hcju/stacksqa huggingface dataset:
(The parenthetical remark in the statement holds by Lemma \ref{lemma-diagonal}.) Observe that $Z$ exists by assumption (1) and Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-diagonal}. By assumption (2) we see that $Z \to V \times_S U$ is locally of finite type. Choose a scheme $W$, a closed point $w_0 \in W$, and an \'etale morphism $W \to Z$ mapping $w_0$ to $z_0$, see Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-finite-type-point}. Then $W$ is locally of finite type over $S$ and $w_0$ is a finite type point of $W$. Let $l = \kappa(z_0)$. Denote $z_{l, 0}$, $v_{l, 0}$, $u_{l, 0}$, and $x_{l, 0}$ the objects of $\mathcal{Z}$, $(\Sch/V)_{fppf}$, $(\Sch/U)_{fppf}$, and $\mathcal{X}$ over $\Spec(l)$ obtained by pullback to $\Spec(l) = w_0$. Consider $$ \xymatrix{ \mathcal{F}_{(\Sch/W)_{fppf}, l, w_0} \ar[r] & \mathcal{F}_{\mathcal{Z}, l, z_{l, 0}} \ar[d] \ar[r] & \mathcal{F}_{(\Sch/U)_{fppf}, l, u_{l, 0}} \ar[d] \\ & \mathcal{F}_{(\Sch/V)_{fppf}, l, v_{l, 0}} \ar[r] & \mathcal{F}_{\mathcal{X}, l, x_{l, 0}} } $$ By Lemma \ref{lemma-fibre-product-deformation-categories} the square is a fibre product of predeformation categories. By Lemma \ref{lemma-versal-change-of-field} we see that the right vertical arrow is smooth. By Formal Deformation Theory, Lemma \ref{formal-defos-lemma-smooth-properties} the left vertical arrow is smooth. By Lemma \ref{lemma-formally-smooth-on-deformation-categories} we see that the left horizontal arrow is smooth. We conclude that the map $$ \mathcal{F}_{(\Sch/W)_{fppf}, l, w_0} \to \mathcal{F}_{(\Sch/V)_{fppf}, l, v_{l, 0}} $$ is smooth by Formal Deformation Theory, Lemma \ref{formal-defos-lemma-smooth-properties}. Thus we conclude that $W \to V$ is smooth at $w_0$ by More on Morphisms, Lemma \ref{more-morphisms-lemma-lifting-along-artinian-at-point}. This exactly means that $Z \to V$ is smooth at $z_0$ and the proof is complete.
Choose an object $x_R$ of $\mathcal{X}$ lying over $\Spec(R)$ whose associated formal object is $\xi$. Let $N = 2$ and apply Lemma \ref{lemma-approximate}. We obtain $A, \mathfrak m_A, x_A, \ldots$. Let $\eta = (A^\wedge, \eta_n, g_n)$ be the formal object associated to $x_A|_{\Spec(A^\wedge)}$. We have a diagram $$ \vcenter{ \xymatrix{ & \eta \ar[d] \\ \xi \ar[r] \ar@{..>}[ru] & \xi_2 = \eta_2 } } \quad\text{lying over}\quad \vcenter{ \xymatrix{ & A^\wedge \ar[d] \\ R \ar[r] \ar@{..>}[ru] & R/\mathfrak m_R^2 = A/\mathfrak m_A^2 } } $$ The versality of $\xi$ means exactly that we can find the dotted arrows in the diagrams, because we can successively find morphisms $\xi \to \eta_3$, $\xi \to \eta_4$, and so on by Formal Deformation Theory, Remark \ref{formal-defos-remark-versal-object}. The corresponding ring map $R \to A^\wedge$ is surjective by Formal Deformation Theory, Lemma \ref{formal-defos-lemma-surjective-cotangent-space}. On the other hand, we have $\dim_k \mathfrak m_R^n/\mathfrak m_R^{n + 1} = \dim_k \mathfrak m_A^n/\mathfrak m_A^{n + 1}$ for all $n$ by construction. Hence $R/\mathfrak m_R^n$ and $A/\mathfrak m_A^n$ have the same (finite) length as $\Lambda$-modules by additivity of length and Formal Deformation Theory, Lemma \ref{formal-defos-lemma-length}. It follows that $R/\mathfrak m_R^n \to A/\mathfrak m_A^n$ is an isomorphism for all $n$, hence $R \to A^\wedge$ is an isomorphism. Thus $\eta$ is isomorphic to a versal object, hence versal itself. By Lemma \ref{lemma-versality-matches} we conclude that $x_A$ is versal at the point $u_0$ of $U = \Spec(A)$ corresponding to $\mathfrak m_A$.
