hcju/stacksqa huggingface dataset:
Let $E \subset \Ob(\textit{Ab}(Y_\etale))$ be the class consisting of products of skyscraper sheaves. We claim that \begin{enumerate} \item every $\mathcal{G}$ in $\textit{Ab}(Y_\etale)$ is a subsheaf of an element of $E$, and \item for every $\mathcal{G} \in E$ there exists an object $\mathcal{H}$ of $\textit{Ab}(X_\etale)$ such that $\Hom(f_!\mathcal{F}, \mathcal{G}) = \Hom(\mathcal{F}, \mathcal{H})$ functorially in $\mathcal{F}$. \end{enumerate} Once the claim has been verified, the dual of Homology, Lemma \ref{homology-lemma-partially-defined-adjoint} produces the adjoint functor $f^!$. \medskip\noindent Part (1) is true because we can map $\mathcal{G}$ to the sheaf $\prod \overline{y}_*\mathcal{G}_{\overline{y}}$ where the product is over all geometric points of $Y$. This is an injection by \'Etale Cohomology, Theorem \ref{etale-cohomology-theorem-exactness-stalks}. (This is the first step in the Godement resolution when done in the setting of abelian sheaves on topological spaces.) \medskip\noindent Part (2) and the final statement of the lemma can be seen as follows. Suppose that $\mathcal{G} = \prod \overline{y}_*A_{\overline{y}}$ for some abelian groups $A_{\overline{y}}$. Then $$ \Hom(f_!\mathcal{F}, \mathcal{G}) = \prod \Hom(f_!\mathcal{F}, \overline{y}_*A_{\overline{y}}) $$ Thus it suffices to find abelian sheaves $\mathcal{H}_{\overline{y}}$ on $X_\etale$ representing the functors $\mathcal{F} \mapsto \Hom(f_!\mathcal{F}, \overline{y}_*A_{\overline{y}})$ and to take $\mathcal{H} = \prod \mathcal{H}_{\overline{y}}$. This reduces us to the case $\mathcal{H} = \overline{y}_*A$ for some fixed geometric point $\overline{y} : \Spec(k) \to Y$ and some fixed abelian group $A$. We claim that in this case $\mathcal{H} = \prod_{f(\overline{x}) = \overline{y}} \overline{x}_*A$ works. This will finish the proof of the lemma. Namely, we have $$ \Hom(f_!\mathcal{F}, \overline{y}_*A) = \Hom_{\textit{Ab}}((f_!\mathcal{F})_{\overline{y}}, A) = \Hom_{\textit{Ab}}(\bigoplus\nolimits_{f(\overline{x}) = \overline{y}} \mathcal{F}_{\overline{x}}, A) $$ by the description of stalks in Lemma \ref{lemma-lqf-f-shriek-stalk} on the one hand and on the other hand we have $$ \Hom(\mathcal{F}, \mathcal{H}) = \prod\nolimits_{f(\overline{x}) = \overline{y}} \Hom(\mathcal{F}, \overline{x}_*A) = \prod\nolimits_{f(\overline{x}) = \overline{y}} \Hom_{\textit{Ab}}(\mathcal{F}_{\overline{x}}, A) $$ We leave it to the reader to identify these as functors of $\mathcal{F}$.
This is true because $j_!$ as defined in Section \ref{section-finite-support} agrees with $j_!$ as defined in \'Etale Cohomology, Section \ref{etale-cohomology-section-extension-by-zero}, see Lemma \ref{lemma-finite-support-etale-shriek}. Finally, in \'Etale Cohomology, Section \ref{etale-cohomology-section-extension-by-zero} the functor $j_!$ is defined as the left adjoint of $j^{-1}$ and hence we conclude by uniqueness of adjoint functors.
This follows immediately from Derived Categories, Lemma \ref{derived-lemma-derived-adjoint-functors}, the fact that $f_!$ is exact (Lemma \ref{lemma-lqf-f-shriek-stalk}) and hence $Lf_! = f_!$ and the fact that we have enough K-injective complexes of abelian sheaves on $Y_\etale$ so that $Rf^!$ is defined.
