hcju/stacksqa huggingface dataset:
Topologies, Lemma \ref{topologies-lemma-sheaf-property-fpqc} reduces us to the case of a Zariski covering and a covering $\{\Spec(S) \to \Spec(R)\}$ with $R \to S$ faithfully flat. We observed in the proof of Lemma \ref{lemma-fully-faithful}  that $h_\mathfrak X$ satisfies the sheaf condition for Zariski coverings. \medskip\noindent Suppose that $R \to S$ is a faithfully flat ring map. Denote $\pi : \Spec(S) \to \Spec(R)$ the corresponding morphism of schemes. It is surjective and flat. Let $f : \Spec(S) \to \mathfrak X$ be a morphism such that $f \circ \text{pr}_1 = f \circ \text{pr}_2$ as maps $\Spec(S \otimes_R S) \to \mathfrak X$. By Descent, Lemma \ref{descent-lemma-equiv-fibre-product} we see that as a map on the underlying sets $f$ is of the form $f = g \circ \pi$ for some (set theoretic) map $g : \Spec(R) \to \mathfrak X$. By Morphisms, Lemma \ref{morphisms-lemma-fpqc-quotient-topology} and the fact that $f$ is continuous we see that $g$ is continuous. \medskip\noindent Pick $y \in \Spec(R)$. Choose $\mathfrak U \subset \mathfrak X$ an affine formal open subscheme containing $g(y)$. Say $\mathfrak U = \text{Spf}(A)$ for some admissible topological ring $A$. By the above we may choose an $r \in R$ such that $y \in D(r) \subset g^{-1}(\mathfrak U)$. The restriction of $f$ to $\pi^{-1}(D(r))$ into $\mathfrak U$ corresponds to a continuous ring map $A \to S_r$ by (\ref{equation-morphisms-affine-formal-schemes}). The two induced ring maps $A \to S_r \otimes_{R_r} S_r = (S \otimes_R S)_r$ are equal by assumption on $f$. Note that $R_r \to S_r$ is faithfully flat. By Descent, Lemma \ref{descent-lemma-ff-exact} the equalizer of the two arrows $S_r \to S_r \otimes_{R_r} S_r$ is $R_r$. We conclude that $A \to S_r$ factors uniquely through a map $A \to R_r$ which is also continuous as it has the same (open) kernel as the map $A \to S_r$. This map in turn gives a morphism $D(r) \to \mathfrak U$ by (\ref{equation-morphisms-affine-formal-schemes}). \medskip\noindent What have we proved so far? We have shown that for any $y \in \Spec(R)$ there exists a standard affine open $y \in D(r) \subset \Spec(R)$ such that the morphism $f|_{\pi^{-1}(D(r))} : \pi^{-1}(D(r)) \to \mathfrak X$ factors uniquely though some morphism $D(r) \to \mathfrak X$. We omit the verification that these morphisms glue to the desired morphism $\Spec(R) \to \mathfrak X$.
Let $M_n$, $n \in \mathbf{N}$ be a fundamental system of open submodules of $M$. We may assume $M_{n + 1} \subset M_n$ for all $n$. The $(M_n + N)/N$ is a fundamental system in $M/N$. Hence we have to show that $M/N = \lim M/(M_n + N)$. Consider the short exact sequences $$ 0 \to N/N \cap M_n \to M/M_n \to M/(M_n + N) \to 0 $$ Since the transition maps of the system $\{N/N\cap M_n\}$ are surjective we see that $M = \lim M/M_n$ (by completeness of $M$) surjects onto $\lim M/(M_n + N)$ by Algebra, Lemma \ref{algebra-lemma-ML-exact-sequence}. As $N$ is closed we see that the kernel of $M \to \lim M/(M_n + N)$ is $N$ (see Lemma \ref{lemma-closed}). Finally, $M \to M/N$ is open by definition of the quotient topology.
We have $0 \to N^\wedge \to M^\wedge \to (M/N)^\wedge$ is exact and statement (2) by Lemma \ref{lemma-closure}. This produces a canonical map $c : M^\wedge/N^\wedge \to (M/N)^\wedge$. The module $M^\wedge/N^\wedge$ is complete and $M^\wedge \to M^\wedge/N^\wedge$ is open by Lemma \ref{lemma-quotient-by-closed}. By the universal property of completion we obtain a canonical map $b : (M/N)^\wedge \to M^\wedge/N^\wedge$. Then $b$ and $c$ are mutually inverse as they are on a dense subset.
