hcju/stacksqa huggingface dataset:
Assume $X$ is countably indexed. We write $X = \colim X_n$ as in Lemma \ref{lemma-countable-affine-formal-algebraic-space}. Write $Y = \colim Y_\lambda$ as in Definition \ref{definition-affine-formal-algebraic-space}. For every $n$ we can pick a $\lambda_n$ such that $X_n \to Y$ factors through $Y_{\lambda_n}$, see Lemma \ref{lemma-factor-through-thickening}. On the other hand, for every $\lambda$ the scheme $Y_\lambda \times_Y X$ is affine (Lemma \ref{lemma-affine-representable-by-algebraic-spaces}) and hence $Y_\lambda \times_Y X \to X$ factors through $X_n$ for some $n$ (Lemma \ref{lemma-factor-through-thickening}). Picture $$ \xymatrix{ Y_\lambda \times_Y X \ar[r] \ar[d] & X_n \ar[r] \ar[d] & X \ar[d] \\ Y_\lambda \ar@{..>}[r] \ar@/_1pc/[rr] & Y_{\lambda_n} \ar[r] & Y } $$ If we can show the dotted arrow exists, then we conclude that $Y = \colim Y_{\lambda_n}$ and $Y$ is countably indexed. To do this we pick a $\mu$ with $\mu \geq \lambda$ and $\mu \geq \lambda_n$. Thus both $Y_\lambda \to Y$ and $Y_{\lambda_n} \to Y$ factor through $Y_\mu \to Y$. Say $Y_\mu = \Spec(B_\mu)$, the closed subscheme $Y_\lambda$ corresponds to $J \subset B_\mu$, and the closed subscheme $Y_{\lambda_n}$ corresponds to $J' \subset B_\mu$. We are trying to show that $J' \subset J$. By the diagram above we know $J'A_\mu \subset JA_\mu$ where $Y_\mu \times_Y X = \Spec(A_\mu)$. Since $X \to Y$ is surjective and flat the morphism $Y_\lambda \times_Y X \to Y_\lambda$ is a faithfully flat morphism of affine schemes, hence $B_\mu \to A_\mu$ is faithfully flat. Thus $J' \subset J$ as desired. \medskip\noindent Assume $Y$ is countably indexed. Then $X$ is countably indexed by Lemma \ref{lemma-property-goes-up-affine}.
Observe that a Noetherian affine formal algebraic space is adic*, see Lemma \ref{lemma-implications-between-types}. Thus by Lemma \ref{lemma-iff-adic-star} we may assume that both $X$ and $Y$ are adic*. We will use the criterion of Lemma \ref{lemma-characterize-noetherian-affine} to see that the lemma holds. Namely, write $Y = \colim Y_n$ as in Lemma \ref{lemma-countable-affine-formal-algebraic-space}. For each $n$ set $X_n = Y_n \times_Y X$. Then $X_n$ is an affine scheme (Lemma \ref{lemma-affine-representable-by-algebraic-spaces}) and $X = \colim X_n$. Each of the morphisms $X_n \to Y_n$ is faithfully flat and of finite type. Thus the lemma follows from the fact that in this situation $X_n$ is Noetherian if and only if $Y_n$ is Noetherian, see Algebra, Lemma \ref{algebra-lemma-descent-Noetherian} (to go down) and Algebra, Lemma \ref{algebra-lemma-Noetherian-permanence} (to go up).
