hcju/stacksqa huggingface dataset:
Let $K' \in D_{perf}(\mathcal{O}_{Y \times_S X})$ be the object assumed to exist in Definition \ref{definition-relative-equivalence-kernel}. Since $X \times_S Y = \lim X_i \times_{S_i} Y_i$ there exists an $i$ and objects $K_i$ and $K'_i$ in $D_{perf}(\mathcal{O}_{Y_i \times_{S_i} X_i})$ whose pullbacks to $Y \times_S X$ give $K$ and $K'$. See Derived Categories of Schemes, Lemma \ref{perfect-lemma-descend-perfect}. By Derived Categories of Schemes, Lemma \ref{perfect-lemma-flat-proper-perfect-direct-image-general} the object $$ R\text{pr}_{13, *}(L\text{pr}_{12}^*K_i \otimes_{\mathcal{O}_{X_i \times_{S_i} Y_i \times_{S_i} X_i}}^\mathbf{L} L\text{pr}_{23}^*K_i') $$ is perfect and its pullback to $X \times_S X$ is equal to $$ R\text{pr}_{13, *}(L\text{pr}_{12}^*K \otimes_{\mathcal{O}_{X \times_S Y \times_S X}}^\mathbf{L} L\text{pr}_{23}^*K') \cong \Delta_{X/S, *}\mathcal{O}_X $$ See proof of Lemma \ref{lemma-base-change-rek}. On the other hand, since $X_i \to S$ is smooth and separated the object $$ \Delta_{i, *}\mathcal{O}_{X_i} $$ of $D(\mathcal{O}_{X_i \times_{S_i} X_i})$ is also perfect (by More on Morphisms, Lemmas \ref{more-morphisms-lemma-smooth-diagonal-perfect} and \ref{more-morphisms-lemma-perfect-proper-perfect-direct-image}) and its pullback to $X \times_S X$ is equal to $$ \Delta_{X/S, *}\mathcal{O}_X $$ See proof of Lemma \ref{lemma-base-change-rek}. Thus by Derived Categories of Schemes, Lemma \ref{perfect-lemma-descend-perfect} after increasing $i$ we may assume that $$ \Delta_{i, *}\mathcal{O}_{X_i} \cong R\text{pr}_{13, *}(L\text{pr}_{12}^*K_i \otimes_{\mathcal{O}_{X_i \times_{S_i} Y_i \times_{S_i} X_i}}^\mathbf{L} L\text{pr}_{23}^*K_i') $$ as desired. The same works for the roles of $K$ and $K'$ reversed.
Let us use chain complexes in the proof of this lemma. The Koszul complex $K_\bullet(\overline{f}_1, \ldots, \overline{f}_r)$ is defined in More on Algebra, Definition \ref{more-algebra-definition-koszul-complex}. By More on Algebra, Lemma \ref{more-algebra-lemma-lift-complex-stably-frees} we can represent $K$ by a complex $$ K_\bullet : A \to A^{\oplus r} \to \ldots \to A^{\oplus r} \to A $$ whose tensor product with $A/\mathfrak mA$ is equal (!) to $K_\bullet(\overline{f}_1, \ldots, \overline{f}_r)$. Denote $f_1, \ldots, f_r \in A$ the components of the arrow $A^{\oplus r} \to A$. These $f_i$ are lifts of the $\overline{f}_i$. By Algebra, Lemma \ref{algebra-lemma-grothendieck-regular-sequence-general} $f_1, \ldots, f_r$ form a regular sequence in $A$ and $A/(f_1, \ldots, f_r)$ is flat over $R$. Let $J = (f_1, \ldots, f_r) \subset A$. Consider the diagram $$ \xymatrix{ K_\bullet \ar[rd] \ar@{..>}[rr]_{\varphi_\bullet} & & K_\bullet(f_1, \ldots, f_r) \ar[ld] \\ & A/J } $$ Since $f_1, \ldots, f_r$ is a regular sequence the south-west arrow is a quasi-isomorphism (see More on Algebra, Lemma \ref{more-algebra-lemma-regular-koszul-regular}). Hence we can find the dotted arrow making the diagram commute for example by Algebra, Lemma \ref{algebra-lemma-compare-resolutions}. Reducing modulo $\mathfrak m$ we obtain a commutative diagram $$ \xymatrix{ K_\bullet(\overline{f}_1, \ldots, \overline{f}_r) \ar[rd] \ar[rr]_{\overline{\varphi}_\bullet} & & K_\bullet(\overline{f}_1, \ldots, \overline{f}_r) \ar[ld] \\ & (A/\mathfrak m A)/(\overline{f}_1, \ldots, \overline{f}_r) } $$ by our choice of $K_\bullet$. Thus $\overline{\varphi}$ is an isomorphism in the derived category $D(A/\mathfrak m A)$. It follows that $\overline{\varphi} \otimes_{A/\mathfrak m A}^\mathbf{L} \lambda$ is an isomorphism. Since $\overline{f}_i \in \mathfrak n / \mathfrak m A$ we see that $$ \text{Tor}_i^{A/\mathfrak m A}( K_\bullet(\overline{f}_1, \ldots, \overline{f}_r), \lambda) = K_i(\overline{f}_1, \ldots, \overline{f}_r) \otimes_{A/\mathfrak m A} \lambda $$ Hence $\varphi_i \bmod \mathfrak n$ is invertible. Since $A$ is local this means that $\varphi_i$ is an isomorphism and the proof is complete.
