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\input{preamble}
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\begin{document}
\title{Stacks}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
In this very short chapter we introduce stacks, and
stacks in groupoids. See \cite{DM}, and \cite{Vis2}.
\section{Presheaves of morphisms associated to fibred categories}
\label{section-morphisms}
\noindent
Let $\mathcal{C}$ be a category.
Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category,
see Categories, Section \ref{categories-section-fibred-categories}.
Suppose that $x, y\in \Ob(\mathcal{S}_U)$ are
objects in the fibre category over $U$. We are going to define
a functor
$$
\mathit{Mor}(x, y) : (\mathcal{C}/U)^{opp} \longrightarrow \textit{Sets}.
$$
In other words this will be a presheaf on $\mathcal{C}/U$, see
Sites, Definition \ref{sites-definition-presheaf}.
Make a choice of pullbacks as in
Categories,
Definition \ref{categories-definition-pullback-functor-fibred-category}.
Then, for $f : V \to U$ we set
$$
\mathit{Mor}(x, y)(f : V \to U) =
\Mor_{\mathcal{S}_V}(f^\ast x, f^\ast y).
$$
Let $f' : V' \to U$ be a second object of $\mathcal{C}/U$.
We also have to define the restriction map corresponding to a
morphism $g : V'/U \to V/U$ in $\mathcal{C}/U$,
in other words $g : V' \to V$ and $f' = f \circ g$.
This will be a map
$$
\Mor_{\mathcal{S}_V}(f^\ast x, f^\ast y)
\longrightarrow
\Mor_{\mathcal{S}_{V'}}({f'}^\ast x, {f'}^\ast y), \quad
\phi \longmapsto \phi|_{V'}
$$
This map will basically be $g^\ast$, except that this transforms
an element $\phi$ of the left hand side into an element
$g^\ast \phi$
of $\Mor_{\mathcal{S}_{V'}}(g^\ast f^\ast x, g^\ast f^\ast y)$.
At this point we use the transformation $\alpha_{g, f}$ of
Categories, Lemma \ref{categories-lemma-fibred}.
In a formula, the restriction map is described by
$$
\phi|_{V'} =
(\alpha_{g, f})_y^{-1} \circ
g^\ast \phi \circ
(\alpha_{g, f})_x.
$$
Of course, nobody thinks of this restriction map in this way.
We will only do this once in order to verify the following
lemma.
\begin{lemma}
\label{lemma-painful}
This actually does give a presheaf.
\end{lemma}
\begin{proof}
Let $g : V'/U \to V/U$ be as above and similarly
$g' : V''/U \to V'/U$ be morphisms in $\mathcal{C}/U$.
So $f' = f \circ g$ and $f'' = f' \circ g' = f \circ g \circ g'$.
Let $\phi \in \Mor_{\mathcal{S}_V}(f^\ast x, f^\ast y)$.
Then we have
\begin{eqnarray*}
& &
(\alpha_{g \circ g', f})_y^{-1} \circ
(g \circ g')^\ast \phi \circ
(\alpha_{g \circ g', f})_x
\\
& = &
(\alpha_{g \circ g', f})_y^{-1} \circ
(\alpha_{g', g})_{f^*y}^{-1} \circ
(g')^*g^\ast \phi \circ
(\alpha_{g', g})_{f^*x} \circ
(\alpha_{g \circ g', f})_x
\\
& = &
(\alpha_{g', f'})_y^{-1} \circ
(g')^*(\alpha_{g, f})_y^{-1} \circ
(g')^* g^\ast \phi \circ
(g')^*(\alpha_{g, f})_x
\circ
(\alpha_{g', f'})_x
\\
& = &
(\alpha_{g', f'})_y^{-1} \circ
(g')^*\Big(
(\alpha_{g, f})_y^{-1} \circ
g^\ast \phi \circ
(\alpha_{g, f})_x
\Big) \circ
(\alpha_{g', f'})_x
\end{eqnarray*}
which is what we want, namely $\phi|_{V''} = (\phi|_{V'})|_{V''}$.
The first equality holds because
$\alpha_{g', g}$ is a transformation of functors, and hence
$$
\xymatrix{
(g \circ g')^*f^*x
\ar[rr]_{(g \circ g')^\ast \phi}
\ar[d]_{(\alpha_{g', g})_{f^*x}} & &
(g \circ g')^*f^*y
\ar[d]^{(\alpha_{g', g})_{f^*y}} \\
(g')^*g^*f^*x
\ar[rr]^{(g')^*g^\ast \phi} & &
(g')^*g^*f^*y
}
$$
commutes. The second equality holds because of property (d) of
a pseudo functor since $f' = f \circ g$ (see
Categories, Definition \ref{categories-definition-functor-into-2-category}).
The last equality follows from the fact that $(g')^*$ is a functor.
\end{proof}
\noindent
From now on we often omit mentioning the transformations
$\alpha_{g, f}$ and we simply identify the functors
$g^* \circ f^*$ and $(f \circ g)^*$. In particular,
given $g : V'/U \to V/U$ the restriction
mappings for the presheaf $\mathit{Mor}(x, y)$
will sometimes be denoted $\phi \mapsto g^*\phi$.
We formalize the construction in a definition.
\begin{definition}
\label{definition-mor-presheaf}
Let $\mathcal{C}$ be a category.
Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category,
see Categories, Section \ref{categories-section-fibred-categories}.
Given an object $U$ of $\mathcal{C}$ and objects
$x$, $y$ of the fibre category, the {\it presheaf
of morphisms from $x$ to $y$} is the presheaf
$$
(f : V \to U) \longmapsto \Mor_{\mathcal{S}_V}(f^*x, f^*y)
$$
described above. It is denoted $\mathit{Mor}(x, y)$.
The subpresheaf $\mathit{Isom}(x, y)$ whose values
over $V$ is the set of isomorphisms
$f^*x \to f^*y$ in the fibre category $\mathcal{S}_V$
is called the {\it presheaf of isomorphisms from $x$ to $y$}.
\end{definition}
\noindent
If $\mathcal{S}$ is fibred in groupoids then of course
$\mathit{Isom}(x, y) = \mathit{Mor}(x, y)$, and it is
customary to use the $\mathit{Isom}$ notation.
\begin{lemma}
\label{lemma-presheaf-mor-map-fibred-categories}
Let $F : \mathcal{S}_1 \to \mathcal{S}_2$ be a $1$-morphism of fibred
categories over the category $\mathcal{C}$. Let $U \in \Ob(\mathcal{C})$
and $x, y\in \Ob(\mathcal{S}_U)$. Then $F$ defines a canonical
morphism of presheaves
$$
\mathit{Mor}_{\mathcal{S}_1}(x, y)
\longrightarrow
\mathit{Mor}_{\mathcal{S}_2}(F(x), F(y))
$$
on $\mathcal{C}/U$.
\end{lemma}
\begin{proof}
By
Categories, Definition \ref{categories-definition-fibred-categories-over-C}
the functor $F$ maps strongly cartesian morphisms to strongly cartesian
morphisms. Hence if $f : V \to U$ is a morphism in $\mathcal{C}$, then
there are canonical isomorphisms $\alpha_V : f^*F(x) \to F(f^*x)$,
$\beta_V : f^*F(y) \to F(f^*y)$ such that $f^*F(x) \to F(f^*x) \to F(x)$
is the canonical morphism $f^*F(x) \to F(x)$, and similarly for $\beta_V$.
Thus we may define
$$
\xymatrix{
\mathit{Mor}_{\mathcal{S}_1}(x, y)(f : V \to U) \ar@{=}[r] &
\Mor_{\mathcal{S}_{1, V}}(f^\ast x, f^\ast y) \ar[d] \\
\mathit{Mor}_{\mathcal{S}_2}(F(x), F(y))(f : V \to U) \ar@{=}[r] &
\Mor_{\mathcal{S}_{2, V}}(f^\ast F(x), f^\ast F(y))
}
$$
by $\phi \mapsto \beta_V^{-1} \circ F(\phi) \circ \alpha_V$.
We omit the verification that this is compatible with the restriction
mappings.
\end{proof}
\begin{remark}
\label{remark-alternative}
Suppose that $p : \mathcal{S} \to \mathcal{C}$ is fibred in groupoids.
In this case we can prove
Lemma \ref{lemma-painful}
using
Categories, Lemma \ref{categories-lemma-fibred-strict}
which says that $\mathcal{S} \to \mathcal{C}$ is equivalent to the
category associated to a contravariant functor
$F : \mathcal{C} \to \text{Groupoids}$.
In the case of the fibred category associated to $F$
we have $g^* \circ f^* = (f \circ g)^*$ on the nose
and there is no need to use the maps $\alpha_{g, f}$.
In this case the lemma is (even more) trivial. Of course then
one uses that the $\mathit{Mor}(x, y)$ presheaf is
unchanged when passing to an equivalent fibred category which follows from
Lemma \ref{lemma-presheaf-mor-map-fibred-categories}.
\end{remark}
\begin{lemma}
\label{lemma-isom-as-2-fibre-product}
Let $\mathcal{C}$ be a category.
Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category,
see Categories, Section \ref{categories-section-fibred-categories}.
Let $U \in \Ob(\mathcal{C})$ and let $x, y \in \Ob(\mathcal{S}_U)$.
Denote $x, y : \mathcal{C}/U \to \mathcal{S}$ also the corresponding
$1$-morphisms, see
Categories, Lemma \ref{categories-lemma-yoneda-2category}.
Then
\begin{enumerate}
\item the $2$-fibre product
$\mathcal{S} \times_{\mathcal{S} \times \mathcal{S}, (x, y)} \mathcal{C}/U$
is fibred in setoids over $\mathcal{C}/U$, and
\item $\mathit{Isom}(x, y)$ is the presheaf of sets corresponding
to this category fibred in setoids, see
Categories, Lemma \ref{categories-lemma-2-category-fibred-setoids}.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted. Hint: Objects of the $2$-fibre product are
$(a : V \to U, z, a : V \to U, (\alpha, \beta))$ where
$\alpha : z \to a^*x$ and $\beta : z \to a^*y$ are isomorphisms
in $\mathcal{S}_V$. Thus the relationship with $\mathit{Isom}(x, y)$
comes by assigning to such an object the isomorphism
$\beta \circ \alpha^{-1}$.
\end{proof}
\section{Descent data in fibred categories}
\label{section-descent-data}
\noindent
In this section we define the notion of a descent datum
in the abstract setting of a fibred category. Before we
do so we point out that this is completely analogous to
descent data for quasi-coherent sheaves
(Descent, Section \ref{descent-section-equivalence})
and descent data for schemes over schemes
(Descent, Section \ref{descent-section-descent-datum}).
\medskip\noindent
We will use the convention where the projection maps
$\text{pr}_i : X \times \ldots \times X \to X$
are labeled starting with $i = 0$. Hence we have
$\text{pr}_0, \text{pr}_1 : X \times X \to X$,
$\text{pr}_0, \text{pr}_1, \text{pr}_2 : X \times X \times X \to X$,
etc.
\begin{definition}
\label{definition-descent-data}
Let $\mathcal{C}$ be a category.
Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category.
Make a choice of pullbacks as in Categories,
Definition \ref{categories-definition-pullback-functor-fibred-category}.
Let $\mathcal{U} = \{f_i : U_i \to U\}_{i \in I}$
be a family of morphisms of $\mathcal{C}$. Assume all the fibre products
$U_i \times_U U_j$, and $U_i \times_U U_j \times_U U_k$ exist.
\begin{enumerate}
\item A {\it descent datum $(X_i, \varphi_{ij})$ in $\mathcal{S}$
relative to the family $\{f_i : U_i \to U\}$} is given by an object $X_i$
of $\mathcal{S}_{U_i}$ for each $i \in I$, an isomorphism
$\varphi_{ij} : \text{pr}_0^*X_i \to \text{pr}_1^*X_j$
in $\mathcal{S}_{U_i \times_U U_j}$ for each pair $(i, j) \in I^2$
such that for every triple of indices $(i, j, k) \in I^3$ the
diagram
$$
\xymatrix{
\text{pr}_0^*X_i \ar[rd]_{\text{pr}_{01}^*\varphi_{ij}}
\ar[rr]_{\text{pr}_{02}^*\varphi_{ik}} & &
\text{pr}_2^*X_k \\
& \text{pr}_1^*X_j \ar[ru]_{\text{pr}_{12}^*\varphi_{jk}} &
}
$$
in the category $\mathcal{S}_{U_i \times_U U_j \times_U U_k}$
commutes. This is called the {\it cocycle condition}.
\item A {\it morphism $\psi : (X_i, \varphi_{ij}) \to
(X'_i, \varphi'_{ij})$ of descent data} is given
by a family $\psi = (\psi_i)_{i\in I}$ of morphisms
$\psi_i : X_i \to X'_i$ in $\mathcal{S}_{U_i}$
such that all the diagrams
$$
\xymatrix{
\text{pr}_0^*X_i \ar[r]_{\varphi_{ij}} \ar[d]_{\text{pr}_0^*\psi_i}
& \text{pr}_1^*X_j \ar[d]^{\text{pr}_1^*\psi_j} \\
\text{pr}_0^*X'_i \ar[r]^{\varphi'_{ij}} &
\text{pr}_1^*X'_j \\
}
$$
in the categories $\mathcal{S}_{U_i \times_U U_j}$ commute.
\item The category of descent data relative to
$\mathcal{U}$ is denoted $DD(\mathcal{U})$.
\end{enumerate}
\end{definition}
\noindent
The fibre products $U_i \times_U U_j$ and $U_i \times_U U_j \times_U U_k$
will exist if each of the morphisms $f_i : U_i \to U$ is {\it representable},
see Categories, Definition \ref{categories-definition-representable-morphism}.
Recall that in a site one of the conditions for a covering $\{U_i \to U\}$ is
that each of the morphisms is representable, see
Sites, Definition \ref{sites-definition-site} part (3).
In fact the main interest in the definition above is where $\mathcal{C}$
is a site and $\{U_i \to U\}$ is a covering of $\mathcal{C}$. However,
a descent datum is just an abstract gadget that can be defined as above.
This is useful: for example, given a fibred category over $\mathcal{C}$
one can look at the collection of families with respect to which descent data
are effective, and try to use these as the family of coverings for a site.
\begin{remarks}
\label{remarks-definition-descent-datum}
Two remarks on Definition \ref{definition-descent-data} are in order.
Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category.
Let $\{f_i : U_i \to U\}_{i \in I}$, and $(X_i, \varphi_{ij})$
be as in Definition \ref{definition-descent-data}.
\begin{enumerate}
\item There is a diagonal morphism $\Delta : U_i \to U_i \times_U U_i$.
We can pull back $\varphi_{ii}$ via this morphism to get an automorphism
$\Delta^\ast \varphi_{ii} \in \text{Aut}_{U_i}(x_i)$.
On pulling back the cocycle condition for the triple $(i, i, i)$
by $\Delta_{123} : U_i \to U_i \times_U U_i \times_U U_i$ we deduce that
$\Delta^\ast \varphi_{ii} \circ \Delta^\ast \varphi_{ii} =
\Delta^\ast \varphi_{ii}$; thus $\Delta^\ast \varphi_{ii} =
\text{id}_{x_i}$.
\item There is a morphism
$\Delta_{13}: U_i \times_U U_j \to U_i \times_U U_j \times_U U_i$
and we can pull back the
cocycle condition for the triple $(i, j, i)$ to get the
identity $(\sigma^\ast \varphi_{ji}) \circ \varphi_{ij} =
\text{id}_{\text{pr}_0^\ast x_i}$, where
$\sigma : U_i \times_U U_j \to U_j \times_U U_i$ is the switching morphism.
\end{enumerate}
\end{remarks}
\begin{lemma}
\label{lemma-pullback}
(Pullback of descent data.)
Let $\mathcal{C}$ be a category.
Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category.
Make a choice pullbacks as in Categories,
Definition \ref{categories-definition-pullback-functor-fibred-category}.
Let $\mathcal{U} = \{f_i : U_i \to U\}_{i \in I}$, and
$\mathcal{V} = \{V_j \to V\}_{j \in J}$
be a families of morphisms of $\mathcal{C}$ with fixed target.
Assume all the fibre products
$U_i \times_U U_{i'}$, $U_i \times_U U_{i'} \times_U U_{i''}$,
$V_j \times_V V_{j'}$, and $V_j \times_V V_{j'} \times_V V_{j''}$ exist.
Let $\alpha : I \to J$, $h : U \to V$ and
$g_i : U_i \to V_{\alpha(i)}$ be a morphism of families
of maps with fixed target, see
Sites, Definition \ref{sites-definition-morphism-coverings}.
\begin{enumerate}
\item Let $(Y_j, \varphi_{jj'})$ be a descent datum relative to the
family $\{V_j \to V\}$. The system
$$
\left(
g_i^*Y_{\alpha(i)},
(g_i \times g_{i'})^*\varphi_{\alpha(i)\alpha(i')}
\right)
$$
is a descent datum relative to $\mathcal{U}$.
\item This construction defines a functor between descent data relative
to $\mathcal{V}$ and descent data relative to $\mathcal{U}$.
\item Given a second $\alpha' : I \to J$, $h' : U \to V$ and
$g'_i : U_i \to V_{\alpha'(i)}$ morphism of families
of maps with fixed target, then if $h = h'$ the two resulting functors
between descent data are canonically isomorphic.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-pullback-functor}
With $\mathcal{U} = \{U_i \to U\}_{i \in I}$,
$\mathcal{V} = \{V_j \to V\}_{j \in J}$,
$\alpha : I \to J$, $h : U \to V$,
and $g_i : U_i \to V_{\alpha(i)}$ as in Lemma \ref{lemma-pullback}
the functor
$$
(Y_j, \varphi_{jj'}) \longmapsto
(g_i^*Y_{\alpha(i)}, (g_i \times g_{i'})^*\varphi_{\alpha(i)\alpha(i')})
$$
constructed in that lemma
is called the {\it pullback functor} on descent data.
\end{definition}
\noindent
Given $h : U \to V$, if there exists a morphism
$\tilde h : \mathcal{U} \to \mathcal{V}$ covering $h$
then $\tilde h^*$ is independent of the choice of
$\tilde h$ as we saw in Lemma \ref{lemma-pullback}.
Hence we will sometimes simply write $h^*$ to indicate
the pullback functor.
\begin{definition}
\label{definition-effective-descent-datum}
Let $\mathcal{C}$ be a category.
Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category.
Make a choice of pullbacks as in Categories,
Definition \ref{categories-definition-pullback-functor-fibred-category}.
Let $\mathcal{U} = \{f_i : U_i \to U\}_{i \in I}$ be a family of morphisms
with target $U$. Assume all the fibre products
$U_i \times_U U_j$ and $U_i \times_U U_j \times_U U_k$ exist.
\begin{enumerate}
\item Given an object $X$ of $\mathcal{S}_U$ the {\it trivial descent datum}
is the descent datum $(X, \text{id}_X)$ with respect to the family
$\{\text{id}_U : U \to U\}$.
\item Given an object $X$ of $\mathcal{S}_U$
we have a {\it canonical descent datum} on the family of
objects $f_i^*X$ by pulling back the trivial
descent datum $(X, \text{id}_X)$ via the
obvious map $\{f_i : U_i \to U\} \to \{\text{id}_U : U \to U\}$.
We denote this descent datum $(f_i^*X, can)$.
\item A descent datum $(X_i, \varphi_{ij})$
relative to $\{f_i : U_i \to U\}$ is called {\it effective}
if there exists an object $X$ of $\mathcal{S}_U$ such that
$(X_i, \varphi_{ij})$ is isomorphic to $(f_i^*X, can)$.