Let $M' \subset M$ be an open submodule whose topology is $I$-adic. Then $\{I^nM'\}_{n \geq 1}$ is a fundamental system of open submodules of $M$. Thus $M^\wedge = \lim M/I^nM'$ contains $(M')^\wedge = \lim M'/I^nM'$ as an open submodule and the topology on $(M')^\wedge$ is $I$-adic by Algebra, Lemma \ref{algebra-lemma-hathat-finitely-generated}. Since $I$ is finitely generated, $I^n$ is finitely generated, say by $f_1, \ldots, f_r$. Observe that the surjection $(f_1, \ldots, f_r) : M^{\oplus r} \to I^n M$ is continuous and open by our description of the topology on $M$ above. By Lemma \ref{lemma-ses} applied to this surjection and to the short exact sequence $0 \to I^nM \to M \to M/I^nM \to 0$ we conclude that $$ (f_1, \ldots, f_r) : (M^\wedge)^{\oplus r} \longrightarrow M^\wedge $$ surjects onto the kernel of the surjection $M^\wedge \to M/I^nM$. Since $f_1, \ldots, f_r$ generate $I^n$ we conclude.
Let $A \to A'$ be an \'etale ring map and let $\sigma : A' \to A/I$ be an $A$-algebra map. By More on Algebra, Lemma \ref{more-algebra-lemma-characterize-henselian-pair} it suffices to lift $\sigma$ to an $A$-algebra map $A' \to A$. To do this, as $A$ is complete, it suffices to find, for every open ideal $J \subset I$, a unique $A$-algebra map $A' \to A/J$ lifting $\sigma$. Since $I$ is a weak ideal of definition, the ideal $I/J$ is locally nilpotent. We conclude by More on Algebra, Lemma \ref{more-algebra-lemma-locally-nilpotent-henselian}.
Let $\mathfrak b \subset B$ be the set of topologically nilpotent elements. We omit the proof of the fact that $\mathfrak b$ is a radical ideal (good exercise in the definitions). Let $g$ be an element of the closure of $\mathfrak b$. Our goal is to show that $g$ is topologically nilpotent. Let $J \subset B$ be an open ideal. We have to show $g^e \in J$ for some $e \geq 1$. We have $g \in \mathfrak b + J$ by Lemma \ref{lemma-closed}. Hence $g = f + h$ for some $f \in \mathfrak b$ and $h \in J$. Pick $m \geq 1$ such that $f^m \in J$. Then $g^{m + 1} \in J$ as desired. \medskip\noindent Let $\varphi : A \to B$ be as in the statement of the lemma. Assertion (1) is clear and assertion (2) follows from this and the fact that $\mathfrak b$ is a closed ideal.
Let $J_\mu$ be a fundamental system of open ideals of $B$. The closure of $IB$ is $\bigcap (IB + J_\lambda)$ by Lemma \ref{lemma-closed}. Let $I_\mu$ be a fundamental system of open ideals in $A$. Then $$ B \widehat{\otimes}_A A/I = \lim (B/J_\lambda \otimes_A A/(I_\mu + I)) = \lim B/(J_\lambda + I_\mu B + I B) $$ Since $A \to B$ is continuous, for every $\lambda$ there is a $\mu$ such that $I_\mu B \subset J_\lambda$, see discussion in Example \ref{example-what-does-it-mean}. Hence the limit can be written as $\lim B/(J_\lambda + IB)$ and the result is clear.
Let $J_n$ be the closure of $\varphi(I^n)B$ in $B$. Since $B$ is complete we have $B = \lim B/J_n$. Let $B' = \lim B/I^nB$ be the $I$-adic completion of $B$. By Algebra, Lemma \ref{algebra-lemma-hathat-finitely-generated}, the $I$-adic topology on $B'$ is complete and $B'/I^nB' = B/I^nB$. Thus the ring map $B' \to B$ is continuous and has dense image as $B' \to B/I^nB \to B/J_n$ is surjective for all $n$. Finally, the map $B' \to B$ is taut because $(I^nB')B = I^nB$ and $A \to B$ is taut. By Lemma \ref{lemma-dense-image-surjective} we see that $B' \to B$ is open and surjective which implies the lemma.
