hcju/stacksqa huggingface dataset:
First assume that $X$ is an affine formal algebraic space. Say $X = \colim X_\lambda$ as in Definition \ref{definition-affine-formal-algebraic-space}. Since the transition morphisms are thickenings, the affine schemes $X_\lambda$ all have isomorphic reductions $X_{red}$. The morphism $X_{red} \to X$ is representable and a thickening by Lemma \ref{lemma-covering-by-thickenings} and the fact that compositions of thickenings are thickenings. We omit the verification of the universal property (use Schemes, Definition \ref{schemes-definition-reduced-induced-scheme}, Schemes, Lemma \ref{schemes-lemma-map-into-reduction}, Properties of Spaces, Definition \ref{spaces-properties-definition-reduced-induced-space}, and Properties of Spaces, Lemma \ref{spaces-properties-lemma-map-into-reduction}). \medskip\noindent Let $X$ and $\{X_i \to X\}_{i \in I}$ be as in Definition \ref{definition-formal-algebraic-space}. For each $i$ let $X_{i, red} \to X_i$ be the reduction as constructed above. For $i, j \in I$ the projection $X_{i, red} \times_X X_j \to X_{i, red}$ is an \'etale (by assumption) morphism of schemes (by Lemma \ref{lemma-presentation-representable}). Hence $X_{i, red} \times_X X_j$ is reduced (see Descent, Lemma \ref{descent-lemma-reduced-local-smooth}). Thus the projection $X_{i, red} \times_X X_j \to X_j$ factors through $X_{j, red}$ by the universal property. We conclude that $$ R_{ij} = X_{i, red} \times_X X_j = X_{i, red} \times_X X_{j, red} = X_i \times_X X_{j, red} $$ because the morphisms $X_{i, red} \to X_i$ are injections of sheaves. Set $U = \coprod X_{i, red}$, set $R = \coprod R_{ij}$, and denote $s, t : R \to U$ the two projections. As a sheaf $R = U \times_X U$ and $s$ and $t$ are \'etale. Then $(t, s) : R \to U$ defines an \'etale equivalence relation by our observations above. Thus $X_{red} = U/R$ is an algebraic space by Spaces, Theorem \ref{spaces-theorem-presentation}. By construction the diagram $$ \xymatrix{ \coprod X_{i, red} \ar[r] \ar[d] & \coprod X_i \ar[d] \\ X_{red} \ar[r] & X } $$ is cartesian. Since the right vertical arrow is \'etale surjective and the top horizontal arrow is representable and a thickening we conclude that $X_{red} \to X$ is representable by Bootstrap, Lemma \ref{bootstrap-lemma-after-fppf-sep-lqf} (to verify the assumptions of the lemma use that a surjective \'etale morphism is surjective, flat, and locally of finite presentation and use that thickenings are separated and locally quasi-finite). Then we can use Spaces, Lemma \ref{spaces-lemma-descent-representable-transformations-property} to conclude that $X_{red} \to X$ is a thickening (use that being a thickening is equivalent to being a surjective closed immersion). \medskip\noindent Finally, suppose that $U \to X$ is a morphism with $U$ a reduced algebraic space over $S$. Then each $X_i \times_X U$ is \'etale over $U$ and therefore reduced (by our definition of reduced algebraic spaces in Properties of Spaces, Section \ref{spaces-properties-section-types-properties}). Then $X_i \times_X U \to X_i$ factors through $X_{i, red}$. Hence $U \to X$ factors through $X_{red}$ because $\{X_i \times_X U \to U\}$ is an \'etale covering.
(The \'etale case follows directly from the construction of the underlying reduced algebraic space in the proof of Lemma \ref{lemma-reduction-formal-algebraic-space}.) Assume $f$ is smooth. Observe that $X \times_Y Y_{red} \to Y_{red}$ is a smooth morphism of algebraic spaces. Hence $X \times_Y Y_{red}$ is a reduced algebraic space by Descent on Spaces, Lemma \ref{spaces-descent-lemma-reduced-local-smooth}. Then the univeral property of reduction shows that the canonical morphism $X_{red} \to X \times_Y Y_{red}$ is an isomorphism.