\end{enumerate}
\end{definition}
\noindent
Note that the rule that associates to $X \in \mathcal{S}_U$ its
canonical descent datum relative to $\mathcal{U}$ defines a
functor
$$
\mathcal{S}_U \longrightarrow DD(\mathcal{U}).
$$
A descent datum is effective if and only if it is in the essential
image of this functor.
Let us make explicit the canonical descent datum as follows.
\begin{lemma}
\label{lemma-trivial-cocycle}
In the situation of
Definition \ref{definition-effective-descent-datum} part (2) the maps
$can_{ij} : \text{pr}_0^*f_i^*X \to \text{pr}_1^*f_j^*X$ are equal to
$(\alpha_{\text{pr}_1, f_j})_X \circ (\alpha_{\text{pr}_0, f_i})_X^{-1}$
where $\alpha_{\cdot, \cdot}$ is as in
Categories, Lemma \ref{categories-lemma-fibred}
and where we
use the equality $f_i \circ \text{pr}_0 = f_j \circ \text{pr}_1$
as maps $U_i \times_U U_j \to U$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Stacks}
\label{section-definition}
\noindent
Here is the definition of a stack. It mixes the notion of a fibred
category with the notion of descent.
\begin{definition}
\label{definition-stack}
Let $\mathcal{C}$ be a site. A {\it stack} over $\mathcal{C}$
is a category $p : \mathcal{S} \to \mathcal{C}$ over $\mathcal{C}$ which
satisfies the following conditions:
\begin{enumerate}
\item $p : \mathcal{S} \to \mathcal{C}$ is a fibred category, see
Categories, Definition \ref{categories-definition-fibred-category},
\item for any $U \in \Ob(\mathcal{C})$ and any $x, y \in \mathcal{S}_U$
the presheaf $\mathit{Mor}(x, y)$ (see
Definition \ref{definition-mor-presheaf}) is a sheaf on
the site $\mathcal{C}/U$, and
\item for any covering $\mathcal{U} = \{f_i : U_i \to U\}_{i \in I}$
of the site $\mathcal{C}$, any descent datum in $\mathcal{S}$
relative to $\mathcal{U}$ is effective.
\end{enumerate}
\end{definition}
\noindent
We find the formulation above the most convenient way to think about
a stack. Namely, given a category over $\mathcal{C}$ in order to verify
that it is a stack you proceed to check properties (1), (2) and
(3) in that order. Certainly properties (2) and (3) do not make sense
if the category isn't fibred. Without (2) we cannot prove that the
descent in (3) is unique up to unique isomorphism and functorial.
\medskip\noindent
The following lemma provides an alternative definition.
\begin{lemma}
\label{lemma-stack-equivalences}
Let $\mathcal{C}$ be a site.
Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category
over $\mathcal{C}$. The following are equivalent
\begin{enumerate}
\item $\mathcal{S}$ is a stack over $\mathcal{C}$, and
\item for any covering $\mathcal{U} = \{f_i : U_i \to U\}_{i \in I}$
of the site $\mathcal{C}$ the functor
$$
\mathcal{S}_U \longrightarrow DD(\mathcal{U})
$$
which associates to an
object its canonical descent datum is an equivalence.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-substack}
Let $p : \mathcal{S} \to \mathcal{C}$ be a stack over the site $\mathcal{C}$.
Let $\mathcal{S}'$ be a subcategory of $\mathcal{S}$.
Assume
\begin{enumerate}
\item if $\varphi : y \to x$ is a strongly cartesian
morphism of $\mathcal{S}$ and
$x$ is an object of $\mathcal{S}'$, then $y$ is isomorphic to an
object of $\mathcal{S}'$,
\item $\mathcal{S}'$ is a full subcategory of $\mathcal{S}$, and
\item if $\{f_i : U_i \to U\}$ is a covering of $\mathcal{C}$,
and $x$ an object of $\mathcal{S}$ over $U$ such that $f_i^*x$
is isomorphic to an object of $\mathcal{S}'$ for each $i$,
then $x$ is isomorphic to an object of $\mathcal{S}'$.
\end{enumerate}
Then $\mathcal{S}' \to \mathcal{C}$ is a stack.
\end{lemma}
\begin{proof}
Omitted. Hints:
The first condition guarantees that $\mathcal{S}'$ is a fibred category.
The second condition guarantees that the $\mathit{Isom}$-presheaves
of $\mathcal{S}'$ are sheaves (as they are identical to their counter parts
in $\mathcal{S}$). The third condition guarantees that the descent condition
holds in $\mathcal{S}'$ as we can first descend in $\mathcal{S}$ and
then (3) implies the resulting object is isomorphic to an object of
$\mathcal{S}'$.
\end{proof}
\begin{lemma}
\label{lemma-stack-equivalent}
Let $\mathcal{C}$ be a site.
Let $\mathcal{S}_1$, $\mathcal{S}_2$ be categories over $\mathcal{C}$.
Suppose that $\mathcal{S}_1$ and $\mathcal{S}_2$ are equivalent
as categories over $\mathcal{C}$.
Then $\mathcal{S}_1$ is a stack over $\mathcal{C}$ if and only if
$\mathcal{S}_2$ is a stack over $\mathcal{C}$.
\end{lemma}
\begin{proof}
Let $F : \mathcal{S}_1 \to \mathcal{S}_2$,
$G : \mathcal{S}_2 \to \mathcal{S}_1$ be functors over $\mathcal{C}$, and let
$i : F \circ G \to \text{id}_{\mathcal{S}_2}$,
$j : G \circ F \to \text{id}_{\mathcal{S}_1}$ be isomorphisms of
functors over $\mathcal{C}$. By
Categories, Lemma \ref{categories-lemma-fibred-equivalent}
we see that $\mathcal{S}_1$ is fibred if and only if $\mathcal{S}_2$
is fibred over $\mathcal{C}$. Hence we may assume that both
$\mathcal{S}_1$ and $\mathcal{S}_2$ are fibred. Moreover, the proof of
Categories, Lemma \ref{categories-lemma-fibred-equivalent}
shows that $F$ and $G$ map strongly cartesian morphisms to strongly
cartesian morphisms, i.e., $F$ and $G$ are $1$-morphisms of fibred
categories over $\mathcal{C}$. This means that given
$U \in \Ob(\mathcal{C})$, and $x, y \in \mathcal{S}_{1, U}$ then
the presheaves
$$
\mathit{Mor}_{\mathcal{S}_1}(x, y),
\mathit{Mor}_{\mathcal{S}_1}(F(x), F(y)) :
(\mathcal{C}/U)^{opp} \longrightarrow \textit{Sets}.
$$
are identified, see
Lemma \ref{lemma-presheaf-mor-map-fibred-categories}. Hence
the first is a sheaf if and only if the second is a sheaf.
Finally, we have to show that if every descent datum in $\mathcal{S}_1$
is effective, then so is every descent datum in $\mathcal{S}_2$.
To do this, let $(X_i, \varphi_{ii'})$ be a descent datum
in $\mathcal{S}_2$ relative the covering $\{U_i \to U\}$ of the site
$\mathcal{C}$. Then $(G(X_i), G(\varphi_{ii'}))$ is a descent datum
in $\mathcal{S}_1$ relative the covering $\{U_i \to U\}$.
Let $X$ be an object of $\mathcal{S}_{1, U}$ such that the
descent datum $(f_i^*X, can)$ is isomorphic to
$(G(X_i), G(\varphi_{ii'}))$. Then $F(X)$ is an object of $\mathcal{S}_{2, U}$
such that the descent datum $(f_i^*F(X), can)$ is isomorphic to
$(F(G(X_i)), F(G(\varphi_{ii'})))$ which in turn is isomorphic to
the original descent datum $(X_i, \varphi_{ii'})$ using $i$.
\end{proof}
\noindent
The $2$-category of stacks over $\mathcal{C}$
is defined as follows.
\begin{definition}
\label{definition-stacks-over-C}
Let $\mathcal{C}$ be a site.
The {\it $2$-category of stacks over $\mathcal{C}$}
is the sub $2$-category of the $2$-category of fibred categories
over $\mathcal{C}$ (see
Categories, Definition \ref{categories-definition-fibred-categories-over-C})
defined as follows:
\begin{enumerate}
\item Its objects will be stacks $p : \mathcal{S} \to \mathcal{C}$.
\item Its $1$-morphisms $(\mathcal{S}, p) \to (\mathcal{S}', p')$
will be functors $G : \mathcal{S} \to \mathcal{S}'$ such that
$p' \circ G = p$ and such that $G$ maps strongly cartesian
morphisms to strongly cartesian morphisms.
\item Its $2$-morphisms $t : G \to H$ for
$G, H : (\mathcal{S}, p) \to (\mathcal{S}', p')$
will be morphisms of functors
such that $p'(t_x) = \text{id}_{p(x)}$
for all $x \in \Ob(\mathcal{S})$.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma-2-product-stacks}
Let $\mathcal{C}$ be a site.
The $(2, 1)$-category of stacks over $\mathcal{C}$
has 2-fibre products, and they are described as in
Categories, Lemma \ref{categories-lemma-2-product-categories-over-C}.
\end{lemma}
\begin{proof}
Let $f : \mathcal{X} \to \mathcal{S}$ and
$g : \mathcal{Y} \to \mathcal{S}$ be
$1$-morphisms of stacks over $\mathcal{C}$
as defined above. The category
$\mathcal{X} \times_\mathcal{S} \mathcal{Y}$
described in
Categories, Lemma \ref{categories-lemma-2-product-categories-over-C} is a
fibred category according to
Categories, Lemma \ref{categories-lemma-2-product-fibred-categories-over-C}.
(This is where we use that $f$ and $g$ preserve strongly cartesian
morphisms.) It remains to show that the morphism presheaves are sheaves
and that descent relative to coverings of $\mathcal{C}$ is effective.
\medskip\noindent
Recall that an object of $\mathcal{X} \times_\mathcal{S} \mathcal{Y}$
is given by a quadruple $(U, x, y, \phi)$.
It lies over the object
$U$ of $\mathcal{C}$. Next, let $(U, x', y', \phi')$ be second
object lying over $U$.
Recall that $\phi : f(x) \to g(y)$, and $\phi' : f(x') \to g(y')$
are isomorphisms in the category $\mathcal{S}_U$. Let us
use these isomorphisms to identify $z = f(x) = g(y)$ and
$z' = f(x') = g(y')$. With this identifications
it is clear that
$$
\mathit{Mor}((U, x, y, \phi), (U, x', y', \phi'))
=
\mathit{Mor}(x, x')
\times_{\mathit{Mor}(z, z')}
\mathit{Mor}(y, y')
$$
as presheaves. However, as the fibred product in the category of
presheaves preserves sheaves (Sites, Lemma \ref{sites-lemma-limit-sheaf})
we see that this is a sheaf.
\medskip\noindent
Let $\mathcal{U} = \{f_i : U_i \to U\}_{i \in I}$ be a covering of the site
$\mathcal{C}$. Let $(X_i, \chi_{ij})$ be a descent datum
in $\mathcal{X} \times_\mathcal{S} \mathcal{Y}$ relative to $\mathcal{U}$.
Write $X_i = (U_i, x_i, y_i, \phi_i)$ as above. Write
$\chi_{ij} = (\varphi_{ij}, \psi_{ij})$ as in the definition of
the category $\mathcal{X} \times_\mathcal{S} \mathcal{Y}$ (see
Categories, Lemma \ref{categories-lemma-2-product-categories-over-C}).
It is clear that $(x_i, \varphi_{ij})$ is a descent datum in
$\mathcal{X}$ and that $(y_i, \psi_{ij})$ is a descent datum in
$\mathcal{Y}$. Since $\mathcal{X}$ and $\mathcal{Y}$ are stacks these
descent data are effective. Thus we get
$x \in \Ob(\mathcal{X}_U)$, and $y \in \Ob(\mathcal{Y}_U)$
with $x_i = x|_{U_i}$, and $y_i = y|_{U_i}$ compatibly with descent data.
Set $z = f(x)$ and $z' = g(y)$ which are both objects of $\mathcal{S}_U$.
The morphisms $\phi_i$ are elements of
$\mathit{Isom}(z, z')(U_i)$ with the property that
$\phi_i|_{U_i \times_U U_j} = \phi_j|_{U_i \times_U U_j}$.
Hence by the sheaf property of $\mathit{Isom}(z, z')$
we obtain an isomorphism $\phi : z = f(x) \to z' = g(y)$.
We omit the verification that the canonical descent datum associated to
the object $(U, x, y, \phi)$ of
$(\mathcal{X} \times_\mathcal{S} \mathcal{Y})_U$ is isomorphic
to the descent datum we started with.
\end{proof}
\begin{lemma}
\label{lemma-characterize-ff}
Let $\mathcal{C}$ be a site.
Let $\mathcal{S}_1$, $\mathcal{S}_2$ be stacks over $\mathcal{C}$.
Let $F : \mathcal{S}_1 \to \mathcal{S}_2$ be a $1$-morphism.
Then the following are equivalent
\begin{enumerate}
\item $F$ is fully faithful,
\item for every $U \in \Ob(\mathcal{C})$ and for every
$x, y \in \Ob(\mathcal{S}_{1, U})$ the map
$$
F :
\mathit{Mor}_{\mathcal{S}_1}(x, y)
\longrightarrow
\mathit{Mor}_{\mathcal{S}_2}(F(x), F(y))
$$
is an isomorphism of sheaves on $\mathcal{C}/U$.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (1). For $U, x, y$ as in (2) the displayed map $F$ evaluates to the map
$F : \Mor_{\mathcal{S}_{1, V}}(x|_V, y|_V) \to
\Mor_{\mathcal{S}_{2, V}}(F(x|_V), F(y|_V))$
on an object $V$ of $\mathcal{C}$ lying over $U$.
Now, since $F$ is fully faithful, the corresponding map
$\Mor_{\mathcal{S}_1}(x|_V, y|_V) \to \Mor_{\mathcal{S}_2}(F(x|_V), F(y|_V))$
is a bijection. Morphisms in the fibre category $\mathcal{S}_{1, V}$ are
exactly those morphisms between $x|_V$ and $y|_V$ in $\mathcal{S}_1$ lying
over $\text{id}_V$. Similarly, morphisms in the fibre category
$\mathcal{S}_{2, V}$ are exactly those morphisms between $F(x|_V)$ and
$F(y|_V)$ in $\mathcal{S}_2$ lying over $\text{id}_V$. Thus we find that $F$
induces a bijection between these also. Hence (2) holds.
\medskip\noindent
Assume (2). Suppose given objects $U$, $V$ of $\mathcal{C}$ and
$x \in \Ob(\mathcal{S}_{1, U})$ and
$y \in \Ob(\mathcal{S}_{1, V})$. To show that $F$ is fully faithful,
it suffices to prove it induces a bijection on
morphisms lying over a fixed $f : U \to V$. Choose a strongly Cartesian
$f^*y \to y$ in $\mathcal{S}_1$ lying above $f$. This results in a
bijection between the set of morphisms $x \to y$ in $\mathcal{S}_1$ lying
over $f$ and $\Mor_{\mathcal{S}_{1, U}}(x, f^*y)$. Since $F$ preserves
strongly Cartesian morphisms as a $1$-morphism in the $2$-category
of stacks over $\mathcal{C}$, we also get a bijection
between the set of morphisms $F(x) \to F(y)$ in $\mathcal{S}_2$ lying
over $f$ and $\Mor_{\mathcal{S}_{2, U}}(F(x), F(f^*y))$.
Since $F$ induces a bijection
$\Mor_{\mathcal{S}_{1, U}}(x, f^*y) \to
\Mor_{\mathcal{S}_{2, U}}(F(x), F(f^*y))$
we conclude (1) holds.
\end{proof}
\begin{lemma}
\label{lemma-characterize-essentially-surjective-when-ff}
Let $\mathcal{C}$ be a site.
Let $\mathcal{S}_1$, $\mathcal{S}_2$ be stacks over $\mathcal{C}$.
Let $F : \mathcal{S}_1 \to \mathcal{S}_2$ be a $1$-morphism which is
fully faithful. Then the following are equivalent
\begin{enumerate}
\item $F$ is an equivalence,
\item for every $U \in \Ob(\mathcal{C})$ and for every
$x \in \Ob(\mathcal{S}_{2, U})$ there exists a covering
$\{f_i : U_i \to U\}$ such that $f_i^*x$ is in the essential image
of the functor $F : \mathcal{S}_{1, U_i} \to \mathcal{S}_{2, U_i}$.
\end{enumerate}
\end{lemma}
\begin{proof}
The implication (1) $\Rightarrow$ (2) is immediate.
To see that (2) implies (1) we have to show that every
$x$ as in (2) is in the essential image of the functor $F$.
To do this choose a covering as in (2),
$x_i \in \Ob(\mathcal{S}_{1, U_i})$, and
isomorphisms $\varphi_i : F(x_i) \to f_i^*x$. Then we get a descent
datum for $\mathcal{S}_1$ relative to $\{f_i : U_i \to U\}$
by taking
$$
\varphi_{ij} :
x_i|_{U_i \times_U U_j}
\longrightarrow
x_j|_{U_i \times_U U_j}
$$
the arrow such that $F(\varphi_{ij}) = \varphi_j^{-1} \circ \varphi_i$.
This descent datum is effective by the axioms of a stack, and hence
we obtain an object $x_1$ of $\mathcal{S}_1$ over $U$. We omit the
verification that $F(x_1)$ is isomorphic to $x$ over $U$.
\end{proof}
\begin{remark}
\label{remark-stack-make-small}
(Cutting down a ``big'' stack to get a stack.)
Let $\mathcal{C}$ be a site. Suppose that $p : \mathcal{S} \to \mathcal{C}$
is functor from a ``big'' category to $\mathcal{C}$, i.e., suppose
that the collection of objects of $\mathcal{S}$ forms a proper class.
Finally, suppose that $p : \mathcal{S} \to \mathcal{C}$ satisfies
conditions (1), (2), (3) of
Definition \ref{definition-stack}.
In general there is no way to replace $p : \mathcal{S} \to \mathcal{C}$
by a equivalent category such that we obtain a stack. The reason is that
it can happen that a fibre categories $\mathcal{S}_U$ may have a proper
class of isomorphism classes of objects.
On the other hand, suppose that
\begin{enumerate}
\item[(4)] for every $U \in \Ob(\mathcal{C})$ there exists a set
$S_U \subset \Ob(\mathcal{S}_U)$ such that every object of
$\mathcal{S}_U$ is isomorphic in $\mathcal{S}_U$ to an element of $S_U$.
\end{enumerate}
In this case we can find a full subcategory $\mathcal{S}_{small}$
of $\mathcal{S}$ such that, setting $p_{small} = p|_{\mathcal{S}_{small}}$,
we have
\begin{enumerate}
\item[(a)] the functor $p_{small} : \mathcal{S}_{small} \to \mathcal{C}$
defines a stack, and
\item[(b)] the inclusion $\mathcal{S}_{small} \to \mathcal{S}$
is fully faithful and essentially surjective.
\end{enumerate}
(Hint: For every $U \in \Ob(\mathcal{C})$
let $\alpha(U)$ denote the smallest ordinal such that
$\Ob(\mathcal{S}_U) \cap V_{\alpha(U)}$ surjects onto the set
of isomorphism classes of $\mathcal{S}_U$, and set
$\alpha = \sup_{U \in \Ob(\mathcal{C})} \alpha(U)$.
Then take
$\Ob(\mathcal{S}_{small}) = \Ob(\mathcal{S}) \cap V_\alpha$.