We can find $g \in S$, $g \not \in \mathfrak q$ with property (1) by the definition of localizations. After replacing $g$ by $gg'$ for some $g' \in S$, $g' \not \in \mathfrak q$ we may assume (2) holds, see Algebra, Lemma \ref{algebra-lemma-regular-sequence-in-neighbourhood}. By Algebra, Theorem \ref{algebra-theorem-openness-flatness} we find that $S_g/(f'_1, \ldots, f'_r)$ is flat over $R$ in an open neighbourhood of $\mathfrak q$. Hence after once more replacing $g$ by $gg'$ for some $g' \in S$, $g' \not \in \mathfrak q$ we may assume (3) holds as well. Finally, we get (4) for a further replacement by More on Algebra, Lemma \ref{more-algebra-lemma-colimit-perfect-complexes}.
The morphism $p$ is proper by Morphisms, Lemma \ref{morphisms-lemma-closed-immersion-proper}. By Cohomology of Schemes, Lemma \ref{coherent-lemma-proper-finite-fibre-finite-in-neighbourhood} there is an open $Y_s \subset V \subset Y$ such that $p|_{p^{-1}(V)} : p^{-1}(V) \to V$ is finite. By More on Morphisms, Theorem \ref{more-morphisms-theorem-criterion-flatness-fibre-Noetherian} there is an open $X_s \subset U \subset X$ such that $p|_U : U \to Y$ is flat. After removing the images of $X \setminus U$ and $Y \setminus V$ (which are closed subsets not containing $s$) we may assume $p$ is flat and finite. Then $p$ is open (Morphisms, Lemma \ref{morphisms-lemma-fppf-open}) and $Y_s \subset p(X) \subset Y$ hence after shrinking $S$ we may assume $p$ is surjective. As $p_s : X_s \to Y_s$ is an isomorphism, the map $$ p^\sharp : \mathcal{O}_Y \longrightarrow p_*\mathcal{O}_X $$ of coherent $\mathcal{O}_Y$-modules ($p$ is finite) becomes an isomorphism after pullback by $i : Y_s \to Y$ (by Cohomology of Schemes, Lemma \ref{coherent-lemma-affine-base-change} for example). By Nakayama's lemma, this implies that $\mathcal{O}_{Y, y} \to (p_*\mathcal{O}_X)_y$ is surjective for all $y \in Y_s$. Hence there is an open $Y_s \subset V \subset Y$ such that $p^\sharp|_V$ is surjective (Modules, Lemma \ref{modules-lemma-finite-type-surjective-on-stalk}). Hence after shrinking $S$ once more we may assume $p^\sharp$ is surjective which means that $p$ is a closed immersion (as $p$ is already finite). Thus now $p$ is a surjective flat closed immersion of Noetherian schemes and hence an isomorphism, see Morphisms, Section \ref{morphisms-section-flat-closed-immersions}.