It is clear that (1) implies (2). Assume (2) and let $Y \to X$ be as in (1). Since the fibre products $X_i \times_X Y$ are formal algebraic spaces (Lemma \ref{lemma-fibre-products-general}) we can pick coverings $\{X_{ij} \to X_i \times_X Y\}$ as in Definition \ref{definition-formal-algebraic-space}. Since $Y$ is quasi-compact, there exist $(i_1, j_1), \ldots, (i_n, j_n)$ such that $$ X_{i_1 j_1} \amalg \ldots \amalg X_{i_n j_n} \longrightarrow Y $$ is surjective and \'etale. Then $X_{i_kj_k} \to X_{i_k}$ is representable by algebraic spaces and \'etale hence $X_{i_kj_k}$ has property $P$ by Lemma \ref{lemma-property-goes-up-affine}. Then $X_{i_1 j_1} \amalg \ldots \amalg X_{i_n j_n}$ is an affine formal algebraic space with property $P$ (small detail omitted on finite disjoint unions of affine formal algebraic spaces). Hence we conclude by applying one of Lemmas \ref{lemma-iff-countably-indexed}, \ref{lemma-iff-adic-star}, and \ref{lemma-iff-noetherian}.
Choose a surjective \'etale morphism $U \to X$ with $U = \coprod U_i$ a disjoint union of affine schemes, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-cover-by-union-affines}. Let $T_i \subset U_i$ be the inverse image of $T$. We have $X_{/T} \times_X U_i = (U_i)_{/T_i}$ (Lemma \ref{lemma-map-completions-representable}). Hence $\{(U_i)_{/T_i} \to X_{/T}\}$ is a covering as in Definition \ref{definition-formal-algebraic-space}. Moreover, if $X \setminus T \to X$ is quasi-compact, so is $U_i \setminus T_i \to U_i$ and if $X$ is locally Noetherian, so is $U_i$. Thus the lemma follows from the affine case which is Lemma \ref{lemma-affine-formal-completion-types}.
Choose a covering $\{Y_j \to Y\}$ as in Definition \ref{definition-formal-algebraic-space}. For each $j$ choose a covering $\{X_{ji} \to Y_j \times_Y X\}$ as in Definition \ref{definition-formal-algebraic-space}. For each $j$ choose a covering $\{Z_{jk} \to Y_j \times_Y Z\}$ as in Definition \ref{definition-formal-algebraic-space}. Observe that $X_{ji} \times_{Y_j} Z_{jk}$ is an affine formal algebraic space by Lemma \ref{lemma-fibre-product-affines-over-separated}. Hence $$ \{X_{ji} \times_{Y_j} Z_{jk} \to X \times_Y Z\} $$ is a covering as in Definition \ref{definition-formal-algebraic-space}. Thus it suffices to prove (1) and (2) in case $X$, $Y$, and $Z$ are affine formal algebraic spaces. \medskip\noindent Assume $X$, $Y$, and $Z$ are affine formal algebraic spaces and McQuillan. Then we can write $X = \text{Spf}(A)$, $Y = \text{Spf}(B)$, $Z = \text{Spf}(C)$ for some weakly admissible topological rings $A$, $B$, and $C$ and the morphsms $X \to Y$ and $Z \to Y$ are given by continuous ring maps $B \to A$ and $B \to C$, see Definition \ref{definition-types-affine-formal-algebraic-space} and Lemma \ref{lemma-morphism-between-formal-spectra} By Lemma \ref{lemma-fibre-product-affines-over-separated} we see that $X \times_Y Z = \text{Spf}(A \widehat{\otimes}_B C)$ and that $A \widehat{\otimes}_B C$ is a weakly admissible topological ring. This reduces cases (1) and (2) of our lemma to parts (3) and (4) of Lemma \ref{lemma-completed-tensor-product}. \medskip\noindent To deduce case (3) from Lemma \ref{lemma-completed-tensor-product} part (5) we need to match the hypotheses. First, we observe that taking the reduction corresponds to dividing by the ideal of topologically nilpotent elements (Example \ref{example-reduction-affine-formal-spectrum}). Second, if $X_{red} \to Y_{red}$ is locally of finite type, then $(X_{ji})_{red} \to (Y_j)_{red}$ is locally of finite type (and hence correspond to finite type homomorphisms of rings). This follows Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-finite-type-local} and the fact that in the commutative diagram $$ \xymatrix{ (X_{ji})_{red} \ar[d] \ar[r] & (Y_j)_{red} \ar[d] \\ X_{red} \ar[r] & Y_{red} } $$ the vertical morphisms are \'etale. Namely, we have $(X_{ji})_{red} = X_{ij} \times_X X_{red}$ and $(Y_j)_{red} = Y_j \times_Y Y_{red}$ by Lemma \ref{lemma-reduction-smooth}.