For notation used see Sets, Section \ref{sets-section-sets-hierarchy}.)
\end{remark}
\section{Stacks in groupoids}
\label{section-stacks-in-groupoids}
\noindent
Among stacks those which are fibred in groupoids are somewhat easier
to comprehend. We redefine them as follows.
\begin{definition}
\label{definition-stack-in-groupoids}
A {\it stack in groupoids} over a site $\mathcal{C}$ is a
category $p : \mathcal{S} \to \mathcal{C}$ over $\mathcal{C}$
such that
\begin{enumerate}
\item $p : \mathcal{S} \to \mathcal{C}$ is fibred
in groupoids over $\mathcal{C}$ (see
Categories, Definition \ref{categories-definition-fibred-groupoids}),
\item for all $U \in \Ob(\mathcal{C})$,
for all $x, y\in \Ob(\mathcal{S}_U)$ the presheaf
$\mathit{Isom}(x, y)$ is a sheaf on the site $\mathcal{C}/U$, and
\item for all coverings $\mathcal{U} = \{U_i \to U\}$ in $\mathcal{C}$,
all descent data $(x_i, \phi_{ij})$ for $\mathcal{U}$ are effective.
\end{enumerate}
\end{definition}
\noindent
Usually the hardest part to check is the third condition.
Here is the lemma comparing this with the notion of a stack.
\begin{lemma}
\label{lemma-stack-in-groupoids-stack}
Let $\mathcal{C}$ be a site.
Let $p : \mathcal{S} \to \mathcal{C}$ be a category over $\mathcal{C}$.
The following are equivalent
\begin{enumerate}
\item $\mathcal{S}$ is a stack in groupoids over $\mathcal{C}$,
\item $\mathcal{S}$ is a stack over $\mathcal{C}$ and all
fibre categories are groupoids, and
\item $\mathcal{S}$ is fibred in groupoids over $\mathcal{C}$
and is a stack over $\mathcal{C}$.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted, but see Categories, Lemma \ref{categories-lemma-fibred-groupoids}.
\end{proof}
\begin{lemma}
\label{lemma-stack-gives-stack-groupoids}
Let $\mathcal{C}$ be a site.
Let $p : \mathcal{S} \to \mathcal{C}$ be a stack.
Let $p' : \mathcal{S}' \to \mathcal{C}$
be the category fibred in groupoids associated to $\mathcal{S}$
constructed in
Categories, Lemma \ref{categories-lemma-fibred-gives-fibred-groupoids}.
Then $p' : \mathcal{S}' \to \mathcal{C}$ is a stack in groupoids.
\end{lemma}
\begin{proof}
Recall that the morphisms in $\mathcal{S}'$ are exactly the
strongly cartesian morphisms of $\mathcal{S}$, and that any isomorphism of
$\mathcal{S}$ is such a morphism. Hence descent data in $\mathcal{S}'$
are exactly the same thing as descent data in $\mathcal{S}$. Now apply
Lemma \ref{lemma-stack-equivalences}. Some details omitted.
\end{proof}
\begin{lemma}
\label{lemma-stack-in-groupoids-equivalent}
Let $\mathcal{C}$ be a site.
Let $\mathcal{S}_1$, $\mathcal{S}_2$ be categories over $\mathcal{C}$.
Suppose that $\mathcal{S}_1$ and $\mathcal{S}_2$ are equivalent
as categories over $\mathcal{C}$.
Then $\mathcal{S}_1$ is a stack in groupoids over $\mathcal{C}$ if and only if
$\mathcal{S}_2$ is a stack in groupoids over $\mathcal{C}$.
\end{lemma}
\begin{proof}
Follows by combining
Lemmas \ref{lemma-stack-in-groupoids-stack} and \ref{lemma-stack-equivalent}.
\end{proof}
\noindent
The $2$-category of stacks in groupoids over $\mathcal{C}$
is defined as follows.
\begin{definition}
\label{definition-stacks-in-groupoids-over-C}
Let $\mathcal{C}$ be a site.
The {\it $2$-category of stacks in groupoids over $\mathcal{C}$}
is the sub $2$-category of the $2$-category of stacks
over $\mathcal{C}$ (see Definition \ref{definition-stacks-over-C})
defined as follows:
\begin{enumerate}
\item Its objects will be stacks in groupoids
$p : \mathcal{S} \to \mathcal{C}$.
\item Its $1$-morphisms $(\mathcal{S}, p) \to (\mathcal{S}', p')$
will be functors $G : \mathcal{S} \to \mathcal{S}'$ such that
$p' \circ G = p$. (Since every morphism is strongly cartesian
every functor preserves them.)
\item Its $2$-morphisms $t : G \to H$ for
$G, H : (\mathcal{S}, p) \to (\mathcal{S}', p')$
will be morphisms of functors
such that $p'(t_x) = \text{id}_{p(x)}$
for all $x \in \Ob(\mathcal{S})$.
\end{enumerate}
\end{definition}
\noindent
Note that any $2$-morphism is automatically an isomorphism, so
that in fact the $2$-category of stacks in groupoids over $\mathcal{C}$
is a (strict) $(2, 1)$-category.
\begin{lemma}
\label{lemma-2-product-stacks-in-groupoids}
Let $\mathcal{C}$ be a category.
The $2$-category of stacks in groupoids over $\mathcal{C}$
has 2-fibre products, and they are described as in
Categories, Lemma \ref{categories-lemma-2-product-categories-over-C}.
\end{lemma}
\begin{proof}
This is clear from
Categories, Lemma \ref{categories-lemma-2-product-fibred-categories}
and Lemmas \ref{lemma-stack-in-groupoids-stack}
and \ref{lemma-2-product-stacks}.
\end{proof}
\section{Stacks in setoids}
\label{section-stacks-in-setoids}
\noindent
This is just a brief section saying that a stack in sets
is the same thing as a sheaf of sets. Please consult
Categories, Section \ref{categories-section-fibred-in-setoids}
for notation.
\begin{definition}
\label{definition-stack-in-sets}
Let $\mathcal{C}$ be a site.
\begin{enumerate}
\item A {\it stack in setoids} over $\mathcal{C}$
is a stack over $\mathcal{C}$ all of whose fibre categories are
setoids.
\item A {\it stack in sets}, or a {\it stack in discrete categories}
is a stack over $\mathcal{C}$ all of whose fibre categories are discrete.
\end{enumerate}
\end{definition}
\noindent
From the discussion in
Section \ref{section-stacks-in-groupoids}
this is the same thing as a stack in groupoids whose fibre categories
are setoids (resp.\ discrete). Moreover, it is also the same thing
as a category fibred in setoids (resp.\ sets) which is a stack.
\begin{lemma}
\label{lemma-when-stack-in-sets}
Let $\mathcal{C}$ be a site. Under the equivalence
$$
\left\{
\begin{matrix}
\text{the category of presheaves}\\
\text{of sets over }\mathcal{C}
\end{matrix}
\right\}
\leftrightarrow
\left\{
\begin{matrix}
\text{the category of categories}\\
\text{fibred in sets over }\mathcal{C}
\end{matrix}
\right\}
$$
of
Categories, Lemma \ref{categories-lemma-2-category-fibred-sets}
the stacks in sets correspond precisely to the sheaves.
\end{lemma}
\begin{proof}
Omitted. Hint: Show that effectivity of descent corresponds exactly to
the sheaf condition.
\end{proof}
\begin{lemma}
\label{lemma-stack-in-setoids-characterize}
Let $\mathcal{C}$ be a site.
Let $\mathcal{S}$ be a category fibred in setoids over $\mathcal{C}$.
Then $\mathcal{S}$ is a stack in setoids if and only if the unique
equivalent category $\mathcal{S}'$ fibred in sets (see
Categories, Lemma \ref{categories-lemma-setoid-fibres})
is a stack in sets. In other words, if and only if the presheaf
$$
U \longmapsto \Ob(\mathcal{S}_U)/\!\!\cong
$$
is a sheaf.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-stack-in-setoids-equivalent}
Let $\mathcal{C}$ be a site.
Let $\mathcal{S}_1$, $\mathcal{S}_2$ be categories over $\mathcal{C}$.
Suppose that $\mathcal{S}_1$ and $\mathcal{S}_2$ are equivalent
as categories over $\mathcal{C}$.
Then $\mathcal{S}_1$ is a stack in setoids over $\mathcal{C}$ if and only if
$\mathcal{S}_2$ is a stack in setoids over $\mathcal{C}$.
\end{lemma}
\begin{proof}
By
Categories, Lemma \ref{categories-lemma-setoid-fibres}
we see that a category $\mathcal{S}$ over $\mathcal{C}$ is fibred in setoids
over $\mathcal{C}$ if and only if it is equivalent over $\mathcal{C}$
to a category fibred in sets.
Hence we see that $\mathcal{S}_1$ is fibred in setoids over $\mathcal{C}$
if and only if $\mathcal{S}_2$ is fibred in setoids over $\mathcal{C}$.
Hence now the lemma follows from
Lemma \ref{lemma-stack-in-setoids-characterize}.
\end{proof}
\noindent
The $2$-category of stacks in setoids over $\mathcal{C}$
is defined as follows.
\begin{definition}
\label{definition-stacks-in-setoids-over-C}
Let $\mathcal{C}$ be a site.
The {\it $2$-category of stacks in setoids over $\mathcal{C}$}
is the sub $2$-category of the $2$-category of stacks
over $\mathcal{C}$ (see Definition \ref{definition-stacks-over-C})
defined as follows:
\begin{enumerate}
\item Its objects will be stacks in setoids
$p : \mathcal{S} \to \mathcal{C}$.
\item Its $1$-morphisms $(\mathcal{S}, p) \to (\mathcal{S}', p')$
will be functors $G : \mathcal{S} \to \mathcal{S}'$ such that
$p' \circ G = p$. (Since every morphism is strongly cartesian
every functor preserves them.)
\item Its $2$-morphisms $t : G \to H$ for
$G, H : (\mathcal{S}, p) \to (\mathcal{S}', p')$
will be morphisms of functors
such that $p'(t_x) = \text{id}_{p(x)}$
for all $x \in \Ob(\mathcal{S})$.
\end{enumerate}
\end{definition}
\noindent
Note that any $2$-morphism is automatically an isomorphism, so
that in fact the $2$-category of stacks in setoids over $\mathcal{C}$
is a (strict) $(2, 1)$-category.
\begin{lemma}
\label{lemma-2-product-stacks-in-setoids}
Let $\mathcal{C}$ be a site.
The $2$-category of stacks in setoids over $\mathcal{C}$
has 2-fibre products, and they are described as in
Categories, Lemma \ref{categories-lemma-2-product-categories-over-C}.
\end{lemma}
\begin{proof}
This is clear from
Categories, Lemmas \ref{categories-lemma-2-product-fibred-categories} and
\ref{categories-lemma-2-product-categories-fibred-setoids}
and
Lemmas \ref{lemma-stack-in-groupoids-stack} and
\ref{lemma-2-product-stacks}.
\end{proof}
\begin{lemma}
\label{lemma-2-fibre-product-gives-stack-in-setoids}
Let $\mathcal{C}$ be a site.
Let $\mathcal{S}, \mathcal{T}$ be stacks in groupoids over $\mathcal{C}$
and let $\mathcal{R}$ be a stack in setoids over $\mathcal{C}$.
Let $f : \mathcal{T} \to \mathcal{S}$ and $g : \mathcal{R} \to \mathcal{S}$
be $1$-morphisms. If $f$ is faithful, then the $2$-fibre product
$$
\mathcal{T} \times_{f, \mathcal{S}, g} \mathcal{R}
$$
is a stack in setoids over $\mathcal{C}$.
\end{lemma}
\begin{proof}
Immediate from the explicit description of the $2$-fibre product in
Categories, Lemma \ref{categories-lemma-2-product-categories-over-C}.
\end{proof}
\begin{lemma}
\label{lemma-2-fibre-product-stacks-in-setoids-over-stack-in-groupoids}
Let $\mathcal{C}$ be a site.
Let $\mathcal{S}$ be a stack in groupoids over $\mathcal{C}$ and
let $\mathcal{S}_i$, $i = 1, 2$ be stacks in setoids over $\mathcal{C}$.
Let $f_i : \mathcal{S}_i \to \mathcal{S}$ be $1$-morphisms.
Then the $2$-fibre product
$$
\mathcal{S}_1 \times_{f_1, \mathcal{S}, f_2} \mathcal{S}_2
$$
is a stack in setoids over $\mathcal{C}$.
\end{lemma}
\begin{proof}
This is a special case of
Lemma \ref{lemma-2-fibre-product-gives-stack-in-setoids}
as $f_2$ is faithful.
\end{proof}
\begin{lemma}
\label{lemma-faithful-descent}
Let $\mathcal{C}$ be a site. Let
$$
\xymatrix{
\mathcal{T}_2 \ar[r] \ar[d]_{G'} & \mathcal{T}_1 \ar[d]^G \\
\mathcal{S}_2 \ar[r]^F & \mathcal{S}_1
}
$$
be a $2$-cartesian diagram of stacks in groupoids over $\mathcal{C}$.
Assume
\begin{enumerate}
\item for every $U \in \Ob(\mathcal{C})$ and
$x \in \Ob((\mathcal{S}_1)_U)$ there exists a covering
$\{U_i \to U\}$ such that $x|_{U_i}$ is in the essential
image of $F : (\mathcal{S}_2)_{U_i} \to (\mathcal{S}_1)_{U_i}$, and
\item $G'$ is faithful,
\end{enumerate}
then $G$ is faithful.
\end{lemma}
\begin{proof}
We may assume that $\mathcal{T}_2$ is the category
$\mathcal{S}_2 \times_{\mathcal{S}_1} \mathcal{T}_1$
described in
Categories, Lemma \ref{categories-lemma-2-product-categories-over-C}.
By
Categories, Lemma \ref{categories-lemma-equivalence-fibred-categories}
the faithfulness of $G, G'$ can be checked on fibre categories.
Suppose that $y, y'$ are objects of $\mathcal{T}_1$ over the object $U$
of $\mathcal{C}$. Let $\alpha, \beta : y \to y'$ be morphisms of
$(\mathcal{T}_1)_U$ such that $G(\alpha) = G(\beta)$. Our object is to
show that $\alpha = \beta$. Considering instead
$\gamma = \alpha^{-1} \circ \beta$ we see that $G(\gamma) = \text{id}_{G(y)}$
and we have to show that $\gamma = \text{id}_y$. By assumption we
can find a covering $\{U_i \to U\}$ such that $G(y)|_{U_i}$ is in the
essential image of $F :(\mathcal{S}_2)_{U_i} \to (\mathcal{S}_1)_{U_i}$.
Since it suffices to show that $\gamma|_{U_i} = \text{id}$ for each $i$,
we may therefore assume that we have
$f : F(x) \to G(y)$ for some object $x$ of $\mathcal{S}_2$ over $U$
and morphisms $f$ of $(\mathcal{S}_1)_U$. In this case we get
a morphism
$$
(1, \gamma) : (U, x, y, f) \longrightarrow (U, x, y, f)
$$
in the fibre category of $\mathcal{S}_2 \times_{\mathcal{S}_1} \mathcal{T}_1$
over $U$ whose image under $G'$ in $\mathcal{S}_1$ is $\text{id}_x$.
As $G'$ is faithful we conclude that $\gamma = \text{id}_y$ and we win.
\end{proof}
\begin{lemma}
\label{lemma-stack-in-setoids-descent}
Let $\mathcal{C}$ be a site.
Let
$$
\xymatrix{
\mathcal{T}_2 \ar[r] \ar[d] & \mathcal{T}_1 \ar[d]^G \\
\mathcal{S}_2 \ar[r]^F & \mathcal{S}_1
}
$$
be a $2$-cartesian diagram of stacks in groupoids over $\mathcal{C}$.
If
\begin{enumerate}
\item $F : \mathcal{S}_2 \to \mathcal{S}_1$ is fully faithful,
\item for every $U \in \Ob(\mathcal{C})$ and
$x \in \Ob((\mathcal{S}_1)_U)$ there exists a covering
$\{U_i \to U\}$ such that $x|_{U_i}$ is in the essential
image of $F : (\mathcal{S}_2)_{U_i} \to (\mathcal{S}_1)_{U_i}$, and
\item $\mathcal{T}_2$ is a stack in setoids.
\end{enumerate}
then $\mathcal{T}_1$ is a stack in setoids.
\end{lemma}
\begin{proof}
We may assume that $\mathcal{T}_2$ is the category
$\mathcal{S}_2 \times_{\mathcal{S}_1} \mathcal{T}_1$
described in
Categories, Lemma \ref{categories-lemma-2-product-categories-over-C}.
Pick $U \in \Ob(\mathcal{C})$ and
$y \in \Ob((\mathcal{T}_1)_U)$.
We have to show that the sheaf $\textit{Aut}(y)$ on $\mathcal{C}/U$
is trivial. To to this we may replace $U$ by the members of
a covering of $U$. Hence by assumption (2) we may assume that
there exists an object $x \in \Ob((\mathcal{S}_2)_U)$
and an isomorphism $f : F(x) \to G(y)$.
Then $y' = (U, x, y, f)$ is an object of $\mathcal{T}_2$ over $U$
which is mapped to $y$ under the projection $\mathcal{T}_2 \to \mathcal{T}_1$.
Because $F$ is fully faithful by (1) the map
$\textit{Aut}(y') \to \textit{Aut}(y)$ is surjective, use the explicit
description of morphisms in $\mathcal{T}_2$ in
Categories, Lemma \ref{categories-lemma-2-product-categories-over-C}.
Since by (3) the sheaf $\textit{Aut}(y')$ is trivial
we get the result of the lemma.
\end{proof}
\begin{lemma}
\label{lemma-relative-sheaf-over-stack-is-stack}
Let $\mathcal{C}$ be a site. Let $F : \mathcal{S} \to \mathcal{T}$
be a $1$-morphism of categories fibred in groupoids over $\mathcal{C}$.
Assume that
\begin{enumerate}
\item $\mathcal{T}$ is a stack in groupoids over $\mathcal{C}$,
\item for every $U \in \Ob(\mathcal{C})$ the functor
$\mathcal{S}_U \to \mathcal{T}_U$ of fibre categories is faithful,
\item for each $U$ and each $y \in \Ob(\mathcal{T}_U)$ the presheaf
$$
(h : V \to U)
\longmapsto
\{(x, f) \mid x \in \Ob(\mathcal{S}_V), f : F(x) \to f^*y\text{ over }V\}/\cong
$$
is a sheaf on $\mathcal{C}/U$.
\end{enumerate}
Then $\mathcal{S}$ is a stack in groupoids over $\mathcal{C}$.
\end{lemma}
\begin{proof}
We have to prove descent for morphisms and descent for objects.
\medskip\noindent
Descent for morphisms. Let $\{U_i \to U\}$ be a covering of $\mathcal{C}$.
Let $x, x'$ be objects of $\mathcal{S}$ over $U$. For each $i$
let $\alpha_i : x|_{U_i} \to x'|_{U_i}$ be a morphism over $U_i$
such that $\alpha_i$ and $\alpha_j$ restrict to the same morphism
$x|_{U_i \times_U U_j} \to x'|_{U_i \times_U U_j}$.
Because $\mathcal{T}$ is a stack in groupoids, there is a morphism
$\beta : F(x) \to F(x')$ over $U$ whose restriction to $U_i$ is $F(\alpha_i)$.
