## 8.12 Functoriality for stacks

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.

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

1. 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 $\mathcal{S}_{u(U)}$.

2. 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 $p(\beta ) = u(a)$.

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 $u(U)$.

Lemma 8.12.1. In the situation above, if $\mathcal{S}$ is a fibred category over $\mathcal{D}$ then $u^ p\mathcal{S}$ is a fibred category over $\mathcal{C}$.

Proof. Please take a look at the discussion surrounding Categories, Definitions 4.33.1 and 4.33.5 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*} & \mathop{Mor}\nolimits _{u^ p\mathcal{S}}((U_1, y_1), (U, y)) \\ & = \mathop{Mor}\nolimits _\mathcal {C}(U_1, U) \times _{\mathop{Mor}\nolimits _\mathcal {D}(u(U_1), u(U))} \mathop{Mor}\nolimits _\mathcal {S}(y_1, y) \\ & = \mathop{Mor}\nolimits _\mathcal {C}(U_1, U) \times _{\mathop{Mor}\nolimits _\mathcal {D}(u(U_1), u(U))} \mathop{Mor}\nolimits _\mathcal {S}(y_1, y') \times _{\mathop{Mor}\nolimits _\mathcal {D}(u(U_1), u(U'))} \mathop{Mor}\nolimits _\mathcal {D}(u(U_1), u(U)) \\ & = \mathop{Mor}\nolimits _\mathcal {S}(y_1, y') \times _{\mathop{Mor}\nolimits _\mathcal {D}(u(U_1), u(U'))} \mathop{Mor}\nolimits _\mathcal {C}(U_1, U) \\ & = \mathop{Mor}\nolimits _{u^ p\mathcal{S}}((U_1, y_1), (U', y')) \times _{\mathop{Mor}\nolimits _\mathcal {C}(U_1, U')} \mathop{Mor}\nolimits _\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 4.33) 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 4.33.2.

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 $u(U) \to u(U')$. $\square$

Lemma 8.12.2. 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 $\mathcal{C}$.

Proof. We have seen in Lemma 8.12.1 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 $a^*(U', y') = (U, u(a)^*y')$.

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 $u^ p\mathcal{S}$. $\square$

Lemma 8.12.3. 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 $\mathcal{C}$.

Proof. This follows immediately from Lemma 8.12.2 and the fact that all fibre categories are groupoids. $\square$

Definition 8.12.4. 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 $f_*\mathcal{S}$ for the fibred category $u^ p\mathcal{S}$ defined above. We say that $f_*\mathcal{S}$ is the pushforward of $\mathcal{S}$ along $f$.

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.

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:

1. An object of $u_{pp}\mathcal{S}$ is a triple $(U, \phi : V \to u(U), x)$ where $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, the map $\phi : V \to u(U)$ is a morphism in $\mathcal{D}$, and $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$.

2. 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 $\mathcal{D}$.

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.

Lemma 8.12.5. In the situation above assume

1. $p : \mathcal{S} \to \mathcal{C}$ is a fibred category,

2. $\mathcal{C}$ has nonempty finite limits, and

3. $u : \mathcal{C} \to \mathcal{D}$ commutes with nonempty finite limits.

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.

Proof. According to Categories, Definition 4.27.1 we have to check RMS1, RMS2, RMS3. Condition RMS1 holds as a composition of strongly cartesian morphisms is strongly cartesian, see Categories, Lemma 4.33.2.

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.

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 4.18.3). 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. $\square$

Lemma 8.12.6. With notation and assumptions as in Lemma 8.12.5. Set $u_ p\mathcal{S} = R^{-1}u_{pp}\mathcal{S}$, see Categories, Section 4.27. Then $u_ p\mathcal{S}$ is a fibred category over $\mathcal{D}$.

Proof. We use the description of $u_ p\mathcal{S}$ given just above Categories, Lemma 4.27.11. 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 4.27.16 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 $\mathcal{D}$.

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 4.33.2. 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 $\mathcal{S}$.

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.

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 4.27.13 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 4.27.14 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.

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.)

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. $\square$

Lemma 8.12.7. With notation and assumptions as in Lemma 8.12.6. If $\mathcal{S}$ is fibred in groupoids, then $u_ p\mathcal{S}$ is fibred in groupoids.

