## 97.8 Algebraic morphisms

The following notion is occasionally useful.

Definition 97.8.1. Let $S$ be a scheme. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. We say that $F$ is *algebraic* if for every scheme $T$ and every object $\xi $ of $\mathcal{Y}$ over $T$ the $2$-fibre product

\[ (\mathit{Sch}/T)_{fppf} \times _{\xi , \mathcal{Y}} \mathcal{X} \]

is an algebraic stack over $S$.

With this terminology in place we have the following result that generalizes Algebraic Stacks, Lemma 94.15.4.

Lemma 97.8.2. Let $S$ be a scheme. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. If

$\mathcal{Y}$ is an algebraic stack, and

$F$ is algebraic (see above),

then $\mathcal{X}$ is an algebraic stack.

**Proof.**
By assumption (1) there exists a scheme $T$ and an object $\xi $ of $\mathcal{Y}$ over $T$ such that the corresponding $1$-morphism $\xi : (\mathit{Sch}/T)_{fppf} \to \mathcal{Y}$ is smooth an surjective. Then $\mathcal{U} = (\mathit{Sch}/T)_{fppf} \times _{\xi , \mathcal{Y}} \mathcal{X}$ is an algebraic stack by assumption (2). Choose a scheme $U$ and a surjective smooth $1$-morphism $(\mathit{Sch}/U)_{fppf} \to \mathcal{U}$. The projection $\mathcal{U} \longrightarrow \mathcal{X}$ is, as the base change of the morphism $\xi : (\mathit{Sch}/T)_{fppf} \to \mathcal{Y}$, surjective and smooth, see Algebraic Stacks, Lemma 94.10.6. Then the composition $(\mathit{Sch}/U)_{fppf} \to \mathcal{U} \to \mathcal{X}$ is surjective and smooth as a composition of surjective and smooth morphisms, see Algebraic Stacks, Lemma 94.10.5. Hence $\mathcal{X}$ is an algebraic stack by Algebraic Stacks, Lemma 94.15.3.
$\square$

Lemma 97.8.3. Let $S$ be a scheme. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. If $\mathcal{X}$ is an algebraic stack and $\Delta : \mathcal{Y} \to \mathcal{Y} \times \mathcal{Y}$ is representable by algebraic spaces, then $F$ is algebraic.

**Proof.**
Choose a representable stack in groupoids $\mathcal{U}$ and a surjective smooth $1$-morphism $\mathcal{U} \to \mathcal{X}$. Let $T$ be a scheme and let $\xi $ be an object of $\mathcal{Y}$ over $T$. The morphism of $2$-fibre products

\[ (\mathit{Sch}/T)_{fppf} \times _{\xi , \mathcal{Y}} \mathcal{U} \longrightarrow (\mathit{Sch}/T)_{fppf} \times _{\xi , \mathcal{Y}} \mathcal{X} \]

is representable by algebraic spaces, surjective, and smooth as a base change of $\mathcal{U} \to \mathcal{X}$, see Algebraic Stacks, Lemmas 94.9.7 and 94.10.6. By our condition on the diagonal of $\mathcal{Y}$ we see that the source of this morphism is representable by an algebraic space, see Algebraic Stacks, Lemma 94.10.11. Hence the target is an algebraic stack by Algebraic Stacks, Lemma 94.15.3.
$\square$

Lemma 97.8.4. Let $S$ be a scheme. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. If $F$ is algebraic and $\Delta : \mathcal{Y} \to \mathcal{Y} \times \mathcal{Y}$ is representable by algebraic spaces, then $\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is representable by algebraic spaces.

**Proof.**
Assume $F$ is algebraic and $\Delta : \mathcal{Y} \to \mathcal{Y} \times \mathcal{Y}$ is representable by algebraic spaces. Take a scheme $U$ over $S$ and two objects $x_1, x_2$ of $\mathcal{X}$ over $U$. We have to show that $\mathit{Isom}(x_1, x_2)$ is an algebraic space over $U$, see Algebraic Stacks, Lemma 94.10.11. Set $y_ i = F(x_ i)$. We have a morphism of sheaves of sets

\[ f : \mathit{Isom}(x_1, x_2) \to \mathit{Isom}(y_1, y_2) \]

and the target is an algebraic space by assumption. Thus it suffices to show that $f$ is representable by algebraic spaces, see Bootstrap, Lemma 80.3.6. Thus we can choose a scheme $V$ over $U$ and an isomorphism $\beta : y_{1, V} \to y_{2, V}$ and we have to show the functor

\[ (\mathit{Sch}/V)_{fppf} \to \textit{Sets},\quad T/V \mapsto \{ \alpha : x_{1, T} \to x_{2, T} \text{ in }\mathcal{X}_ T \mid F(\alpha ) = \beta |_ T\} \]

is an algebraic space. Consider the objects $z_1 = (V, x_{1, V}, \text{id})$ and $z_2 = (V, x_{2, V}, \beta )$ of

\[ \mathcal{Z} = (\mathit{Sch}/V)_{fppf} \times _{y_{1, V}, \mathcal{Y}} \mathcal{X} \]

Then it is straightforward to verify that the functor above is equal to $\mathit{Isom}(z_1, z_2)$ on $(\mathit{Sch}/V)_{fppf}$. Hence this is an algebraic space by our assumption that $F$ is algebraic (and the definition of algebraic stacks).
$\square$

## Comments (0)