Since we take the colimit in the category of fppf sheaves, we see that $X$ is a sheaf. Choose and fix $\lambda \in \Lambda$. Choose an \'etale covering $\{X_{i, \lambda} \to X_\lambda\}$ where $X_i$ is an affine scheme over $S$, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-cover-by-union-affines}. For each $\mu \geq \lambda$ there exists a cartesian diagram $$ \xymatrix{ X_{i, \lambda} \ar[r] \ar[d] & X_{i, \mu} \ar[d] \\ X_\lambda \ar[r] & X_\mu } $$ with \'etale vertical arrows, see More on Morphisms of Spaces, Theorem \ref{spaces-more-morphisms-theorem-topological-invariance} (this also uses that a thickening is a surjective closed immersion which satisfies the conditions of the theorem). Moreover, these diagrams are unique up to unique isomorphism and hence $X_{i, \mu} = X_\mu \times_{X_{\mu'}} X_{i, \mu'}$ for $\mu' \geq \mu$. The morphisms $X_{i, \mu} \to X_{i, \mu'}$ is a thickening as a base change of a thickening. Each $X_{i, \mu}$ is an affine scheme by Limits of Spaces, Proposition \ref{spaces-limits-proposition-affine} and the fact that $X_{i, \lambda}$ is affine. Set $X_i = \colim_{\mu \geq \lambda} X_{i, \mu}$. Then $X_i$ is an affine formal algebraic space. The morphism $X_i \to X$ is \'etale because given an affine scheme $U$ any $U \to X$ factors through $X_\mu$ for some $\mu \geq \lambda$ (details omitted). In this way we see that $X$ is a formal algebraic space.
Say $X = \Spec(A)$ and $T$ corresponds to the radical ideal $I \subset A$. Let $U = \Spec(B)$ be an affine scheme over $S$ and let $f : U \to X$ be an element of $F(U)$. Then $f$ corresponds to a ring map $\varphi : A \to B$ such that every prime of $B$ contains $\varphi(I) B$. Thus every element of $\varphi(I)$ is nilpotent in $B$, see Algebra, Lemma \ref{algebra-lemma-Zariski-topology}. Setting $J = \Ker(\varphi)$ we conclude that $I/J$ is a locally nilpotent ideal in $A/J$. Equivalently, $V(J) = V(I) = T$. In other words, the functor of the lemma equals $\colim \Spec(A/J)$ where the colimit is over the collection of ideals $J$ with $V(J) = T$. Thus our functor is an affine formal algebraic space. It is McQuillan (Definition \ref{definition-types-affine-formal-algebraic-space}) because the maps $A \to A/J$ are surjective and hence $A^\wedge = \lim A/J \to A/J$ is surjective, see Lemma \ref{lemma-mcquillan-affine-formal-algebraic-space}.
Denote $F$ the functor. Let $\{U_i \to U\}$ be an fppf covering. Then $\coprod |U_i| \to |U|$ is surjective. Since $X$ is an fppf sheaf, it follows that $F$ is an fppf sheaf. \medskip\noindent Let $\{g_i : X_i \to X\}$ be an \'etale covering such that $X_i$ is affine for all $i$, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-cover-by-union-affines}. The morphisms $F \times_X X_i \to F$ are \'etale (see Spaces, Lemma \ref{spaces-lemma-base-change-representable-transformations-property}) and the map $\coprod F \times_X X_i \to F$ is a surjection of sheaves. Thus it suffices to prove that $F \times_X X_i$ is an affine formal algebraic space. A $U$-valued point of $F \times_X X_i$ is a morphism $U \to X_i$ whose image is contained in the closed subset $g_i^{-1}(T) \subset |X_i|$. Thus this follows from Lemma \ref{lemma-completion-affine-is-affine-formal-algebraic-space}.