Let $I_\lambda \subset A$, $\lambda \in \Lambda$ and $J_\mu \subset C$, $\mu \in M$ be fundamental systems of open ideals, then by definition $$ A \widehat{\otimes}_B C = \lim_{\lambda, \mu} A/I_\lambda \otimes_B C/J_\mu $$ with the limit topology. Thus a fundamental system of open ideals is given by the kernels $K_{\lambda, \mu}$ of the maps $A \widehat{\otimes}_B C \to A/I_\lambda \otimes_B C/J_\mu$. Note that $K_{\lambda, \mu}$ is the closure of the ideal $I_\lambda(A \widehat{\otimes}_B C) + J_\mu(A \widehat{\otimes}_B C)$. Finally, we have a ring homomorphism $\tau : A \otimes_B C \to A \widehat{\otimes}_B C$ with dense image. \medskip\noindent Proof of (1). If $I_\lambda$ and $J_\mu$ consist of topologically nilpotent elements, then so does $K_{\lambda, \mu}$ by Lemma \ref{lemma-topologically-nilpotent}. Hence  $A \widehat{\otimes}_B C$ is weakly admissible by definition. \medskip\noindent Proof of (2). Assume for some $\lambda_0$ and $\mu_0$ the ideals $I = I_{\lambda_0} \subset A$ and $J_{\mu_0} \subset C$ are ideals of definition. Thus for every $\lambda$ there exists an $n$ such that $I^n \subset I_\lambda$. For every $\mu$ there exists an $m$ such that $J^m \subset J_\mu$. Then $$ \left(I(A \widehat{\otimes}_B C) + J(A \widehat{\otimes}_B C)\right)^{n + m} \subset I_\lambda(A \widehat{\otimes}_B C) + J_\mu(A \widehat{\otimes}_B C) $$ It follows that the open ideal $K = K_{\lambda_0, \mu_0}$ satisfies $K^{n + m} \subset K_{\lambda, \mu}$. Hence $K$ is an ideal of definition of $A \widehat{\otimes}_B C$ and $A \widehat{\otimes}_B C$ is admissible by definition. \medskip\noindent Proof of (3). If $\Lambda$ and $M$ are countable, so is $\Lambda \times M$. \medskip\noindent Proof of (4). Assume $\Lambda = \mathbf{N}$ and $M = \mathbf{N}$ and we have finitely generated ideals $I \subset A$ and $J \subset C$ such that $I_n = I^n$ and $J_n = J^n$. Then $$ I(A \widehat{\otimes}_B C) + J(A \widehat{\otimes}_B C) $$ is a finitely generated ideal and it is easily seen that $A \widehat{\otimes}_B C$ is the completion of $A \otimes_B C$ with respect to this ideal. Hence (4) follows from Algebra, Lemma \ref{algebra-lemma-hathat-finitely-generated}. \medskip\noindent Proof of (5). Let $\mathfrak c \subset C$ be the ideal of topologically nilpotent elements. Since $A$ and $C$ are adic Noetherian, we see that $\mathfrak a$ and $\mathfrak c$ are ideals of definition (details omitted). From part (4) we already know that $A \widehat{\otimes}_B C$ is adic and that $\mathfrak a(A \widehat{\otimes}_B C) + \mathfrak c(A \widehat{\otimes}_B C)$ is a finitely generated ideal of definition. Since $$ A \widehat{\otimes}_B C / \left(\mathfrak a(A \widehat{\otimes}_B C) + \mathfrak c(A \widehat{\otimes}_B C)\right) = A/\mathfrak a \otimes_{B/\mathfrak b} C/\mathfrak c $$ is Noetherian as a finite type algebra over the Noetherian ring $C/\mathfrak c$ we conclude by Algebra, Lemma \ref{algebra-lemma-completion-Noetherian}.
Suppose given $U \to X$ and $V \to X$ where $U, V$ are schemes over $S$. Let us show that $U \times_X V$ is representable. Write $X = \colim X_\lambda$ as in Definition \ref{definition-affine-formal-algebraic-space}. The discussion above shows that Zariski locally on $U$ and $V$ the morphisms factors through some $X_\lambda$. In this case $U \times_X V = U \times_{X_\lambda} V$ which is a scheme. Thus the diagonal is representable, see Spaces, Lemma \ref{spaces-lemma-representable-diagonal}. Given $(a, b) : W \to X \times_S X$ where $W$ is a scheme over $S$ consider the map $X \times_{\Delta, X \times_S X, (a, b)} W \to W$. As before locally on $W$ the morphisms $a$ and $b$ map into the affine scheme $X_\lambda$ for some $\lambda$ and then we get the morphism $X_\lambda \times_{\Delta_\lambda, X_\lambda \times_S X_\lambda, (a, b)} W \to W$. This is the base change of $\Delta_\lambda : X_\lambda \to X_\lambda \times_S X_\lambda$ which is a closed immersion as $X_\lambda \to S$ is separated (because $X_\lambda$ is affine). Thus $X \to X \times_S X$ is a closed immersion.