Then we can think of $\xi = (x, \beta)$ and $\xi' = (x', \text{id}_{F(x')})$
as sections of the presheaf associated to $y = F(x')$ over $U$
in assumption (3). On the other hand, the restrictions of
$\xi$ and $\xi'$ to $U_i$ are $(x|_{U_i}, F(\alpha_i))$
and $(x'|_{U_i}, \text{id}_{F(x'|_{U_i})})$.
These are isomorphic to each other by the morphism $\alpha_i$.
Thus $\xi$ and $\xi'$ are isomorphic by assumption (3). This means there is a
morphism $\alpha : x \to x'$ over $U$ with $F(\alpha) = \beta$.
Since $F$ is faithful on fibre categories we obtain
$\alpha|_{U_i} = \alpha_i$.
\medskip\noindent
Descent of objects. Let $\{U_i \to U\}$ be a covering of $\mathcal{C}$.
Let $(x_i, \varphi_{ij})$ be a descent datum for $\mathcal{S}$ with
respect to the given covering. Because $\mathcal{T}$ is a stack in groupoids,
there is an object $y$ in $\mathcal{T}_U$ and isomorphisms
$\beta_i : F(x_i) \to y|_{U_i}$ such that
$F(\varphi_{ij}) = \beta_j|_{U_i \times_U U_j} \circ
(\beta_i|_{U_i \times_U U_j})^{-1}$.
Then $(x_i, \beta_i)$ are sections of the presheaf associated to
$y$ over $U$ defined in assumption (3).
Moreover, $\varphi_{ij}$ defines an isomorphism
from the pair $(x_i, \beta_i)|_{U_i \times_U U_j}$ to
the pair $(x_j, \beta_j)|_{U_i \times_U U_j}$.
Hence by assumption (3) there exists a pair $(x, \beta)$ over $U$
whose restriction to $U_i$ is isomorphic to $(x_i, \beta_i)$.
This means there are morphisms $\alpha_i : x_i \to x|_{U_i}$
with $\beta_i = \beta|_{U_i} \circ F(\alpha_i)$.
Since $F$ is faithful on fibre categories a calculation shows
that $\varphi_{ij} = \alpha_j|_{U_i \times_U U_j} \circ
(\alpha_i|_{U_i \times_U U_j})^{-1}$. This finishes the proof.
\end{proof}
\section{The inertia stack}
\label{section-the-inertia-stack}
\noindent
Let
$p : \mathcal{S} \to \mathcal{C}$ and
$p' : \mathcal{S}' \to \mathcal{C}$
be fibred categories over the category $\mathcal{C}$.
Let $F : \mathcal{S} \to \mathcal{S}'$ be a $1$-morphism of
fibred categories over $\mathcal{C}$.
Recall that we have defined in
Categories, Definition \ref{categories-definition-inertia-fibred-category}
an {\it relative inertia fibred category}
$\mathcal{I}_{\mathcal{S}/\mathcal{S}'} \to \mathcal{C}$ as the
category whose objects are pairs $(x , \alpha)$ where
$x \in \Ob(\mathcal{S})$ and $\alpha : x \to x$ with
$F(\alpha) = \text{id}_{F(x)}$. There is also an absolute version,
namely the {\it inertia} $\mathcal{I}_\mathcal{S}$ of $\mathcal{S}$.
These inertia categories are actually stacks over $\mathcal{C}$ provided
that $\mathcal{S}$ and $\mathcal{S}'$ are stacks.
\begin{lemma}
\label{lemma-inertia}
Let $\mathcal{C}$ be a site. Let
$p : \mathcal{S} \to \mathcal{C}$ and
$p' : \mathcal{S}' \to \mathcal{C}$
be stacks over the site $\mathcal{C}$.
Let $F : \mathcal{S} \to \mathcal{S}'$ be a $1$-morphism of
stacks over $\mathcal{C}$.
\begin{enumerate}
\item The inertia $\mathcal{I}_{\mathcal{S}/\mathcal{S}'}$ and
$\mathcal{I}_\mathcal{S}$ are stacks over $\mathcal{C}$.
\item If $\mathcal{S}, \mathcal{S}'$ are stacks in groupoids over
$\mathcal{S}$, then so are $\mathcal{I}_{\mathcal{S}/\mathcal{S}'}$ and
$\mathcal{I}_\mathcal{S}$.
\item If $\mathcal{S}, \mathcal{S}'$ are stacks in setoids over $\mathcal{S}$,
then so are $\mathcal{I}_{\mathcal{S}/\mathcal{S}'}$ and
$\mathcal{I}_\mathcal{S}$.
\end{enumerate}
\end{lemma}
\begin{proof}
The first three assertions follow from
Lemmas \ref{lemma-2-product-stacks},
\ref{lemma-2-product-stacks-in-groupoids}, and
\ref{lemma-2-product-stacks-in-setoids}
and the equivalence in
Categories, Lemma \ref{categories-lemma-inertia-fibred-category} part (1).
\end{proof}
\begin{lemma}
\label{lemma-characterize-stack-in-setoids}
Let $\mathcal{C}$ be a site.
If $\mathcal{S}$ is a stack in groupoids, then the
canonical $1$-morphism $\mathcal{I}_\mathcal{S} \to \mathcal{S}$
is an equivalence if and only if $\mathcal{S}$ is a stack in setoids.
\end{lemma}
\begin{proof}
Follows directly from
Categories, Lemma \ref{categories-lemma-characterize-fibred-setoids-inertia}.
\end{proof}
\section{Stackification of fibred categories}
\label{section-stackify}
\noindent
Here is the result.
\begin{lemma}
\label{lemma-stackify}
Let $\mathcal{C}$ be a site.
Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category over $\mathcal{C}$.
There exists a stack $p' : \mathcal{S}' \to \mathcal{C}$ and a
$1$-morphism $G : \mathcal{S} \to \mathcal{S}'$
of fibred categories over $\mathcal{C}$ (see
Categories, Definition \ref{categories-definition-fibred-categories-over-C})
such that
\begin{enumerate}
\item for every $U \in \Ob(\mathcal{C})$, and any
$x, y \in \Ob(\mathcal{S}_U)$ the map
$$
\mathit{Mor}(x, y) \longrightarrow \mathit{Mor}(G(x), G(y))
$$
induced by $G$ identifies the right hand side with the sheafification
of the left hand side, and
\item for every $U \in \Ob(\mathcal{C})$, and any
$x' \in \Ob(\mathcal{S}'_U)$ there exists a covering
$\{U_i \to U\}_{i \in I}$ such that for every $i \in I$ the
object $x'|_{U_i}$ is in the essential image of the
functor $G : \mathcal{S}_U \to \mathcal{S}'_U$.
\end{enumerate}
Moreover the stack $\mathcal{S}'$ is determined up to unique
$2$-isomorphism by these conditions.
\end{lemma}
\begin{proof}[Proof by naive method]
In this proof method we proceed in stages:
\medskip\noindent
First, given $x$ lying over $U$ and any object $y$ of
$\mathcal{S}$, we say that two morphisms
$a, b : x \to y$ of $\mathcal{S}$
lying over the same arrow of $\mathcal{C}$
are {\it locally equal}
if there exists a covering $\{f_i : U_i \to U\}$ of $\mathcal{C}$
such that the compositions
$$
f_i^*x \to x \xrightarrow{a} y,
\quad
f_i^*x \to x \xrightarrow{b} y
$$
are equal. This gives an equivalence relation $\sim$
on arrows of $\mathcal{S}$. If $b \sim b'$ then
$a \circ b \circ c \sim a \circ b' \circ c$ (verification omitted).
Hence we can quotient out by this equivalence relation to
obtain a new category $\mathcal{S}^1$ over $\mathcal{C}$
together with a morphism $G^1 : \mathcal{S} \to \mathcal{S}^1$.
\medskip\noindent
One checks that $G^1$ preserves strongly cartesian morphisms
and that $\mathcal{S}^1$ is a fibred category over $\mathcal{C}$.
Checks omitted. Thus we reduce to the case where locally equal
morphisms are equal.
\medskip\noindent
Next, we add morphisms as follows. Given
$x$ lying over $U$ and any object $y$ of lying over $V$
a {\it locally defined morphism from $x$ to $y$} is given by
\begin{enumerate}
\item a morphism $f : U \to V$,
\item a covering $\{f_i : U_i \to U\}$ of $U$, and
\item morphisms $a_i : f_i^*x \to y$ with $p(a_i) = f \circ f_i$
\end{enumerate}
with the property that the compositions
$$
(f_i \times f_j)^*x \to f_i^*x \xrightarrow{a_i} y,
\quad
(f_i \times f_j)^*x \to f_j^*x \xrightarrow{a_j} y
$$
are equal. Note that a usual morphism $a : x \to y$ gives a locally
defined morphism $(p(a) : U \to V, \{\text{id}_U\}, a)$.
We say two locally defined morphisms
$(f, \{f_i : U_i \to U\}, a_i)$ and $(g, \{g_j : U'_j \to U\}, b_j)$
are {\it equal} if $f = g$ and the compositions
$$
(f_i \times g_j)^*x \to f_i^*x \xrightarrow{a_i} y,
\quad
(f_i \times g_j)^*x \to g_j^*x \xrightarrow{b_j} y
$$
are equal (this is the right condition since we are in the
situation where locally equal morphisms are equal).
To compose locally defined morphisms
$(f, \{f_i : U_i \to U\}, a_i)$ from $x$ to $y$ and
$(g, \{g_j : V_j \to V\}, b_j)$ from $y$ to $z$ lying over $W$,
just take $g \circ f : U \to W$, the covering
$\{U_i \times_V V_j \to U\}$, and as maps the compositions
$$
x|_{U_i \times_V V_j}
\xrightarrow{\text{pr}_0^*a_i}
y|_{V_j}
\xrightarrow{b_j}
z
$$
We omit the verification that this is a locally defined morphism.
\medskip\noindent
One checks that $\mathcal{S}^2$ with the same objects as
$\mathcal{S}$ and with locally defined morphisms as morphisms
is a category over $\mathcal{C}$, that there is a functor
$G^2 : \mathcal{S} \to \mathcal{S}^2$ over $\mathcal{C}$,
that this functor preserves strongly cartesian objects,
and that $\mathcal{S}^2$ is a fibred category over $\mathcal{C}$.
Checks omitted. This reduces one to the case where the
morphism presheaves of $\mathcal{S}$ are all sheaves, by
checking that the effect of using locally defined morphisms
is to take the sheafification of the (separated) morphisms
presheaves.
\medskip\noindent
Finally, in the case where the morphism presheaves are all sheaves
we have to add objects in order to make sure descent conditions are
effective in the end result. The simplest way to do this is to
consider the category $\mathcal{S}'$ whose objects are
pairs $(\mathcal{U}, \xi)$ where
$\mathcal{U} = \{U_i \to U\}$ is a covering of $\mathcal{C}$ and
$\xi = (X_i, \varphi_{ii'})$ is a descent datum relative $\mathcal{U}$.
Suppose given two such data
$(\mathcal{U}, \xi) = (\{f_i : U_i \to U\}, x_i, \varphi_{ii'})$ and
$(\mathcal{V}, \eta) = (\{g_j : V_j \to V\}, y_j, \psi_{jj'})$.
We define
$$
\Mor_{\mathcal{S}'}((\mathcal{U}, \xi), (\mathcal{V}, \eta))
$$
as the set of $(f, a_{ij})$, where $f : U \to V$ and
$$
a_{ij} :
x_i|_{U_i \times_V V_j}
\longrightarrow
y_j
$$
are morphisms of $\mathcal{S}$ lying over $U_i \times_V V_j \to V_j$.
These have to satisfy the following condition: for any
$i, i' \in I$ and $j, j' \in J$ set
$W = (U_i \times_U U_{i'}) \times_V (V_j \times_V V_{j'})$. Then
$$
\xymatrix{
x_i|_W \ar[r]_{a_{ij}|_W} \ar[d]_{\varphi_{ii'}|_W} &
y_j|_W \ar[d]^{\psi_{jj'}|_W} \\
x_{i'}|_W \ar[r]^{a_{i'j'}|_W} &
y_{j'}|_W
}
$$
commutes. At this point you have to verify the following things:
\begin{enumerate}
\item there is a well defined composition on morphisms as above,
\item this turns $\mathcal{S}'$ into a category over $\mathcal{C}$,
\item there is a functor $G : \mathcal{S} \to \mathcal{S}'$ over $\mathcal{C}$,
\item for $x, y$ objects of $\mathcal{S}$ we have
$\Mor_\mathcal{S}(x, y) = \Mor_{\mathcal{S}'}(G(x), G(y))$,
\item any object of $\mathcal{S}'$ locally comes from an object of
$\mathcal{S}$, i.e., part (2) of the lemma holds,
\item $G$ preserves strongly cartesian morphisms,
\item $\mathcal{S}'$ is a fibred category over $\mathcal{C}$, and
\item $\mathcal{S}'$ is a stack over $\mathcal{C}$.
\end{enumerate}
This is all not hard but there is a lot of it. Details omitted.
\end{proof}
\begin{proof}[Less naive proof]
Here is a less naive proof.
By Categories, Lemma \ref{categories-lemma-fibred-strict}
there exists an equivalence of
fibred categories $\mathcal{S} \to \mathcal{S}'$ where $\mathcal{S}'$
is a split fibred category, i.e., one in which the pullback
functors compose on the nose. Obviously the lemma for $\mathcal{S}'$
implies the lemma for $\mathcal{S}$. Hence we may think of $\mathcal{S}$
as a presheaf in categories.
\medskip\noindent
Consider the $2$-category $\textit{Cat}$ temporarily as a
category by forgetting about $2$-morphisms.
Let us think of a category as a quintuple
$(\text{Ob}, \text{Arrows}, s, t, \circ)$ as in
Categories, Section \ref{categories-section-definition-categories}.
Consider the forgetful functor
$$
forget : \textit{Cat} \to \textit{Sets}, \quad
(\text{Ob}, \text{Arrows}, s, t, \circ)
\mapsto
\text{Ob} \amalg \text{Arrows}.
$$
Then $forget$ is faithful, $\textit{Cat}$ has limits and
$forget$ commutes with them, $\textit{Cat}$ has directed colimits and
$forget$ commutes with them, and $forget$ reflects isomorphisms.
Hence, according to the first part of
Sites, Section \ref{sites-section-sheaves-algebraic-structures}
we can sheafify presheaves with values in $\textit{Cat}$, and
the result commutes with $forget$. Applying this to
$\mathcal{S}$ we obtain a sheafification $\mathcal{S}^\#$
which has a sheaf of objects and a sheaf of morphisms
both of which are the sheafifications of the corresponding
presheaves for $\mathcal{S}$. In this case it is quite
easy to see that the map $\mathcal{S} \to \mathcal{S}^\#$
has the properties (1) and (2) of the lemma.
\medskip\noindent
However, the category $\mathcal{S}^\#$ may not yet be a
stack since, although the presheaf of objects is a sheaf,
the descent condition may not yet be satisfied.
To remedy this we have to add more objects. But the argument
above does reduce us to the case where $\mathcal{S} = \mathcal{S}_F$
for some sheaf(!) $F : \mathcal{C}^{opp} \to \textit{Cat}$ of
categories. In this case consider the functor
$F' : \mathcal{C}^{opp} \to \textit{Cat}$ defined by
\begin{enumerate}
\item The set $\Ob(F'(U))$ is the set of pairs
$(\mathcal{U}, \xi)$ where $\mathcal{U} = \{U_i \to U\}$
is a covering of $U$ and $\xi = (x_i, \varphi_{ii'})$ is
a descent datum relative to $\mathcal{U}$.
\item A morphism in $F'(U)$ from
$(\mathcal{U}, \xi)$ to $(\mathcal{V}, \eta)$
is an element of
$$
\colim \Mor_{DD(\mathcal{W})}(a^*\xi, b^*\eta)
$$
where the colimit is over all common refinements
$a : \mathcal{W} \to \mathcal{U}$, $b : \mathcal{W} \to \mathcal{V}$.
This colimit is filtered (verification omitted).
Hence composition of morphisms in $F(U)$ is defined by
finding a common refinement and composing in $DD(\mathcal{W})$.
\item Given $h : V \to U$ and an object
$(\mathcal{U}, \xi)$ of $F'(U)$ we set $F'(h)(\mathcal{U}, \xi)$
equal to $(V \times_U \mathcal{U}, \text{pr}_1^*\xi)$.
More precisely, if $\mathcal{U} = \{U_i \to U\}$
and $\xi = (x_i, \varphi_{ii'})$, then
$V \times_U \mathcal{U} = \{V \times_U U_i \to V\}$
which comes with a canonical morphism
$\text{pr}_1 : V \times_U \mathcal{U} \to \mathcal{U}$ and
$\text{pr}_1^*\xi$ is the pullback of $\xi$ with respect to
this morphism (see Definition \ref{definition-pullback-functor}).
\item Given $h : V \to U$, objects $(\mathcal{U}, \xi)$
and $(\mathcal{V}, \eta)$ and a morphism between them, represented by
$a : \mathcal{W} \to \mathcal{U}$, $b : \mathcal{W} \to \mathcal{V}$,
and $\alpha : a^*\xi \to b^*\eta$, then $F'(h)(\alpha)$ is
represented by
$a' : V \times_U\mathcal{W} \to V \times_U\mathcal{U}$,
$b' : V \times_U\mathcal{W} \to V \times_U\mathcal{V}$,
and the pullback $\alpha'$ of the morphism $\alpha$ via
the map $V \times_U \mathcal{W} \to \mathcal{W}$. This works
since pullbacks in $\mathcal{S}_F$ commute on the nose.
\end{enumerate}
There is a map $F \to F'$ given by associating to
an object $x$ of $F(U)$ the object $(\{U \to U\}, (x, triv))$ of
$F'(U)$. At this point you have to check that the corresponding
functor $\mathcal{S}_F \to \mathcal{S}_{F'}$ has properties (1)
and (2) of the lemma, and finally that $\mathcal{S}_{F'}$ is
a stack. Details omitted.
\end{proof}
\begin{lemma}
\label{lemma-stackify-universal-property}
Let $\mathcal{C}$ be a site.
Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category over $\mathcal{C}$.
Let $p' : \mathcal{S}' \to \mathcal{C}$ and $G : \mathcal{S} \to \mathcal{S}'$
the stack and $1$-morphism constructed in Lemma \ref{lemma-stackify}.
This construction has the following universal property: Given a stack
$q : \mathcal{X} \to \mathcal{C}$ and a $1$-morphism
$F : \mathcal{S} \to \mathcal{X}$ of fibred categories over $\mathcal{C}$
there exists a $1$-morphism $H : \mathcal{S}' \to \mathcal{X}$
such that the diagram
$$
\xymatrix{
\mathcal{S} \ar[rr]_F \ar[rd]_G & & \mathcal{X} \\
& \mathcal{S}' \ar[ru]_H
}
$$
is $2$-commutative.
\end{lemma}
\begin{proof}
Omitted. Hint: Suppose that $x' \in \Ob(\mathcal{S}'_U)$.
By the result of Lemma \ref{lemma-stackify}
there exists a covering $\{U_i \to U\}_{i \in I}$
such that $x'|_{U_i} = G(x_i)$ for some $x_i \in \Ob(\mathcal{S}_{U_i})$.
Moreover, there exist coverings $\{U_{ijk} \to U_i \times_U U_j\}$
and isomorphisms $\alpha_{ijk} : x_i|_{U_{ijk}} \to x_j|_{U_{ijk}}$
with $G(\alpha_{ijk}) = \text{id}_{x'|_{U_{ijk}}}$. Set $y_i = F(x_i)$.