Let $\Spec(l) \to U$ be a morphism with $l$ of finite type over $S$. Then the image $u_0 \in U$ is a finite type point of $U$ and $\kappa(u_0) \subset l$ is a finite extension, see discussion in Morphisms, Section \ref{morphisms-section-points-finite-type}. Hence we see that $\mathcal{F}_{(\Sch/U)_{fppf}, l, u_{l, 0}} \to \mathcal{F}_{\mathcal{X}, l, x_{l, 0}}$ is smooth by Lemma \ref{lemma-versal-change-of-field}.
The key ingredient of the proof is More on Morphisms, Lemma \ref{more-morphisms-lemma-lifting-along-artinian-at-point} which (almost) says that a morphism of schemes of finite type over $S$ satisfying (\ref{equation-smooth}) is a smooth morphism. The other arguments of the proof are essentially bookkeeping. \medskip\noindent Let $V$ be a scheme over $S$ and let $y$ be an object of $\mathcal{Y}$ over $V$. Let $Z$ be an algebraic space representing the $2$-fibre product $\mathcal{Z} = \mathcal{X} \times_{f, \mathcal{X}, y} (\Sch/V)_{fppf}$. We have to show that the projection morphism $Z \to V$ is smooth, see Algebraic Stacks, Definition \ref{algebraic-definition-relative-representable-property}. In fact, it suffices to do this when $V$ is an affine scheme locally of finite presentation over $S$, see Criteria for Representability, Lemma \ref{criteria-lemma-check-property-limit-preserving}. Then $(\Sch/V)_{fppf}$ is limit preserving by Lemma \ref{lemma-limit-preserving-algebraic-space}. Hence $Z \to S$ is locally of finite presentation by Lemmas \ref{lemma-fibre-product-limit-preserving} and \ref{lemma-limit-preserving-algebraic-space}. Choose a scheme $W$ and a surjective \'etale morphism $W \to Z$. Then $W$ is locally of finite presentation over $S$. \medskip\noindent Since $f$ satisfies (\ref{equation-smooth}) we see that so does $\mathcal{Z} \to (\Sch/V)_{fppf}$, see Lemma \ref{lemma-base-change-smooth}. Next, we see that $(\Sch/W)_{fppf} \to \mathcal{Z}$ satisfies (\ref{equation-smooth}) by Lemma \ref{lemma-smooth-smooth}. Thus the composition $$ (\Sch/W)_{fppf} \to \mathcal{Z} \to (\Sch/V)_{fppf} $$ satisfies (\ref{equation-smooth}) by Lemma \ref{lemma-composition-smooth}. More on Morphisms, Lemma \ref{more-morphisms-lemma-lifting-along-artinian-at-point} shows that the composition $W \to Z \to V$ is smooth at every finite type point $w_0$ of $W$. Since the smooth locus is open we conclude that $W \to V$ is a smooth morphism of schemes by Morphisms, Lemma \ref{morphisms-lemma-enough-finite-type-points}. Thus we conclude that $Z \to V$ is a smooth morphism of algebraic spaces by definition.
