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Tag 04SX

84.9. Morphisms representable by algebraic spaces

In analogy with Categories, Definition 4.40.5 we make the following definition.

Definition 84.9.1. Let $S$ be a scheme contained in $\textit{Sch}_{fppf}$. A $1$-morphism $f : \mathcal{X} \to \mathcal{Y}$ of categories fibred in groupoids over $(\textit{Sch}/S)_{fppf}$ is called representable by algebraic spaces if for any $U \in \mathop{\rm Ob}\nolimits((\textit{Sch}/S)_{fppf})$ and any $y : (\textit{Sch}/U)_{fppf} \to \mathcal{Y}$ the category fibred in groupoids $$ (\textit{Sch}/U)_{fppf} \times_{y, \mathcal{Y}} \mathcal{X} $$ over $(\textit{Sch}/U)_{fppf}$ is representable by an algebraic space over $U$.

Choose an algebraic space $F_y$ over $U$ which represents $(\textit{Sch}/U)_{fppf} \times_{y, \mathcal{Y}} \mathcal{X}$. We may think of $F_y$ as an algebraic space over $S$ which comes equipped with a canonical morphism $f_y : F_y \to U$ over $S$, see Spaces, Section 56.16. Here is the diagram \begin{equation} \tag{84.9.1.1} \vcenter{ \xymatrix{ F_y \ar[d]_{f_y} & (\textit{Sch}/U)_{fppf} \times_{y, \mathcal{Y}} \mathcal{X} \ar@{~>}[l] \ar[d]_{\text{pr}_0} \ar[r]_-{\text{pr}_1} & \mathcal{X} \ar[d]^f \\ U & (\textit{Sch}/U)_{fppf} \ar@{~>}[l] \ar[r]^-y & \mathcal{Y} } } \end{equation} where the squiggly arrows represent the construction which associates to a stack fibred in setoids its associated sheaf of isomorphism classes of objects. The right square is $2$-commutative, and is a $2$-fibre product square.

Here is the analogue of Categories, Lemma 4.40.7.

Lemma 84.9.2. Let $S$ be a scheme contained in $\textit{Sch}_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $(\textit{Sch}/S)_{fppf}$. The following are necessary and sufficient conditions for $f$ to be representable by algebraic spaces:

  1. for each scheme $U/S$ the functor $f_U : \mathcal{X}_U \longrightarrow \mathcal{Y}_U$ between fibre categories is faithful, and
  2. for each $U$ and each $y \in \mathop{\rm Ob}\nolimits(\mathcal{Y}_U)$ the presheaf $$ (h : V \to U) \longmapsto \{(x, \phi) \mid x \in \mathop{\rm Ob}\nolimits(\mathcal{X}_V), \phi : h^*y \to f(x)\}/\cong $$ is an algebraic space over $U$.

Here we have made a choice of pullbacks for $\mathcal{Y}$.

Proof. This follows from the description of fibre categories of the $2$-fibre products $(\textit{Sch}/U)_{fppf} \times_{y, \mathcal{Y}} \mathcal{X}$ in Categories, Lemma 4.40.3 combined with Lemma 84.8.2. $\square$

Here are some lemmas about this notion that work in great generality.

Lemma 84.9.3. Let $S$ be an object of $\textit{Sch}_{fppf}$. Consider a $2$-commutative diagram $$ \xymatrix{ \mathcal{X}' \ar[r] \ar[d]_{f'} & \mathcal{X} \ar[d]^f \\ \mathcal{Y}' \ar[r] & \mathcal{Y} } $$ of $1$-morphisms of categories fibred in groupoids over $(\textit{Sch}/S)_{fppf}$. Assume the horizontal arrows are equivalences. Then $f$ is representable by algebraic spaces if and only if $f'$ is representable by algebraic spaces.

Proof. Omitted. $\square$

Lemma 84.9.4. Let $S$ be an object of $\textit{Sch}_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $S$. If $\mathcal{X}$ and $\mathcal{Y}$ are representable by algebraic spaces over $S$, then the $1$-morphism $f$ is representable by algebraic spaces.

Proof. Omitted. This relies only on the fact that the category of algebraic spaces over $S$ has fibre products, see Spaces, Lemma 56.7.3. $\square$

Lemma 84.9.5. Let $S$ be an object of $\textit{Sch}_{fppf}$. Let $a : F \to G$ be a map of presheaves of sets on $(\textit{Sch}/S)_{fppf}$. Denote $a' : \mathcal{S}_F \to \mathcal{S}_G$ the associated map of categories fibred in sets. Then $a$ is representable by algebraic spaces (see Bootstrap, Definition 71.3.1) if and only if $a'$ is representable by algebraic spaces.

