## 93.14 2-Fibre products of algebraic stacks

The $2$-category of algebraic stacks has products and $2$-fibre products. The first lemma is really a special case of Lemma 93.14.3 but its proof is slightly easier.

Lemma 93.14.1. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}$, $\mathcal{Y}$ be algebraic stacks over $S$. Then $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ is an algebraic stack, and is a product in the $2$-category of algebraic stacks over $S$.

Proof. An object of $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ over $T$ is just a pair $(x, y)$ where $x$ is an object of $\mathcal{X}_ T$ and $y$ is an object of $\mathcal{Y}_ T$. Hence it is immediate from the definitions that $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ is a stack in groupoids. If $(x, y)$ and $(x', y')$ are two objects of $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ over $T$, then

$\mathit{Isom}((x, y), (x', y')) = \mathit{Isom}(x, x') \times \mathit{Isom}(y, y').$

Hence it follows from the equivalences in Lemma 93.10.11 and the fact that the category of algebraic spaces has products that the diagonal of $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ is representable by algebraic spaces. Finally, suppose that $U, V \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$, and let $x, y$ be surjective smooth morphisms $x : (\mathit{Sch}/U)_{fppf} \to \mathcal{X}$, $y : (\mathit{Sch}/V)_{fppf} \to \mathcal{Y}$. Note that

$(\mathit{Sch}/U \times _ S V)_{fppf} = (\mathit{Sch}/U)_{fppf} \times _{(\mathit{Sch}/S)_{fppf}} (\mathit{Sch}/V)_{fppf}.$

The object $(\text{pr}_ U^*x, \text{pr}_ V^*y)$ of $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ over $(\mathit{Sch}/U \times _ S V)_{fppf}$ thus defines a $1$-morphism

$(\mathit{Sch}/U \times _ S V)_{fppf} \longrightarrow \mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$

which is the composition of base changes of $x$ and $y$, hence is surjective and smooth, see Lemmas 93.10.6 and 93.10.5. We conclude that $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ is indeed an algebraic stack. We omit the verification that it really is a product. $\square$

Lemma 93.14.2. 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 93.12.1. 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

$\xymatrix{ \mathit{Isom}(u, v) \ar[d] \ar[rr] & & \mathit{Isom}(y, y') \ar[d]^{\phi \mapsto g(\phi ) \circ \alpha } \\ \mathit{Isom}(x, x') \ar[rr]^-{\psi \mapsto \alpha ' \circ f(\psi )} & & \mathit{Isom}(f(x), g(y')) }$

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 93.10.11) 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

$(\mathit{Sch}/U)_{fppf} \times _{f \circ x, \mathcal{Z}, g \circ y} (\mathit{Sch}/V)_{fppf} \longrightarrow \mathcal{X} \times _{f, \mathcal{Z}, g} \mathcal{Y}.$

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 93.10.11. Moreover, the morphism is the composition of base changes of $x$ and $y$, hence surjective and smooth, see Lemmas 93.10.6 and 93.10.5. 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

$(\mathit{Sch}/W)_{fppf} \longrightarrow (\mathit{Sch}/U)_{fppf} \times _{f \circ x, \mathcal{Z}, g \circ y} (\mathit{Sch}/V)_{fppf}$

is surjective and smooth which proves the last condition. $\square$

Lemma 93.14.3. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ be algebraic stacks over $S$. Let $f : \mathcal{X} \to \mathcal{Z}$, $g : \mathcal{Y} \to \mathcal{Z}$ be $1$-morphisms of algebraic stacks. Then the $2$-fibre product $\mathcal{X} \times _{f, \mathcal{Z}, g} \mathcal{Y}$ is an algebraic stack. It is also the $2$-fibre product in the $2$-category of algebraic stacks over $(\mathit{Sch}/S)_{fppf}$.

Proof. The fact that $\mathcal{X} \times _{f, \mathcal{Z}, g} \mathcal{Y}$ is an algebraic stack follows from the stronger Lemma 93.14.2. The fact that $\mathcal{X} \times _{f, \mathcal{Z}, g} \mathcal{Y}$ is a $2$-fibre product in the $2$-category of algebraic stacks over $S$ follows formally from the fact that the $2$-category of algebraic stacks over $S$ is a full sub $2$-category of the $2$-category of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. $\square$

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