The statement means that if we write $R(g \circ f)_* = Rg_* \circ Rf_*$ and $R(g' \circ f')_* = Rg'_* \circ Rf'_*$, then the isomorphism $m_! \circ Rg'_* \circ Rf'_* \to Rg_* \circ Rf_* \circ k_!$ of the outer rectangle is equal to the composition $$ m_! \circ Rg'_* \circ Rf'_* \to Rg_* \circ l_! \circ Rf'_* \to Rg_* \circ Rf_* \circ k_! $$ of the two maps of the squares in the diagram. To prove this choose a K-injective complex $\mathcal{J}^\bullet$ of $\Lambda$-modules on $X'_\etale$ and a quasi-isomorphism $k_!\mathcal{J}^\bullet \to \mathcal{I}^\bullet$ to a K-injective complex $\mathcal{I}^\bullet$ of $\Lambda$-modules on $X_\etale$. The proof of Lemma \ref{lemma-shriek-proper-and-open} shows that the canonical map $$ a : l_!f'_*\mathcal{J}^\bullet \to f_*\mathcal{I}^\bullet $$ is a quasi-isomorphism and this quasi-isomorphism produces the second arrow on applying $Rg_*$. By Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-K-injective-flat} the complex $f_*\mathcal{I}^\bullet$, resp.\ $f'_*\mathcal{J}^\bullet$ is a K-injective complex of $\Lambda$-modules on  $Y_\etale$, resp.\ $Y'_\etale$. (Using this is cheating and could be avoided.) In particular, the same reasoning gives that the canonical map $$ b : m_!g'_*f'_*\mathcal{J}^\bullet \to g_*f_*\mathcal{I}^\bullet $$ is a quasi-isomorphism and this quasi-isomorphism represents the first arrow. Finally, the proof of Lemma \ref{lemma-shriek-proper-and-open} show that $g_*l_!f'_!\mathcal{J}^\bullet$ represents $Rg_*(l_!f'_*\mathcal{J}^\bullet)$ because $f'_*\mathcal{J}^\bullet$ is K-injective. Hence $Rg_*(a) = g_*(a)$ and the composition $g_*(a) \circ b$ is the arrow of Lemma \ref{lemma-shriek-proper-and-open} for the rectangle.
The statement means that if we write $(h \circ h')_! = h_! \circ h'_!$ and $(g \circ g')_! = g_! \circ g'_!$ using the equalities of Lemma \ref{lemma-f-shriek-composition}, then the isomorphism $h_! \circ h'_! \circ Rf''_* \to Rf_* \circ g_! \circ g'_!$ of the outer rectangle is equal to the composition $$ h_! \circ h'_! \circ Rf''_* \to h_! \circ Rf'_* \circ g'_! \to Rf_* \circ g_! \circ g'_! $$ of the two maps of the squares in the diagram. To prove this choose a K-injective complex $\mathcal{I}^\bullet$ of $\Lambda$-modules on $X''_\etale$ and a quasi-isomorphism $g'_!\mathcal{I}^\bullet \to \mathcal{J}^\bullet$ to a K-injective complex $\mathcal{J}^\bullet$ of $\Lambda$-modules on $X'_\etale$. Next, choose a quasi-isomorphism $g_!\mathcal{J}^\bullet \to \mathcal{K}^\bullet$ to a K-injective complex $\mathcal{K}^\bullet$ of $\Lambda$-modules on $X_\etale$. The proof of Lemma \ref{lemma-shriek-proper-and-open} shows that the canonical maps $$ h'_!f''_*\mathcal{I}^\bullet \to f'_*\mathcal{J}^\bullet \quad\text{and}\quad h_!f'_*\mathcal{J}^\bullet \to f_*\mathcal{K}^\bullet $$ are quasi-isomorphisms and these quasi-isomorphisms define the first and second arrow above. Since $g_!$ is an exact functor (Lemma \ref{lemma-lqf-f-shriek-separated-colimits}) we find that $g_!g'_!\mathcal{I}^\bullet \to \mathcal{K}^\bullet$ is a quasi-ismorphism and hence the canonical map $$ h_!h'_!f''_*\mathcal{I}^\bullet \to f_*\mathcal{K}^\bullet $$ is a quasi-isomorphism and represents the map for the outer rectangle in the derived category. Clearly this map is the composition of the other two and the proof is complete.
