Lemma 99.9.13. Let $\mathcal X$ be an algebraic stack. Let $U$ be an algebraic space and $U \to \mathcal X$ a surjective smooth morphism. For an open immersion $V \hookrightarrow U$, there exists an algebraic stack $\mathcal Y$, an open immersion $\mathcal Y \to \mathcal X$, and a surjective smooth morphism $V \to \mathcal Y$.

**Proof.**
We define a category fibred in groupoids $\mathcal Y$ by letting the fiber category $\mathcal{Y}_ T$ over an object $T$ of $(\mathit{Sch}/S)_{fppf}$ be the full subcategory of $\mathcal{X}_ T$ consisting of all $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ T)$ such that the projection morphism $V \times _{\mathcal X, y} T \to T$ surjective. Now for any morphism $x : T \to \mathcal X$, the $2$-fibred product $T \times _{x, \mathcal X} \mathcal Y$ has fiber category over $T'$ consisting of triples $(f : T' \to T, y \in \mathcal{X}_{T'}, f^*x \simeq y)$ such that $V \times _{\mathcal X, y} T' \to T'$ is surjective. Note that $T \times _{x, \mathcal X} \mathcal Y$ is fibered in setoids since $\mathcal Y \to \mathcal X$ is faithful (see Stacks, Lemma 8.6.7). Now the isomorphism $f^*x \simeq y$ gives the diagram

where both squares are cartesian. The morphism $V \times _{\mathcal X, x} T \to T$ is smooth by base change, and hence open. Let $T_0 \subset T$ be its image. From the cartesian squares we deduce that $V \times _{\mathcal X, y} T' \to T'$ is surjective if and only if $f$ lands in $T_0$. Therefore $T \times _{x, \mathcal X} \mathcal Y$ is representable by $T_0$, so the inclusion $\mathcal Y \to \mathcal X$ is an open immersion. By Algebraic Stacks, Lemma 93.15.5 we conclude that $\mathcal{Y}$ is an algebraic stack. Lastly if we denote the morphism $V \to \mathcal X$ by $g$, we have $V \times _{\mathcal X} V \to V$ is surjective (the diagonal gives a section). Hence $g$ is in the image of $\mathcal{Y}_ V \to \mathcal{X}_ V$, i.e., we obtain a morphism $g' : V \to \mathcal{Y}$ fitting into the commutative diagram

Since $V \times _{g, \mathcal X} \mathcal Y \to V$ is a monomorphism, it is in fact an isomorphism since $(1, g')$ defines a section. Therefore $g' : V \to \mathcal Y$ is a smooth morphism, as it is the base change of the smooth morphism $g : V \to \mathcal{X}$. It is surjective by our construction of $\mathcal{Y}$ which finishes the proof of the lemma. $\square$

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