The statement makes sense by the discussion in Spaces, Section \ref{spaces-section-representable} and \ref{spaces-section-representable-properties}. By Lemma \ref{lemma-diagonal-affine-formal-algebraic-space} the morphisms $X_\lambda \to X$ are representable. Given $U \to X$ where $U$ is a scheme, then the discussion following Definition \ref{definition-affine-formal-algebraic-space} shows that Zariski locally on $U$ the morphism factors through some $X_\mu$ with $\lambda \leq \mu$. In this case $U \times_X X_\lambda = U \times_{X_\mu} X_\lambda$ so that $U \times_X X_\lambda \to U$ is a base change of the thickening $X_\lambda \to X_\mu$.
It follows from Lemmas \ref{lemma-covering-by-thickenings} and \ref{lemma-factor-through-thickening} that an affine formal algebraic space satisfies (1) and (2). In order to prove the converse we may assume $X$ is not empty. Let $\Lambda$ be the category of representable morphisms $T \to X$ which are thickenings where $T$ is an affine scheme over $S$. This category is directed. Since $X$ is not empty, $\Lambda$ contains at least one object. If $T \to X$ and $T' \to X$ are in $\Lambda$, then we can factor $T \amalg T' \to X$ through $T'' \to X$ in $\Lambda$. Between any two objects of $\Lambda$ there is a unique arrow or none. Thus $\Lambda$ is a directed set and by assumption $X = \colim_{T \to X\text{ in }\Lambda} T$. To finish the proof we need to show that any arrow $T \to T'$ in $\Lambda$ is a thickening. This is true because $T' \to X$ is a monomorphism of sheaves, so that $T = T \times_{T'} T' = T \times_X T'$ and hence the morphism $T \to T'$ equals the projection $T \times_X T' \to T'$ which is a thickening because $T \to X$ is a thickening.
It is clear that (5) implies (4). \medskip\noindent Assume (4) for $X = \colim X_\lambda$ as in Definition \ref{definition-affine-formal-algebraic-space}. Set $A = \lim \Gamma(X_\lambda, \mathcal{O}_{X_\lambda})$. Let $T \to X$ be a closed immersion with $T$ a scheme (note that $T \to X$ is representable by Lemma \ref{lemma-diagonal-affine-formal-algebraic-space}). Since $X_\lambda \to X$ is a thickening, so is $X_\lambda \times_X T \to T$. On the other hand, $X_\lambda \times_X T \to X_\lambda$ is a closed immersion, hence $X_\lambda \times_X T$ is affine. Hence $T$ is affine by Limits, Proposition \ref{limits-proposition-affine}. Then $T \to X$ factors through $X_\lambda$ for some $\lambda$ by Lemma \ref{lemma-factor-through-thickening}. Thus $A \to \Gamma(X_\lambda, \mathcal{O}) \to \Gamma(T, \mathcal{O})$ is surjective. In this way we see that (3) holds. \medskip\noindent It is clear that (3) implies (2). \medskip\noindent Assume (2) for $A$ and $X \to \Spec(A)$. Write $X = \colim X_\lambda$ as in Definition \ref{definition-affine-formal-algebraic-space}. Then $A_\lambda = \Gamma(X_\lambda, \mathcal{O})$ is a quotient of $A$ by assumption (2). Hence $A^\wedge = \lim A_\lambda$ is a complete topological ring, see discussion in More on Algebra, Section \ref{more-algebra-section-topological-ring}. The maps $A^\wedge \to A_\lambda$ are surjective as $A \to A_\lambda$ is. We claim that for any $\lambda$ the kernel $I_\lambda \subset A^\wedge$ of $A^\wedge \to A_\lambda$ is a weak ideal of definition. Namely, it is open by definition