Choose a closed point $s \in S$. Since $k$ is algebraically closed this is a $k$-rational point. Set $Y_0 = Y_s$. The restriction $K_0$ of $K$ to $X \times Y_0$ is the Fourier-Mukai kernel of a relative equivalence from $X$ to $Y_0$ over $\Spec(k)$ by Lemma \ref{lemma-base-change-rek}. Let $K'_0$ in $D_{perf}(\mathcal{O}_{Y_0 \times X})$ be the  object assumed to exist in Definition \ref{definition-relative-equivalence-kernel}. Then $K'_0$ is the Fourier-Mukai kernel of a relative equivalence from $Y_0$ to $X$ over $\Spec(k)$ by the symmetry inherent in Definition \ref{definition-relative-equivalence-kernel}. Hence by Lemma \ref{lemma-base-change-rek} we see that the pullback $$ M = (Y_0 \times X \times S \to Y_0 \times X)^*K'_0 $$ on $(Y_0 \times S) \times_S (X \times S) = Y_0 \times X \times S$ is the Fourier-Mukai kernel of a relative equivalence from $Y_0 \times S$ to $X \times S$ over $S$. Now consider the kernel $$ K_{new} = R\text{pr}_{13, *}(L\text{pr}_{12}^*M \otimes_{\mathcal{O}_{(Y_0 \times S) \times_S (X \times S) \times_S Y}}^\mathbf{L} L\text{pr}_{23}^*K) $$ on $(Y_0 \times S) \times_S Y$. This is the Fourier-Mukai kernel of a relative equivalence from $Y_0 \times S$ to $Y$ over $S$ since it is the composition of two invertible arrows in the category constructed in Section \ref{section-category-Fourier-Mukai-kernels}. Moreover, this composition passes through base change (Lemma \ref{lemma-base-change-is-functor}). Hence we see that the pullback of $K_{new}$ to $((Y_0 \times S) \times_S Y)_s = Y_0 \times Y_0$ is equal to the composition of $K_0$ and $K'_0$ and hence equal to the identity in this category. In other words, we have $$ L(Y_0 \times Y_0 \to (Y_0 \times S) \times_S Y)^*K_{new} \cong \Delta_{Y_0/k, *}\mathcal{O}_{Y_0} $$ Thus by Lemma \ref{lemma-no-deformations} we conclude that $Y \to S$ is isomorphic to $Y_0 \times S$ in an open neighbourhood of $s$. This finishes the proof.
It suffices to prove the statement for $D(\mathcal{A})$ as the others are full subcategories of this one. Since every object in $D(\mathcal{A})$ is a complex of objects of $\mathcal{A}$ it is immediate that the set of isomorphism classes of objects of $D^b(\mathcal{A})$ is countable. Moreover, for bounded complexes $A^\bullet$ and $B^\bullet$ of $\mathcal{A}$ it is clear that $\Hom_{K^b(\mathcal{A})}(A^\bullet, B^\bullet)$ is countable. We have $$ \Hom_{D^b(\mathcal{A})}(A^\bullet, B^\bullet) = \colim_{s : (A')^\bullet \to A^\bullet \text{ qis and }(A')^\bullet\text{ bounded}} \Hom_{K^b(\mathcal{A})}((A')^\bullet, B^\bullet) $$ by Derived Categories, Lemma \ref{derived-lemma-bounded-derived}. Thus this is a countable set as a countable colimit of
Observe that $X$ is Noetherian by Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}. Hence $D_{perf}(\mathcal{O}_X)$ is a full subcategory of $D^b_{\textit{Coh}}(\mathcal{O}_X)$ by Derived Categories of Schemes, Lemma \ref{perfect-lemma-perfect-on-noetherian}. Thus it suffices to prove the result for $D^b_{\textit{Coh}}(\mathcal{O}_X)$. Recall that $D^b_{\textit{Coh}}(\mathcal{O}_X) = D^b(\textit{Coh}(\mathcal{O}_X))$ by Derived Categories of Schemes, Proposition \ref{perfect-proposition-DCoh}. Hence by Lemma \ref{lemma-countable-abelian} it suffices to prove that $\textit{Coh}(\mathcal{O}_X)$ is countable. This we omit.