It follows from Lemma \ref{lemma-reduction-formal-algebraic-space}, Properties of Spaces, Definition \ref{spaces-properties-definition-reduced-induced-space} (which uses Properties of Spaces, Lemma \ref{spaces-properties-lemma-reduced-closed-subspace} to construct $Z$), and the definition of $X_{/T}$ that $Z$ and $(X_{/T})_{red}$ are reduced algebraic spaces characterized the same mapping property: a morphism $g : Y \to X$ whose source is a reduced algebraic space factors through them if and only if $|Y|$ maps into $T \subset |X|$.
By Algebra, Lemma \ref{algebra-lemma-qc-open} if (1) holds, then we can find an ideal $I \subset A$ as in (2). If (3) holds then we can find an ideal $I \subset A$ as in (2). Moreover, completions of Noetherian rings are Noetherian by Algebra, Lemma \ref{algebra-lemma-completion-Noetherian-Noetherian}. All in all we see that it suffices to prove (2). \medskip\noindent Proof of (2). Let $I = (f_1, \ldots, f_r) \subset A$ cut out $T$. If $Z = \Spec(B)$ is an affine scheme and $g : Z \to X$ is a morphism with $g(Z) \subset T$ (set theoretically), then $g^\sharp(f_i)$ is nilpotent in $B$ for each $i$. Thus $I^n$ maps to zero in $B$ for some $n$. Hence we see that $X_{/T} = \colim \Spec(A/I^n) = \text{Spf}(A^\wedge)$.
Choose $\{X_i \to X\}$ and $\{Y_j \to Y\}$ as in Definition \ref{definition-formal-algebraic-space}. Then $\{X_i \times_Z Y_j \to X \times_Z Y\}$ is a family of maps which are representable by algebraic spaces and \'etale. Thus Lemma \ref{lemma-etale-covering-by-formal-algebraic-spaces} tells us it suffices to show that $X \times_Z Y$ is a formal algebraic space when $X$ and $Y$ are affine formal algebraic spaces. \medskip\noindent Assume $X$ and $Y$ are affine formal algebraic spaces. Write $X = \colim X_\lambda$ and $Y = \colim Y_\mu$ as in Definition \ref{definition-affine-formal-algebraic-space}. Then $X \times_Z Y = \colim X_\lambda \times_Z Y_\mu$. Each $X_\lambda \times_Z Y_\mu$ is an algebraic space. For $\lambda \leq \lambda'$ and $\mu \leq \mu'$ the morphism $$ X_\lambda \times_Z Y_\mu \to X_\lambda \times_Z Y_{\mu'} \to X_{\lambda'} \times_Z Y_{\mu'} $$ is a thickening as a composition of base changes of thickenings. Thus we conclude by applying Lemma \ref{lemma-colimit-is-formal}.
Observe that $U \times_X V$ is a scheme by Lemma \ref{lemma-diagonal-formal-algebraic-space}. Let $U_{red}, V_{red}, X_{red}$ be the reduction of $U, V, X$. Then $$ U_{red} \times_{X_{red}} V_{red} = U_{red} \times_X V_{red} \to U \times_X V $$ is a thickening of schemes. From this the equivalence of (1) and (2) is clear, keeping in mind the analogous lemma for algebraic spaces, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-characterize-quasi-separated}. We omit the proof of the equivalence of (2) and (3).