This is a special case of Lemma \ref{lemma-shriek-proper-and-open} if $f$ and $g$ are separated. We urge the reader to skip the proof in the general case as we'll mainly use the case where $f$ and $g$ are separated. \medskip\noindent Represent $K$ by a complex $\mathcal{K}^\bullet$ of sheaves of $\Lambda$-modules on $X_\etale$. Choose a quasi-isomorphism $f_!\mathcal{K}^\bullet \to \mathcal{I}^\bullet$ into a K-injective complex $\mathcal{I}^\bullet$ of sheaves of $\Lambda$-modules on $Y_\etale$. Consider the map $$ g_!\mathcal{K}^\bullet = h_!f_!\mathcal{K}^\bullet = h_*f_!\mathcal{K}^\bullet \longrightarrow h_*\mathcal{I}^\bullet $$ where the equalities are Lemmas \ref{lemma-lqf-separated-shriek-composition} and \ref{lemma-proper-f-shriek}. This map of complexes determines the map $g_!K \to Rh_*(f_!K)$ of the statement of the lemma. To check the map is an isomorphism we may work locally on $Z$. Hence we may assume that the dimension of fibres of $h$ is bounded, see Morphisms, Lemma \ref{morphisms-lemma-morphism-finite-type-bounded-dimension}. Then we see that $Rh_*$ has finite cohomological dimension, see \'Etale Cohomology, Lemma \ref{etale-cohomology-lemma-cohomological-dimension-proper}. Hence by Derived Categories, Lemma \ref{derived-lemma-unbounded-right-derived}, if we show that $R^qh_*(f_!\mathcal{F}) = 0$ for $q > 0$ and any sheaf $\mathcal{F}$ of $\Lambda$-modules on $X_\etale$, then $h_*f_!\mathcal{K}^\bullet \to h_*\mathcal{I}^\bullet$ is a quasi-isomorphism. \medskip\noindent Observe that $\mathcal{G} = f_!\mathcal{F}$ is a sheaf of $\Lambda$-modules on $Y$ whose stalks are nonzero only at points $y \in Y$ such that $\kappa(y)/\kappa(h(y))$ is a finite extension. This follows from the description of stalks of $f_!\mathcal{F}$ in Lemma \ref{lemma-lqf-f-shriek-stalk} and the fact that both $f$ and $g$ are locally quasi-finite. Hence by the proper base change theorem (\'Etale Cohomology, Lemma \ref{etale-cohomology-lemma-proper-base-change-stalk}) it suffices to show that $H^q(Y_{\overline{z}}, \mathcal{H}) = 0$ where $\mathcal{H}$ is a sheaf on the proper scheme $Y_{\overline{z}}$ over $\kappa(\overline{z})$ whose support is contained in the set of closed points. Thus the required vanishing by \'Etale Cohomology, Lemma \ref{etale-cohomology-lemma-supported-in-closed-points}.