By axiom [2], Lemma \ref{lemma-deformation-category}, and Remark \ref{remark-deformation-category-implies} we see that $\mathcal{F}_{\mathcal{X}, k, x_0}$ satisfies (S1) and (S2). Since also the tangent space has finite dimension by axiom [3] we deduce from Formal Deformation Theory, Lemma \ref{formal-defos-lemma-versal-object-existence} that $\mathcal{F}_{\mathcal{X}, k, x_0}$ has a versal formal object $\xi$. Assumption (3) says $\xi$ is effective. By axiom [1] and Lemma \ref{lemma-approximate-versal} there exists a morphism of finite type $U \to S$, an object $x$ of $\mathcal{X}$ over $U$, and a finite type point $u_0$ of $U$ with residue field $k$ such that $x$ is versal at $u_0$ and such that $x|_{\Spec(k)} \cong x_0$. By openness of versality we may shrink $U$ and assume that $x$ is versal at every finite type point of $U$. We claim that $$ x : (\Sch/U)_{fppf} \longrightarrow \mathcal{X} $$ is smooth which proves the lemma. Namely, by Lemma \ref{lemma-versal-smooth} $x$ satisfies (\ref{equation-smooth}) whereupon Lemma \ref{lemma-implies-smooth} finishes the proof.
By Lemma \ref{lemma-finite-dimension} the functor $G$ satisfies [3]. As $F \to G$ is injective, we conclude that $F$ also satisfies [3]. Moreover, as $F \to G$ is injective, we see that given schemes $U$, $V$ and morphisms $U \to F$ and $V \to F$, then $U \times_F V = U \times_G V$. Hence $\Delta : F \to F \times F$ is representable (by schemes) as this holds for $G$ by assumption. Thus Proposition \ref{proposition-spaces-diagonal-representable} applies\footnote{The set theoretic condition [-1] holds for $F$ as it holds for $G$. Details omitted.}.
Lemma \ref{lemma-get-smooth} applies to $\mathcal{X}$. Using this we choose, for every finite type field $k$ over $S$ and every isomorphism class of object $x_0 \in \Ob(\mathcal{X}_{\Spec(k)})$, an affine scheme $U_{k, x_0}$ of finite type over $S$ and a smooth morphism $(\Sch/U_{k, x_0})_{fppf} \to \mathcal{X}$ such that there exists a finite type point $u_{k, x_0} \in U_{k, x_0}$ with residue field $k$ such that $x_0$ is the image of $u_{k, x_0}$. Then $$ (\Sch/U)_{fppf} \to \mathcal{X}, \quad\text{with}\quad U = \coprod\nolimits_{k, x_0} U_{k, x_0} $$ is smooth\footnote{Set theoretical remark: This coproduct is (isomorphic to) an object of $(\Sch/S)_{fppf}$ as we have a bound on the index set by axiom [-1], see Sets, Lemma \ref{sets-lemma-what-is-in-it}.}. To finish the proof it suffices to show this map is surjective, see Criteria for Representability, Lemma \ref{criteria-lemma-stacks-etale} (this is where we use axiom [0]). By Criteria for Representability, Lemma \ref{criteria-lemma-check-property-limit-preserving} it suffices to show that $(\Sch/U)_{fppf} \times_\mathcal{X} (\Sch/V)_{fppf} \to (\Sch/V)_{fppf}$ is surjective for those $y : (\Sch/V)_{fppf} \to \mathcal{X}$ where $V$ is an affine scheme locally of finite presentation over $S$. By assumption (1) the fibre product $(\Sch/U)_{fppf} \times_\mathcal{X} (\Sch/V)_{fppf}$ is representable by an algebraic space $W$. Then $W \to V$ is smooth, hence the image is open. Hence it suffices to show that the image of $W \to V$ contains all finite type points of $V$, see Morphisms, Lemma \ref{morphisms-lemma-enough-finite-type-points}. Let $v_0 \in V$ be a finite type point. Then $k = \kappa(v_0)$ is a finite type field over $S$. Denote $x_0 = y|_{\Spec(k)}$ the pullback of $y$ by $v_0$. Then $(u_{k, x_0}, v_0)$ will give a morphism $\Spec(k) \to W$ whose composition with $W \to V$ is $v_0$ and we win.