If (2) holds, then $X_{red}$ is a separated algebraic space by applying Properties of Spaces, Lemma \ref{spaces-properties-lemma-characterize-quasi-separated} to morphisms $U \to X_{red}$ and $V \to X_{red}$ with $U, V$ affine and using that $U \times_{X_{red}} V = U \times_X V$. \medskip\noindent Assume (1). Let $U \to X$ and $V \to X$ be as in (2). Observe that $U \times_X V$ is a scheme by Lemma \ref{lemma-diagonal-formal-algebraic-space}. Let $U_{red}, V_{red}, X_{red}$ be the reduction of $U, V, X$. Then $$ U_{red} \times_{X_{red}} V_{red} = U_{red} \times_X V_{red} \to U \times_X V $$ is a thickening of schemes. It follows that $(U \times_X V)_{red} = (U_{red} \times_{X_{red}} V_{red})_{red}$. In particular, we see that $(U \times_X V)_{red}$ is an affine scheme and that $$ \mathcal{O}(U) \otimes_\mathbf{Z} \mathcal{O}(V) \longrightarrow \mathcal{O}((U \times_X V)_{red}) $$ is surjective, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-characterize-quasi-separated}. Then $U \times_X V$ is affine by Limits of Spaces, Proposition \ref{spaces-limits-proposition-affine}. On the other hand, the morphism $U \times_X V \to U \times V$ of affine schemes is the composition $$ U \times_X V = X \times_{(X \times_S X)} (U \times_S V) \to U \times_S V \to U \times V $$ The first morphism is a monomorphism and locally of finite type (Lemma \ref{lemma-diagonal-formal-algebraic-space}). The second morphism is an immersion (Schemes, Lemma \ref{schemes-lemma-fibre-product-after-map}). Hence the composition is a monomorphism which is locally of finite type. On the other hand, the composition is integral as the map on underlying reduced affine schemes is a closed immersion by the above and hence universally closed (use Morphisms, Lemma \ref{morphisms-lemma-integral-universally-closed}). Thus the ring map $$ \mathcal{O}(U) \otimes_\mathbf{Z} \mathcal{O}(V) \longrightarrow \mathcal{O}(U \times_X V) $$ is an epimorphism which is integral of finite type hence finite hence surjective (use Morphisms, Lemma \ref{morphisms-lemma-finite-integral} and Algebra, Lemma \ref{algebra-lemma-finite-epimorphism-surjective}).
Write $X = \colim X_\lambda$ and $Y = \colim Y_\mu$ as in Definition \ref{definition-affine-formal-algebraic-space}. Then $X \times_Z Y = \colim X_\lambda \times_Z Y_\mu$. Since $Z$ is separated the fibre products are affine, hence we see that (1) holds. Assume $X$ and $Y$ corresponds to the weakly admissible topological $S$-algebras $A$ and $B$ and $X_\lambda = \Spec(A/I_\lambda)$ and $Y_\mu = \Spec(B/J_\mu)$. Then $$ X_\lambda \times_Z Y_\mu \to X_\lambda \times Y_\mu \to \Spec(A \otimes B) $$ is a closed immersion. Thus one of the conditions of Lemma \ref{lemma-mcquillan-affine-formal-algebraic-space} holds and we conclude that $X \times_Z Y$ is McQuillan. If also $Z$ is McQuillan corresponding to $C$, then $$ X_\lambda \times_Z Y_\mu = \Spec(A/I_\lambda \otimes_C B/J_\mu) $$ hence we see that the weakly admissible topological ring corresponding to $X \times_Z Y$ is the completed tensor product (see Definition \ref{definition-toplogy-tensor-product}).
The statement makes sense because $U \to X$ is representable by algebraic spaces (Lemma \ref{lemma-space-to-formal-space}). Let $T$ be a scheme and $T \to X$ a morphism. We have to show that $U \times_X T \to T$ is separated. Since $U \times_X T \to U \times_S T$ is a monomorphism, it suffices to show that $U \times_S T \to T$ is separated. As this is the base change of $U \to S$ this follows. We used in the argument above: Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-base-change-separated}, \ref{spaces-morphisms-lemma-composition-separated}, \ref{spaces-morphisms-lemma-monomorphism-separated}, and \ref{spaces-morphisms-lemma-separated-implies-morphism-separated}.