Consider the category of compactifications of $X$ over $Y$, which is cofiltered according to More on Flatness, Theorem \ref{flat-theorem-nagata} and Lemmas \ref{flat-lemma-compactifications-cofiltered} and \ref{flat-lemma-compactifyable}. To every choice of a compactification $$ j : X \to \overline{X},\quad \overline{f} : \overline{X} \to Y $$ the construction above associates the functor $R\overline{f}_* \circ j_! : D(X_\etale, \Lambda) \to D(Y_\etale, \Lambda)$. Let's be a little more explicit. Given a complex $\mathcal{K}^\bullet$ of sheaves of $\Lambda$-modules on $X_\etale$, we choose a quasi-isomorphism $j_!\mathcal{K}^\bullet \to \mathcal{I}^\bullet$ into a K-injective complex of sheaves of $\Lambda$-modules on $\overline{X}_\etale$. Then our functor sends $\mathcal{K}^\bullet$ to $\overline{f}_*\mathcal{I}^\bullet$. \medskip\noindent Suppose given a morphism $g : \overline{X}_1 \to \overline{X}_2$ between compactifications $j_i : X \to \overline{X}_i$ over $Y$. Then we get an isomorphism $$ R\overline{f}_{2, *} \circ j_{2, !} = R\overline{f}_{2, *} \circ Rg_* \circ j_{1, !} = R\overline{f}_{1, *} \circ j_{1, !} $$ using Lemma \ref{lemma-shriek-lqf-and-proper} in the first equality. \medskip\noindent To finish the proof, since the category of compactifications of $X$ over $Y$ is cofiltered, it suffices to show compositions of morphisms of compactifications of $X$ over $Y$ are turned into compositions of isomorphisms of functors\footnote{Namely, if $\alpha, \beta : F \to G$ are morphisms of functors and $\gamma : G \to H$ is an isomorphism of functors such that $\gamma \circ \alpha = \gamma \circ \beta$, then we conclude $\alpha = \beta$.}. To do this, suppose that $j_3 : X \to \overline{X}_3$ is a third compactification and that $h : \overline{X}_2 \to \overline{X}_3$ is a morphism of compactifications. Then we have to show that the composition $$ R\overline{f}_{3, *} \circ j_{3, !} = R\overline{f}_{3, *} \circ Rh_* \circ j_{2, !} = R\overline{f}_{2, *} \circ j_{2, !} = R\overline{f}_{2, *} \circ Rg_* \circ j_{1, !} = R\overline{f}_{1, *} \circ j_{1, !} $$ is equal to the isomorphism of functors constructed using simply $j_3$, $g \circ h$, and $j_1$. A calculation shows that it suffices to prove that the composition of the maps $$ j_{3, !} \to Rh_* \circ j_{2, !} \to Rh_* \circ Rg_* \circ j_{1, !} $$ of Lemma \ref{lemma-shriek-lqf-and-proper} agrees with the corresponding map $j_{3, !} \to R(h \circ g)_* \circ j_{1, !}$ via the identification $R(h \circ g)_* = Rh_* \circ Rg_*$. Since the map of Lemma \ref{lemma-shriek-lqf-and-proper} is a special case of the map of Lemma \ref{lemma-shriek-proper-and-open} (as $j_1$ and $j_2$ are separated) this follows immediately from Lemma \ref{lemma-shriek-proper-and-open-compose}.
To do this we choose a compactification $\overline{Z}$ of $Z$ over $T$, then a compactification $\overline{Y}$ of $Y$ over $\overline{Z}$, and then a compactification $\overline{X}$ of $X$ over $\overline{Y}$. This uses More on Flatness, Theorem \ref{flat-theorem-nagata} and Lemma \ref{flat-lemma-compactifyable}. Let $W \subset \overline{Y}$ be the inverse image of $Z$ under $\overline{Y} \to \overline{Z}$ and let $U \subset V \subset \overline{X}$ be the inverse images of $Y \subset W$ under $\overline{X} \to \overline{Y}$. This produces the following diagram $$ \xymatrix{ X \ar[d]_f \ar[r] & U \ar[r] \ar[d] \ar@{}[dr]|A & V \ar[d] \ar[r] \ar@{}[rd]|B & \overline{X} \ar[d] \\ Y \ar[d]_g \ar[r] & Y \ar[r] \ar[d] & W \ar[r] \ar[d] \ar@{}[rd]|C & \overline{Y} \ar[d] \\ Z \ar[d]_h \ar[r] & Z \ar[d] \ar[r] & Z \ar[d] \ar[r] & \overline{Z} \ar[d] \\ T \ar[r] & T \ar[r] & T \ar[r] & T } $$ Without introducing tons of notation but arguing exactly as in the proof of Lemma \ref{lemma-shriek-composition} we see that the maps in the first displayed diagram use the maps of Lemma \ref{lemma-shriek-proper-and-open} for the rectangles $A + B$, $B + C$, $A$, and $C$ as indicated in the diagram in the statement of the lemma. Since by Lemmas \ref{lemma-shriek-proper-and-open-compose} and \ref{lemma-shriek-proper-and-open-compose-horizontal} we have $\gamma_{A + B} = \gamma_B \circ \gamma_A$ and $\gamma_{B + C} = \gamma_B \circ \gamma_C$ we conclude that the desired equality holds provided $\gamma_A \circ \gamma_C = \gamma_C \circ \gamma_A$. This is true because the two squares $A$ and $C$ only intersect in one point (similar to the last argument in Remark \ref{remark-going-around}).
