategory of $\mathcal{X}_ T$ consisting of all $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ T)$ such that the morphism $\coprod _{i \in I} (\mathcal{X}_ i \times _{\mathcal X} T) \to T$ is surjective. Let $x_ i \in \mathop{\mathrm{Ob}}\nolimits ((\mathcal{X}_ i)_ T)$. Then $(x_ i, 1)$ gives a section of $\mathcal{X}_ i \times _{\mathcal X} T \to T$, so we have an isomorphism. Thus $\mathcal{X}_ i \subset \mathcal{X}'$ is a full subcategory. Now let $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ T)$. Then $\mathcal{X}_ i \times _{\mathcal X} T$ is representable by an open subscheme $T_ i \subset T$. The $2$-fibred product $\mathcal{X}' \times _{\mathcal X} T$ has fiber over $T'$ consisting of $(y \in \mathcal{X}_{T'}, f : T' \to T, f^*x \simeq y)$ such that $\coprod (\mathcal{X}_ i \times _{\mathcal X, y} T') \to T'$ is surjective. The isomorphism $f^*x \simeq y$ induces an isomorphism $\mathcal{X}_ i \times _{\mathcal X, y} T' \simeq T_ i \times _ T T'$. Then the $T_ i \times _ T T'$ cover $T'$ if and only if $f$ lands in $\bigcup T_ i$. Therefore we have a diagram

$\xymatrix{ T_ i \ar[r] \ar[d] & \bigcup T_ i \ar[r] \ar[d] & T \ar[d] \\ \mathcal{X}_ i \ar[r] & \mathcal{X}' \ar[r] & \mathcal{X} }$

with both squares cartesian. By Algebraic Stacks, Lemma 94.15.5 we conclude that $\mathcal{X'} \subset \mathcal{X}$ is algebraic and an open substack. It is also clear from the cartesian squares above that the morphism $\coprod _{i \in I} \mathcal{X}_ i \to \mathcal{X}'$ which finishes the proof of the lemma. $\square$

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