Proof. Omitted. $\square$

Lemma 84.9.6. Let $S$ be an object of $\textit{Sch}_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in setoids over $(\textit{Sch}/S)_{fppf}$. Let $F$, resp. $G$ be the presheaf which to $T$ associates the set of isomorphism classes of objects of $\mathcal{X}_T$, resp. $\mathcal{Y}_T$. Let $a : F \to G$ be the map of presheaves corresponding to $f$. Then $a$ is representable by algebraic spaces (see Bootstrap, Definition 71.3.1) if and only if $f$ is representable by algebraic spaces.

Proof. Omitted. Hint: Combine Lemmas 84.9.3 and 84.9.5. $\square$

Lemma 84.9.7. Let $S$ be a scheme contained in $\textit{Sch}_{fppf}$. Let $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ be categories fibred in groupoids over $(\textit{Sch}/S)_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism representable by algebraic spaces. Let $g : \mathcal{Z} \to \mathcal{Y}$ be any $1$-morphism. Consider the fibre product diagram $$ \xymatrix{ \mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X} \ar[r]_-{g'} \ar[d]_{f'} & \mathcal{X} \ar[d]^f \\ \mathcal{Z} \ar[r]^g & \mathcal{Y} } $$ Then the base change $f'$ is a $1$-morphism representable by algebraic spaces.

Proof. This is formal. $\square$

Lemma 84.9.8. Let $S$ be a scheme contained in $\textit{Sch}_{fppf}$. Let $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ be categories fibred in groupoids over $(\textit{Sch}/S)_{fppf}$ Let $f : \mathcal{X} \to \mathcal{Y}$, $g : \mathcal{Z} \to \mathcal{Y}$ be $1$-morphisms. Assume

  1. $f$ is representable by algebraic spaces, and
  2. $\mathcal{Z}$ is representable by an algebraic space over $S$.

Then the $2$-fibre product $\mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X}$ is representable by an algebraic space.

Proof. This is a reformulation of Bootstrap, Lemma 71.3.6. First note that $\mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X}$ is fibred in setoids over $(\textit{Sch}/S)_{fppf}$. Hence it is equivalent to $\mathcal{S}_F$ for some presheaf $F$ on $(\textit{Sch}/S)_{fppf}$, see Categories, Lemma 4.38.5. Moreover, let $G$ be an algebraic space which represents $\mathcal{Z}$. The $1$-morphism $\mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X} \to \mathcal{Z}$ is representable by algebraic spaces by Lemma 84.9.7. And $\mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X} \to \mathcal{Z}$ corresponds to a morphism $F \to G$ by Categories, Lemma 4.38.6. Then $F \to G$ is representable by algebraic spaces by Lemma 84.9.6. Hence Bootstrap, Lemma 71.3.6 implies that $F$ is an algebraic space as desired. $\square$

Let $S$, $\mathcal{X}$, $\mathcal{Y}$, $\mathcal{Z}$, $f$, $g$ be as in Lemma 84.9.8. Let $F$ and $G$ be algebraic spaces over $S$ such that $F$ represents $\mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X}$ and $G$ represents $\mathcal{Z}$. The $1$-morphism $f' : \mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X} \to \mathcal{Z}$ corresponds to a morphism $f' : F \to G$ of algebraic spaces by (84.8.2.1). Thus we have the following diagram \begin{equation} \tag{84.9.8.1} \vcenter{ \xymatrix{ F \ar[d]_{f'} & \mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X} \ar@{~>}[l] \ar[d] \ar[r] & \mathcal{X} \ar[d]^f \\ G & \mathcal{Z} \ar@{~>}[l] \ar[r]^-g & \mathcal{Y} } } \end{equation} where the squiggly arrows represent the construction which associates to a stack fibred in setoids its associated sheaf of isomorphism classes of objects.

Lemma 84.9.9. Let $S$ be a scheme contained in $\textit{Sch}_{fppf}$. Let $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ be categories fibred in groupoids over $(\textit{Sch}/S)_{fppf}$. If $f : \mathcal{X} \to \mathcal{Y}$, $g : \mathcal{Y} \to \mathcal{Z}$ are $1$-morphisms representable by algebraic spaces, then $$ g \circ f : \mathcal{X} \longrightarrow \mathcal{Z} $$ is a $1$-morphism representable by algebraic spaces.