Then you can check that
$$
F(\alpha_{ijk}) : y_i|_{U_{ijk}} \to y_j|_{U_{ijk}}
$$
agree on overlaps and therefore (as $\mathcal{X}$ is a stack) define
a morphism $\beta_{ij} : y_i|_{U_i \times_U U_j} \to y_j|_{U_i \times_U U_j}$.
Next, you check that the $\beta_{ij}$ define a descent datum. Since
$\mathcal{X}$ is a stack these descent data are effective and we find
an object $y$ of $\mathcal{X}_U$ agreeing with $G(x_i)$ over $U_i$.
The hint is to set $H(x') = y$.
\end{proof}
\begin{lemma}
\label{lemma-stackify-universal-property-more}
Notation and assumptions as in
Lemma \ref{lemma-stackify-universal-property}.
There is a canonical equivalence of categories
$$
\Mor_{\textit{Fib}/\mathcal{C}}(\mathcal{S}, \mathcal{X})
=
\Mor_{\textit{Stacks}/\mathcal{C}}(\mathcal{S}', \mathcal{X})
$$
given by the constructions in the proof of the aforementioned lemma.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-stackification-fibre-product-fibred-categories}
Let $\mathcal{C}$ be a site.
Let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Z} \to \mathcal{Y}$
be morphisms of fibred categories over $\mathcal{C}$.
In this case the stackification of the $2$-fibre product is the $2$-fibre
product of the stackifications.
\end{lemma}
\begin{proof}
Let us denote $\mathcal{X}', \mathcal{Y}', \mathcal{Z}'$ the stackifications
and $\mathcal{W}$ the stackification of
$\mathcal{X} \times_\mathcal{Y} \mathcal{Z}$. By construction of $2$-fibre
products there is a canonical $1$-morphism
$\mathcal{X} \times_\mathcal{Y} \mathcal{Z} \to
\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{Z}'$.
As the second $2$-fibre product is a stack (see
Lemma \ref{lemma-2-product-stacks})
this $1$-morphism induces a $1$-morphism
$h : \mathcal{W} \to \mathcal{X}' \times_{\mathcal{Y}'} \mathcal{Z}'$
by the universal property of stackification, see
Lemma \ref{lemma-stackify-universal-property}.
Now $h$ is a morphism of stacks, and we may check that it is an
equivalence using
Lemmas \ref{lemma-characterize-ff} and
\ref{lemma-characterize-essentially-surjective-when-ff}.
\medskip\noindent
Thus we first prove that $h$ induces isomorphisms of $\mathit{Mor}$-sheaves.
Let $\xi, \xi'$ be objects of $\mathcal{W}$ over
$U \in \Ob(\mathcal{C})$. We want to show that
$$
h : \mathit{Mor}(\xi, \xi') \longrightarrow \mathit{Mor}(h(\xi), h(\xi'))
$$
is an isomorphism. To do this we may work locally on $U$ (see
Sites, Section \ref{sites-section-glueing-sheaves}).
Hence by construction of $\mathcal{W}$ (see
Lemma \ref{lemma-stackify})
we may assume that $\xi, \xi'$
actually come from objects $(x, y, \alpha)$ and $(x', y', \alpha')$
of $\mathcal{X} \times_\mathcal{Y} \mathcal{Z}$ over $U$.
By the same lemma once more we see that in this case
$\mathit{Mor}(\xi, \xi')$ is the sheafification of
$$
V/U \longmapsto
\Mor_{\mathcal{X}_V}(x|_V, x'|_V)
\times_{\Mor_{\mathcal{Z}_V}(f(x)|_V, f(x')|_V)}
\Mor_{\mathcal{Y}_V}(y|_V, y'|_V)
$$
and that $\mathit{Mor}(h(\xi), h(\xi'))$ is equal to the fibre product
$$
\mathit{Mor}(i(x), i(x'))
\times_{\mathit{Mor}(k(f(x)), k(f(x'))}
\mathit{Mor}(j(x), j(x'))
$$
where $i : \mathcal{X} \to \mathcal{X}'$,
$j : \mathcal{Y} \to \mathcal{Y}'$, and
$k : \mathcal{Z} \to \mathcal{Z}'$ are the canonical functors.
Thus the first displayed map of this paragraph is an isomorphism as
sheafification is exact (and hence the sheafification of a fibre product
of presheaves is the fibre product of the sheafifications).
\medskip\noindent
Finally, we have to check that any object of
$\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{Z}'$
over $U$ is locally on $U$ in the essential image of $h$.
Write such an object as a triple $(x', y', \alpha)$.
Then $x'$ locally comes from an object of $\mathcal{X}$,
$y'$ locally comes from an object of $\mathcal{Y}$, and
having made suitable replacements for $x'$, $y'$ the morphism
$\alpha$ of $\mathcal{Z}'_U$ locally comes from a morphism of
$\mathcal{Z}$. In other words, we have shown that any object of
$\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{Z}'$
over $U$ is locally on $U$ in the essential image of
$\mathcal{X} \times_\mathcal{Y} \mathcal{Z} \to
\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{Z}'$, hence
a fortiori it is locally in the essential image of $h$.
\end{proof}
\begin{lemma}
\label{lemma-stackification-inertia}
Let $\mathcal{C}$ be a site.
Let $\mathcal{X}$ be a fibred category over $\mathcal{C}$.
The stackification of the inertia fibred category $\mathcal{I}_\mathcal{X}$
is inertia of the stackification of $\mathcal{X}$.
\end{lemma}
\begin{proof}
This follows from the fact that stackification is compatible
with $2$-fibre products by
Lemma \ref{lemma-stackification-fibre-product-fibred-categories}
and the fact that there is a formula for the inertia in terms of
$2$-fibre products of categories over $\mathcal{C}$, see
Categories, Lemma \ref{categories-lemma-inertia-fibred-category}.
\end{proof}
\section{Stackification of categories fibred in groupoids}
\label{section-stackify-groupoids}
\noindent
Here is the result.
\begin{lemma}
\label{lemma-stackify-groupoids}
Let $\mathcal{C}$ be a site.
Let $p : \mathcal{S} \to \mathcal{C}$ be a category
fibred in groupoids over $\mathcal{C}$.
There exists a stack in groupoids
$p' : \mathcal{S}' \to \mathcal{C}$ and a
$1$-morphism $G : \mathcal{S} \to \mathcal{S}'$
of categories fibred in groupoids over $\mathcal{C}$ (see
Categories, Definition
\ref{categories-definition-categories-fibred-in-groupoids-over-C})
such that
\begin{enumerate}
\item for every $U \in \Ob(\mathcal{C})$, and any
$x, y \in \Ob(\mathcal{S}_U)$ the map
$$
\mathit{Mor}(x, y) \longrightarrow \mathit{Mor}(G(x), G(y))
$$
induced by $G$ identifies the right hand side with the sheafification
of the left hand side, and
\item for every $U \in \Ob(\mathcal{C})$, and any
$x' \in \Ob(\mathcal{S}'_U)$ there exists a covering
$\{U_i \to U\}_{i \in I}$ such that for every $i \in I$ the
object $x'|_{U_i}$ is in the essential image of the
functor $G : \mathcal{S}_{U_i} \to \mathcal{S}'_{U_i}$.
\end{enumerate}
Moreover the stack in groupoids $\mathcal{S}'$ is determined up to unique
$2$-isomorphism by these conditions.
\end{lemma}
\begin{proof}
Apply Lemma \ref{lemma-stackify}. The result will be a
stack in groupoids by applying Lemma \ref{lemma-stack-in-groupoids-stack}.
\end{proof}
\begin{lemma}
\label{lemma-stackify-groupoids-universal-property}
Let $\mathcal{C}$ be a site.
Let $p : \mathcal{S} \to \mathcal{C}$ be a category fibred in groupoids
over $\mathcal{C}$. Let $p' : \mathcal{S}' \to \mathcal{C}$ and
$G : \mathcal{S} \to \mathcal{S}'$
the stack in groupoids and $1$-morphism constructed in
Lemma \ref{lemma-stackify-groupoids}.
This construction has the following universal property: Given a stack
in groupoids $q : \mathcal{X} \to \mathcal{C}$ and a $1$-morphism
$F : \mathcal{S} \to \mathcal{X}$ of categories over $\mathcal{C}$
there exists a $1$-morphism $H : \mathcal{S}' \to \mathcal{X}$
such that the diagram
$$
\xymatrix{
\mathcal{S} \ar[rr]_F \ar[rd]_G & & \mathcal{X} \\
& \mathcal{S}' \ar[ru]_H
}
$$
is $2$-commutative.
\end{lemma}
\begin{proof}
This is a special case of
Lemma \ref{lemma-stackify-universal-property}.
\end{proof}
\begin{lemma}
\label{lemma-stackification-fibre-product-categories-fibred-in-groupoids}
Let $\mathcal{C}$ be a site.
Let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Y} \to \mathcal{Z}$
be morphisms of categories fibred in groupoids over $\mathcal{C}$.
In this case the stackification of the $2$-fibre product is the $2$-fibre
product of the stackifications.
\end{lemma}
\begin{proof}
This is a special case of
Lemma \ref{lemma-stackification-fibre-product-fibred-categories}.
\end{proof}
\section{Inherited topologies}
\label{section-topology}
\noindent
It turns out that a fibred category over a site inherits a canonical
topology from the underlying site.
\begin{lemma}
\label{lemma-topology-inherited}
Let $\mathcal{C}$ be a site. Let $p : \mathcal{S} \to \mathcal{C}$
be a fibred category. Let $\text{Cov}(\mathcal{S})$
be the set of families $\{x_i \to x\}_{i \in I}$ of morphisms in $\mathcal{S}$
with fixed target such that (a) each $x_i \to x$ is strongly cartesian,
and (b) $\{p(x_i) \to p(x)\}_{i \in I}$ is a covering of $\mathcal{C}$.
Then $(\mathcal{S}, \text{Cov}(\mathcal{S}))$ is a site.
\end{lemma}
\begin{proof}
We have to check the three conditions of
Sites, Definition \ref{sites-definition-site}.
\begin{enumerate}
\item If $x \to y$ is an isomorphism of $\mathcal{S}$, then
it is strongly cartesian by
Categories, Lemma \ref{categories-lemma-composition-cartesian}
and $p(x) \to p(y)$ is an isomorphism of $\mathcal{C}$. Thus
$\{p(x) \to p(y)\}$ is a covering of $\mathcal{C}$ whence
$\{x \to y\} \in \text{Cov}(\mathcal{S})$.
\item If $\{x_i \to x\}_{i\in I} \in \text{Cov}(\mathcal{S})$ and for each
$i$ we have $\{y_{ij} \to x_i\}_{j\in J_i} \in \text{Cov}(\mathcal{S})$, then
each composition $p(y_{ij}) \to p(x)$ is strongly cartesian by
Categories, Lemma \ref{categories-lemma-composition-cartesian}
and $\{p(y_{ij}) \to p(x)\}_{i \in I, j\in J_i} \in \text{Cov}(\mathcal{C})$.
Hence also $\{y_{ij} \to x\}_{i \in I, j\in J_i} \in \text{Cov}(\mathcal{S})$.
\item Suppose $\{x_i \to x\}_{i\in I}\in \text{Cov}(\mathcal{S})$
and $y \to x$ is a morphism of $\mathcal{S}$. As $\{p(x_i) \to p(x)\}$
is a covering of $\mathcal{C}$ we see that $p(x_i) \times_{p(x)} p(y)$
exists. Hence
Categories, Lemma \ref{categories-lemma-fibred-category-representable-goes-up}
implies that $x_i \times_x y$ exists, that $p(x_i \times_x y) =
p(x_i) \times_{p(x)} p(y)$, and that $x_i \times_x y \to y$ is
strongly cartesian. Since also
$\{p(x_i) \times_{p(x)} p(y) \to p(y) \}_{i\in I} \in \text{Cov}(\mathcal{C})$
we conclude that
$\{x_i \times_x y \to y \}_{i\in I} \in \text{Cov}(\mathcal{S})$
\end{enumerate}
This finishes the proof.
\end{proof}
\noindent
Note that if $p : \mathcal{S} \to \mathcal{C}$ is fibred in groupoids, then
the coverings of the site $\mathcal{S}$ in
Lemma \ref{lemma-topology-inherited}
are characterized by
$$
\{x_i \to x\} \in \text{Cov}(\mathcal{S})
\Leftrightarrow
\{p(x_i) \to p(x)\} \in \text{Cov}(\mathcal{C})
$$
because every morphism of $\mathcal{S}$ is strongly cartesian.
\begin{definition}
\label{definition-topology-inherited}
Let $\mathcal{C}$ be a site. Let $p : \mathcal{S} \to \mathcal{C}$ be a
fibred category. We say $(\mathcal{S}, \text{Cov}(\mathcal{S}))$ as in
Lemma \ref{lemma-topology-inherited}
is the {\it structure of site on $\mathcal{S}$ inherited from $\mathcal{C}$}.
We sometimes indicate this by saying that
{\it $\mathcal{S}$ is endowed with the topology inherited from $\mathcal{C}$}.
\end{definition}
\noindent
In particular we obtain a topos of sheaves $\Sh(\mathcal{S})$
in this situation. It turns out that this topos is functorial with respect
to $1$-morphisms of fibred categories.
\begin{lemma}
\label{lemma-topology-inherited-functorial}
Let $\mathcal{C}$ be a site. Let $F : \mathcal{X} \to \mathcal{Y}$
be a $1$-morphism of fibred categories over $\mathcal{C}$.
Then $F$ is a continuous and cocontinuous functor between the structure
of sites inherited from $\mathcal{C}$. Hence $F$ induces a morphism of topoi
$f : \Sh(\mathcal{X}) \to \Sh(\mathcal{Y})$ with
$f_* = {}_sF = {}_pF$ and $f^{-1} = F^s = F^p$. In particular
$f^{-1}(\mathcal{G})(x) = \mathcal{G}(F(x))$
for a sheaf $\mathcal{G}$ on $\mathcal{Y}$ and object $x$ of $\mathcal{X}$.
\end{lemma}
\begin{proof}
We first prove that $F$ is continuous.
Let $\{x_i \to x\}_{i \in I}$ be a covering of $\mathcal{X}$. By
Categories, Definition \ref{categories-definition-fibred-categories-over-C}
the functor $F$ transforms strongly cartesian morphisms into strongly
cartesian morphisms, hence $\{F(x_i) \to F(x)\}_{i \in I}$ is a covering
of $\mathcal{Y}$. This proves part (1) of
Sites, Definition \ref{sites-definition-continuous}.
Moreover, let $x' \to x$ be a morphism of $\mathcal{X}$. By
Categories, Lemma \ref{categories-lemma-fibred-category-representable-goes-up}
the fibre product $x_i \times_x x'$ exists and $x_i \times_x x' \to x'$
is strongly cartesian. Hence $F(x_i \times_x x') \to F(x')$ is strongly
cartesian. By
Categories, Lemma \ref{categories-lemma-fibred-category-representable-goes-up}
applied to $\mathcal{Y}$ this means that
$F(x_i \times_x x') = F(x_i) \times_{F(x)} F(x')$.
This proves part (2) of
Sites, Definition \ref{sites-definition-continuous}
and we conclude that $F$ is continuous.
\medskip\noindent
Next we prove that $F$ is cocontinuous.
Let $x \in \Ob(\mathcal{X})$ and let $\{y_i \to F(x)\}_{i \in I}$
be a covering in $\mathcal{Y}$. Denote $\{U_i \to U\}_{i \in I}$ the
corresponding covering of $\mathcal{C}$. For each $i$ choose a strongly
cartesian morphism $x_i \to x$ in $\mathcal{X}$ lying over $U_i \to U$.
Then $F(x_i) \to F(x)$ and $y_i \to F(x)$ are both a strongly cartesian
morphisms in $\mathcal{Y}$ lying over $U_i \to U$. Hence there exists
a unique isomorphism $F(x_i) \to y_i$ in $\mathcal{Y}_{U_i}$ compatible
with the maps to $F(x)$. Thus $\{x_i \to x\}_{i \in I}$ is a covering of
$\mathcal{X}$ such that $\{F(x_i) \to F(x)\}_{i \in I}$ is isomorphic to
$\{y_i \to F(x)\}_{i \in I}$. Hence $F$ is cocontinuous, see
Sites, Definition \ref{sites-definition-cocontinuous}.
\medskip\noindent
The final assertion follows from the first two, see
Sites, Lemmas
\ref{sites-lemma-cocontinuous-morphism-topoi},
\ref{sites-lemma-pu-sheaf}, and
\ref{sites-lemma-when-shriek}.
\end{proof}
\begin{lemma}
\label{lemma-localizing}
Let $\mathcal{C}$ be a site. Let $p : \mathcal{X} \to \mathcal{C}$
be a category fibred in groupoids. Let $x \in \Ob(\mathcal{X})$
lying over $U = p(x)$. The functor $p$ induces an equivalence of sites
$\mathcal{X}/x \to \mathcal{C}/U$ where $\mathcal{X}$ is endowed with
the topology inherited from $\mathcal{C}$.
\end{lemma}
\begin{proof}
Here $\mathcal{C}/U$ is the localization of the site
$\mathcal{C}$ at the object $U$ and similarly for $\mathcal{X}/x$.
It follows from
Categories, Definition \ref{categories-definition-fibred-groupoids}
that the rule $x'/x \mapsto p(x')/p(x)$ defines an equivalence of
categories $\mathcal{X}/x \to \mathcal{C}/U$. Whereupon it follows from
Definition \ref{definition-topology-inherited}
that coverings of $x'$ in $\mathcal{X}/x$ are in bijective correspondence
with coverings of $p(x')$ in $\mathcal{C}/U$.
\end{proof}
\begin{lemma}
\label{lemma-stack-in-groupoids-over-stack-in-groupoids}
Let $\mathcal{C}$ be a site. Let $p : \mathcal{X} \to \mathcal{C}$
and $q : \mathcal{Y} \to \mathcal{C}$
be stacks in groupoids. Let $F : \mathcal{X} \to \mathcal{Y}$
be a $1$-morphism of categories over $\mathcal{C}$. If $F$ turns
$\mathcal{X}$ into a category fibred in groupoids over $\mathcal{Y}$,
then $\mathcal{X}$ is a stack in groupoids over $\mathcal{Y}$ (with
topology inherited from $\mathcal{C}$).
\end{lemma}
\begin{proof}
Let us prove descent for objects. Let $\{y_i \to y\}$ be a covering
of $\mathcal{Y}$. Let $(x_i, \varphi_{ij})$ be a descent datum in $\mathcal{X}$
with respect to this covering. Then $(x_i, \varphi_{ij})$ is also a descent
datum with respect to the covering $\{q(y_i) \to q(y)\}$ of
$\mathcal{C}$. As $\mathcal{X}$ is a stack in groupoids we obtain an
object $x$ over $q(y)$ and isomorphisms $\psi_i : x|_{q(y_i)} \to x_i$
over $q(y_i)$ compatible with the $\varphi_{ij}$, i.e., such that
$$
\varphi_{ij} = \psi_j|_{q(y_i) \times_{q(y)} q(y_j)}
\circ \psi_i^{-1}|_{q(y_i) \times_{q(y)} q(y_j)}.
$$
Consider the sheaf $\mathit{I} = \mathit{Isom}_\mathcal{Y}(F(x), y)$
on $\mathcal{C}/p(x)$. Note that $s_i = F(\psi_i) \in \mathit{I}(q(x_i))$
because $F(x_i) = y_i$. Because $F(\varphi_{ij}) = \text{id}$ (as we started
with a descent datum over $\{y_i \to y\}$) the displayed formula shows
that $s_i|_{q(y_i) \times_{q(y)} q(y_j)} = s_j|_{q(y_i) \times_{q(y)} q(y_j)}$.