Choose a compactification $j : X \to \overline{X}$ over $Y$ and denote $\overline{f} : \overline{X} \to Y$ the structure morphism. Let $j' : X' \to \overline{X}'$ and $\overline{f}' : \overline{X}' \to Y'$ denote the base changes of $j$ and $\overline{f}$. Since $Rf_! = R\overline{f} \circ j_!$ and $Rf'_! = R\overline{f}' \circ j'_!$ the isomorphism can be constructed via $$ g^{-1} \circ R\overline{f} \circ j_! \to R\overline{f}' \circ (\overline{g}')^{-1} \circ j_! \to R\overline{f}' \circ j'_! \circ (g')^{-1}  $$ where the first arrow is the isomorphism given to us by the proper base change theorem (\'Etale Cohomology, Lemma \ref{etale-cohomology-lemma-proper-base-change-mod-n}) and the second arrow is the isomorphism of Lemma \ref{lemma-base-change-f-shriek-separated}. \medskip\noindent To finish the proof we have to show two things: first we have to show that the isomorphism of functors so obtained does not depend on the choice of the compactification and second we have to show that if we vertically stack two base change diagrams as in the lemma, then these base change isomorphisms are compatible with the isomorphisms of Lemma \ref{lemma-shriek-composition}. A straightforward argument which we omit shows that both follow if we can show that the isomorphisms \begin{enumerate} \item $Rg_* \circ Rf_* = R(g \circ f)_*$ for $f : X \to Y$ and $g : Y \to Z$ proper, \item $g_! \circ f_! = (g \circ f)_!$ for $f : X \to Y$ and $g : Y \to Z$ separated and quasi-finite, and \item $g_! \circ Rf'_* = Rf_* \circ g'_!$ for  $f : X \to Y$ and $f' : X' \to Y'$ proper and $g : Y' \to Y$ and $g' : X' \to X$ separated and quasi-finite with $f \circ g' = g \circ f'$ \end{enumerate} are compatible with base change. This holds for (1) by Cohomology on Sites, Remark \ref{sites-cohomology-remark-compose-base-change}, for (2) by Remark \ref{remark-f-shriek-base-change-composition}, and (3) by Lemma \ref{lemma-shriek-proper-and-open-base-change}.
By construction it suffices to prove this when $f$ is an open immersion and when $f$ is a proper morphism. For any open immersion $j : U \to X$ of schemes, the functor $j_! : D(U_\etale) \to D(X_\etale)$ is a left adjoint to pullback $j^{-1} : D(X_\etale) \to D(U_\etale)$ and hence commutes with direct sums, see Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-adjoint-lower-shriek-restrict}. In the proper case we have $Rf_! = Rf_*$ and we get the result from \'Etale Cohomology, Lemma \ref{etale-cohomology-lemma-proper-mod-n-direct-sums} and the fact that our coefficient ring $\Lambda$ is a torsion ring.
By construction it suffices to prove this when $f$ is an open immersion and when $f$ is a proper morphism. For any open immersion $j : U \to X$ of schemes, the functor $j_! : D(U_\etale) \to D(X_\etale)$ is exact and hence the statement holds with $N = 0$ in this case. If $f$ is proper then $Rf_! = Rf_*$, i.e., it is a right derived functor. Hence the bound on the left by Derived Categories, Lemma \ref{derived-lemma-negative-vanishing}. Moreover in this case $f_* : \textit{Mod}(X_\etale, \Lambda) \to \textit{Mod}(Y_\etale, \Lambda)$ has bounded cohomological dimension by Morphisms, Lemma \ref{morphisms-lemma-morphism-finite-type-bounded-dimension} and \'Etale Cohomology, Lemma \ref{etale-cohomology-lemma-cohomological-dimension-proper}. Thus we conclude by Derived Categories, Lemma \ref{derived-lemma-unbounded-right-derived}.
