hcju/stacksqa huggingface dataset:
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.
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}.
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 $\mathcal{C}/U$.
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}.
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 $(\mathcal{S}_2)_{U_i}$.
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.
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).
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).
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 $\mathcal{X} \to \mathcal{Y}$.
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.
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 $u(U) \to u(U')$.
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 $u^p\mathcal{S}$.
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.
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.