The Stacks project

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

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

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

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

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

of $u_{pp}\mathcal{S}$ and a morphism

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

is a morphism and that

is an element of $R$ which form a solution of the existence problem posed by RMS2.

Finally, suppose that

are two morphisms of $u_{pp}\mathcal{S}$ and suppose that

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

is an element of $R$ and equalizes $(a, b, \alpha )$ and $(a', b', \alpha ')$. Hence $R$ satisfies RMS3 as well. $\square$

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

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

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

works, in fact the diagram

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

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

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

is strongly cartesian by the result of the preceding paragraphs and we win. $\square$

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. $\square$

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

on objects and by

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

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

on objects and we let

be the unique morphism $\beta : \phi ^*\text{pr}(H(x)) \to (\phi ')^*\text{pr}(H(x'))$ such that $q(\beta ) = b$ and the diagram

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

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

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. $\square$

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

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 $u_ p\mathcal{S}$. $\square$

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

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

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

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

is the sheafification of the presheaf

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

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

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

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). $\square$