Proof. By Lemma 8.12.6 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 4.35.2. 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 4.35.2 hence we see that $(a, b, \alpha ) \in R$ is invertible in $u_ p\mathcal{S}$ as desired. $\square$

Lemma 8.12.8. 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

1. $p : \mathcal{S} \to \mathcal{C}$ is a fibred category,

2. $q : \mathcal{T} \to \mathcal{D}$ is a fibred category,

3. $\mathcal{C}$ has nonempty finite limits, and

4. $u : \mathcal{C} \to \mathcal{D}$ commutes with nonempty finite limits.

Then we have a canonical equivalence of categories

$\mathop{Mor}\nolimits _{\textit{Fib}/\mathcal{C}}(\mathcal{S}, u^ p\mathcal{T}) = \mathop{Mor}\nolimits _{\textit{Fib}/\mathcal{D}}(u_ p\mathcal{S}, \mathcal{T})$

of morphism categories.

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 $\mathcal{C}$.

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 8.12.6 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

$A : \mathop{Mor}\nolimits _{\textit{Fib}/\mathcal{D}}(u_ p\mathcal{S}, \mathcal{T}) \longrightarrow \mathop{Mor}\nolimits _{\textit{Fib}/\mathcal{C}}(\mathcal{S}, u^ p\mathcal{T})$

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 $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T}_{u(U)})$. 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.

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 8.12.5 then $H(\alpha )$ is strongly cartesian, and $\text{pr}(H(\alpha ))$ is strongly cartesian, see proof of Lemma 8.12.1. Hence in this case the morphism $\beta$ has $q(\beta ) = \text{id}_ V$ and is strongly cartesian. Hence $\beta$ is an isomorphism by Categories, Lemma 4.33.2. Thus by Categories, Lemma 4.27.16 we obtain a canonical extension $G : u_ p\mathcal{S} \to \mathcal{T}$.

Next, let us prove that $G$ transforms strongly cartesian morphisms into strongly cartesian morphisms. Suppose that $f : X \to Y$ is a strongly cartesian. By the characterization of strongly cartesian morphisms in $u_ p\mathcal{S}$ we can write $f$ as $((a, b, \alpha ) : X' \to Y, r : X' \to Y)$ where $r \in R$ and $\alpha$ strongly cartesian in $\mathcal{S}$. By the above it suffices to show that $G'(a, b \alpha )$ is strongly cartesian. As before the condition that $\alpha$ is strongly cartesian implies that $\text{pr}(H(a, \alpha )) : \text{pr}(H(x)) \to \text{pr}(H(x'))$ is strongly cartesian in $\mathcal{T}$. Since in the commutative square above now all arrows except possibly $\beta$ is strongly cartesian it follows that also $\beta$ is strongly cartesian as desired. Clearly the construction $H \mapsto G$ is functorial in $H$ and we obtain a functor

$B : \mathop{Mor}\nolimits _{\textit{Fib}/\mathcal{C}}(\mathcal{S}, u^ p\mathcal{T}) \longrightarrow \mathop{Mor}\nolimits _{\textit{Fib}/\mathcal{D}}(u_ p\mathcal{S}, \mathcal{T})$

To finish the proof of the lemma we have to show that the functors $A$ and $B$ are mutually quasi-inverse. We omit the verifications. $\square$

Definition 8.12.9. Let $f : \mathcal{D} \to \mathcal{C}$ be a morphism of sites given by a continuous functor $u : \mathcal{C} \to \mathcal{D}$ satisfying the hypotheses and conclusions of Sites, Proposition 7.14.7. Let $\mathcal{S}$ be a stack over $\mathcal{C}$. In this setting we write $f^{-1}\mathcal{S}$ for the stackification of the fibred category $u_ p\mathcal{S}$ over $\mathcal{D}$ constructed above. We say that $f^{-1}\mathcal{S}$ is the pullback of $\mathcal{S}$ along $f$.

Of course, if $\mathcal{S}$ is a stack in groupoids, then $f^{-1}\mathcal{S}$ is a stack in groupoids by Lemmas 8.9.1 and 8.12.7.

Lemma 8.12.10. Let $f : \mathcal{D} \to \mathcal{C}$ be a morphism of sites given by a continuous functor $u : \mathcal{C} \to \mathcal{D}$ satisfying the hypotheses and conclusions of Sites, Proposition 7.14.7. Let $p : \mathcal{S} \to \mathcal{C}$ and $q : \mathcal{T} \to \mathcal{D}$ be stacks. Then we have a canonical equivalence of categories

$\mathop{Mor}\nolimits _{\textit{Stacks}/\mathcal{C}}(\mathcal{S}, f_*\mathcal{T}) = \mathop{Mor}\nolimits _{\textit{Stacks}/\mathcal{D}}(f^{-1}\mathcal{S}, \mathcal{T})$

of morphism categories.