By Lemmas \ref{lemma-characterize-quasi-separated} and \ref{lemma-characterize-quasi-compact} and Definitions \ref{definition-separated} and \ref{definition-quasi-compact} we see that $X$ is quasi-compact and quasi-separated. By Yasuda's lemma (Lemma \ref{lemma-structure-quasi-compact-quasi-separated}) we can write $X = \colim X_\lambda$ as a filtered colimit of thickenings of algebraic spaces. However, each $X_\lambda$ is affine by Limits of Spaces, Lemma \ref{spaces-limits-lemma-reduction-scheme} because $(X_\lambda)_{red} = X_{red}$. Hence $X$ is an affine formal algebraic space by definition.
It is clear that (1) implies (2) because we can take $U = X$ and $V = Y$. Conversely, (2) implies (1) by Bootstrap, Lemma \ref{bootstrap-lemma-representable-by-spaces-cover} applied to $U \to X \to Y$. \medskip\noindent Assume (1) is true and consider a diagram as in (3). Then $U \to Y$ is representable by algebraic spaces (as the composition $U \to X \to Y$, see Bootstrap, Lemma \ref{bootstrap-lemma-composition-transformation}) and factors through $V$. Thus $U \to V$ is representable by algebraic spaces by Lemma \ref{lemma-permanence-representable}. \medskip\noindent It is clear that (3) implies (2). Thus now (1) -- (3) are equivalent. \medskip\noindent Observe that the condition in (4) makes sense as the fibre product $Y_j \times_Y X$ is a formal algebraic space by Lemma \ref{lemma-fibre-products}. It is clear that (4) implies (5). \medskip\noindent Assume $X_i \to Y_i \to Y$ as in (5). Then we set $V = \coprod Y_i$ and $U = \coprod X_i$ to see that (5) implies (2). \medskip\noindent Finally, assume (1) -- (3) are true. Thus we can choose any covering $\{Y_j \to Y\}$ as in Definition \ref{definition-formal-algebraic-space} and for each $j$ any covering $\{X_{ji} \to Y_j \times_Y X\}$ as in Definition \ref{definition-formal-algebraic-space}. Then $X_{ij} \to Y_j$ is representable by algebraic spaces by (3) and we see that (4) is true. This concludes the proof.
Let $\{Y_i \to Y\}$ be as in Definition \ref{definition-formal-algebraic-space}. Then $X \times_Y Y_i \to X$ is a family of morphisms representable by algebraic spaces, \'etale, and jointly surjective. Thus it suffices to show that $X \times_Y Y_i$ is a formal algebraic space, see Lemma \ref{lemma-etale-covering-by-formal-algebraic-spaces}. This follows from Lemma \ref{lemma-algebraic-space-over-affine-formal}.
We will show that $f$ is affine; it will then follow that $f$ is representable and affine by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-affine-local}. Write $Y = \colim Y_\mu$ and $X = \colim X_\lambda$ as in Definition \ref{definition-affine-formal-algebraic-space}. Let $T \to Y$ be a morphism where $T$ is a scheme over $S$. We have to show that $X \times_Y T \to T$ is affine, see Bootstrap, Definition \ref{bootstrap-definition-property-transformation}. To do this we may assume that $T$ is affine and we have to prove that $X \times_Y T$ is affine. In this case $T \to Y$ factors through $Y_\mu \to Y$ for some $\mu$, see Lemma \ref{lemma-factor-through-thickening}. Since $f$ is quasi-compact we see that $X \times_Y T$ is quasi-compact (Lemma \ref{lemma-characterize-quasi-compact-morphism}). Hence $X \times_Y T \to X$ factors through $X_\lambda$ for some $\lambda$. Similarly $X_\lambda \to Y$ factors through $Y_\mu$ after increasing $\mu$. Then $X \times_Y T = X_\lambda \times_{Y_\mu} T$. We conclude as fibre products of affine schemes are affine.