Proof. This follows from Lemma 84.9.8. Details omitted. $\square$

Lemma 84.9.10. Let $S$ be a scheme contained in $\textit{Sch}_{fppf}$. Let $\mathcal{X}_i, \mathcal{Y}_i$ be categories fibred in groupoids over $(\textit{Sch}/S)_{fppf}$, $i = 1, 2$. Let $f_i : \mathcal{X}_i \to \mathcal{Y}_i$, $i = 1, 2$ be $1$-morphisms representable by algebraic spaces. Then $$ f_1 \times f_2 : \mathcal{X}_1 \times \mathcal{X}_2 \longrightarrow \mathcal{Y}_1 \times \mathcal{Y}_2 $$ is a $1$-morphism representable by algebraic spaces.

Proof. Write $f_1 \times f_2$ as the composition $\mathcal{X}_1 \times \mathcal{X}_2 \to \mathcal{Y}_1 \times \mathcal{X}_2 \to \mathcal{Y}_1 \times \mathcal{Y}_2$. The first arrow is the base change of $f_1$ by the map $\mathcal{Y}_1 \times \mathcal{X}_2 \to \mathcal{Y}_1$, and the second arrow is the base change of $f_2$ by the map $\mathcal{Y}_1 \times \mathcal{Y}_2 \to \mathcal{Y}_2$. Hence this lemma is a formal consequence of Lemmas 84.9.9 and 84.9.7. $\square$

Lemma 84.9.11. Let $S$ be a scheme contained in $\textit{Sch}_{fppf}$. Let $\mathcal{X} \to \mathcal{Z}$ and $\mathcal{Y} \to \mathcal{Z}$ be $1$-morphisms of categories fibred in groupoids over $(\textit{Sch}/S)_{fppf}$. If $\mathcal{X} \to \mathcal{Z}$ is representable by algebraic spaces and $\mathcal{Y}$ is a stack in groupoids, then $\mathcal{X} \times_\mathcal{Z} \mathcal{Y}$ is a stack in groupoids.

Proof. The property of a morphism being representable by algebraic spaces is preserved under base-change (Lemma 84.9.8), and so, passing to the base-change $\mathcal{X} \times_\mathcal{Z} \mathcal{Y}$ over $\mathcal{Y}$, we may reduce to the case of a morphism of categories fibred in groupoids $\mathcal{X} \to \mathcal{Y}$ which is representable by algebraic spaces, and whose target is a stack in groupoids; our goal is then to prove that $\mathcal{X}$ is also a stack in groupoids. This follows from Stacks, Lemma 8.6.11 whose assumptions are satisfied as a result of Lemma 84.9.2. $\square$

    The code snippet corresponding to this tag is a part of the file algebraic.tex and is located in lines 495–835 (see updates for more information).

    \section{Morphisms representable by algebraic spaces}
    \label{section-morphisms-representable-by-algebraic-spaces}
    
    \noindent
    In analogy with Categories, Definition
    \ref{categories-definition-representable-map-categories-fibred-in-groupoids}
    we make the following definition.
    
    \begin{definition}
    \label{definition-representable-by-algebraic-spaces}
    Let $S$ be a scheme contained in $\Sch_{fppf}$.
    A $1$-morphism $f : \mathcal{X} \to \mathcal{Y}$ of
    categories fibred in groupoids over $(\Sch/S)_{fppf}$
    is called {\it representable by algebraic spaces} if
    for any $U \in \Ob((\Sch/S)_{fppf})$
    and any $y : (\Sch/U)_{fppf} \to \mathcal{Y}$
    the category fibred in groupoids
    $$
    (\Sch/U)_{fppf} \times_{y, \mathcal{Y}} \mathcal{X}
    $$
    over $(\Sch/U)_{fppf}$
    is representable by an algebraic space over $U$.
    \end{definition}
    
    \noindent
    Choose an algebraic space $F_y$ over $U$ which represents
    $(\Sch/U)_{fppf} \times_{y, \mathcal{Y}} \mathcal{X}$.
    We may think of $F_y$ as an algebraic space over $S$
    which comes equipped with a canonical morphism $f_y : F_y \to U$
    over $S$, see
    Spaces, Section \ref{spaces-section-change-base-scheme}.
    Here is the diagram
    \begin{equation}
    \label{equation-representable-by-algebraic-spaces}
    \vcenter{
    \xymatrix{
    F_y \ar[d]_{f_y} &
    (\Sch/U)_{fppf} \times_{y, \mathcal{Y}} \mathcal{X}
    \ar@{~>}[l] \ar[d]_{\text{pr}_0} \ar[r]_-{\text{pr}_1} &
    \mathcal{X} \ar[d]^f \\
    U &
    (\Sch/U)_{fppf} \ar@{~>}[l] \ar[r]^-y &
    \mathcal{Y}
    }
    }
    \end{equation}
    where the squiggly arrows represent the construction which associates
    to a stack fibred in setoids its associated sheaf of isomorphism classes
    of objects. The right square is
    $2$-commutative, and is a $2$-fibre product square.
    