Proof. For $i = 1, 2$ an $i$-morphism of stacks is the same thing as a $i$-morphism of fibred categories, see Definition 8.4.5. By Lemma 8.12.8 we have already

$\mathop{Mor}\nolimits _{\textit{Fib}/\mathcal{C}}(\mathcal{S}, u^ p\mathcal{T}) = \mathop{Mor}\nolimits _{\textit{Fib}/\mathcal{D}}(u_ p\mathcal{S}, \mathcal{T})$

Hence the result follows from Lemma 8.8.3 as $u^ p\mathcal{T} = f_*\mathcal{T}$ and $f^{-1}\mathcal{S}$ is the stackification of $u_ p\mathcal{S}$. $\square$

Lemma 8.12.11. Let $f : \mathcal{D} \to \mathcal{C}$ be a morphism of sites given by a continuous functor $u : \mathcal{C} \to \mathcal{D}$ satisfying the hypotheses and conclusions of Sites, Proposition 7.14.7. Let $\mathcal{S} \to \mathcal{C}$ be a fibred category, and let $\mathcal{S} \to \mathcal{S}'$ be the stackification of $\mathcal{S}$. Then $f^{-1}\mathcal{S}'$ is the stackification of $u_ p\mathcal{S}$.

Proof. Omitted. Hint: This is the analogue of Sites, Lemma 7.13.4. $\square$

The following lemma tells us that the $2$-category of stacks over $\mathit{Sch}_{fppf}$ is a “full 2-sub category” of the $2$-category of stacks over $\mathit{Sch}'_{fppf}$ provided that $\mathit{Sch}'_{fppf}$ contains $\mathit{Sch}_{fppf}$ (see Topologies, Section 34.12).

Lemma 8.12.12. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor satisfying the assumptions of Sites, Lemma 7.21.8. Let $f : \mathcal{D} \to \mathcal{C}$ be the corresponding morphism of sites. Then

1. for every stack $p : \mathcal{S} \to \mathcal{C}$ the canonical functor $\mathcal{S} \to f_*f^{-1}\mathcal{S}$ is an equivalence of stacks,

2. given stacks $\mathcal{S}, \mathcal{S}'$ over $\mathcal{C}$ the construction $f^{-1}$ induces an equivalence

$\mathop{Mor}\nolimits _{\textit{Stacks}/\mathcal{C}}(\mathcal{S}, \mathcal{S}') \longrightarrow \mathop{Mor}\nolimits _{\textit{Stacks}/\mathcal{D}}(f^{-1}\mathcal{S}, f^{-1}\mathcal{S}')$

of morphism categories.

Proof. Note that by Lemma 8.12.10 we have an equivalence of categories

$\mathop{Mor}\nolimits _{\textit{Stacks}/\mathcal{D}}(f^{-1}\mathcal{S}, f^{-1}\mathcal{S}') = \mathop{Mor}\nolimits _{\textit{Stacks}/\mathcal{C}}(\mathcal{S}, f_*f^{-1}\mathcal{S}')$

Hence (2) follows from (1).

To prove (1) we are going to use Lemma 8.4.8. This lemma tells us that we have to show that $can : \mathcal{S} \to f_*f^{-1}\mathcal{S}$ is fully faithful and that all objects of $f_*f^{-1}\mathcal{S}$ are locally in the essential image.

We quickly describe the functor $can$, see proof of Lemma 8.12.8. To do this we introduce the functor $c'' : \mathcal{S} \to u_{pp}\mathcal{S}$ defined by $c''(x/U) = (U, \text{id} : u(U) \to u(U), x)$, and $c''(\alpha /a) = (a, u(a), \alpha )$. We set $c' : \mathcal{S} \to u_ p\mathcal{S}$ equal to the composition of $c''$ and the canonical functor $u_{pp}\mathcal{S} \to u_ p\mathcal{S}$. We set $c : \mathcal{S} \to f^{-1}\mathcal{S}$ equal to the composition of $c'$ and the canonical functor $u_ p\mathcal{S} \to f^{-1}\mathcal{S}$. Then $can : \mathcal{S} \to f_*f^{-1}\mathcal{S}$ is the functor which to $x/U$ associates the pair $(U, c(x))$ and to $\alpha /a$ the morphism $(a, c(\alpha ))$.