The field $K$ is the filtered union of its countable subfields. Dually, $\Spec(K)$ is the cofiltered limit of the spectra of the countable subfields of $K$. Hence Limits, Lemma \ref{limits-lemma-descend-finite-presentation} guarantees that we can find a countable subfield $k$ and morphisms $X_0 \to S_0$ and $Y_0 \to S_0$ of schemes of finite type over $k$ such that $X \to S$ and $Y \to S$ are the base changes of these. \medskip\noindent By Lemma \ref{lemma-countable-finite-type} there is a countable set $I$ and pairs $(S_{0, i} \to S_0, h_{0, i})$ such that \begin{enumerate} \item $S_{0, i} \to S_0$ is a morphism of finite type, set $X_{0, i} = X_0 \times_{S_0} S_{0, i}$ and $Y_{0, i} = Y_0 \times_{S_0} S_{0, i}$, \item $h_{0, i} : X_{0, i} \to Y_{0, i}$ is an isomorphism over $S_{0, i}$. \end{enumerate} such that every pair $(T \to S_0, h_T)$ with $T \to S_0$ of finite type and $h_T : X_0 \times_{S_0} T \to Y_0 \times_{S_0} T$ an isomorphism is isomorphic to one of these. Denote $(S_i \to S, h_i)$ the base change of $(S_{0, i} \to S_0, h_{0, i})$ by $\Spec(K) \to \Spec(k)$. We claim this works. \medskip\noindent Let $s \in S(K)$ and let $h_s : X_s \to Y_s$ be an isomorphism over $K = \kappa(s)$. We can write $K$ as the filtered union of its finitely generated $k$-subalgebras. Hence by Limits, Proposition \ref{limits-proposition-characterize-locally-finite-presentation} and Lemma \ref{limits-lemma-descend-finite-presentation} we can find such a finitely generated $k$-subalgebra $K \supset A \supset k$ such that \begin{enumerate} \item there is a commutative diagram $$ \xymatrix{ \Spec(K) \ar[d]_s \ar[r] & \Spec(A) \ar[d]^{s'} \\ S \ar[r] & S_0} $$ for some morphism $s' : \Spec(A) \to S_0$ over $k$, \item $h_s$ is the base change of an isomorphism $h_{s'} : X_0 \times_{S_0, s'} \Spec(A) \to X_0 \times_{S_0, s'} \Spec(A)$ over $A$. \end{enumerate} Of course, then $(s' : \Spec(A) \to S_0, h_{s'})$ is isomorphic to the pair $(S_{0, i} \to S_0, h_{0, i})$ for some $i \in I$. This concludes the proof because the commutative diagram in (1) shows that $s$ is in the image of the base change of $s'$ to $\Spec(K)$.
By More on Morphisms of Spaces, Lemmas \ref{spaces-more-morphisms-lemma-descending-property-thickening} and \ref{spaces-more-morphisms-lemma-base-change-thickening} the property $P$ that a morphism of algebraic spaces is a (first order) thickening is fpqc local on the base and stable under base change. Thus the discussion in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} indeed applies. Having said this the equivalence of (1) and (2) follows from the fact that $P = P_1 + P_2$ where $P_1$ is the property of being a closed immersion and $P_2$ is the property of being surjective. (Strictly speaking, the reader should also consult More on Morphisms of Spaces, Definition \ref{spaces-more-morphisms-definition-thickening}, Properties of Stacks, Definition \ref{stacks-properties-definition-immersion} and the discussion following, Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-surjective-representable}, Properties of Stacks, Section \ref{stacks-properties-section-surjective} to see that all the concepts all match up.) The final assertion is clear from the foregoing.
Since $\mathcal{X} \to \mathcal{Y}'$ factors through the closed substack $\mathcal{Y}$ we see that $\mathcal{X} \times_\mathcal{Y} \mathcal{X} = \mathcal{X} \times_{\mathcal{Y}'} \mathcal{X}$. Hence $\mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}'$ is isomorphic to the composition $$ \mathcal{X} \times_{\mathcal{Y}'} \mathcal{X} \to \mathcal{X} \times_{\mathcal{Y}'} \mathcal{X}' \to \mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}' $$ both of which are thickenings as base changes of thickenings (Lemma \ref{lemma-base-change-thickening}). Hence so is the composition (Lemma \ref{lemma-composition-thickening}). Since $\mathcal{X} \to \mathcal{X}'$ is a monomorphism, the final statement of the lemma follows from Properties of Stacks, Lemma \ref{stacks-properties-lemma-monomorphism-diagonal} applied to $\mathcal{X} \to \mathcal{X}' \to \mathcal{Y}'$.