Let $R = \mathcal{O}_{U, u}^\wedge$. Let $k = \kappa(u)$ be the residue field of $R$. Let $\xi$ be the formal object of $\mathcal{X}$ over $R$ associated to $x$. Since $x$ is not versal at $u$, we see that $\xi$ is not versal, see Lemma \ref{lemma-versality-matches}. By the discussion following Definition \ref{definition-versal-formal-object} this means we can find morphisms $\xi_1 \to x_A \to x_B$ of $\mathcal{X}$ lying over closed immersions $\Spec(k) \to \Spec(A) \to \Spec(B)$ where $A, B$ are Artinian local rings with residue field $k$, an $n \geq 1$ and a commutative diagram $$ \vcenter{ \xymatrix{ & x_A \ar[ld] \\ \xi_n & \xi_1 \ar[u] \ar[l] } } \quad\text{lying over}\quad \vcenter{ \xymatrix{ & \Spec(A) \ar[ld] \\ \Spec(R/\mathfrak m^n) & \Spec(k) \ar[u] \ar[l] } } $$ such that there does {\bf not} exist an $m \geq n$ and a commutative diagram $$ \vcenter{ \xymatrix{ & & x_B \ar[lldd] \\ & & x_A \ar[ld] \ar[u] \\ \xi_m & \xi_n \ar[l] & \xi_1 \ar[u] \ar[l] } } \quad\text{lying over} \vcenter{ \xymatrix{ & & \Spec(B) \ar[lldd] \\ & & \Spec(A) \ar[ld] \ar[u] \\ \Spec(R/\mathfrak m^m) & \Spec(R/\mathfrak m^n) \ar[l] & \Spec(k) \ar[u] \ar[l] } } $$ We may moreover assume that $B \to A$ is a small extension, i.e., that the kernel $I$ of the surjection $B \to A$ is isomorphic to $k$ as an $A$-module. This follows from Formal Deformation Theory, Remark \ref{formal-defos-remark-versal-object}. Then we simply define $$ T = U \amalg_{\Spec(A)} \Spec(B) $$ By property (RS*) we find $y$ over $T$ whose restriction to $\Spec(B)$ is $x_B$ and whose restriction to $U$ is $x$ (this gives the arrow $x \to y$ lying over $U \to T$). To finish the proof we verify conditions (1), (2), and (3). \medskip\noindent By the construction of the pushout we have a commutative diagram $$ \xymatrix{ 0 \ar[r] & I \ar[r] & B \ar[r] & A \ar[r] & 0 \\ 0 \ar[r] & I \ar[r] \ar[u] & \Gamma(T, \mathcal{O}_T) \ar[r] \ar[u] & \Gamma(U, \mathcal{O}_U) \ar[r] \ar[u] & 0 } $$ with exact rows. This immediately proves (1) and (2). To finish the proof we will argue by contradiction. Assume we have a pair $(W, \beta)$ as in (3). Since $\Spec(B) \to T$ factors through $W$ we get the morphism $$ x_B \to y|_W \xrightarrow{\beta} x $$ Since $B$ is Artinian local with residue field $k = \kappa(u)$ we see that $x_B \to x$ lies over a morphism $\Spec(B) \to U$ which factors through $\Spec(\mathcal{O}_{U, u}/\mathfrak m_u^m)$ for some $m \geq n$. In other words, $x_B \to x$ factors through $\xi_m$ giving a map $x_B \to \xi_m$. The compatibility condition on the morphism $\alpha$ in condition (3) translates into the condition that $$ \xymatrix{ x_B \ar[d] & x_A \ar[d] \ar[l] \\ \xi_m & \xi_n \ar[l] } $$ is commutative. This gives the contradiction we were looking for.