The implication (1) $\Rightarrow$ (2) is Algebraic Stacks, Lemma \ref{algebraic-lemma-base-change-by-space-representable-by-space}. Conversely, let $W \to \mathcal{Y}$ be as in (2). To prove (1) it suffices to show that $f$ is faithful on fibre categories, see Algebraic Stacks, Lemma \ref{algebraic-lemma-characterize-representable-by-algebraic-spaces}. Assumption (2) implies in particular that $W \times_\mathcal{Y} \mathcal{X} \to W$ is faithful. Hence the faithfulness of $f$ follows from Stacks, Lemma \ref{stacks-lemma-faithful-descent}.
The implication from left to right follows from Lemma \ref{lemma-property-spaces-too}. Assume $V \to W$ has $P$. Let $T$ be a scheme, and let $T \to \mathcal{Y}$ be a morphism. Consider the commutative diagram $$ \xymatrix{ T \times_\mathcal{Y} \mathcal{X} \ar[d] & T \times_\mathcal{Y} W \ar[d] \ar[l] \ar[r] & W \ar[d] \\ T & T \times_\mathcal{Y} V \ar[l] \ar[r] & V } $$ of algebraic spaces. The squares are cartesian. The bottom left morphism is a surjective, flat morphism which is locally of finite presentation, hence $\{T \times_\mathcal{Y} V \to T\}$ is an fppf covering. Hence the fact that the right vertical arrow has property $P$ implies that the left vertical arrow has property $P$.
Choose an algebraic space $W$ and a morphism $W \to \mathcal{Z}$ which is surjective, flat, and locally of finite presentation. By the discussion above the composition $W \to \mathcal{Y}$ is also surjective, flat, and locally of finite presentation. Denote $V = W \times_\mathcal{Z} \mathcal{W} = V \times_\mathcal{Y} \mathcal{X}$. By Lemma \ref{lemma-check-property-covering} we see that $f$ has $\mathcal{P}$ if and only if $V \to W$ does and that $\mathcal{W} \to \mathcal{Z}$ has $\mathcal{P}$ if and only if $V \to W$ does. The lemma follows.
Since $U \to \mathcal{Y}$ is surjective and smooth, see Algebraic Stacks, Lemma \ref{algebraic-lemma-smooth-quotient-smooth-presentation} the base change $U' \to \mathcal{X}'$ is also surjective and smooth. Hence, by Algebraic Stacks, Lemma \ref{algebraic-lemma-stack-presentation} it suffices to show that $R' = U' \times_{\mathcal{X}'} U'$ in order to get a smooth groupoid $(U', R', s', t', c')$ and a presentation $[U'/R'] \to \mathcal{X}'$. Using that $R = V \times_\mathcal{Y} V$ (see Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-2-cartesian}) this follows from $$ R' = U \times_\mathcal{X} U \times_\mathcal{X} \mathcal{X}' = (U \times_\mathcal{X} \mathcal{X}') \times_{\mathcal{X}'} (U \times_\mathcal{X} \mathcal{X}') $$ see Categories, Lemmas \ref{categories-lemma-associativity-2-fibre-product} and \ref{categories-lemma-2-fibre-product-erase-factor}. Clearly the projection morphisms $U' \to U$ and $R' \to R$ give the desired morphism of groupoids $\text{pr} : (U', R', s', t', c') \to (U, R, s, t, c)$. Hence the morphism $[\text{pr}]$ of quotient stacks by Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-functorial}. \medskip\noindent We still have to show that the diagram $2$-commutes. It is clear that the diagram $$ \xymatrix{ U' \ar[d]_{\text{pr}_U} \ar[r]_{f'} & \mathcal{X}' \ar[d]^g \\ U \ar[r]^f & \mathcal{X} } $$ $2$-commutes where $\text{pr}_U : U' \to U$ is the projection. There is a canonical $2$-arrow $\tau : f \circ t \to f \circ s$ in $\Mor(R, \mathcal{X})$ coming from $R = U \times_\mathcal{X} U$, $t = \text{pr}_0$, and $s = \text{pr}_1$. Using the isomorphism $R' \to U' \times_{\mathcal{X}'} U'$ we get similarly an isomorphism $\tau' : f' \circ t' \to f' \circ s'$. Note that $g \circ f' \circ t' = f \circ t \circ \text{pr}_R$ and $g \circ f' \circ s' = f \circ s \circ \text{pr}_R$, where $\text{pr}_R : R' \to R$ is the projection. Thus it makes sense to ask if \begin{equation} \label{equation-verify} \tau \star \text{id}_{\text{pr}_R} = \text{id}_g \star \tau'. \end{equation} Now we make two claims: (1) if Equation (\ref{equation-verify}) holds, then the diagram $2$-commutes, and (2) Equation (\ref{equation-verify}) holds. We omit the proof of both claims. Hints: part (1) follows from the construction of $f = f_{can}$ and $f' = f'_{can}$ in Algebraic Stacks, Lemma \ref{algebraic-lemma-map-space-into-stack}. Part (2) follows by carefuly working through the definitions.