Part (1) is clear from the construction. Assume $A$ has a finitely generated ideal of definition $I \subset A$. Then $I^n(A')^\wedge =  \Ker((A')^\wedge \to A'/I^nA')$ by Algebra, Lemma \ref{algebra-lemma-hathat-finitely-generated}. Thus $I(A')^\wedge$ is a finitely generated ideal of definition and we see that (2) holds. Finally, assume that $A$ is Noetherian and adic. By (2) we know that $(A')^\wedge$ is adic. By Algebra, Lemma \ref{algebra-lemma-completion-Noetherian-Noetherian} we see that $(A')^\wedge$ is Noetherian. Hence (3) holds.
It is clear that (1) implies (2) and that (2) implies (3). Assume $\{X_i \to X\}$ and $X_i \to Y_i \to Y$ as in (3) and let a diagram as in (1) be given. Since $Y_i \times_Y V$ is a formal algebraic space (Lemma \ref{lemma-fibre-products-general}) we may pick coverings $\{Y_{ij} \to Y_i \times_Y V\}$ as in Definition \ref{definition-formal-algebraic-space}. For each $(i, j)$ we may similarly choose coverings $\{X_{ijk} \to Y_{ij} \times_{Y_i} X_i \times_X U\}$ as in Definition \ref{definition-formal-algebraic-space}. Since $U$ is quasi-compact we can choose $(i_1, j_1, k_1), \ldots, (i_n, j_n, k_n)$ such that $$ X_{i_1 j_1 k_1} \amalg \ldots \amalg X_{i_n j_n k_n} \longrightarrow U $$ is surjective. For $s = 1, \ldots, n$ consider the commutative diagram $$ \xymatrix{ & & & X_{i_s j_s k_s} \ar[ld] \ar[d] \ar[rd] \\ X \ar[d] & X_{i_s} \ar[l] \ar[d] & X_{i_s} \times_X U \ar[l] \ar[d] & Y_{i_s j_s} \ar[ld] \ar[rd] & X_{i_s} \times_X U \ar[d] \ar[r] & U \ar[d] \ar[r] & X \ar[d] \\ Y & Y_{i_s} \ar[l] & Y_{i_s} \times_Y V \ar[l] & & Y_{i_s} \times_Y V \ar[r] & V \ar[r] & Y } $$ Let us say that $P$ holds for a morphism of countably indexed affine formal algebraic spaces if it holds for the corresponding morphism of $\textit{WAdm}^{count}$. Observe that the maps $X_{i_s j_s k_s} \to X_{i_s}$, $Y_{i_s j_s} \to Y_{i_s}$ are given by completions of \'etale ring maps, see Lemma \ref{lemma-etale}. Hence we see that $P(X_{i_s} \to Y_{i_s})$ implies $P(X_{i_s j_s k_s} \to Y_{i_s j_s})$ by axiom (\ref{item-axiom-1}). Observe that the maps $Y_{i_s j_s} \to V$ are given by completions of \'etale rings maps (same lemma as before). By axiom (\ref{item-axiom-2}) applied to the diagram $$ \xymatrix{ X_{i_s j_s k_s} \ar@{=}[r] \ar[d] & X_{i_s j_s k_s} \ar[d] \\ Y_{i_s j_s} \ar[r] & V } $$ (this is permissible as identities are faithfully flat ring maps) we conclude that $P(X_{i_s j_s k_s} \to V)$ holds. By axiom (\ref{item-axiom-3}) we find that $P(\coprod_{s = 1, \ldots, n} X_{i_s j_s k_s} \to V)$ holds. Since the morphism $\coprod X_{i_s j_s k_s} \to U$ is surjective by construction, the corresponding morphism of $\textit{WAdm}^{count}$ is the completion of a faithfully flat \'etale ring map, see Lemma \ref{lemma-etale-surjective}. One more application of axiom (\ref{item-axiom-2}) (with $B' = B$) implies that $P(U \to V)$ is true as desired.