of the limit topology. If $f \in I_\lambda$, then for any $\mu \in \Lambda$ the image of $f$ in $A_\mu$ is zero in all the residue fields of the points of $X_\mu$. Hence it is a nilpotent element of $A_\mu$. Hence some power $f^n \in I_\mu$. Thus $f^n \to 0$ as $n \to 0$. Thus $A^\wedge$ is weakly admissible. Finally, suppose that $I \subset A^\wedge$ is a weak ideal of definition. Then $I \subset A^\wedge$ is open and hence there exists some $\lambda$ such that $I \supset I_\lambda$. Thus we obtain a morphism $\Spec(A^\wedge/I) \to \Spec(A_\lambda) \to X$. Then it follows that $X = \colim \Spec(A^\wedge/I)$ where now the colimit is over all weak ideals of definition. Thus (1) holds. \medskip\noindent Assume (1). In this case it is clear that $X$ is an affine formal algebraic space. Let $X = \colim X_\lambda$ be any presentation as in Definition \ref{definition-affine-formal-algebraic-space}. For each $\lambda$ we can find a weak ideal of definition $I \subset A$ such that $X_\lambda \to X$ factors through $\Spec(A/I) \to X$, see Lemma \ref{lemma-factor-through-thickening}. Then $X_\lambda = \Spec(A/I_\lambda)$ with $I \subset I_\lambda$. Conversely, for any weak ideal of definition $I \subset A$ the morphism $\Spec(A/I) \to X$ factors through $X_\lambda$ for some $\lambda$, i.e., $I_\lambda \subset I$. It follows that each $I_\lambda$ is a weak ideal of definition and that they form a cofinal subset of the set of weak ideals of definition. Hence $A = \lim A/I = \lim A/I_\lambda$ and we see that (5) is true and moreover that $A = \lim \Gamma(X_\lambda, \mathcal{O}_{X_\lambda})$.
Let $f : \text{Spf}(B) \to \text{Spf}(A)$ be as in the lemma. Let $J \subset B$ be a weak ideal of definition. By Lemma \ref{lemma-factor-through-thickening} there exists a weak ideal of definition $I \subset A$ such that $\Spec(B/J) \to \text{Spf}(B) \to \text{Spf}(A)$ factors through $\Spec(A/I)$. By Schemes, Lemma \ref{schemes-lemma-morphism-into-affine} we obtain an $S$-algebra map $A/I \to B/J$. These maps are compatible for varying $J$ and define the map $f^\sharp : A \to B$. This map is continuous because for every weak ideal of definition $J \subset B$ there is a weak ideal of definition $I \subset A$ such that $f^\sharp(I) \subset J$. The equality $f = \text{Spf}(f^\sharp)$ holds by our choice of the ring maps $A/I \to B/J$ which make up $f^\sharp$.
Let $T$ be a scheme over $S$ and $T \to Y$ a map. We have to show that the algebraic space $X \times_Y T$ is a scheme. Write $X = \colim X_\lambda$ as in Definition \ref{definition-affine-formal-algebraic-space}. Let $W \subset X \times_Y T$ be a quasi-compact open subspace. The restriction of the projection $X \times_Y T \to X$ to $W$ factors through $X_\lambda$ for some $\lambda$. Then $$ W \to X_\lambda \times_S T $$ is a monomorphism (hence separated) and locally quasi-finite (because $W \to X \times_Y T \to T$ is locally quasi-finite by our assumption on $X \to Y$, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-quasi-finite}). Hence $W$ is a scheme by Morphisms of Spaces, Proposition \ref{spaces-morphisms-proposition-locally-quasi-finite-separated-over-scheme}. Thus $X \times_Y T$ is a scheme by Properties of Spaces, Lemma \ref{spaces-properties-lemma-subscheme}.