After shrinking $U$ we may assume $U$ is affine and $U$ maps into an affine open $\Spec(\Lambda)$ of $S$. If $x$ is not versal at $u$ then we may pick $x \to y$ lying over $U \to T$ as in Lemma \ref{lemma-single-point}. Write $U = \Spec(R_0)$ and $T = \Spec(R)$. The morphism $U \to T$ corresponds to a surjective ring map $R \to R_0$ whose kernel is an ideal of square zero. By assumption (3) we get that $y$ comes from an object $x'$ over $U' = \Spec(R')$ for some finite type $\Lambda$-subalgebra $R' \subset R$. After increasing $R'$ we may and do assume that $R' \to R_0$ is surjective, so that $U \subset U'$ is a first order thickening. Thus we now have $$ x \to y \to x' \text{ lying over } U \to T \to U' $$ By assumption (1) there is an algebraic space $Z$ over $S$ representing $$ (\Sch/U)_{fppf} \times_{x, \mathcal{X}, x'} (\Sch/U')_{fppf} $$ see Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-diagonal}. By construction of $2$-fibre products, a $V$-valued point of $Z$ corresponds to a triple $(a, a', \alpha)$ consisting of morphisms $a : V \to U$, $a' : V \to U'$ and a morphism $\alpha : a^*x \to (a')^*x'$. We obtain a commutative diagram $$ \xymatrix{ U \ar[rd] \ar[rdd] \ar[rrd] \\ & Z \ar[r]_{p'} \ar[d]^p & U' \ar[d] \\ & U \ar[r] & S } $$ The morphism $i : U \to Z$ comes the isomorphism $x \to x'|_U$. Let $z_0 = i(u_0) \in Z$. By Lemma \ref{lemma-base-change-versal} we see that $Z \to U'$ is smooth at $z_0$. After replacing $U$ by an affine open neighbourhood of $u_0$, replacing $U'$ by the corresponding open, and replacing $Z$ by the intersection of the inverse images of these opens by $p$ and $p'$, we reach the situation where $Z \to U'$ is smooth along $i(U)$. Since $u \leadsto u_0$ the point $u$ is in this open. Condition (3) of Lemma \ref{lemma-single-point} is clearly preserved by shrinking $U$ (all of the schemes $U$, $T$, $U'$ have the same underlying topological space). Since $U \to U'$ is a first order thickening of affine schemes, we can choose a morphism $i' : U' \to Z$ such that $p' \circ i' = \text{id}_{U'}$ and whose restriction to $U$ is $i$ (More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-smooth-formally-smooth}). Pulling back the universal morphism $p^*x \to (p')^*x'$ by $i'$ we obtain a morphism $$ x' \to x $$ lying over $p \circ i' : U' \to U$ such that the composition $$ x \to x' \to x $$ is the identity. Recall that we have $y \to x'$ lying over the morphism $T \to U'$. Composing we get a morphism $y \to x$ whose existence contradicts condition (3) of Lemma \ref{lemma-single-point}. This contradiction finishes the proof.
Since there are no specializations among the points $u_n$ (and in particular the $u_n$ are pairwise distinct), for every $n$ we can find an open $U' \subset U$ such that $u_n \in U'$ and $u_i \not \in U'$ for $i = 1, \ldots, n - 1$. By Lemma \ref{lemma-single-point} for each $n \geq 1$ we can find $$ x \to y_n \quad\text{in }\mathcal{X}\text{ lying over}\quad U \to T_n $$ such that \begin{enumerate} \item the morphism $U \to T_n$ is a first order thickening, \item we have a short exact sequence $$ 0 \to \kappa(u_n) \to \mathcal{O}_{T_n} \to \mathcal{O}_U \to 0 $$ \item there does {\bf not} exist a pair $(W, \alpha)$ consisting of an open neighbourhood $W \subset T_n$ of $u_n$ and a morphism $\beta : y_n|_W \to x$ such that the composition $$ x|_{U \cap W} \xrightarrow{\text{restriction of }x \to y_n} y_n|_W \xrightarrow{\beta} x $$ is the canonical morphism $x|_{U \cap W} \to x$. \end{enumerate} Thus we can define inductively $$ U_1 = T_1, \quad U_{n + 1} = U_n \amalg_U T_{n + 1} $$ Setting $x_1 = y_1$ and using (RS*) we find inductively $x_{n + 1}$ over $U_{n + 1}$ restricting to $x_n$ over $U_n$ and $y_{n + 1}$ over $T_{n + 1}$. Property (1) for $U \to U_n$ follows from the construction of the pushout in More on Morphisms, Lemma \ref{more-morphisms-lemma-pushout-along-thickening}. Property (2) for $U_n$ similarly follows from property (2) for $T_n$ by the construction of the pushout. After shrinking to an open neighbourhood $U'$ of $u_n$ as discussed above, property (3) for $(U_n, x_n)$ follows from property (3) for $(T_n, y_n)$ simply because the corresponding open subschemes of $T_n$ and $U_n$ are isomorphic. Some details omitted.