Choose a covering $\{Y_j \to Y\}$ as in Definition \ref{definition-formal-algebraic-space}. For each $j$ choose a covering $\{X_{ji} \to Y_j \times_Y X\}$ as in Definition \ref{definition-formal-algebraic-space}. For each $j$ choose a covering $\{Z_{jk} \to Y_j \times_Y Z\}$ as in Definition \ref{definition-formal-algebraic-space}. Observe that $X_{ji} \times_{Y_j} Z_{jk}$ is an affine formal algebraic space which is countably indexed, see Lemma \ref{lemma-types-fibre-products}. Then we see that $$ \{X_{ji} \times_{Y_j} Z_{jk} \to X \times_Y Z\} $$ is a covering as in Definition \ref{definition-formal-algebraic-space}. Moreover, the morphisms $X_{ji} \times_{Y_j} Z_{jk} \to Z$ factor through $Z_{jk}$. By assumption we know that $X_{ji} \to Y_j$ corresponds to a morphism $B_j \to A_{ji}$ of $\text{WAdm}^{count}$ having property $P$. The morphisms $Z_{jk} \to Y_j$ correspond to morphisms $B_j \to C_{jk}$ in $\text{WAdm}^{count}$. Since $X_{ji} \times_{Y_j} Z_{jk} = \text{Spf}(A_{ji} \widehat{\otimes}_{B_j} C_{jk})$ by Lemma \ref{lemma-fibre-product-affines-over-separated} we see that it suffices to show that $C_{jk} \to A_{ji} \widehat{\otimes}_{B_j} C_{jk}$ has property $P$ which is exactly what the condition that $P$ is stable under base change guarantees.
Choose a covering $\{Z_k \to Z\}$ as in Definition \ref{definition-formal-algebraic-space}. For each $k$ choose a covering $\{Y_{kj} \to Z_k \times_Z Y\}$ as in Definition \ref{definition-formal-algebraic-space}. For each $k$ and $j$ choose a covering $\{X_{kji} \to Y_{kj} \times_Y X\}$ as in Definition \ref{definition-formal-algebraic-space}. If $f$ and $g$ satisfies the equivalent conditions of Lemma \ref{lemma-property-defines-property-morphisms} then $X_{kji} \to Y_{jk}$ and $Y_{jk} \to Z_k$ correspond to arrows $B_{kj} \to A_{kji}$ and $C_k \to B_{kj}$ of $\text{WAdm}^{count}$ having property $P$. Hence the compositions do too and we conclude.
Choose a covering $\{Z_k \to Z\}$ as in Definition \ref{definition-formal-algebraic-space}. For each $k$ choose a covering $\{Y_{kj} \to Z_k \times_Z Y\}$ as in Definition \ref{definition-formal-algebraic-space}. For each $k$ and $j$ choose a covering $\{X_{kji} \to Y_{kj} \times_Y X\}$ as in Definition \ref{definition-formal-algebraic-space}. Let $X_{kji} \to Y_{jk}$ and $Y_{jk} \to Z_k$ correspond to arrows $B_{kj} \to A_{kji}$ and $C_k \to B_{kj}$ of $\text{WAdm}^{count}$. If $g \circ f$ and $g$ satisfies the equivalent conditions of Lemma \ref{lemma-property-defines-property-morphisms} then $C_k \to B_{kj}$ and $C_k \to A_{kji}$ satisfy $P$. Hence $B_{kj} \to A_{kji}$ does too and we conclude.