Observe that $$ (\Delta, \Delta') : (\mathcal{X} \subset \mathcal{X}') \longrightarrow (\mathcal{X} \times_\mathcal{Y} \mathcal{X} \subset \mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}') $$ is a morphism of thickenings (Lemma \ref{lemma-reduced-diagonal}). Moreover $\Delta$ and $\Delta'$ are representable by algebraic spaces by Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-properties-diagonal}. Hence, via the discussion in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} the lemma follows for cases (a), (b), (c), (d), (e), (f), (g), (h), (i), and (j) by using More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-thicken-property-morphisms}. \medskip\noindent Lemma \ref{lemma-reduced-diagonal} tells us that $\mathcal{X} = (\mathcal{X} \times_\mathcal{Y} \mathcal{X}) \times_{(\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}')} \mathcal{X}'$. Moreover, $\Delta$ and $\Delta'$ are locally of finite type by the aforementioned Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-properties-diagonal}. Hence the result for cases (k), (l), (m), (n), (o), (p), and (q) by using More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-properties-that-extend-over-thickenings}.
In each case we reduce to a question about the diagonal and then we use Lemma \ref{lemma-thickening-diagonals} applied to the morphism of thickenings $$ (\mathcal{X} \subset \mathcal{X}') \to \left(\Spec(\mathbf{Z}) \subset \Spec(\mathbf{Z})\right) $$ We do this after viewing $\mathcal{X} \subset \mathcal{X}'$ as a thickening of algebraic stacks over $\Spec(\mathbf{Z})$ via Algebraic Stacks, Definition \ref{algebraic-definition-viewed-as}. \medskip\noindent Case (1). An algebraic stack is an algebraic space if and only if its diagonal is a monomorphism, see Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-hierarchy} (this also follows immediately from Algebraic Stacks, Proposition \ref{algebraic-proposition-algebraic-stack-no-automorphisms}). \medskip\noindent Case (2). By (1) we may assume that $\mathcal{X}$ and $\mathcal{X}'$ are algebraic spaces and then we can use More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-thickening-scheme}. \medskip\noindent Case (3) -- (6). Each of these cases corresponds to a condition on the diagonal, see Morphisms of Stacks, Definitions \ref{stacks-morphisms-definition-separated} and \ref{stacks-morphisms-definition-absolute-separated}.
By Lemma \ref{lemma-thickening} the morphisms $\mathcal{X} \to \mathcal{X}'$ and $\mathcal{Y} \to \mathcal{Y}'$ are universal homeomorphisms. Thus any condition on $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ is equivalent with the corresponding condition on $|f'| : |\mathcal{X}'| \to |\mathcal{Y}'|$ and the same is true after arbitrary base change by a morphism $\mathcal{Z}' \to \mathcal{Y}'$. This proves that (2), (3), (4), (6), (7) hold. \medskip\noindent In cases (8), (9), (10), (12) we can translate the conditions on $f$ and $f'$ into conditions on the diagonals $\Delta$ and $\Delta'$ as in Lemma \ref{lemma-thickening-diagonals}. See Morphisms of Stacks, Definition \ref{stacks-morphisms-definition-separated} and Lemma \ref{stacks-morphisms-lemma-hierarchy}. Hence these cases follow from Lemma \ref{lemma-thickening-diagonals}. \medskip\noindent Proof of (11). If $f'$ is representable, then so is $f$, because for a scheme $T$ and a morphism $T \to \mathcal{Y}$ we have $\mathcal{X} \times_\mathcal{Y} T = \mathcal{X} \times_{\mathcal{X}'} (\mathcal{X}' \times_{\mathcal{Y}'} T)$ and $\mathcal{X} \to \mathcal{X}'$ is a closed immersion (hence representable). Conversely, assume $f$ is representable, and let $T' \to \mathcal{Y}'$ be a morphism where $T'$ is a scheme. Then $$ \mathcal{X} \times_{\mathcal{Y}} (\mathcal{Y} \times_{\mathcal{Y}'} T') = \mathcal{X} \times_{\mathcal{X}'} (\mathcal{X}' \times_{\mathcal{Y}'} T') \to \mathcal{X}' \times_{\mathcal{Y}'} T' $$ is a thickening (by Lemma \ref{lemma-base-change-thickening}) and the source is a scheme. Hence the target is a scheme by Lemma \ref{lemma-thickening-properties}. \medskip\noindent In cases (1) and (5) if either $f$ or $f'$ has the stated property, then both $f$ and $f'$ are representable by (11). In this case choose an algebraic space $V'$ and a surjective smooth morphism $V' \to \mathcal{Y}'$. Set $V = \mathcal{Y} \times_{\mathcal{Y}'} V'$, $U' = \mathcal{X}' \times_{\mathcal{Y}'} V'$, and $U = \mathcal{X} \times_{\mathcal{Y}'} V'$. Then the desired results follow from the corresponding results for the morphism $(U \subset U') \to (V \subset V')$ of thickenings of algebraic spaces via the principle of Properties of Stacks, Lemma \ref{stacks-properties-lemma-check-property-covering}. See More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-thicken-property-morphisms} for the corresponding results in the case of algebraic spaces.