    \medskip\noindent
    Here is the analogue of Categories,
    Lemma \ref{categories-lemma-criterion-representable-map-stack-in-groupoids}.
    
    \begin{lemma}
    \label{lemma-criterion-map-representable-spaces-fibred-in-groupoids}
    Let $S$ be a scheme contained in $\Sch_{fppf}$.
    Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism
    of categories fibred in groupoids over $(\Sch/S)_{fppf}$.
    The following are necessary and sufficient conditions for
    $f$ to be representable by algebraic spaces:
    \begin{enumerate}
    \item for each scheme $U/S$ the
    functor $f_U : \mathcal{X}_U \longrightarrow \mathcal{Y}_U$
    between fibre categories is faithful, and
    \item for each $U$ and each $y \in \Ob(\mathcal{Y}_U)$ the presheaf
    $$
    (h : V \to U)
    \longmapsto
    \{(x, \phi) \mid x \in \Ob(\mathcal{X}_V), \phi : h^*y \to f(x)\}/\cong
    $$
    is an algebraic space over $U$.
    \end{enumerate}
    Here we have made a choice of pullbacks for $\mathcal{Y}$.
    \end{lemma}
    
    \begin{proof}
    This follows from the description of fibre categories of the $2$-fibre products
    $(\Sch/U)_{fppf} \times_{y, \mathcal{Y}} \mathcal{X}$ in
    Categories, Lemma \ref{categories-lemma-identify-fibre-product}
    combined with
    Lemma \ref{lemma-characterize-representable-by-space}.
    \end{proof}
    
    \noindent
    Here are some lemmas about this notion that work in great generality.
    
    \begin{lemma}
    \label{lemma-representable-by-spaces-morphism-equivalent}
    Let $S$ be an object of $\Sch_{fppf}$.
    Consider a $2$-commutative diagram
    $$
    \xymatrix{
    \mathcal{X}' \ar[r] \ar[d]_{f'} & \mathcal{X} \ar[d]^f \\
    \mathcal{Y}' \ar[r] & \mathcal{Y}
    }
    $$
    of $1$-morphisms of categories fibred in groupoids over
    $(\Sch/S)_{fppf}$.
    Assume the horizontal arrows are equivalences.
    Then $f$ is representable by algebraic spaces
    if and only if $f'$ is representable by algebraic spaces.
    \end{lemma}
    
    \begin{proof}
    Omitted.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-morphism-spaces-gives-representable-by-spaces}
    Let $S$ be an object of $\Sch_{fppf}$.
    Let $f : \mathcal{X} \to \mathcal{Y}$
    be a $1$-morphism of categories fibred in groupoids over $S$.
    If $\mathcal{X}$ and $\mathcal{Y}$ are representable by
    algebraic spaces over $S$, then the $1$-morphism $f$
    is representable by algebraic spaces.
    \end{lemma}
    
    \begin{proof}
    Omitted. This relies only on the fact that
    the category of algebraic spaces over $S$ has fibre products,
    see Spaces, Lemma \ref{spaces-lemma-fibre-product-spaces}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-map-presheaves-representable-by-algebraic-spaces}
    Let $S$ be an object of $\Sch_{fppf}$.
    Let $a : F \to G$ be a map of presheaves of sets on $(\Sch/S)_{fppf}$.
    Denote $a' : \mathcal{S}_F  \to \mathcal{S}_G$ the associated
    map of categories fibred in sets.
    Then $a$ is representable by algebraic spaces (see
    Bootstrap,
    Definition \ref{bootstrap-definition-morphism-representable-by-spaces})
    if and only if $a'$ is representable by algebraic spaces.
    \end{lemma}
    
