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The corresponding content:
Lemma 65.12.5. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids over $(\textit{Sch}/S)_{fppf}$. 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 (63.17.2.1) 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} \tag{65.12.5.1} (\textit{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 (65.12.5.1) 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 (65.12.5.1) is representable by $\text{Res}_{Z/U}(I)$ which is an algebraic space by Proposition 65.11.5. $\square$
\begin{lemma}
\label{lemma-relative-hilbert}
Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids
over $(\Sch/S)_{fppf}$. 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
(\ref{examples-stacks-equation-diagram-hilbert-d-stack})
is representable by algebraic spaces.
\end{lemma}
\begin{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{equation-res-isom}
(\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 (\ref{equation-res-isom}) 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 (\ref{equation-res-isom})
is representable by $\text{Res}_{Z/U}(I)$ which is an algebraic space by
Proposition \ref{proposition-restriction-of-scalars-algebraic-space}.
\end{proof}
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