The equivalence of (a) and (b) is immediate. Below we will give an algebraic proof of the axioms, but it turns out we've already proven them. Namely, using Lemma \ref{lemma-representable-affine} the equivalent conditions (a) and (b) translate to saying the corresponding morphism of affine formal algebraic spaces is representable. Since this condition is ``\'etale local on the source and target'' by Lemma \ref{lemma-representable-by-algebraic-spaces-local} we immediately get axioms (\ref{item-axiom-1}), (\ref{item-axiom-2}), and (\ref{item-axiom-3}). \medskip\noindent Direct algebraic proof of (\ref{item-axiom-1}), (\ref{item-axiom-2}), (\ref{item-axiom-3}). Let a diagram (\ref{equation-localize}) as in Situation \ref{situation-local-property} be given. By Example \ref{example-representable-morphism-from-completion} the maps $A \to (A')^\wedge$ and $B \to (B')^\wedge$ satisfy (a) and (b). \medskip\noindent Assume (a) and (b) hold for $\varphi$. Let $J \subset B$ be a weak ideal of definition. Then the closure of $JA$, resp.\ $J(B')^\wedge$ is a weak ideal of definition $I \subset A$, resp.\ $J' \subset (B')^\wedge$. Then the closure of $I(A')^\wedge$ is a weak ideal of definition $I' \subset (A')^\wedge$. A topological argument shows that $I'$ is also the closure of $J(A')^\wedge$ and of $J'(A')^\wedge$. Finally, as $J$ runs over a fundamental system of weak ideals of definition of $B$ so do the ideals $I$ and $I'$ in $A$ and $(A')^\wedge$. It follows that (a) holds for $\varphi'$. This proves (\ref{item-axiom-1}). \medskip\noindent Assume $A \to A'$ is faithfully flat and that (a) and (b) hold for $\varphi'$. Let $J \subset B$ be a weak ideal of definition. Using (a) and (b) for the maps $B \to (B')^\wedge \to (A')^\wedge$ we find that the closure $I'$ of $J(A')^\wedge$ is a weak ideal of definition. In particular, $I'$ is open and hence the inverse image of $I'$ in $A$ is open. Now we have (explanation below) \begin{align*} A \cap I' & = A \cap \bigcap (J(A')^\wedge + \Ker((A')^\wedge \to A'/I_0A')) \\ & = A \cap \bigcap \Ker((A')^\wedge \to A'/JA' + I_0 A') \\ & = \bigcap (JA + I_0) \end{align*} which is the closure of $JA$ by Lemma \ref{lemma-closed}. The intersections are over weak ideals of definition $I_0 \subset A$. The first equality because a fundamental system of neighbourhoods of $0$ in $(A')^\wedge$ are the kernels of the maps $(A')^\wedge \to A'/I_0A'$. The second equality is trivial. The third equality because $A \to A'$ is faithfully flat, see Algebra, Lemma \ref{algebra-lemma-faithfully-flat-universally-injective}. Thus the closure of $JA$ is open. By Lemma \ref{lemma-topologically-nilpotent} the closure of $JA$ is a weak ideal of definition of $A$. Finally, given a weak ideal of definition $I \subset A$ we can find $J$ such that $J(A')^\wedge$ is contained in the closure of $I(A')^\wedge$ by property (a) for $B \to (B')^\wedge$ and $\varphi'$. Thus we see that (a) holds for $\varphi$. This proves (\ref{item-axiom-2}). \medskip\noindent We omit the proof of (\ref{item-axiom-3}).
The property of a map in $\textit{WAdm}^{count}$ being ``taut'' is a local property by Lemma \ref{lemma-representable-property-rings}. Thus Lemma \ref{lemma-property-defines-property-morphisms} exactly tells us that (1), (2), and (3) are equivalent. On the other hand, by Lemma \ref{lemma-representable-affine} being ``taut'' on maps in $\textit{WAdm}^{count}$ corresponds exactly to being ``representable by algebraic spaces'' for the corresponding morphisms of countably indexed affine formal algebraic spaces. Thus the implication (1) $\Rightarrow$ (2) of Lemma \ref{lemma-representable-by-algebraic-spaces-local} shows that (4) implies (1) of the current lemma. Similarly, the implication (4) $\Rightarrow$ (1) of Lemma \ref{lemma-representable-by-algebraic-spaces-local} shows that (2) implies (4) of the current lemma.