We define $\text{Inf}_x$ as the functor $$ \text{Mod}_A \longrightarrow \textit{Sets},\quad M \longrightarrow \text{Inf}(x'_M/x) = \text{Lift}(\text{id}_x, A[M]) $$ mapping $M$ to the group of infinitesimal automorphisms of the trivial deformation $x'_M$ of $x$ to $\Spec(A[M])$ or equivalently the group of lifts of $\text{id}_x$ in $\mathit{Aut}_\mathcal{X}(x'_M)$. We define $T_x$ as the functor $$ \text{Mod}_A \longrightarrow \textit{Sets},\quad M \longrightarrow \text{Lift}(x, A[M]) $$ of isomorphism classes of infintesimal deformations of $x$ to $\Spec(A[M])$. We apply Formal Deformation Theory, Lemma \ref{formal-defos-lemma-linear-functor} to $\text{Inf}_x$ and $T_x$. This lemma is applicable, since (RS*) tells us that $$ \textit{Lift}(x, A[M \times N]) = \textit{Lift}(x, A[M]) \times \textit{Lift}(x, A[N]) $$ as categories (and trivial deformations match up too). \medskip\noindent Let $(x, A' \to A)$ be a deformation situation. Consider the ring map $g : A' \times_A A' \to A[I]$ defined by the rule $g(a_1, a_2) = \overline{a_1} \oplus a_2 - a_1$. There is an isomorphism $$ A' \times_A A' \longrightarrow A' \times_A A[I] $$ given by $(a_1, a_2) \mapsto (a_1, g(a_1, a_2))$. This isomorphism commutes with the projections to $A'$ on the first factor, and hence with the projections to $A$. Thus applying (RS*) twice we find equivalences of categories \begin{align*} \textit{Lift}(x, A') \times \textit{Lift}(x, A') & = \textit{Lift}(x, A' \times_A A') \\ & = \textit{Lift}(x, A' \times_A A[I]) \\ & = \textit{Lift}(x, A') \times \textit{Lift}(x, A[I]) \end{align*} Using these maps and projection onto the last factor of the last product we see that we obtain ``difference maps'' $$ \text{Inf}(x'/x) \times  \text{Inf}(x'/x) \longrightarrow \text{Inf}_x(I) \quad\text{and}\quad \text{Lift}(x, A') \times \text{Lift}(x, A') \longrightarrow T_x(I) $$ These difference maps satisfy the transitivity rule ``$(x'_1 - x'_2) + (x'_2 - x'_3) = x'_1 - x'_3$'' because $$ \xymatrix{ A' \times_A A' \times_A A' \ar[rrrrr]_-{(a_1, a_2, a_3) \mapsto (g(a_1, a_2), g(a_2, a_3))} \ar[rrrrrd]_{(a_1, a_2, a_3) \mapsto g(a_1, a_3)} & & & & & A[I] \times_A A[I] = A[I \times I] \ar[d]^{+} \\ & & & & & A[I] } $$ is commutative. Inverting the string of equivalences above we obtain an action which is free and transitive provided $\text{Inf}(x'/x)$, resp.\ $\text{Lift}(x, A')$ is nonempty. Note that $\text{Inf}(x'/x)$ is always nonempty as it is a group.