Hence the local sections $s_i$ glue to $s : F(x) \to y$. As $F$ is fibred in
groupoids we see that $x$ is isomorphic to an object $x'$ with $F(x') = y$.
We omit the verification that $x'$ in the fibre category of $\mathcal{X}$
over $y$ is a solution to the problem of descent posed by the descent datum
$(x_i, \varphi_{ij})$. We also omit the proof of the sheaf property
of the $\mathit{Isom}$-presheaves of $\mathcal{X}/\mathcal{Y}$.
\end{proof}
\begin{lemma}
\label{lemma-stack-over-stack}
Let $\mathcal{C}$ be a site. Let $p : \mathcal{X} \to \mathcal{C}$
be a stack. Endow $\mathcal{X}$ with the topology inherited from
$\mathcal{C}$ and let $q : \mathcal{Y} \to \mathcal{X}$ be a stack.
Then $\mathcal{Y}$ is a stack over $\mathcal{C}$.
If $p$ and $q$ define stacks in groupoids, then
$\mathcal{Y}$ is a stack in groupoids over $\mathcal{C}$.
\end{lemma}
\begin{proof}
We check the three conditions in Definition \ref{definition-stack}
to prove that $\mathcal{Y}$ is a stack over $\mathcal{C}$.
By Categories, Lemma \ref{categories-lemma-fibred-over-fibred}
we find that $\mathcal{Y}$ is a fibred category over $\mathcal{C}$.
Thus condition (1) holds.
\medskip\noindent
Let $U$ be an object of $\mathcal{C}$ and let $y_1, y_2$ be objects
of $\mathcal{Y}$ over $U$. Denote $x_i = q(y_i)$ in $\mathcal{X}$.
Consider the map of presheaves
$$
q : \mathit{Mor}_{\mathcal{Y}/\mathcal{C}}(y_1, y_2)
\longrightarrow
\mathit{Mor}_{\mathcal{X}/\mathcal{C}}(x_1, x_2)
$$
on $\mathcal{C}/U$, see Lemma \ref{lemma-presheaf-mor-map-fibred-categories}.
Let $\{U_i \to U\}$ be a covering and let $\varphi_i$ be a section
of the presheaf on the left over $U_i$ such that $\varphi_i$ and
$\varphi_j$ restrict to the same section over $U_i \times_U U_j$.
We have to find a morphism $\varphi : x_1 \to x_2$ restricting to $\varphi_i$.
Note that $q(\varphi_i) = \psi|_{U_i}$ for some morphism
$\psi : x_1 \to x_2$ over $U$ because the second presheaf is a sheaf
(by assumption). Let $y_{12} \to y_2$ be the stronly $\mathcal{X}$-cartesian
morphism of $\mathcal{Y}$ lying over $\psi$. Then $\varphi_i$ corresponds
to a morphism $\varphi'_i : y_1|_{U_i} \to y_{12}|_{U_i}$ over $x_1|_{U_i}$.
In other words, $\varphi'_i$ now define local sections of the presheaf
$$
\mathit{Mor}_{\mathcal{Y}/\mathcal{X}}(y_1, y_{12})
$$
over the members of the covering $\{x_1|_{U_i} \to x_1\}$. By assumption these
glue to a unique morphism $y_1 \to y_{12}$ which composed with the given
morphism $y_{12} \to y_2$ produces the desired morphism $y_1 \to y_2$.
\medskip\noindent
Finally, we show that descent data are effective. Let $\{f_i : U_i \to U\}$
be a covering of $\mathcal{C}$ and let $(y_i, \varphi_{ij})$ be a descent
datum relative to this covering (Definition \ref{definition-descent-data}).
Setting $x_i = q(y_i)$ and $\psi_{ij} = q(\varphi_{ij})$
we obtain a descent datum $(x_i, \psi_{ij})$ for the covering in $\mathcal{X}$.
By assumption on $\mathcal{X}$ we may assume $x_i = x|_{U_i}$
and the $\psi_{ij}$ equal to the canonical descent datum
(Definition \ref{definition-effective-descent-datum}).
In this case $\{x|_{U_i} \to x\}$ is a covering and we can view
$(y_i, \varphi_{ij})$ as a descent datum relative to this covering.
By our assumption that $\mathcal{Y}$ is a stack over $\mathcal{C}$
we see that it is effective which finishes the proof of condition (3).
\medskip\noindent
The final assertion follows because $\mathcal{Y}$ is a stack over
$\mathcal{C}$ and is fibred in groupoids by
Categories, Lemma
\ref{categories-lemma-fibred-in-groupoids-over-fibred-in-groupoids}.
\end{proof}
\section{Gerbes}
\label{section-gerbes}
\noindent
Gerbes are a special kind of stacks in groupoids.
\begin{definition}
\label{definition-gerbe}
A {\it gerbe} over a site $\mathcal{C}$ is a category
$p : \mathcal{S} \to \mathcal{C}$ over $\mathcal{C}$ such that
\begin{enumerate}
\item $p : \mathcal{S} \to \mathcal{C}$ is a stack
in groupoids over $\mathcal{C}$ (see
Definition \ref{definition-stack-in-groupoids}),
\item for $U \in \Ob(\mathcal{C})$ there exists
a covering $\{U_i \to U\}$ in $\mathcal{C}$ such that
$\mathcal{S}_{U_i}$ is nonempty, and
\item for $U \in \Ob(\mathcal{C})$ and
$x, y \in \Ob(\mathcal{S}_U)$ there exists
a covering $\{U_i \to U\}$ in $\mathcal{C}$ such that
$x|_{U_i} \cong y|_{U_i}$ in $\mathcal{S}_{U_i}$.
\end{enumerate}
\end{definition}
\noindent
In other words, a gerbe is a stack in groupoids such that any two objects
are locally isomorphic and such that objects exist locally.
\begin{lemma}
\label{lemma-gerbe-equivalent}
Let $\mathcal{C}$ be a site.
Let $\mathcal{S}_1$, $\mathcal{S}_2$ be categories over $\mathcal{C}$.
Suppose that $\mathcal{S}_1$ and $\mathcal{S}_2$ are equivalent
as categories over $\mathcal{C}$.
Then $\mathcal{S}_1$ is a gerbe over $\mathcal{C}$ if and only if
$\mathcal{S}_2$ is a gerbe over $\mathcal{C}$.
\end{lemma}
\begin{proof}
Assume $\mathcal{S}_1$ is a gerbe over $\mathcal{C}$. By
Lemma \ref{lemma-stack-in-groupoids-equivalent}
we see $\mathcal{S}_2$ is a stack in groupoids over $\mathcal{C}$.
Let $F : \mathcal{S}_1 \to \mathcal{S}_2$,
$G : \mathcal{S}_2 \to \mathcal{S}_1$ be equivalences of categories over
$\mathcal{C}$. Given $U \in \Ob(\mathcal{C})$ we see that there exists
a covering $\{U_i \to U\}$ such that $(\mathcal{S}_1)_{U_i}$ is
nonempty. Applying $F$ we see that $(\mathcal{S}_2)_{U_i}$ is nonempty.
Given $U \in \Ob(\mathcal{C})$ and
$x, y \in \Ob((\mathcal{S}_2)_U)$ there exists
a covering $\{U_i \to U\}$ in $\mathcal{C}$ such that
$G(x)|_{U_i} \cong G(y)|_{U_i}$ in $(\mathcal{S}_1)_{U_i}$. By
Categories, Lemma \ref{categories-lemma-equivalence-fibred-categories}
this implies $x|_{U_i} \cong y|_{U_i}$ in $(\mathcal{S}_2)_{U_i}$.
\end{proof}
\noindent
We want to generalize the definition of gerbes a bit. Namely, let
$F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids
over a site $\mathcal{C}$. We want to say what it
means for $\mathcal{X}$ to be a gerbe over $\mathcal{Y}$. By
Section \ref{section-topology}
the category $\mathcal{Y}$ inherits the structure of a site from
$\mathcal{C}$. A naive guess is: Just require that
$\mathcal{X} \to \mathcal{Y}$ {\it is} a gerbe in the sense above. Except
the notion so obtained is not invariants under replacing $\mathcal{X}$
by an equivalent stack in groupoids over $\mathcal{C}$; this is even the
case for the property of being fibred in groupoids over $\mathcal{Y}$.
However, it turns out that we can replace $\mathcal{X}$ by an equivalent
stack in groupoids over $\mathcal{Y}$ which is fibred in groupoids over
$\mathcal{Y}$, and then the property of being a gerbe over $\mathcal{Y}$
is independent of this choice. Here is the precise formulation.
\begin{lemma}
\label{lemma-when-gerbe}
Let $\mathcal{C}$ be a site. Let $p : \mathcal{X} \to \mathcal{C}$
and $q : \mathcal{Y} \to \mathcal{C}$ be stacks in groupoids.
Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories
over $\mathcal{C}$. The following are equivalent
\begin{enumerate}
\item For some (equivalently any) factorization $F = F' \circ a$ where
$a : \mathcal{X} \to \mathcal{X}'$ is an equivalence of categories over
$\mathcal{C}$ and $F'$ is fibred in groupoids, the map
$F' : \mathcal{X}' \to \mathcal{Y}$ is a gerbe (with the topology
on $\mathcal{Y}$ inherited from $\mathcal{C}$).
\item The following two conditions are satisfied
\begin{enumerate}
\item for $y \in \Ob(\mathcal{Y})$ lying over
$U \in \Ob(\mathcal{C})$ there exists a covering
$\{U_i \to U\}$ in $\mathcal{C}$ and objects $x_i$ of $\mathcal{X}$
over $U_i$ such that $F(x_i) \cong y|_{U_i}$ in $\mathcal{Y}_{U_i}$, and
\item for $U \in \Ob(\mathcal{C})$,
$x, x' \in \Ob(\mathcal{X}_U)$, and $b : F(x) \to F(x')$ in
$\mathcal{Y}_U$ there exists
a covering $\{U_i \to U\}$ in $\mathcal{C}$ and morphisms
$a_i : x|_{U_i} \to x'|_{U_i}$ in $\mathcal{X}_{U_i}$ with
$F(a_i) = b|_{U_i}$.
\end{enumerate}
\end{enumerate}
\end{lemma}
\begin{proof}
By
Categories, Lemma
\ref{categories-lemma-ameliorate-morphism-categories-fibred-groupoids}
there exists a factorization $F = F' \circ a$ where
$a : \mathcal{X} \to \mathcal{X}'$ is an equivalence of categories over
$\mathcal{C}$ and $F'$ is fibred in groupoids. By
Categories, Lemma \ref{categories-lemma-amelioration-unique}
given any two such factorizations $F = F' \circ a = F'' \circ b$
we have that $\mathcal{X}'$ is equivalent to $\mathcal{X}''$ as
categories over $\mathcal{Y}$. Hence
Lemma \ref{lemma-gerbe-equivalent}
guarantees that the condition (1) is independent of the choice of the
factorization. Moreover, this means that we may assume
$\mathcal{X}' = \mathcal{X} \times_{F, \mathcal{Y}, \text{id}} \mathcal{Y}$
as in the proof of
Categories, Lemma
\ref{categories-lemma-ameliorate-morphism-categories-fibred-groupoids}
\medskip\noindent
Let us prove that (a) and (b) imply that $\mathcal{X}' \to \mathcal{Y}$
is a gerbe. First of all, by
Lemma \ref{lemma-stack-in-groupoids-over-stack-in-groupoids}
we see that $\mathcal{X}' \to \mathcal{Y}$ is a stack in groupoids.
Next, let $y$ be an object of $\mathcal{Y}$ lying over
$U \in \Ob(\mathcal{C})$. By (a) we can find a covering
$\{U_i \to U\}$ in $\mathcal{C}$ and objects $x_i$ of $\mathcal{X}$
over $U_i$ and isomorphisms $f_i : F(x_i) \to y|_{U_i}$ in
$\mathcal{Y}_{U_i}$. Then $(U_i, x_i, y|_{U_i}, f_i)$ are objects
of $\mathcal{X}'_{U_i}$, i.e., the second condition of
Definition \ref{definition-gerbe}
holds. Finally, let $(U, x, y, f)$ and $(U, x', y, f')$ be objects
of $\mathcal{X}'$ lying over the same object $y \in \Ob(\mathcal{Y})$.
Set $b = (f')^{-1} \circ f$. By condition (b) we can find a covering
$\{U_i \to U\}$ and isomorphisms $a_i : x|_{U_i} \to x'|_{U_i}$
in $\mathcal{X}_{U_i}$ with $F(a_i) = b|_{U_i}$. Then
$$
(a_i, \text{id}) : (U, x, y, f)|_{U_i} \to (U, x', y, f')|_{U_i}
$$
is a morphism in $\mathcal{X}'_{U_i}$ as desired. This proves that
(2) implies (1).
\medskip\noindent
To prove that (1) implies (2) one reads the arguments
in the preceding paragraph backwards. Details omitted.
\end{proof}
\begin{definition}
\label{definition-gerbe-over-stack-in-groupoids}
Let $\mathcal{C}$ be a site. Let $\mathcal{X}$
and $\mathcal{Y}$ be stacks in groupoids over $\mathcal{C}$.
Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories
over $\mathcal{C}$. We say $\mathcal{X}$ is a {\it gerbe over} $\mathcal{Y}$
if the equivalent conditions of
Lemma \ref{lemma-when-gerbe}
are satisfied.
\end{definition}
\noindent
This definition does not conflict with
Definition \ref{definition-gerbe}
when $\mathcal{Y} = \mathcal{C}$ because in this case we may take
$\mathcal{X}' = \mathcal{X}$ in part (1) of
Lemma \ref{lemma-when-gerbe}.
Note that conditions (2)(a) and (2)(b) of
Lemma \ref{lemma-when-gerbe}
are quite close in spirit to conditions (2) and (3) of
Definition \ref{definition-gerbe}.
Namely, (2)(a) says that the map of presheaves of isomorphism
classes of objects becomes a surjection after sheafification.
Moreover, (2)(b) says that
$$
\mathit{Isom}_\mathcal{X}(x, x')
\longrightarrow
\mathit{Isom}_\mathcal{Y}(F(x), F(x'))
$$
is a surjection of sheaves on $\mathcal{C}/U$ for any $U$ and
$x, x' \in \Ob(\mathcal{X}_U)$.
\begin{lemma}
\label{lemma-base-change-gerbe}
Let $\mathcal{C}$ be a site. Let
$$
\xymatrix{
\mathcal{X}' \ar[r]_{G'} \ar[d]_{F'} & \mathcal{X} \ar[d]^F \\
\mathcal{Y}' \ar[r]^G & \mathcal{Y}
}
$$
be a $2$-fibre product of stacks in groupoids over $\mathcal{C}$.
If $\mathcal{X}$ is a gerbe over $\mathcal{Y}$, then
$\mathcal{X}'$ is a gerbe over $\mathcal{Y}'$.
\end{lemma}
\begin{proof}
By the uniqueness property of a $2$-fibre product may assume that
$\mathcal{X}' = \mathcal{Y}' \times_\mathcal{Y} \mathcal{X}$
as in
Categories, Lemma \ref{categories-lemma-2-product-categories-over-C}.
Let us prove properties (2)(a) and (2)(b) of
Lemma \ref{lemma-when-gerbe}
for $\mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'$.
\medskip\noindent
Let $y'$ be an object of $\mathcal{Y}'$ lying over the object $U$
of $\mathcal{C}$. By assumption there exists
a covering $\{U_i \to U\}$ of $U$ and objects
$x_i \in \mathcal{X}_{U_i}$ with isomorphisms
$\alpha_i : G(y')|_{U_i} \to F(x_i)$.
Then $(U_i, y'|_{U_i}, x_i, \alpha_i)$ is an object of
$\mathcal{Y}' \times_\mathcal{Y} \mathcal{X}$ over $U_i$
whose image in $\mathcal{Y}'$ is $y'|_{U_i}$. Thus (2)(a) holds.
\medskip\noindent
Let $U \in \Ob(\mathcal{C})$, let $x'_1, x'_2$ be objects of
$\mathcal{Y}' \times_\mathcal{Y} \mathcal{X}$ over $U$, and let
$b' : F'(x'_1) \to F'(x'_2)$ be a morphism in $\mathcal{Y}'_U$.
Write $x'_i = (U, y'_i, x_i, \alpha_i)$. Note that $F'(x'_i) = x_i$
and $G'(x'_i) = y'_i$. By assumption there exists
a covering $\{U_i \to U\}$ in $\mathcal{C}$ and morphisms
$a_i : x_1|_{U_i} \to x_2|_{U_i}$ in $\mathcal{X}_{U_i}$ with
$F(a_i) = G(b')|_{U_i}$. Then $(b'|_{U_i}, a_i)$ is a morphism
$x'_1|_{U_i} \to x'_2|_{U_i}$ as required in (2)(b).
\end{proof}
\begin{lemma}
\label{lemma-composition-gerbe}
Let $\mathcal{C}$ be a site. Let
$F : \mathcal{X} \to \mathcal{Y}$ and $G : \mathcal{Y} \to \mathcal{Z}$
be $1$-morphisms of stacks in groupoids over $\mathcal{C}$.
If $\mathcal{X}$ is a gerbe over $\mathcal{Y}$ and
$\mathcal{Y}$ is a gerbe over $\mathcal{Z}$, then
$\mathcal{X}$ is a gerbe over $\mathcal{Z}$.
\end{lemma}
\begin{proof}
Let us prove properties (2)(a) and (2)(b) of
Lemma \ref{lemma-when-gerbe}
for $\mathcal{X} \to \mathcal{Z}$.
\medskip\noindent
Let $z$ be an object of $\mathcal{Z}$ lying over the object $U$
of $\mathcal{C}$. By assumption on $G$ there exists
a covering $\{U_i \to U\}$ of $U$ and objects
$y_i \in \mathcal{Y}_{U_i}$ such that $G(y_i) \cong z|_{U_i}$.
By assumption on $F$ there exist coverings $\{U_{ij} \to U_i\}$
and objects $x_{ij} \in \mathcal{X}_{U_{ij}}$ such that
$F(x_{ij}) \cong y_i|_{U_{ij}}$.
Then $\{U_{ij} \to U\}$ is a covering of $\mathcal{C}$
and $(G \circ F)(x_{ij}) \cong z|_{U_{ij}}$. Thus (2)(a) holds.
\medskip\noindent
Let $U \in \Ob(\mathcal{C})$, let $x_1, x_2$ be objects of
$\mathcal{X}$ over $U$, and let
$c : (G \circ F)(x_1) \to (G \circ F)(x_2)$ be a morphism in
$\mathcal{Z}_U$. By assumption on $G$ there exists
a covering $\{U_i \to U\}$ of $U$ and morphisms
$b_i : F(x_1)|_{U_i} \to F(x_2)|_{U_i}$ in $\mathcal{Y}_{U_i}$
such that $G(b_i) = c|_{U_i}$.
By assumption on $F$ there exist coverings $\{U_{ij} \to U_i\}$
and morphisms $a_{ij} : x_1|_{U_{ij}} \to x_2|_{U_{ij}}$ in
$\mathcal{X}_{U_{ij}}$ such that $F(a_{ij}) = b_i|_{U_{ij}}$.
Then $\{U_{ij} \to U\}$ is a covering of $\mathcal{C}$
and $(G \circ F)(a_{ij}) = c|_{U_{ij}}$ as required in (2)(b).