Part (1) follows immediately from the definitions, please see Definition \ref{definition-taut} for the definition of a taut ring maps. Conversely, assume $B$ is complete, $I \subset A$ is a finitely generated ideal of definition, and $\varphi$ is taut. Then Lemma \ref{lemma-taut-is-adic} tells us the topology on $B$ is the $I$-adic topology as desired. This proves (2). The final statement is a trivial consequence of (1) and (2).
By Lemma \ref{lemma-permanence-representable} we see that $f$ is representable by algebraic spaces. Let $T$ be a scheme and let $T \to Z$ be a morphism. Then we can apply Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-finite-type} to the morphisms $T \times_Z X \to T \times_Z Y \to T$ of algebraic spaces to conclude.
In each of the 5 cases the morphism $f : X \to Y$ is representable by algebraic spaces, see Lemma \ref{lemma-representable-by-algebraic-spaces-local}. We will use this below without further mention. \medskip\noindent It is clear that (1) implies (2) because we can take $U = X$ and $V = Y$. Conversely, assume given a diagram as in (2). Let $T$ be a scheme and let $T \to Y$ be a morphism. Then we can consider $$ \xymatrix{ U \times_Y T \ar[d] \ar[r] & V \times_Y T \ar[d] \\ X \times_Y T \ar[r] & T } $$ The vertical arrows are \'etale and the top horizontal arrow is locally of finite type as base changes of such morphisms. Hence by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-finite-type-local} we conclude that $X \times_Y T \to T$ is locally of finite type. In other words (1) holds. \medskip\noindent Assume (1) is true and consider a diagram as in (3). Then $U \to Y$ is locally of finite type (as the composition $U \to X \to Y$, see Bootstrap, Lemma \ref{bootstrap-lemma-composition-transformation-property}). Let $T$ be a scheme and let $T \to V$ be a morphism. Then the projection $T \times_V U \to T$ factors as $$ T \times_V U = (T \times_Y U) \times_{(V \times_Y V)} V \to T \times_Y U \to T $$ The second arrow is locally of finite type (as a base change of the composition $U \to X \to Y$) and the first is the base change of the diagonal $V \to V \times_Y V$ which is locally of finite type by Lemma \ref{lemma-diagonal-morphism-formal-algebraic-spaces}. \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 locally of finite type by (3) and we see that (4) is true. This concludes the proof.
Write $X = \text{Spf}(A)$ where $A$ is a weakly admissible topological ring. Let $I_\lambda$ be a fundamental system of weakly admissible ideals of definition in $A$. Then $Y \times_X \Spec(A/I_\lambda)$ is a closed subscheme of $\Spec(A/I_\lambda)$ and hence affine (Definition \ref{definition-closed-immersion}). Say $Y \times_X \Spec(A/I_\lambda) = \Spec(B_\lambda)$. The ring map $A/I_\lambda \to B_\lambda$ is surjective. Hence the projections $$ B = \lim B_\lambda \longrightarrow B_\lambda $$ are surjective as the compositions $A \to B \to B_\lambda$ are surjective. It follows that $Y$ is McQuillan by Lemma \ref{lemma-mcquillan-affine-formal-algebraic-space}. The ring map $A \to B$ is taut by Lemma \ref{lemma-representable-affine}. The kernel is closed because $B$ is complete and $A \to B$ is continuous. Finally, as $A \to B_\lambda$ is surjective for all $\lambda$ we see that the image of $A$ in $B$ is dense.