    \begin{proof}
    Omitted.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-map-fibred-setoids-representable-algebraic-spaces}
    Let $S$ be an object of $\Sch_{fppf}$.
    Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of
    categories fibred in setoids over $(\Sch/S)_{fppf}$.
    Let $F$, resp.\ $G$ be the presheaf which to $T$ associates
    the set of isomorphism classes of objects of
    $\mathcal{X}_T$, resp.\ $\mathcal{Y}_T$.
    Let $a : F \to G$ be the map of presheaves corresponding to $f$.
    Then $a$ is representable by algebraic spaces (see
    Bootstrap,
    Definition \ref{bootstrap-definition-morphism-representable-by-spaces})
    if and only if $f$ is representable by algebraic spaces.
    \end{lemma}
    
    \begin{proof}
    Omitted. Hint: Combine
    Lemmas \ref{lemma-representable-by-spaces-morphism-equivalent}
    and \ref{lemma-map-presheaves-representable-by-algebraic-spaces}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-base-change-representable-by-spaces}
    Let $S$ be a scheme contained in $\Sch_{fppf}$.
    Let $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$
    be categories fibred in groupoids over $(\Sch/S)_{fppf}$.
    Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism
    representable by algebraic spaces.
    Let $g : \mathcal{Z} \to \mathcal{Y}$ be any $1$-morphism.
    Consider the fibre product diagram
    $$
    \xymatrix{
    \mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X} \ar[r]_-{g'} \ar[d]_{f'} &
    \mathcal{X} \ar[d]^f \\
    \mathcal{Z} \ar[r]^g & \mathcal{Y}
    }
    $$
    Then the base change $f'$ is a $1$-morphism representable by
    algebraic spaces.
    \end{lemma}
    
    \begin{proof}
    This is formal.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-base-change-by-space-representable-by-space}
    Let $S$ be a scheme contained in $\Sch_{fppf}$.
    Let $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$
    be categories fibred in groupoids over $(\Sch/S)_{fppf}$
    Let $f : \mathcal{X} \to \mathcal{Y}$,
    $g : \mathcal{Z} \to \mathcal{Y}$ be $1$-morphisms.
    Assume
    \begin{enumerate}
    \item $f$ is representable by algebraic spaces, and
    \item $\mathcal{Z}$ is representable by an algebraic space over $S$.
    \end{enumerate}
    Then the $2$-fibre product
    $\mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X}$
    is representable by an algebraic space.
    \end{lemma}
    
    \begin{proof}
    This is a reformulation of
    Bootstrap, Lemma \ref{bootstrap-lemma-representable-by-spaces-over-space}.
    First note that
    $\mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X}$
    is fibred in setoids over $(\Sch/S)_{fppf}$.
    Hence it is equivalent to $\mathcal{S}_F$ for some presheaf
    $F$ on $(\Sch/S)_{fppf}$, see
    Categories, Lemma \ref{categories-lemma-setoid-fibres}.
    Moreover, let $G$ be an algebraic space which represents
    $\mathcal{Z}$. The $1$-morphism
    $\mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X} \to \mathcal{Z}$
    is representable by algebraic spaces by
    Lemma \ref{lemma-base-change-representable-by-spaces}.
    And $\mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X} \to \mathcal{Z}$
    corresponds to a morphism $F \to G$ by
    Categories, Lemma \ref{categories-lemma-2-category-fibred-setoids}.
    Then $F \to G$ is representable by algebraic spaces by
    Lemma \ref{lemma-map-fibred-setoids-representable-algebraic-spaces}.
    Hence
    Bootstrap, Lemma \ref{bootstrap-lemma-representable-by-spaces-over-space}
    implies that $F$ is an algebraic space as desired.
    \end{proof}
    
    \noindent
    Let $S$, $\mathcal{X}$, $\mathcal{Y}$, $\mathcal{Z}$, $f$, $g$ be as in
    Lemma \ref{lemma-base-change-by-space-representable-by-space}.
    Let $F$ and $G$ be algebraic spaces over $S$ such that
    $F$ represents $\mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X}$
    and $G$ represents $\mathcal{Z}$. The $1$-morphism
    $f' : \mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X} \to \mathcal{Z}$
    corresponds to a morphism $f' : F \to G$ of algebraic spaces
    by (\ref{equation-morphisms-spaces}).
    Thus we have the following diagram
    \begin{equation}
    \label{equation-representable-by-algebraic-spaces-on-space}
    \vcenter{
    \xymatrix{
    F \ar[d]_{f'} &
    \mathcal{Z} \times_{g, \mathcal{Y}, f} \mathcal{X}
    \ar@{~>}[l] \ar[d] \ar[r] &
    \mathcal{X} \ar[d]^f \\
    G &
    \mathcal{Z} \ar@{~>}[l] \ar[r]^-g &
    \mathcal{Y}
    }
    }
    \end{equation}
    where the squiggly arrows represent the construction which associates
    to a stack fibred in setoids its associated sheaf of isomorphism classes
    of objects.
    
