The Stacks project

Lemma 96.12.5. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. Assume that $\Delta : \mathcal{Y} \to \mathcal{Y} \times \mathcal{Y}$ is representable by algebraic spaces. Then

\[ \mathcal{H}_ d(\mathcal{X}/\mathcal{Y}) \longrightarrow \mathcal{H}_ d(\mathcal{X}) \times \mathcal{Y} \]

see Examples of Stacks, Equation ( is representable by algebraic spaces.

Proof. Let $U$ be a scheme and let $\xi = (U, Z, p, x, 1)$ be an object of $\mathcal{H}_ d(\mathcal{X}) = \mathcal{H}_ d(\mathcal{X}/S)$ over $U$. Here $p$ is just the structure morphism of $U$. The fifth component $1$ exists and is unique since everything is over $S$. Also, let $y$ be an object of $\mathcal{Y}$ over $U$. We have to show the $2$-fibre product
\begin{equation} \label{criteria-equation-res-isom} (\mathit{Sch}/U)_{fppf} \times _{\xi \times y, \mathcal{H}_ d(\mathcal{X}) \times \mathcal{Y}} \mathcal{H}_ d(\mathcal{X}/\mathcal{Y}) \end{equation}

is representable by an algebraic space. To explain why this is so we introduce

\[ I = \mathit{Isom}_\mathcal {Y}(y|_ Z, F(x)) \]

which is an algebraic space over $Z$ by assumption. Let $a : U' \to U$ be a scheme over $U$. What does it mean to give an object of the fibre category of ( over $U'$? Well, it means that we have an object $\xi ' = (U', Z', y', x', \alpha ')$ of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ over $U'$ and isomorphisms $(U', Z', p', x', 1) \cong (U, Z, p, x, 1)|_{U'}$ and $y' \cong y|_{U'}$. Thus $\xi '$ is isomorphic to $(U', U' \times _{a, U} Z, a^*y, x|_{U' \times _{a, U} Z}, \alpha )$ for some morphism

\[ \alpha : a^*y|_{U' \times _{a, U} Z} \longrightarrow F(x|_{U' \times _{a, U} Z}) \]

in the fibre category of $\mathcal{Y}$ over $U' \times _{a, U} Z$. Hence we can view $\alpha $ as a morphism $b : U' \times _{a, U} Z \to I$. In this way we see that ( is representable by $\text{Res}_{Z/U}(I)$ which is an algebraic space by Proposition 96.11.5. $\square$

Comments (2)

Comment #28 by David Zureick-Brown on

Typo: representably.

Comment #29 by Johan on

Fixed. Thanks!

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