Assumptions (2) and (3) imply that $X_{red} = X \times_Y Y_{red} = Y_{red}$. We will use this without further mention. \medskip\noindent If $Y' \to Y$ is an \'etale morphism of formal algebraic spaces over $S$, then the base change $f' : X \times_Y Y' \to Y'$ satisfies conditions (1) -- (4). Hence by Lemma \ref{lemma-characterize-closed-immersion} we may assume $Y$ is an affine formal algebraic space. \medskip\noindent Say $Y = \colim_{\lambda \in \Lambda} Y_\lambda$ as in Definition \ref{definition-affine-formal-algebraic-space}. Then $X_\lambda = X \times_Y Y_\lambda$ is an algebraic space endowed with a monomorphism $f_\lambda : X_\lambda \to Y_\lambda$ which induces an isomorphism $X_{\lambda, red} \to Y_{\lambda, red}$. Thus $X_\lambda$ is an affine scheme by Limits of Spaces, Proposition \ref{spaces-limits-proposition-affine} (as $X_{\lambda, red} \to X_\lambda$ is surjective and integral). To finish the proof it suffices to show that $X_\lambda \to Y_\lambda$ is a closed immersion which we will do in the next paragraph. \medskip\noindent Let $X \to Y$ be a monomorphism of affine schemes such that $X_{red} = X \times_Y Y_{red} = Y_{red}$. In general, this does not imply that $X \to Y$ is a closed immersion, see Examples, Section \ref{examples-section-epimorphism-not-surjective}. However, under our assumption (4) we know that in the previous parapgrah either $X_\lambda \to Y_\lambda$ is of finite type or $Y_\lambda$ is Noetherian. This means that $X \to Y$ corresponds to a ring map $R \to A$ such that $R/I \to A/IA$ is an isomorphism where $I \subset R$ is the nil radical (ie., the maximal locally nilpotent ideal of $R$) and either $R \to A$ is of finite type or $R$ is Noetherian. In the first case $R \to A$ is surjective by Algebra, Lemma \ref{algebra-lemma-surjective-mod-locally-nilpotent} and in the second case $I$ is finitely generated, hence nilpotent, hence $R \to A$ is surjective by Nakayama's lemma, see Algebra, Lemma \ref{algebra-lemma-NAK} part (11).
We can use Lemma \ref{lemma-representable-affine} to relate tautness of $\varphi$ to representability of $\text{Spf}(\varphi)$. We will use this without further mention below. It follows that $X = \colim \Spec(A/I)$ and $Y = \colim \Spec(B/J(I))$ where $I \subset A$ runs over the weak ideals of definition of $A$ and $J(I)$ is the closure of $IB$ in $B$. \medskip\noindent Assume (2). Choose a ring map $A[x_1, \ldots, x_r] \to B$ whose image is dense. Then $A[x_1, \ldots, x_r] \to B \to B/J(I)$ has dense image too which means that it is surjective. Therefore $B/J(I)$ is of finite type over $A/I$. Let $T \to X$ be a morphism with $T$ a quasi-compact scheme. Then $T \to X$ factors through $\Spec(A/I)$ for some $I$ (Lemma \ref{lemma-factor-through-thickening}). Then $T \times_X Y = T \times_{\Spec(A/I)} \Spec(B/J(I))$, see proof of Lemma \ref{lemma-representable-affine}. Hence $T \times_Y X \to T$ is of finite type as the base change of the morphism $\Spec(B/J(I)) \to \Spec(A/I)$ which is of finite type. Thus (1) is true. \medskip\noindent Assume (1). Pick any $I \subset A$ as above. Since $\Spec(A/I) \times_X Y = \Spec(B/J(I))$ we see that $A/I \to B/J(I)$ is of finite type. Choose $b_1, \ldots, b_r \in B$ mapping to generators of $B/J(I)$ over $A/I$. We claim that the image of the ring map $A[x_1, \ldots, x_r] \to B$ sending $x_i$ to $b_i$ is dense. To prove this, let $I' \subset I$ be a second weak ideal of definition. Then we have $$ B/(J(I') + IB) = B/J(I) $$ because $J(I)$ is the closure of $IB$ and because $J(I')$ is open. Hence we may apply Algebra, Lemma \ref{algebra-lemma-surjective-mod-locally-nilpotent} to see that $(A/I')[x_1, \ldots, x_r] \to B/J(I')$ is surjective. Thus (2) is true, concluding the proof.