\end{proof}
\begin{lemma}
\label{lemma-gerbe-descent}
Let $\mathcal{C}$ be a site. Let
$$
\xymatrix{
\mathcal{X}' \ar[r]_{G'} \ar[d]_{F'} & \mathcal{X} \ar[d]^F \\
\mathcal{Y}' \ar[r]^G & \mathcal{Y}
}
$$
be a $2$-cartesian diagram of stacks in groupoids over $\mathcal{C}$.
If for every $U \in \Ob(\mathcal{C})$ and
$x \in \Ob(\mathcal{Y}_U)$ there exists a covering
$\{U_i \to U\}$ such that $x|_{U_i}$ is in the essential
image of $G : \mathcal{Y}'_{U_i} \to \mathcal{Y}_{U_i}$ and
$\mathcal{X}'$ is a gerbe over $\mathcal{Y}'$, then
$\mathcal{X}$ is a gerbe over $\mathcal{Y}$.
\end{lemma}
\begin{proof}
By the uniqueness property of a $2$-fibre product may assume that
$\mathcal{X}' = \mathcal{Y}' \times_\mathcal{Y} \mathcal{X}$
as in
Categories, Lemma \ref{categories-lemma-2-product-categories-over-C}.
Let us prove properties (2)(a) and (2)(b) of
Lemma \ref{lemma-when-gerbe}
for $\mathcal{X} \to \mathcal{Y}$.
\medskip\noindent
Let $y$ be an object of $\mathcal{Y}$ lying over the object $U$
of $\mathcal{C}$. By assumption there exists
a covering $\{U_i \to U\}$ of $U$ and objects
$y'_i \in \mathcal{Y}'_{U_i}$ with $G(y'_i) \cong y|_{U_i}$.
By (2)(a) for $\mathcal{X}' \to \mathcal{Y}'$ there exist
coverings $\{U_{ij} \to U_i\}$ and objects $x'_{ij}$ of
$\mathcal{X}'$ over $U_{ij}$ with $F'(x'_{ij})$ isomorphic
to the restriction of $y'_i$ to $U_{ij}$. Then
$\{U_{ij} \to U\}$ is a covering of $\mathcal{C}$ and
$G'(x'_{ij})$ are objects of $\mathcal{X}$ over $U_{ij}$
whose images in $\mathcal{Y}$ are isomorphic to the restrictions
$y|_{U_{ij}}$. This proves (2)(a) for $\mathcal{X} \to \mathcal{Y}$.
\medskip\noindent
Let $U \in \Ob(\mathcal{C})$, let $x_1, x_2$ be objects of
$\mathcal{X}$ over $U$, and let $b : F(x_1) \to F(x_2)$ be a morphism
in $\mathcal{Y}_U$. By assumption we may choose a covering
$\{U_i \to U\}$ and objects $y'_i$ of $\mathcal{Y}'$
over $U_i$ such that there exist isomorphisms
$\alpha_i : G(y'_i) \to F(x_1)|_{U_i}$.
Then we get objects
$$
x'_{1i} = (U_i, y'_i, x_1|_{U_i}, \alpha_i)
\quad\text{and}\quad
x'_{2i} = (U_i, y'_i, x_2|_{U_i}, b|_{U_i} \circ \alpha_i)
$$
of $\mathcal{X}'$ over $U_i$. The identity morphism on $y'_i$ is a
morphism $F'(x'_{1i}) \to F'(x'_{2i})$. By (2)(b) for
$\mathcal{X}' \to \mathcal{Y}'$ there exist coverings
$\{U_{ij} \to U_i\}$ and morphisms
$a'_{ij} : x'_{1i}|_{U_{ij}} \to x'_{2i}|_{U_{ij}}$
such that $F'(a'_{ij}) = \text{id}_{y'_i}|_{U_{ij}}$. Unwinding the definition
of morphisms in $\mathcal{Y}' \times_\mathcal{Y} \mathcal{X}$
we see that $G'(a'_{ij}) : x_1|_{U_{ij}} \to x_2|_{U_{ij}}$ are
the morphisms we're looking for, i.e., (2)(b) holds for
$\mathcal{X} \to \mathcal{Y}$.
\end{proof}
\noindent
Gerbes all of whose automorphism sheaves are abelian play an important
role in algebraic geometry.
\begin{lemma}
\label{lemma-gerbe-abelian-auts}
Let $p : \mathcal{S} \to \mathcal{C}$ be a gerbe over a site $\mathcal{C}$.
Assume that for all $U \in \Ob(\mathcal{C})$ and $x \in \Ob(\mathcal{S}_U)$
the sheaf of groups $\mathit{Aut}(x) = \mathit{Isom}(x, x)$ on $\mathcal{C}/U$
is abelian. Then there exist
\begin{enumerate}
\item a sheaf $\mathcal{G}$ of abelian groups on $\mathcal{C}$,
\item for every $U \in \Ob(\mathcal{C})$ and every $x \in \Ob(\mathcal{S}_U)$
an isomorphism $\mathcal{G}|_U \to \mathit{Aut}(x)$
\end{enumerate}
such that for every $U$ and every morphism $\varphi : x \to y$
in $\mathcal{S}_U$ the diagram
$$
\xymatrix{
\mathcal{G}|_U \ar[d] \ar@{=}[rr] & & \mathcal{G}|_U \ar[d] \\
\mathit{Aut}(x)
\ar[rr]^{\alpha \mapsto \varphi \circ \alpha \circ \varphi^{-1}} & &
\mathit{Aut}(y)
}
$$
is commutative.
\end{lemma}
\begin{proof}
Let $x, y$ be two objects of $\mathcal{S}$ with $U = p(x) = p(y)$.
\medskip\noindent
If there is a morphism $\varphi : x \to y$ over $U$, then it is an
isomorphism and then we indeed get an isomorphism
$\mathit{Aut}(x) \to \mathit{Aut}(y)$ sending
$\alpha$ to $\varphi \circ \alpha \circ \varphi^{-1}$.
Moreover, since we are assuming $\mathit{Aut}(x)$ is
commutative, this isomorphism is independent of the choice
of $\varphi$ by a simple computation: namely, if $\psi$ is
a second such map, then
$$
\varphi \circ \alpha \circ \varphi^{-1} =
\psi \circ \psi^{-1} \circ \varphi \circ \alpha \circ \varphi^{-1} =
\psi \circ \alpha \circ \psi^{-1} \circ \varphi \circ \varphi^{-1} =
\psi \circ \alpha \circ \psi^{-1}
$$
The upshot is a canonical isomorphism of sheaves
$\mathit{Aut}(x) \to \mathit{Aut}(y)$. Furthermore, if there
is a third object $z$ and a morphism $y \to z$ (and hence
also a morphism $x \to z$), then the
canonical isomorphisms $\mathit{Aut}(x) \to \mathit{Aut}(y)$,
$\mathit{Aut}(y) \to \mathit{Aut}(z)$, and
$\mathit{Aut}(x) \to \mathit{Aut}(z)$ are compatible in the sense
that
$$
\xymatrix{
\mathit{Aut}(x) \ar[rd] \ar[rr] & & \mathit{Aut}(z) \\
& \mathit{Aut}(y) \ar[ru]
}
$$
commutes.
\medskip\noindent
If there is no morphism from $x$ to $y$ over $U$, then we can
choose a covering $\{U_i \to U\}$ such that there exist morphisms
$x|_{U_i} \to y|_{U_i}$. This gives canonical isomorphisms
$$
\mathit{Aut}(x)|_{U_i} \longrightarrow \mathit{Aut}(y)|_{U_i}
$$
which agree over $U_i \times_U U_j$ (by canonicity). By glueing of
sheaves (Sites, Lemma \ref{sites-lemma-glue-maps}) we get a unique
isomorphism $\mathit{Aut}(x) \to \mathit{Aut}(y)$ whose restriction
to any $U_i$ is the canonical isomorphism of the previous paragraph.
Similarly to the above these canonical isomorphisms satisfy
a compatibility if we have a third object over $U$.
\medskip\noindent
What if the fibre category of $\mathcal{S}$ over $U$ is empty?
Well, in this case we can find a covering $\{U_i \to U\}$
and objects $x_i$ of $\mathcal{S}$ over $U_i$. Then we set
$\mathcal{G}_i = \mathit{Aut}(x_i)$. By the above we obtain
canonical isomorphisms
$$
\varphi_{ij} :
\mathcal{G}_i|_{U_i \times_U U_j}
\longrightarrow
\mathcal{G}_j|_{U_i \times_U U_j}
$$
whose restrictions to $U_i \times_U U_j \times_U U_k$ satisfy
the cocycle condition explained in Sites, Section
\ref{sites-section-glueing-sheaves}.
By Sites, Lemma \ref{sites-lemma-glue-sheaves}
we obtain a sheaf $\mathcal{G}$ over $U$ whose
restriction to $U_i$ gives $\mathcal{G}_i$
in a manner compatible with the glueing maps $\varphi_{ij}$.
\medskip\noindent
If $\mathcal{C}$ has a final object $U$, then this finishes the proof
as we can take $\mathcal{G}$ equal to the sheaf we just constructed.
In the general case we need to verify that the sheaves $\mathcal{G}$
constructed over varying $U$ are compatible in a canonical manner.
This is omitted.
\end{proof}
\section{Functoriality for stacks}
\label{section-inverse-image}
\noindent
In this section we study what happens if we want to change the base site
of a stack. This section can be skipped on a first reading.
\medskip\noindent
Let $u : \mathcal{C} \to \mathcal{D}$ be a functor between categories.
Let $p : \mathcal{S} \to \mathcal{D}$ be a category over $\mathcal{D}$.
In this situation we denote $u^p\mathcal{S}$ the category over $\mathcal{C}$
defined as follows
\begin{enumerate}
\item An object of $u^p\mathcal{S}$ is a pair $(U, y)$ consisting
of an object $U$ of $\mathcal{C}$ and an object $y$ of $\mathcal{S}_{u(U)}$.
\item A morphism $(a, \beta) : (U, y) \to (U', y')$ is given by
a morphism $a : U \to U'$ of $\mathcal{C}$ and a morphism $\beta : y \to y'$
of $\mathcal{S}$ such that $p(\beta) = u(a)$.
\end{enumerate}
Note that with these definitions the fibre category of
$u^p\mathcal{S}$ over $U$ is equal to the fibre category of
$\mathcal{S}$ over $u(U)$.
\begin{lemma}
\label{lemma-fibred-category-pushforward}
In the situation above, if $\mathcal{S}$ is a fibred category over
$\mathcal{D}$ then $u^p\mathcal{S}$ is a fibred category over $\mathcal{C}$.
\end{lemma}
\begin{proof}
Please take a look at the discussion surrounding
Categories, Definitions \ref{categories-definition-cartesian-over-C} and
\ref{categories-definition-fibred-category}
before reading this proof.
Let $(a, \beta) : (U, y) \to (U', y')$ be a morphism of $u^p\mathcal{S}$.
We claim that $(a, \beta)$ is strongly cartesian if and only if
$\beta$ is strongly cartesian. First, assume $\beta$ is strongly cartesian.
Consider any second morphism
$(a_1, \beta_1) : (U_1, y_1) \to (U', y')$ of $u^p\mathcal{S}$.
Then
\begin{align*}
& \Mor_{u^p\mathcal{S}}((U_1, y_1), (U, y)) \\
& =
\Mor_\mathcal{C}(U_1, U)
\times_{\Mor_\mathcal{D}(u(U_1), u(U))}
\Mor_\mathcal{S}(y_1, y) \\
& =
\Mor_\mathcal{C}(U_1, U)
\times_{\Mor_\mathcal{D}(u(U_1), u(U))}
\Mor_\mathcal{S}(y_1, y')
\times_{\Mor_\mathcal{D}(u(U_1), u(U'))}
\Mor_\mathcal{D}(u(U_1), u(U)) \\
& =
\Mor_\mathcal{S}(y_1, y')
\times_{\Mor_\mathcal{D}(u(U_1), u(U'))}
\Mor_\mathcal{C}(U_1, U) \\
& =
\Mor_{u^p\mathcal{S}}((U_1, y_1), (U', y'))
\times_{\Mor_\mathcal{C}(U_1, U')}
\Mor_\mathcal{C}(U_1, U)
\end{align*}
the second equality as $\beta$ is strongly cartesian. Hence we see that
indeed $(a, \beta)$ is strongly cartesian. Conversely, suppose that
$(a, \beta)$ is strongly cartesian. Choose a strongly cartesian morphism
$\beta' : y'' \to y'$ in $\mathcal{S}$ with $p(\beta') = u(a)$.
Then bot $(a, \beta) : (U, y) \to (U, y')$ and
$(a, \beta') : (U, y'') \to (U, y)$ are strongly cartesian and
lift $a$. Hence, by the uniqueness of strongly cartesian morphisms
(see discussion in
Categories, Section \ref{categories-section-fibred-categories})
there exists an isomorphism $\iota : y \to y''$ in $\mathcal{S}_{u(U)}$
such that $\beta = \beta' \circ \iota$, which implies that
$\beta$ is strongly cartesian in $\mathcal{S}$ by
Categories, Lemma \ref{categories-lemma-composition-cartesian}.
\medskip\noindent
Finally, we have to show
that given $(U', y')$ and $U \to U'$ we can find a strongly
cartesian morphism $(U, y) \to (U', y')$ in $u^p\mathcal{S}$
lifting the morphism $U \to U'$. This follows from the above as by assumption
we can find a strongly cartesian morphism $y \to y'$ lifting the
morphism $u(U) \to u(U')$.
\end{proof}
\begin{lemma}
\label{lemma-stack-pushforward}
Let $u : \mathcal{C} \to \mathcal{D}$ be a continuous functor of sites.
Let $p : \mathcal{S} \to \mathcal{D}$ be a stack over $\mathcal{D}$.
Then $u^p\mathcal{S}$ is a stack over $\mathcal{C}$.
\end{lemma}
\begin{proof}
We have seen in
Lemma \ref{lemma-fibred-category-pushforward}
that $u^p\mathcal{S}$ is a fibred category over $\mathcal{C}$.
Moreover, in the proof of that lemma we have seen that a morphism
$(a, \beta)$ of $u^p\mathcal{S}$ is strongly cartesian if and
only $\beta$ is strongly cartesian in $\mathcal{S}$. Hence,
given a morphism $a : U \to U'$ of $\mathcal{C}$, not only
do we have the equalities
$(u^p\mathcal{S})_U = \mathcal{S}_U$ and
$(u^p\mathcal{S})_{U'} = \mathcal{S}_{U'}$, but via these
equalities the pullback functors agree; in a formula
$a^*(U', y') = (U, u(a)^*y')$.
\medskip\noindent
Having said this, let $\mathcal{U} = \{U_i \to U\}$ be a covering
of $\mathcal{C}$. As $u$ is continuous we see that
$\mathcal{V} = \{u(U_i) \to u(U)\}$ is a covering of $\mathcal{D}$,
and that $u(U_i \times_U U_j) = u(U_i) \times_{u(U)} u(U_j)$ and
similarly for the triple fibre products $U_i \times_U U_j \times_U U_k$.
As we have the identifications of fibre categories and pullbacks
we see that descend data relative to $\mathcal{U}$ are identical
to descend data relative to $\mathcal{V}$. Since by assumption we have
effective descent in $\mathcal{S}$ we conclude the same holds for
$u^p\mathcal{S}$.
\end{proof}
\begin{lemma}
\label{lemma-stack-in-groupoids-pushforward}
Let $u : \mathcal{C} \to \mathcal{D}$ be a continuous functor of sites.
Let $p : \mathcal{S} \to \mathcal{D}$ be a stack in groupoids
over $\mathcal{D}$. Then $u^p\mathcal{S}$ is a stack in groupoids
over $\mathcal{C}$.
\end{lemma}
\begin{proof}
This follows immediately from
Lemma \ref{lemma-stack-pushforward}
and the fact that all fibre categories are groupoids.
\end{proof}
\begin{definition}
\label{definition-pushforward-stack}
Let $f : \mathcal{D} \to \mathcal{C}$ be a morphism of sites
given by the continuous functor $u : \mathcal{C} \to \mathcal{D}$.
Let $\mathcal{S}$ be a fibred category over $\mathcal{D}$.
In this setting we write {\it $f_*\mathcal{S}$} for the fibred
category $u^p\mathcal{S}$ defined above. We say that
$f_*\mathcal{S}$ is the {\it pushforward of $\mathcal{S}$ along $f$}.
\end{definition}
\noindent
By the results above we know that $f_*\mathcal{S}$ is a stack (in groupoids)
if $\mathcal{S}$ is a stack (in groupoids).
It is harder to define the pullback of a stack (and we'll need additional
assumptions for our particular construction -- feel free to write up and
submit a more general construction). We do this in several steps.
\medskip\noindent
Let $u : \mathcal{C} \to \mathcal{D}$ be a functor between categories.
Let $p : \mathcal{S} \to \mathcal{C}$ be a category over $\mathcal{C}$.
In this setting we define a category $u_{pp}\mathcal{S}$ as follows:
\begin{enumerate}
\item An object of $u_{pp}\mathcal{S}$ is a triple
$(U, \phi : V \to u(U), x)$ where $U \in \Ob(\mathcal{C})$,
the map $\phi : V \to u(U)$ is a morphism in $\mathcal{D}$,
and $x \in \Ob(\mathcal{S}_U)$.
\item A morphism
$$
(U_1, \phi_1 : V_1 \to u(U_1), x_1)
\longrightarrow
(U_2, \phi_2 : V_2 \to u(U_2), x_2)
$$
of $u_{pp}\mathcal{S}$ is given by a
$(a, b, \alpha)$ where $a : U_1 \to U_2$ is a morphism of $\mathcal{C}$,
$b : V_1 \to V_2$ is a morphism of $\mathcal{D}$, and
$\alpha : x_1 \to x_2$ is morphism of $\mathcal{S}$,
such that $p(\alpha) = a$ and the diagram
$$
\xymatrix{
V_1 \ar[d]_{\phi_1} \ar[r]_b & V_2 \ar[d]^{\phi_2} \\
u(U_1) \ar[r]^{u(a)} & u(U_2)
}
$$
commutes in $\mathcal{D}$.
\end{enumerate}
We think of $u_{pp}\mathcal{S}$ as a category over $\mathcal{D}$ via
$$
p_{pp} : u_{pp}\mathcal{S} \longrightarrow \mathcal{D},
\quad
(U, \phi : V \to u(U), x) \longmapsto V.
$$
The fibre category of $u_{pp}\mathcal{S}$ over an object $V$ of $\mathcal{D}$
does not have a simple description.
\begin{lemma}
\label{lemma-right-multiplicative-system}
In the situation above assume
\begin{enumerate}
\item $p : \mathcal{S} \to \mathcal{C}$ is a fibred category,
\item $\mathcal{C}$ has nonempty finite limits, and
\item $u : \mathcal{C} \to \mathcal{D}$ commutes with nonempty finite limits.
\end{enumerate}
Consider the set $R \subset \text{Arrows}(u_{pp}\mathcal{S})$ of morphisms
of the form
$$
(a, \text{id}_V, \alpha) :
(U', \phi' : V \to u(U'), x')
\longrightarrow
(U, \phi : V \to u(U), x)
$$
with $\alpha$ strongly cartesian. Then $R$ is a right multiplicative system.
\end{lemma}
\begin{proof}
According to
Categories, Definition \ref{categories-definition-multiplicative-system}
we have to check RMS1, RMS2, RMS3.