Proof of (1). Write $Y = \colim_{\lambda \in \Lambda} Y_\lambda$ as in Definition \ref{definition-affine-formal-algebraic-space}. Since $f$ is representable and affine, the fibre products $X_\lambda = Y_\lambda \times_Y X$ are affine. And $X = \colim Y_\lambda \times_Y X$. Thus $X$ is an affine formal algebraic space. \medskip\noindent Proof of (2). If $Y$ is countably indexed, then in the argument above we may assume $\Lambda$ is countable. Then we immediately see that $X$ is countably indexed too. \medskip\noindent Proof of (3). Assume $Y$ is adic*. Then $Y = \text{Spf}(B)$ for some adic topological ring $B$ which has a finitely generated ideal $J$ such that $\{J^n\}$ is a fundamental system of open ideals. Of course, then $Y = \colim \Spec(B/J^n)$. The schemes $X \times_Y \Spec(B/J^n)$ are affine and we can write $X \times_Y \Spec(B/J^n) = \Spec(A_n)$. Then $X = \colim \Spec(A_n)$. The $B$-algebra maps $A_{n + 1} \to A_n$ are surjective and induce isomorphisms $A_{n + 1}/J^nA_{n + 1} \to A_n$. By Algebra, Lemma \ref{algebra-lemma-limit-complete} the ring $A = \lim A_n$ is $J$-adically complete and $A/J^n A = A_n$. Hence $X = \text{Spf}(A^\wedge)$ is adic*. \medskip\noindent Proof of (4). Combining (3) with Lemma \ref{lemma-implications-between-types} we see that $X$ is adic*. Thus we can use the criterion of Lemma \ref{lemma-characterize-noetherian-affine}. First, it tells us the affine schemes $Y_\lambda$ are Noetherian. Then $X_\lambda \to Y_\lambda$ is of finite type, hence $X_\lambda$ is Noetherian too (Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}). Then the criterion tells us $X$ is Noetherian and the proof is complete.
Parts (1) and (2) are equivalent by Lemma \ref{lemma-affine-representable-by-algebraic-spaces}. \medskip\noindent Assume the equivalent conditions (1) and (2) hold. If $I \subset A$ is a weak ideal of definition, then $\Spec(A/I) \to \text{Spf}(A)$ is representable and a thickening (this is clear from the construction of the formal spectrum but it also follows from Lemma \ref{lemma-mcquillan-affine-formal-algebraic-space}). Then $\Spec(A/I) \times_{\text{Spf}(A)} \text{Spf}(B) \to \text{Spf}(B)$ is representable and a thickening as a base change. Hence by Lemma \ref{lemma-mcquillan-affine-formal-algebraic-space} there is a weak ideal of definition $J(I) \subset B$ such that $\Spec(A/I) \times_{\text{Spf}(A)} \text{Spf}(B) = \Spec(B/J(I))$ as subfunctors of $\text{Spf}(B)$. We obtain a cartesian diagram $$ \xymatrix{ \Spec(B/J(I)) \ar[d] \ar[r] & \Spec(A/I) \ar[d] \\ \text{Spf}(B) \ar[r] & \text{Spf}(A) } $$ By Lemma \ref{lemma-fibre-product-affines-over-separated} we see that $B/J(I) = B \widehat{\otimes}_A A/I$. It follows that $J(I)$ is the closure of the ideal $\varphi(I)B$, see Lemma \ref{lemma-closure-image-ideal}. Since $\text{Spf}(A) = \colim \Spec(A/I)$ with $I$ as above, we find that $\text{Spf}(B) = \colim \Spec(B/J(I))$. Thus the ideals $J(I)$ form a fundamental system of weak ideals of definition (see Lemma \ref{lemma-mcquillan-affine-formal-algebraic-space}). Hence (3) holds. \medskip\noindent Assume (3) holds. We are essentially just going to reverse the arguments given in the previous paragraph. Let $I \subset A$ be a weak ideal of definition. By Lemma \ref{lemma-fibre-product-affines-over-separated} we get a cartesian diagram $$ \xymatrix{ \text{Spf}(B \widehat{\otimes}_A A/I) \ar[d] \ar[r] & \Spec(A/I) \ar[d] \\ \text{Spf}(B) \ar[r] & \text{Spf}(A) } $$ If $J(I)$ is the closure of $IB$, then $J(I)$ is open in $B$ by tautness of $\varphi$. Hence if $J$ is open in $B$ and $J \subset J(B)$, then $B/J \otimes_A A/I = B/(IB + J) = B/J(I)$ because $J(I) = \bigcap_{J \subset B\text{ open}} (IB + J)$ by Lemma \ref{lemma-closed}. Hence the limit defining the completed tensor product collapses to give $B \widehat{\otimes}_A A/I = B/J(I)$. Thus $\text{Spf}(B \widehat{\otimes}_A A/I) = \Spec(B/J(I))$. This proves that $\text{Spf}(B) \times_{\text{Spf}(A)} \Spec(A/I)$ is representable for every weak ideal of definition $I \subset A$. Since every morphism $T \to \text{Spf}(A)$ with $T$ quasi-compact factors through $\Spec(A/I)$ for some weak ideal of definition $I$ (Lemma \ref{lemma-factor-through-thickening}) we conclude that $\text{Spf}(\varphi)$ is representable, i.e., (2) holds. This finishes the proof.
