Lemma 8.10.1. 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.

## 8.10 Inherited topologies

It turns out that a fibred category over a site inherits a canonical topology from the underlying site.

**Proof.**
We have to check the three conditions of Sites, Definition 7.6.2.

If $x \to y$ is an isomorphism of $\mathcal{S}$, then it is strongly cartesian by Categories, Lemma 4.33.2 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 $\{ x \to y\} \in \text{Cov}(\mathcal{S})$.

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 4.33.2 and $\{ p(y_{ij}) \to p(x)\} _{i \in I, j\in J_ i} \in \text{Cov}(\mathcal{C})$. Hence also $\{ y_{ij} \to x\} _{i \in I, j\in J_ i} \in \text{Cov}(\mathcal{S})$.

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 4.33.13 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 $\{ x_ i \times _ x y \to y \} _{i\in I} \in \text{Cov}(\mathcal{S})$

This finishes the proof. $\square$

Note that if $p : \mathcal{S} \to \mathcal{C}$ is fibred in groupoids, then the coverings of the site $\mathcal{S}$ in Lemma 8.10.1 are characterized by

because every morphism of $\mathcal{S}$ is strongly cartesian.

Definition 8.10.2. 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 8.10.1 is the *structure of site on $\mathcal{S}$ inherited from $\mathcal{C}$*. We sometimes indicate this by saying that *$\mathcal{S}$ is endowed with the topology inherited from $\mathcal{C}$*.

In particular we obtain a topos of sheaves $\mathop{\mathit{Sh}}\nolimits (\mathcal{S})$ in this situation. It turns out that this topos is functorial with respect to $1$-morphisms of fibred categories.

Lemma 8.10.3. 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 $f : \mathop{\mathit{Sh}}\nolimits (\mathcal{X}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{Y})$ 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 $\mathcal{X}$.

**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 4.33.9 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 7.13.1. Moreover, let $x' \to x$ be a morphism of $\mathcal{X}$. By Categories, Lemma 4.33.13 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 4.33.13 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 7.13.1 and we conclude that $F$ is continuous.

Next we prove that $F$ is cocontinuous. Let $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X})$ 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 7.20.1.

The final assertion follows from the first two, see Sites, Lemmas 7.21.1, 7.20.2, and 7.21.5. $\square$

Lemma 8.10.4. Let $\mathcal{C}$ be a site. Let $p : \mathcal{X} \to \mathcal{C}$ be a category fibred in groupoids. Let $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X})$ 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 $\mathcal{C}$.

**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 4.35.1 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 8.10.2 that coverings of $x'$ in $\mathcal{X}/x$ are in bijective correspondence with coverings of $p(x')$ in $\mathcal{C}/U$.
$\square$

Lemma 8.10.5. 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 $\mathcal{C}$).

**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

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 $\mathcal{X}/\mathcal{Y}$. $\square$

Lemma 8.10.6. 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 $\mathcal{C}$.

**Proof.**
We check the three conditions in Definition 8.4.1 to prove that $\mathcal{Y}$ is a stack over $\mathcal{C}$. By Categories, Lemma 4.33.12 we find that $\mathcal{Y}$ is a fibred category over $\mathcal{C}$. Thus condition (1) holds.

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

on $\mathcal{C}/U$, see Lemma 8.2.3. 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

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 $y_1 \to y_2$.

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 8.3.1). 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 8.3.5). 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).

The final assertion follows because $\mathcal{Y}$ is a stack over $\mathcal{C}$ and is fibred in groupoids by Categories, Lemma 4.35.13. $\square$

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