Condition RMS1 holds as a composition of strongly cartesian morphisms is
strongly cartesian, see
Categories, Lemma \ref{categories-lemma-composition-cartesian}.
\medskip\noindent
To check RMS2 suppose we have a morphism
$$
(a, b, \alpha) :
(U_1, \phi_1 : V_1 \to u(U_1), x_1)
\longrightarrow
(U, \phi : V \to u(U), x)
$$
of $u_{pp}\mathcal{S}$ and a morphism
$$
(c, \text{id}_V, \gamma) :
(U', \phi' : V \to u(U'), x')
\longrightarrow
(U, \phi : V \to u(U), x)
$$
with $\gamma$ strongly cartesian from $R$. In this situation set
$U'_1 = U_1 \times_U U'$, and denote $a' : U'_1 \to U'$ and
$c' : U'_1 \to U_1$ the projections.
As $u(U'_1) = u(U_1) \times_{u(U)} u(U')$
we see that $\phi'_1 = (\phi_1, \phi') : V_1 \to u(U'_1)$ is
a morphism in $\mathcal{D}$. Let $\gamma_1 : x_1' \to x_1$
be a strongly cartesian morphism of $\mathcal{S}$ with
$p(\gamma_1) = \phi'_1$ (which exists because $\mathcal{S}$ is a
fibred category over $\mathcal{C}$). Then as $\gamma : x' \to x$ is
strongly cartesian there exists a unique morphism
$\alpha' : x'_1 \to x'$ with $p(\alpha') = a'$.
At this point we see that
$$
(a', b, \alpha') :
(U_1, \phi_1 : V_1 \to u(U'_1), x'_1)
\longrightarrow
(U, \phi : V \to u(U'), x')
$$
is a morphism and that
$$
(c', \text{id}_{V_1}, \gamma_1) :
(U'_1, \phi'_1 : V_1 \to u(U'_1), x'_1)
\longrightarrow
(U_1, \phi : V_1 \to u(U_1), x_1)
$$
is an element of $R$ which form a solution of the existence problem
posed by RMS2.
\medskip\noindent
Finally, suppose that
$$
(a, b, \alpha), (a', b', \alpha') :
(U_1, \phi_1 : V_1 \to u(U_1), x_1)
\longrightarrow
(U, \phi : V \to u(U), x)
$$
are two morphisms of $u_{pp}\mathcal{S}$ and suppose that
$$
(c, \text{id}_V, \gamma) :
(U, \phi : V \to u(U), x)
\longrightarrow
(U', \phi : V \to u(U'), x')
$$
is an element of $R$ which equalizes the morphisms
$(a, b, \alpha)$ and $(a', b', \alpha')$. This implies in particular
that $b = b'$. Let $d : U_2 \to U_1$ be the equalizer of $a, a'$ which
exists (see
Categories, Lemma \ref{categories-lemma-almost-finite-limits-exist}).
Moreover, $u(d) : u(U_2) \to u(U_1)$ is the equalizer of $u(a), u(a')$
hence (as $b = b'$) there is a morphism $\phi_2 : V_1 \to u(U_2)$ such that
$\phi_1 = u(d) \circ \phi_1$. Let $\delta : x_2 \to x_1$ be a strongly
cartesian morphism of $\mathcal{S}$ with $p(\delta) = u(d)$.
Now we claim that $\alpha \circ \delta = \alpha' \circ \delta$.
This is true because
$\gamma$ is strongly cartesian,
$\gamma \circ \alpha \circ \delta = \gamma \circ \alpha' \circ \delta$, and
$p(\alpha \circ \delta) = p(\alpha' \circ \delta)$.
Hence the arrow
$$
(d, \text{id}_{V_1}, \delta) :
(U_2, \phi_2 : V_1 \to u(U_2), x_2)
\longrightarrow
(U_1, \phi_1 : V_1 \to u(U_1), x_1)
$$
is an element of $R$ and equalizes $(a, b, \alpha)$ and $(a', b', \alpha')$.
Hence $R$ satisfies RMS3 as well.
\end{proof}
\begin{lemma}
\label{lemma-fibred-category-pullback}
With notation and assumptions as in
Lemma \ref{lemma-right-multiplicative-system}.
Set $u_p\mathcal{S} = R^{-1}u_{pp}\mathcal{S}$, see
Categories, Section \ref{categories-section-localization}.
Then $u_p\mathcal{S}$ is a fibred category over $\mathcal{D}$.
\end{lemma}
\begin{proof}
We use the description of $u_p\mathcal{S}$ given just above
Categories, Lemma \ref{categories-lemma-right-localization}.
Note that the functor $p_{pp} : u_{pp}\mathcal{S} \to \mathcal{D}$
transforms every element of $R$ to an identity morphism.
Hence by
Categories, Lemma \ref{categories-lemma-properties-right-localization}
we obtain a canonical functor $p_p : u_p\mathcal{S} \to \mathcal{D}$
extending the given functor. This is how we think of
$u_p\mathcal{S}$ as a category over $\mathcal{D}$.
\medskip\noindent
First we want to characterize the $\mathcal{D}$-strongly cartesian morphisms
in $u_p\mathcal{S}$.
A morphism $f : X \to Y$ of $u_p\mathcal{S}$ is the equivalence class of
a pair $(f' : X' \to Y, r : X' \to X)$ with $r \in R$.
In fact, in $u_p\mathcal{S}$ we have $f = (f', 1) \circ (r, 1)^{-1}$
with obvious notation.
Note that an isomorphism is always strongly cartesian, as are
compositions of strongly cartesian morphisms, see
Categories, Lemma \ref{categories-lemma-composition-cartesian}.
Hence $f$ is strongly cartesian if and only if $(f', 1)$ is so.
Thus the following claim completely characterizes strongly cartesian
morphisms. Claim: A morphism
$$
(a, b, \alpha) :
X_1 = (U_1, \phi_1 : V_1 \to u(U_1), x_1)
\longrightarrow
(U_2, \phi_2 : V_2 \to u(U_2), x_2) = X_2
$$
of $u_{pp}\mathcal{S}$ has image $f = ((a, b, \alpha), 1)$
strongly cartesian in $u_p\mathcal{S}$ if and only if $\alpha$
is a strongly cartesian morphism of $\mathcal{S}$.
\medskip\noindent
Assume $\alpha$ strongly cartesian.
Let $X = (U, \phi : V \to u(U), x)$ be another object, and let
$f_2 : X \to X_2$ be a morphism of $u_p\mathcal{S}$
such that $p_p(f_2) = b \circ b_1$ for some $b_1 : U \to U_1$.
To show that $f$ is strongly cartesian we have to show
that there exists a unique morphism $f_1 : X \to X_1$ in $u_p\mathcal{S}$
such that $p_p(f_1) = b_1$ and $f_2 = f \circ f_1$ in $u_p\mathcal{S}$.
Write $f_2 = (f'_2 : X' \to X_2, r : X' \to X)$. Again we can write
$f_2 = (f'_2, 1) \circ (r, 1)^{-1}$ in $u_p\mathcal{S}$. Since $(r, 1)$
is an isomorphism whose image in $\mathcal{D}$ is an identity we
see that finding a morphism $f_1 : X \to X_1$ with the required
properties is the same thing as finding a morphism $f'_1 : X' \to X_1$
in $u_p\mathcal{S}$ with $p(f'_1) = b_1$ and $f'_2 = f \circ f'_1$.
Hence we may assume that $f_2$ is of the form
$f_2 = ((a_2, b_2, \alpha_2), 1)$ with $b_2 = b \circ b_1$. Here is
a picture
$$
\xymatrix{
& & (U_1, V_1 \to u(U_1), x_1) \ar[d]^{(a, b, \alpha)} \\
(U, V \to u(U), x) \ar[rr]^{(a_2, b_2, \alpha_2)} & &
(U_2, V_2 \to u(U_2), x_2)
}
$$
Now it is clear how to construct the morphism $f_1$.
Namely, set $U' = U \times_{U_2} U_1$ with projections
$c : U' \to U$ and $a_1 : U' \to U_1$.
Pick a strongly cartesian morphism $\gamma : x' \to x$
lifting the morphism $c$. Since $b_2 = b \circ b_1$,
and since $u(U') = u(U) \times_{u(U_2)} u(U_1)$ we see that
$\phi' = (\phi, \phi_1 \circ b_1) : V \to u(U')$. Since
$\alpha$ is strongly cartesian, and
$a \circ a_1 = a_2 \circ c = p(\alpha_2 \circ \gamma)$
there exists a morphism
$\alpha_1 : x' \to x_1$ lifting $a_1$ such that
$\alpha \circ \alpha_1 = \alpha_2 \circ \gamma$.
Set $X' = (U', \phi' : V \to u(U'), x')$.
Thus we see that
$$
f_1 =
((a_1, b_1, \alpha_1) : X' \to X_1, (c, \text{id}_V, \gamma) : X' \to X) :
X \longrightarrow X_1
$$
works, in fact the diagram
$$
\xymatrix{
(U', \phi' : V \to u(U'), x')
\ar[d]_{(c, \text{id}_V, \gamma)}
\ar[rr]_{(a_1, b_1, \alpha_1)}
& & (U_1, V_1 \to u(U_1), x_1) \ar[d]^{(a, b, \alpha)} \\
(U, V \to u(U), x) \ar[rr]^{(a_2, b_2, \alpha_2)} & &
(U_2, V_2 \to u(U_2), x_2)
}
$$
is commutative by construction. This proves existence.
\medskip\noindent
Next we prove uniqueness, still in the special case
$f = ((a, b, \alpha), 1)$ and $f_2 = ((a_2, b_2, \alpha_2), 1)$.
We strongly advise the reader to skip this part.
Suppose that $g_1, g'_1 : X \to X_1$ are two morphisms of
$u_p\mathcal{S}$ such that $p_p(g_1) = p_p(g'_1) = b_1$ and
$f_2 = f \circ g_1 = f \circ g'_1$. Our goal is to show that
$g_1 = g'_1$. By
Categories, Lemma \ref{categories-lemma-morphisms-right-localization}
we may represent $g_1$ and $g'_1$ as the equivalence classes of
$(f_1 : X' \to X_1, r : X' \to X)$ and
$(f'_1 : X' \to X_1, r : X' \to X)$ for some $r \in R$. By
Categories, Lemma \ref{categories-lemma-equality-morphisms-right-localization}
we see that $f_2 = f \circ g_1 = f \circ g'_1$ means that there
exists a morphism $r' : X'' \to X'$ in $u_{pp}\mathcal{S}$ such that
$r' \circ r \in R$ and
$$
(a, b, \alpha) \circ f_1 \circ r' =
(a, b, \alpha) \circ f'_1 \circ r' =
(a_2, b_2, \alpha_2) \circ r'
$$
in $u_{pp}\mathcal{S}$. Note that
now $g_1$ is represented by $(f_1 \circ r', r \circ r')$ and
similarly for $g'_1$. Hence we may assume that
$$
(a, b, \alpha) \circ f_1 =
(a, b, \alpha) \circ f'_1 =
(a_2, b_2, \alpha_2).
$$
Write $r = (c, \text{id}_V, \gamma) : (U', \phi' : V \to u(U'), x')$,
$f_1 = (a_1, b_1, \alpha_1)$, and $f'_1 = (a'_1, b_1, \alpha'_1)$.
Here we have used the condition that $p_p(g_1) = p_p(g'_1)$.
The equalities above are now equivalent to
$a \circ a_1 = a \circ a'_1 = a_2 \circ c$ and
$\alpha \circ \alpha_1 = \alpha \circ \alpha'_1 = \alpha_2 \circ \gamma$.
It need not be the case that $a_1 = a'_1$ in this situation.
Thus we have to precompose by one more morphism from $R$.
Namely, let $U'' = \text{Eq}(a_1, a'_1)$ be the equalizer of
$a_1$ and $a'_1$ which is a subobject of $U'$. Denote
$c' : U'' \to U'$ the canonical monomorphism.
Because of the relations among the morphisms
above we see that $V \to u(U')$ maps into
$u(U'') = u(\text{Eq}(a_1, a'_1)) = \text{Eq}(u(a_1), u(a'_1))$.
Hence we get a new object $(U'', \phi'' : V \to u(U''), x'')$, where
$\gamma' : x'' \to x'$ is a strongly cartesian morphism lifting $\gamma$.
Then we see that we may precompose $f_1$ and $f'_1$ with the element
$(c', \text{id}_V, \gamma')$ of $R$. After doing this, i.e., replacing
$(U', \phi' : V \to u(U'), x')$ with
$(U'', \phi'' : V \to u(U''), x'')$, we get back to the previous
situation where in addition we now have that $a_1 = a'_1$.
In this case it follows formally from the fact that $\alpha$
is strongly cartesian (!) that $\alpha_1 = \alpha'_1$.
This shows that $g_1 = g'_1$ as desired.
\medskip\noindent
We omit the proof of the fact that for any strongly cartesian morphism
of $u_p\mathcal{S}$ of the form $((a, b, \alpha), 1)$ the morphism
$\alpha$ is strongly cartesian in $\mathcal{S}$.
(We do not need the characterization of strongly cartesian morphisms
in the rest of the proof, although we do use it later in this section.)
\medskip\noindent
Let $(U, \phi : V \to u(U), x)$ be an object of $u_p\mathcal{S}$.
Let $b : V' \to V$ be a morphism of $\mathcal{D}$. Then the morphism
$$
(\text{id}_U, b, \text{id}_x) :
(U, \phi \circ b : V' \to u(U), x)
\longrightarrow
(U, \phi : V \to u(U), x)
$$
is strongly cartesian by the result of the preceding paragraphs and we win.
\end{proof}
\begin{lemma}
\label{lemma-fibred-groupoids-category-pullback}
With notation and assumptions as in
Lemma \ref{lemma-fibred-category-pullback}.
If $\mathcal{S}$ is fibred in groupoids, then $u_p\mathcal{S}$ is fibred
in groupoids.
\end{lemma}
\begin{proof}
By
Lemma \ref{lemma-fibred-category-pullback}
we know that $u_p\mathcal{S}$ is a fibred category.
Let $f : X \to Y$ be a morphism of $u_p\mathcal{S}$ with
$p_p(f) = \text{id}_V$. We are done if we can show that $f$ is invertible, see
Categories, Lemma \ref{categories-lemma-fibred-groupoids}.
Write $f$ as the equivalence class of
a pair $((a, b, \alpha), r)$ with $r \in R$. Then $p_p(r) = \text{id}_V$,
hence $p_{pp}((a, b, \alpha)) = \text{id}_V$. Hence $b = \text{id}_V$.
But any morphism of $\mathcal{S}$ is strongly cartesian, see
Categories, Lemma \ref{categories-lemma-fibred-groupoids}
hence we see that $(a, b, \alpha) \in R$ is invertible
in $u_p\mathcal{S}$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-adjointness-pullback-pushforward}
Let $u : \mathcal{C} \to \mathcal{D}$ be a functor.
Let $p : \mathcal{S} \to \mathcal{C}$ and $q : \mathcal{T} \to \mathcal{D}$
be categories over $\mathcal{C}$ and $\mathcal{D}$. Assume that
\begin{enumerate}
\item $p : \mathcal{S} \to \mathcal{C}$ is a fibred category,
\item $q : \mathcal{T} \to \mathcal{D}$ is a fibred category,
\item $\mathcal{C}$ has nonempty finite limits, and
\item $u : \mathcal{C} \to \mathcal{D}$ commutes with nonempty finite limits.
\end{enumerate}
Then we have a canonical equivalence of categories
$$
\Mor_{\textit{Fib}/\mathcal{C}}(\mathcal{S}, u^p\mathcal{T})
=
\Mor_{\textit{Fib}/\mathcal{D}}(u_p\mathcal{S}, \mathcal{T})
$$
of morphism categories.
\end{lemma}
\begin{proof}
In this proof we use the notation $x/U$ to denote an object
$x$ of $\mathcal{S}$ which lies over $U$ in $\mathcal{C}$.
Similarly $y/V$ denotes an object $y$ of $\mathcal{T}$
which lies over $V$ in $\mathcal{D}$. In the same vein
$\alpha/a : x/U \to x'/U'$ denotes the morphism
$\alpha : x \to x'$ with image $a : U \to U'$ in $\mathcal{C}$.
\medskip\noindent
Let $G : u_p\mathcal{S} \to \mathcal{T}$ be a $1$-morphism of fibred
categories over $\mathcal{D}$. Denote $G' : u_{pp}\mathcal{S} \to \mathcal{T}$
the composition of $G$ with the canonical (localization) functor
$u_{pp}\mathcal{S} \to u_p\mathcal{S}$. Then consider the functor
$H : \mathcal{S} \to u^p\mathcal{T}$ given by
$$
H(x/U) = (U, G'(U, \text{id}_{u(U)} : u(U) \to u(U), x))
$$
on objects and by
$$
H((\alpha, a) : x/U \to x'/U') = G'(a, u(a), \alpha)
$$
on morphisms. Since $G$ transforms strongly cartesian morphisms
into strongly cartesian morphisms, we see that if $\alpha$ is strongly
cartesian, then $H(\alpha)$ is strongly cartesian.
Namely, we've seen in the proof of
Lemma \ref{lemma-fibred-category-pullback}
that in this case the map $(a, u(a), \alpha)$ becomes
strongly cartesian in $u_p\mathcal{S}$. Clearly this construction is
functorial in $G$ and we obtain a functor
$$
A :
\Mor_{\textit{Fib}/\mathcal{D}}(u_p\mathcal{S}, \mathcal{T})
\longrightarrow
\Mor_{\textit{Fib}/\mathcal{C}}(\mathcal{S}, u^p\mathcal{T})
$$
\medskip\noindent
Conversely, let $H : \mathcal{S} \to u^p\mathcal{T}$ be a $1$-morphism
of fibred categories. Recall that an object of $u^p\mathcal{T}$ is
a pair $(U, y)$ with $y \in \Ob(\mathcal{T}_{u(U)})$. We denote
$\text{pr} : u^p\mathcal{T} \to \mathcal{T}$ the functor $(U, y) \mapsto y$.
In this case we define a functor
$G' : u_{pp}\mathcal{S} \to \mathcal{T}$ by the rules
$$
G'(U, \phi : V \to u(U), x) = \phi^*\text{pr}(H(x))
$$
on objects and we let
$$
G'((a, b, \alpha) : (U, \phi : V \to u(U), x)
\to (U', \phi' : V' \to u(U'), x'))
=
\beta
$$
be the unique morphism
$\beta : \phi^*\text{pr}(H(x)) \to (\phi')^*\text{pr}(H(x'))$
such that $q(\beta) = b$ and the diagram
$$
\xymatrix{
\phi^*\text{pr}(H(x)) \ar[d] \ar[r]_-{\beta} &
(\phi')^*\text{pr}(H(x')) \ar[d] \\
\text{pr}(H(x)) \ar[r]^{\text{pr}(H(a, \alpha))} & \text{pr}(H(x'))
}
$$
Such a morphism exists and is unique because $\mathcal{T}$ is a fibred
category.
\medskip\noindent
We check that $G'(r)$ is an isomorphism if $r \in R$.
Namely, if
$$
(a, \text{id}_V, \alpha) :
(U', \phi' : V \to u(U'), x')
\longrightarrow
(U, \phi : V \to u(U), x)
$$
with $\alpha$ strongly cartesian is an element of the right multiplicative
system $R$ of
Lemma \ref{lemma-right-multiplicative-system}
then $H(\alpha)$ is strongly cartesian, and
$\text{pr}(H(\alpha))$ is strongly cartesian, see proof of
Lemma \ref{lemma-fibred-category-pushforward}.
H