Choose an algebraic space $U'$ and a surjective smooth morphism $U' \to \mathcal{B}'$. Choose an algebraic space $V'$ and a surjective smooth morphism $V' \to U' \times_{\mathcal{B}'} \mathcal{Y}'$. Choose an algebraic space $W'$ and a surjective smooth morphism $W' \to V' \times_{\mathcal{Y}'} \mathcal{X}'$. Let $U, V, W$ be the base change of $U', V', W'$ by $\mathcal{B} \to \mathcal{B}'$. Then flatness of $f'$ is equivalent to flatness of $W' \to V'$ and we are given that $W \to V$ is flat. Hence we may apply the lemma in the case of algebraic spaces to the diagram $$ \xymatrix{ (W \subset W') \ar[rr] \ar[rd] & & (V \subset V') \ar[ld] \\ & (U \subset U') } $$ of thickenings of algebraic spaces. See More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-flatness-morphism-thickenings-fp-over-ft}. The statement about flatness of $\mathcal{Y}'/\mathcal{B}'$ at points in the image of $|f'|$ follows in the same manner.
Namely $f : V \to U$ is \'etale as a morphism in $W_{spaces, \etale}$ and we can apply Lemma \ref{lemma-deform-property-fp-over-ft} because $U' \to \mathcal{X}'$ and $V' \to \mathcal{X}'$ are smooth and $U = \mathcal{X} \times_{\mathcal{X}'} U'$ and $V = \mathcal{X} \times_{\mathcal{X}'} V'$.
We claim the fibre categories of $p$ are groupoids. If $(f, f', \gamma')$ as in (\ref{equation-morphism}) is a morphism such that $f : U \to V$ is an isomorphism, then $f'$ is an isomorphism by Lemma \ref{lemma-deform-property-fp-over-ft} and hence $(f, f', \gamma')$ is an isomorphism. \medskip\noindent Consider a morphism $f : V \to U$ in $W_{spaces, \etale}$ and an object $\xi = (U, U', a, i, x', \alpha)$ of $\mathcal{C}$ over $U$. We are going to construct the ``pullback'' $f^*\xi$ over $V$.  Namely, set $b = a \circ f$. Let $f' : V' \to U'$ be the \'etale morphism whose restriction to $V$ is $f$ (More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-topological-invariance}). Denote $j : V \to V'$ the corresponding thickening. Let $y' = x' \circ f'$ and $\gamma = \text{id} : x' \circ f' \to y'$. Set $$ \beta = \alpha \star \text{id}_f : x \circ b = x \circ a \circ f \to x' \circ i \circ f = x' \circ f' \circ j = y' \circ j $$ It is clear that $(f, f', \gamma) : (V, V', b, j, y', \beta) \to (U, U', a, i, x', \alpha)$ is a morphism as in (\ref{equation-morphism}). The morphisms $(f, f', \gamma)$ so constructed are strongly cartesian (Categories, Definition \ref{categories-definition-cartesian-over-C}). We omit the detailed proof, but essentially the reason is that given a morphism $(g, g', \epsilon) : (Y, Y', c, k, z', \delta) \to (U, U', a, i, x', \alpha)$ in $\mathcal{C}$ such that $g$ factors as $g = f \circ h$ for some $h : Y \to V$, then we get a unique factorization $g' = f' \circ h'$ from More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-topological-invariance} and after that one can produce the necessary $\zeta$ such that $(h, h', \zeta) :  (Y, Y', c, k, z', \delta) \to (V, V', b, j, y', \beta)$ is a morphism of $\mathcal{C}$ with $(g, g', \epsilon) = (f, f', \gamma) \circ (h, h', \zeta)$. \medskip\noindent Therefore $p : \mathcal{C} \to W_\etale$ is a fibred category (Categories, Definition \ref{categories-definition-fibred-category}). Combined with the fact that the fibre categories are groupoids seen above we conclude that $p : \mathcal{C} \to W_\etale$ is fibred in groupoids by Categories, Lemma \ref{categories-lemma-fibred-groupoids}.