Lemma 94.14.2. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{Z}$ be a stack in groupoids over $(\mathit{Sch}/S)_{fppf}$ whose diagonal is representable by algebraic spaces. Let $\mathcal{X}$, $\mathcal{Y}$ be algebraic stacks over $S$. Let $f : \mathcal{X} \to \mathcal{Z}$, $g : \mathcal{Y} \to \mathcal{Z}$ be $1$-morphisms of stacks in groupoids. Then the $2$-fibre product $\mathcal{X} \times _{f, \mathcal{Z}, g} \mathcal{Y}$ is an algebraic stack.
Proof. We have to check conditions (1), (2), and (3) of Definition 94.12.1. The first condition follows from Stacks, Lemma 8.5.6.
The second condition we have to check is that the $\mathit{Isom}$-sheaves are representable by algebraic spaces. To do this, suppose that $T$ is a scheme over $S$, and $u, v$ are objects of $(\mathcal{X} \times _{f, \mathcal{Z}, g} \mathcal{Y})_ T$. By our construction of $2$-fibre products (which goes all the way back to Categories, Lemma 4.32.3) we may write $u = (x, y, \alpha )$ and $v = (x', y', \alpha ')$. Here $\alpha : f(x) \to g(y)$ and similarly for $\alpha '$. Then it is clear that
is a cartesian diagram of sheaves on $(\mathit{Sch}/T)_{fppf}$. Since by assumption the sheaves $\mathit{Isom}(y, y')$, $\mathit{Isom}(x, x')$, $\mathit{Isom}(f(x), g(y'))$ are algebraic spaces (see Lemma 94.10.11) we see that $\mathit{Isom}(u, v)$ is an algebraic space.
Let $U, V \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$, and let $x, y$ be surjective smooth morphisms $x : (\mathit{Sch}/U)_{fppf} \to \mathcal{X}$, $y : (\mathit{Sch}/V)_{fppf} \to \mathcal{Y}$. Consider the morphism
As the diagonal of $\mathcal{Z}$ is representable by algebraic spaces the source of this arrow is representable by an algebraic space $F$, see Lemma 94.10.11. Moreover, the morphism is the composition of base changes of $x$ and $y$, hence surjective and smooth, see Lemmas 94.10.6 and 94.10.5. Choosing a scheme $W$ and a surjective étale morphism $W \to F$ we see that the composition of the displayed $1$-morphism with the corresponding $1$-morphism
is surjective and smooth which proves the last condition. $\square$
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