By Lemma \ref{lemma-fibre-product-RS-star} we see that $\mathcal{W}$ satisfies (RS*) and hence $T_w(M)$ and $\text{Inf}_w(M)$ are defined. The horizontal arrows are defined using the functoriality of Lemma \ref{lemma-functoriality}. \medskip\noindent Definition of the ``boundary'' map $\delta : \text{Inf}_y(M) \to T_w(M)$. Choose isomorphisms $p(x) \to y$ and $y \to q(z)$ such that $w = (x, z, p(x) \to y \to q(z))$ in the description of the $2$-fibre product of Categories, Lemma \ref{categories-lemma-2-product-fibred-categories} and more precisely Categories, Lemma \ref{categories-lemma-2-product-categories-over-C}. Let $x', y', z', w'$ denote the trivial deformation of $x, y, z, w$ over $A[M]$. By pullback we get isomorphisms $y' \to p(x')$ and $q(z') \to y'$. An element $\alpha \in \text{Inf}_y(M)$ is the same thing as an automorphism $\alpha : y' \to y'$ over $A[M]$ which restricts to the identity on $y$ over $A$. Thus setting $$ \delta(\alpha) = (x', z', p(x') \to y' \xrightarrow{\alpha} y' \to q(z')) $$ we obtain an object of $T_w(M)$. This is a map of $A$-modules by Formal Deformation Theory, Lemma \ref{formal-defos-lemma-morphism-linear-functors}. \medskip\noindent The rest of the proof is exactly the same as the proof of Formal Deformation Theory, Lemma \ref{formal-defos-lemma-deformation-categories-fiber-product-morphisms}.
We prove this by verifying condition (4) of Lemma \ref{lemma-SGE-implies-openness-versality}. Let $(\xi_n)$ and $(R_n)$ be as in Remark \ref{remark-strong-effectiveness} such that $\Ker(R_m \to R_n)$ is an ideal of square zero for all $m \geq n$. Set $A = R_1$ and $x = \xi_1$. Denote $M_n = \Ker(R_n \to R_1)$. Then $M_n$ is an $A$-module. Set $R = \lim R_n$. Let $$ \tilde R = \{(r_1, r_2, r_3 \ldots) \in \prod R_n \text{ such that all have the same image in }A\} $$ Then $\tilde R \to A$ is surjective with kernel $M = \prod M_n$. There is a map $R \to \tilde R$ and a map $\tilde R \to A[M]$, $(r_1, r_2, r_3, \ldots) \mapsto (r_1, r_2 - r_1, r_3 - r_2, \ldots)$. Together these give a short exact sequence $$ (x, R \to A) \to (x, \tilde R \to A) \to (x, A[M]) $$ of deformation situations, see Remark \ref{remark-short-exact-sequence-thickenings}. The associated sequence of kernels $0 \to \lim M_n \to M \to M \to 0$ is the canonical sequence computing the limit of the system of modules $(M_n)$. \medskip\noindent Let $o_x(\tilde R) \in \mathcal{O}_x(M)$ be the obstruction element. Since we have the lifts $\xi_n$ we see that $o_x(\tilde R)$ maps to zero in $\mathcal{O}_x(M_n)$. By assumption (5)(b) we see that $o_x(\tilde R) = 0$. Choose a lift $\tilde \xi$ of $x$ to $\Spec(\tilde R)$. Let $\tilde \xi_n$ be the restriction of $\tilde \xi$ to $\Spec(R_n)$. There exists elements $t_n \in T_x(M_n)$ such that $t_n \cdot \tilde \xi_n = \xi_n$ by Lemma \ref{lemma-properties-lift-RS-star} part (2)(b). By assumption (5)(a) we can find $t \in T_x(M)$ mapping to $t_n$ in $T_x(M_n)$. After replacing $\tilde \xi$ by $t \cdot \tilde \xi$ we find that $\tilde \xi$ restricts to $\xi_n$ over $\Spec(R_n)$ for all $n$. In particular, since $\xi_{n + 1}$ restricts to $\xi_n$ over $\Spec(R_n)$, the restriction $\overline{\xi}$ of $\tilde \xi$ to $\Spec(A[M])$ has the property that it restricts to the trivial deformation over $\Spec(A[M_n])$ for all $n$. Hence by assumption (5)(a) we find that $\overline{\xi}$ is the trivial deformation of $x$. By axiom (RS*) applied to $R = \tilde R \times_{A[M]} A$ this implies that $\tilde \xi$ is the pullback of a deformation $\xi$ of $x$ over $R$. This finishes the proof.