Assume (1). Let $T \to \mathcal{Y}$ be a morphism whose source is a scheme. To prove (2) we have to show that the morphism of algebraic spaces $T \times_\mathcal{Y} \mathcal{X} \to T$ is surjective. By Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-surjective} this means we have to show that $|T \times_\mathcal{Y} \mathcal{X}| \to |T|$ is surjective. Applying Lemma \ref{lemma-points-cartesian} we see that this follows from (1). \medskip\noindent Conversely, assume (2). Let $y : \Spec(K) \to \mathcal{Y}$ be a morphism from the spectrum of a field into $\mathcal{Y}$. By assumption the morphism $\Spec(K) \times_{y, \mathcal{Y}} \mathcal{X} \to \Spec(K)$ of algebraic spaces is surjective. By Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-surjective} this means there exists a field extension $K \subset K'$ and a morphism $\Spec(K') \to \Spec(K) \times_{y, \mathcal{Y}} \mathcal{X}$ such that the left square of the diagram $$ \xymatrix{ \Spec(K') \ar[r] \ar[d] & \Spec(K) \times_{y, \mathcal{Y}} \mathcal{X} \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ \Spec(K) \ar@{=}[r] & \Spec(K) \ar[r]^-y & \mathcal{Y} } $$ is commutative. This shows that $|X| \to |\mathcal{Y}|$ is surjective.
The assumption means that we have a smooth groupoid $(U, R, s, t, c)$ in algebraic spaces, and an equivalence $f : [U/R] \to \mathcal{X}$. We may assume $\mathcal{X} = [U/R]$. The induced morphism $p : U \to \mathcal{X}$ is smooth and surjective, see Algebraic Stacks, Lemma \ref{algebraic-lemma-smooth-quotient-smooth-presentation}. Hence $|U| \to |\mathcal{X}|$ is surjective by Lemma \ref{lemma-characterize-surjective}. Note that $R = U \times_\mathcal{X} U$, see Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-2-cartesian}. Hence Lemma \ref{lemma-points-cartesian} implies the map $$ |R| \longrightarrow |U| \times_{|\mathcal{X}|} |U| $$ is surjective. Hence the image of $|R| \to |U| \times |U|$ is exactly the set of pairs $(u_1, u_2) \in |U| \times |U|$ such that $u_1$ and $u_2$ have the same image in $|\mathcal{X}|$. Combining these two statements we get the result of the lemma.
This follows formally from the fact that there exists a scheme $U$ and a surjective, open, continuous map $U \to |\mathcal{X}|$ of topological spaces. Namely, if $U \to \mathcal{X}$ is surjective and smooth, then Lemma \ref{lemma-topology-points} guarantees that $|U| \to |\mathcal{X}|$ is continuous, surjective, and open.
We will use Lemma \ref{lemma-characterize-surjective}. Suppose $\mathcal{U}$ and $\mathcal{U} \to \mathcal{X}$ are as in (4). Then since $|\mathcal{U}| \to |\mathcal{X}|$ is surjective and continuous we conclude that $|\mathcal{X}|$ is quasi-compact. Thus (4) implies (1). The implications (2) $\Rightarrow$ (3) $\Rightarrow$ (4) are immediate. Assume (1), i.e., $\mathcal{X}$ is quasi-compact, i.e., that $|\mathcal{X}|$ is quasi-compact. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Then since $|U| \to |\mathcal{X}|$ is open we see that there exists a quasi-compact open $U' \subset U$ such that $|U'| \to |X|$ is surjective (and still smooth). Hence (2) holds.