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
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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