Let $A$ be an \'etale $B$-algebra. Then $B/J \to A/JA$ is \'etale for every open ideal $J \subset B$. Hence the morphism $\text{Spf}(A^\wedge) \to Y$ is representable and \'etale. The functor $\text{Spf}$ is fully faithful by Lemma \ref{lemma-morphism-between-formal-spectra}. To finish the proof we will show in the next paragraph that any $X \to Y$ as in (1) is in the essential image. \medskip\noindent Choose a weak ideal of definition $J_0 \subset B$. Set $Y_0 = \Spec(B/J_0)$ and $X_0 = Y_0 \times_Y X$. Then $X_0 \to Y_0$ is an \'etale morphism of affine schemes (see Lemma \ref{lemma-affine-representable-by-algebraic-spaces}). Say $X_0 = \Spec(A_0)$. By Algebra, Lemma \ref{algebra-lemma-lift-etale} we can find an \'etale algebra map $B \to A$ such that $A_0 \cong A/J_0A$. Consider an ideal of definition $J \subset J_0$. As above we may write $\Spec(B/J) \times_Y X = \Spec(\bar A)$ for some \'etale ring map $B/J \to \bar A$. Then both $B/J \to \bar A$ and $B/J \to A/JA$ are \'etale ring maps lifting the \'etale ring map $B/J_0 \to A_0$. By More on Algebra, Lemma \ref{more-algebra-lemma-locally-nilpotent-henselian} there is a unique $B/J$-algebra isomorphism $\varphi_J : A/JA \to \bar A$ lifting the identification modulo $J_0$. Since the maps $\varphi_J$ are unique they are compatible for varying $J$. Thus $$ X = \colim \Spec(B/J) \times_Y X = \colim \Spec(A/JA) = \text{Spf}(A) $$ and we see that the lemma holds.
Let $J \subset B$ be a weak ideal of definition. As every element of $J$ is topologically nilpotent, we see that every element of $1 + J$ is a unit. It follows that $J$ is contained in the Jacobson radical of $B$ (Algebra, Lemma \ref{algebra-lemma-contained-in-radical}). Hence a flat ring map $B \to A$ is faithfully flat if and only if $B/J \to A/JA$ is faithfully flat (Algebra, Lemma \ref{algebra-lemma-ff-rings}). In this way we see that (2) -- (4) are equivalent. If (1) holds, then for every weak ideal of definition $J \subset B$ the morphism $\Spec(A/JA) = \Spec(B/J) \times_Y X \to \Spec(B/J)$ is surjective which implies (3). Conversely, assume (3). A morphism $T \to Y$ with $T$ quasi-compact factors through $\Spec(B/J)$ for some ideal of definition $J$ of $B$ (Lemma \ref{lemma-factor-through-thickening}). Hence $X \times_Y T = \Spec(A/JA) \times_{\Spec(B/J)} T \to T$ is surjective as a base change of the surjective morphism $\Spec(A/JA) \to \Spec(B/J)$. Thus (1) holds.