    \begin{lemma}
    \label{lemma-composition-representable-by-spaces}
    Let $S$ be a scheme contained in $\Sch_{fppf}$.
    Let $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$
    be categories fibred in groupoids over $(\Sch/S)_{fppf}$.
    If $f : \mathcal{X} \to \mathcal{Y}$, $g : \mathcal{Y} \to \mathcal{Z}$
    are $1$-morphisms representable by algebraic spaces, then
    $$
    g \circ f : \mathcal{X} \longrightarrow \mathcal{Z}
    $$
    is a $1$-morphism representable by algebraic spaces.
    \end{lemma}
    
    \begin{proof}
    This follows from
    Lemma \ref{lemma-base-change-by-space-representable-by-space}.
    Details omitted.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-product-representable-by-spaces}
    Let $S$ be a scheme contained in $\Sch_{fppf}$.
    Let $\mathcal{X}_i, \mathcal{Y}_i$ be categories fibred in groupoids over
    $(\Sch/S)_{fppf}$, $i = 1, 2$.
    Let $f_i : \mathcal{X}_i \to \mathcal{Y}_i$, $i = 1, 2$
    be $1$-morphisms representable by algebraic spaces.
    Then
    $$
    f_1 \times f_2 :
    \mathcal{X}_1 \times \mathcal{X}_2
    \longrightarrow
    \mathcal{Y}_1 \times \mathcal{Y}_2
    $$
    is a $1$-morphism representable by algebraic spaces.
    \end{lemma}
    
    \begin{proof}
    Write $f_1 \times f_2$ as the composition
    $\mathcal{X}_1 \times \mathcal{X}_2 \to
    \mathcal{Y}_1 \times \mathcal{X}_2 \to
    \mathcal{Y}_1 \times \mathcal{Y}_2$.
    The first arrow is the base change of $f_1$ by the map
    $\mathcal{Y}_1 \times \mathcal{X}_2 \to \mathcal{Y}_1$, and the second arrow
    is the base change of $f_2$ by the map
    $\mathcal{Y}_1 \times \mathcal{Y}_2 \to \mathcal{Y}_2$.
    Hence this lemma is a formal
    consequence of Lemmas \ref{lemma-composition-representable-by-spaces}
    and \ref{lemma-base-change-representable-by-spaces}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-get-a-stack}
    \begin{reference}
    Lemma in an email of Matthew Emerton dated June 15, 2016
    \end{reference}
    Let $S$ be a scheme contained in $\Sch_{fppf}$.
    Let $\mathcal{X} \to \mathcal{Z}$ and $\mathcal{Y} \to \mathcal{Z}$
    be $1$-morphisms of categories fibred in groupoids over $(\Sch/S)_{fppf}$.
    If $\mathcal{X} \to \mathcal{Z}$ is representable by algebraic spaces
    and $\mathcal{Y}$ is a stack in groupoids, then
    $\mathcal{X} \times_\mathcal{Z} \mathcal{Y}$ is a stack in groupoids.
    \end{lemma}
    
    \begin{proof}
    The property of a morphism being representable by algebraic spaces
    is preserved under base-change
    (Lemma \ref{lemma-base-change-by-space-representable-by-space}),
    and so, passing to the base-change
    $\mathcal{X} \times_\mathcal{Z} \mathcal{Y}$ over $\mathcal{Y}$,
    we may reduce to the case of a morphism of categories
    fibred in groupoids $\mathcal{X} \to \mathcal{Y}$
    which is representable by algebraic spaces, and
    whose target is a stack in groupoids; our goal is then to prove
    that $\mathcal{X}$ is also a stack in groupoids.
    This follows from Stacks, Lemma
    \ref{stacks-lemma-relative-sheaf-over-stack-is-stack}
    whose assumptions are satisfied as a result of
    Lemma \ref{lemma-criterion-map-representable-spaces-fibred-in-groupoids}.
    \end{proof}

    Comments (2)

    Comment #2248 by David Hansen on October 6, 2016 a 3:18 pm UTC

    The use of "middle square" in reference to part of diagram 77.9.8.1 is confusing (cf. second line below the diagram) - I only see left and right squares. :)

    Comment #2282 by Johan (site) on November 3, 2016 a 6:41 pm UTC

    Dear David Hansen, indeed! Fixed.

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