Assume (1). By Lemma \ref{lemma-implications-between-types} we can write $X = \text{Spf}(A)$ where $A$ is a weakly admissible topological $S$-algebra. For any presentation $X = \colim X_n$ as in Lemma \ref{lemma-countable-affine-formal-algebraic-space} part (1) we see that $A = \lim A_n$ with $X_n = \Spec(A_n)$ and $A_n = A/I_n$ for some weak ideal of definition $I_n \subset A$. This follows from the final statement of Lemma \ref{lemma-mcquillan-affine-formal-algebraic-space} which moreover implies that $\{I_n\}$ is a fundamental system of open neighbourhoods of $0$. Thus we have a sequence $$ A \supset I_1 \supset I_2 \supset I_3 \supset \ldots $$ of weak ideals of definition with $A = \lim A/I_n$. In this way we see that condition (1) implies each of the conditions (2) -- (5). \medskip\noindent Assume (5). First note that the limit topology on $A = \lim B/J_n$ is a linearly topologized, complete topology, see More on Algebra, Section \ref{more-algebra-section-topological-ring}. If $f \in A$ maps to zero in $B/J_1$, then some power maps to zero in $B/J_2$ as its image in $J_1/J_2$ is nilpotent, then a further power maps to zero in $J_2/J_3$, etc, etc. In this way we see the open ideal $\Ker(A \to B/J_1)$ is a weak ideal of definition. Thus $A$ is weakly admissible. In this way we see that (5) implies (2). \medskip\noindent It is clear that (4) is a special case of (5) by taking $B = A$. It is clear that (3) is a special case of (2). \medskip\noindent Assume $A$ is as in (2). Let $E_n$ be a countable fundamental system of neighbourhoods of $0$ in $A$. Since $A$ is a weakly admissible topological ring we can find open ideals $I_n \subset E_n$. We can also choose a weak ideal of definition $J \subset A$. Then $J \cap I_n$ is a fundamental system of weak ideals of definition of $A$ and we get $X = \text{Spf}(A) = \colim \Spec(A/(J \cap I_n))$ which shows that $X$ is a countably indexed affine formal algebraic space.
Suppose given $U \to X$ and $V \to X$ with $U, V$ schemes over $S$. Then $U \times_X V$ is a sheaf. Choose $\{X_i \to X\}$ as in Definition \ref{definition-formal-algebraic-space}. For every $i$ the morphism $$ (U \times_X X_i) \times_{X_i} (V \times_X X_i) = (U \times_X V) \times_X X_i \to U \times_X V $$ is representable and \'etale as a base change of $X_i \to X$ and its source is a scheme (use Lemmas \ref{lemma-diagonal-affine-formal-algebraic-space} and \ref{lemma-presentation-representable}). These maps are jointly surjective hence $U \times_X V$ is an algebraic space by Bootstrap, Theorem \ref{bootstrap-theorem-final-bootstrap}. The morphism $U \times_X V \to U \times_S V$ is a monomorphism. It is also locally quasi-finite, because on precomposing with the morphism displayed above we obtain the composition $$ (U \times_X X_i) \times_{X_i} (V \times_X X_i) \to (U \times_X X_i) \times_S (V \times_X X_i) \to U \times_S V $$ which is locally quasi-finite as a composition of a closed immersion (Lemma \ref{lemma-diagonal-affine-formal-algebraic-space}) and an \'etale morphism, see Descent on Spaces, Lemma \ref{spaces-descent-lemma-locally-quasi-finite-etale-local-source}. Hence we conclude that $U \times_X V$ is a scheme by Morphisms of Spaces, Proposition \ref{spaces-morphisms-proposition-locally-quasi-finite-separated-over-scheme}. Thus $\Delta$ is representable, see Spaces, Lemma \ref{spaces-lemma-representable-diagonal}. \medskip\noindent In fact, since we've shown above that the morphisms of schemes $U \times_X V \to U \times_S V$ are aways monomorphisms and locally quasi-finite we conclude that $\Delta : X \to X \times_S X$ is a monomorphism and locally quasi-finite, see Spaces, Lemma \ref{spaces-lemma-transformation-diagonal-properties}. Then we can use the principle of Spaces, Lemma \ref{spaces-lemma-representable-transformations-property-implication} to see that $\Delta$ is separated and locally of finite type. Namely, a monomorphism of schemes is separated (Schemes, Lemma \ref{schemes-lemma-monomorphism-separated}) and a locally quasi-finite morphism of schemes is locally of finite type (follows from the definition in Morphisms, Section \ref{morphisms-section-quasi-finite}).
Let $Z \to Y$ be a morphism where $Z$ is a scheme over $S$. We have to show that $X \times_Y Z$ is an algebraic space. Choose a scheme $U$ and a surjective \'etale morphism $U \to X$. Then $U \times_Y Z \to X \times_Y Z$ is representable surjective \'etale (Spaces, Lemma \ref{spaces-lemma-base-change-representable-transformations-property}) and $U \times_Y Z$ is a scheme by Lemma \ref{lemma-diagonal-formal-algebraic-space}. Hence the result by Bootstrap, Theorem \ref{bootstrap-theorem-final-bootstrap}.