Fully faithfulness. To prove this we are going to use Lemma 8.4.7. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Let $x, y \in \mathcal{S}_ U$. First off, as $u$ is fully faithful, we have

$\mathop{Mor}\nolimits _{(f_*f^{-1}\mathcal{S})_ U}(can(x), can(y)) = \mathop{Mor}\nolimits _{(f^{-1}\mathcal{S})_{u(U)}}(c(x), c(y))$

directly from the definition of $f_*$. Similar holds after pulling back to any $U'/U$. Because $f^{-1}\mathcal{S}$ is the stackification of $u_ p\mathcal{S}$, and since $u$ is continuous and cocontinuous the presheaf

$U'/U \longmapsto \mathop{Mor}\nolimits _{(f^{-1}\mathcal{S})_{u(U')}}(c(x|_{U'}), c(y|_{U'}))$

is the sheafification of the presheaf

$U'/U \longmapsto \mathop{Mor}\nolimits _{(u_ p\mathcal{S})_{u(U')}}(c'(x|_{U'}), c'(y|_{U'}))$

Hence to finish the proof of fully faithfulness it suffices to show that for any $U$ and $x, y$ the map

$\mathop{Mor}\nolimits _{\mathcal{S}_ U}(x, y) \longrightarrow \mathop{Mor}\nolimits _{(u_ p\mathcal{S})_ U}(c'(x), c'(y))$

is bijective. A morphism $f : x \to y$ in $u_ p\mathcal{S}$ over $u(U)$ is given by an equivalence class of diagrams

$\xymatrix{ (U', \phi : u(U) \to u(U'), x') \ar[d]_{(c, \text{id}_{u(U)}, \gamma )} \ar[r]_{(a, b, \alpha )} & (U, \text{id} : u(U) \to u(U), y) \\ (U, \text{id} : u(U) \to u(U), x) }$

with $\gamma$ strongly cartesian and $b = \text{id}_{u(U)}$. But since $u$ is fully faithful we can write $\phi = u(c')$ for some morphism $c' : U \to U'$ and then we see that $a \circ c' = \text{id}_ U$ and $c \circ c' = \text{id}_{U'}$. Because $\gamma$ is strongly cartesian we can find a morphism $\gamma ' : x \to x'$ lifting $c'$ such that $\gamma \circ \gamma ' = \text{id}_ x$. By definition of the equivalence classes defining morphisms in $u_ p\mathcal{S}$ it follows that the morphism

$\xymatrix{ (U, \text{id} : u(U) \to u(U), x) \ar[rr]_{(\text{id}, \text{id}, \alpha \circ \gamma ')} & & (U, \text{id} : u(U) \to u(U), y) }$

of $u_{pp}\mathcal{S}$ induces the morphism $f$ in $u_ p\mathcal{S}$. This proves that the map is surjective. We omit the proof that it is injective.

Finally, we have to show that any object of $f_*f^{-1}\mathcal{S}$ locally comes from an object of $\mathcal{S}$. This is clear from the constructions (details omitted). $\square$

Comment #1216 by JuanPablo on

After the defintion 8.12.4 (tag 04WE), after defining $u_{pp}$ it says that if $\mathcal{S}$ is a fibred category over $\mathcal{C}$, then $u_{pp}(\mathcal{S})$ is not necessarily fibred.

I think this is not true, that is even if $\mathcal{S}$ is not fibred then for any object $(V,\phi:V\rightarrow u(U), x)$ in $(u_{pp}\mathcal{S})_V$ and morphism $g:V'\rightarrow V$ in $\mathcal{D}$, the morphism $(g,1,1):(V',\phi g: V'\rightarrow u(U),x)\rightarrow (V,\phi:V\rightarrow u(U),x)$ is a strong cartesian morphism in $u_{pp}\mathcal{S}$ over $g$.

Further in Lemma 8.12.6 the localization $u_{pp}\mathcal{S}\rightarrow u_p\mathcal{S}$ is a 1-morphism of fibred categories over $\mathcal{C}$.

The problem is that for $\mathcal{S}$ fibred the canonical morphism $\mathcal{S}\rightarrow u^pu_{pp} \mathcal{S}$ is a 1-morphism of categories over $\mathcal{C}$, but is not a 1-morphism of fibred categories (does not preserve strongly cartesian morphisms).

Comment #1245 by on

OK, I agree with what you say. For the moment I have just removed the erroneous statement, see here. If I misunderstood and there are more mistakes in this section, please let us know. (The whole section should be rewritten completely methinks.)

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