hcju/stacksqa huggingface dataset:
Assume (1). By Lemma \ref{lemma-RS-fiber-square} we see that every object of $\mathcal{F}(A_1 \times_A A_2)$ is of the form $x_1 \times_x x_2$. Moreover $$ \Mor_{A_1 \times_A A_2}(x_1 \times_x x_2, y_1 \times_y y_2) = \Mor_{A_1}(x_1, y_1) \times_{\Mor_A(x, y)} \Mor_{A_2}(x_2, y_2). $$ Hence we see that $\mathcal{F}(A_1 \times_A A_2)$ is a $2$-fibre product of $\mathcal{F}(A_1)$ with $\mathcal{F}(A_2)$ over $\mathcal{F}(A)$ by Categories, Remark \ref{categories-remark-other-description-2-fibre-product}. In other words, we see that (2) holds. \medskip\noindent The implication (2) $\Rightarrow$ (3) is immediate. \medskip\noindent Assume (3). Let $q_1 : A_1 \to A$ and $q_2 : A_2 \to A$ be given with $q_2$ a small extension. We will use the description of the $2$-fibre product $\mathcal{F}(A_1) \times_{\mathcal{F}(A)} \mathcal{F}(A_2)$ from Categories, Remark \ref{categories-remark-other-description-2-fibre-product}. Hence let $y \in \mathcal{F}(A_1 \times_A A_2)$ correspond to $(x_1, x_2, x, a_1 : x_1 \to x, a_2 : x_2 \to x)$. Let $z$ be an object of $\mathcal{F}$ lying over $C$. Then \begin{align*} \Mor_\mathcal{F}(z, y) & = \{(f, \alpha) \mid f : C \to A_1 \times_A A_2, \alpha : f_*z \to y\} \\ & = \{(f_1, f_2, \alpha_1, \alpha_2) \mid f_i : C \to A_i, \ \alpha_i : f_{i, *}z \to x_i, \\ & \quad\quad\quad\quad q_1 \circ f_1 = q_2 \circ f_2, \ q_{1, *} \alpha_1 = q_{2, *}\alpha_2\} \\ & = \Mor_\mathcal{F}(z, x_1) \times_{\Mor_\mathcal{F}(z, x)} \Mor_\mathcal{F}(z, x_2) \end{align*} whence $y$ is a fibre product of $x_1$ and $x_2$ over $x$. Thus we see that $\mathcal{F}$ satisfies (RS) in case $A_2 \to A$ is a small extension. Hence (RS) holds by Lemma \ref{lemma-RS-small-extension}.
Using the reformulation of Lemma \ref{lemma-RS-2-categorical} and the explanation of (S1) following Definition \ref{definition-S1-S2} it is immediate that (RS) implies (S1). This proves the first part of (S2). The second part of (S2) follows because Lemma \ref{lemma-RS-fiber-square} tells us that $y = x_1 \times_{d, x_0, e} x_2 = y'$ if $y, y'$ are as in the second part of the definition of (S2) in Definition \ref{definition-S1-S2}. (In fact the morphism $y \to y'$ is compatible with both $a, a'$ and $c, c'$!)
We prove that (1) is equivalent to (2) and (2) is equivalent to (3).  The equivalence of (3) and (4) follows from Lemma \ref{lemma-factor-small-extension}. \medskip\noindent Let $f_1: A_1 \to A$ and $f_2: A_2 \to A$ be ring maps in $\mathcal{C}_\Lambda$ with $f_2$ surjective. By Remark \ref{remark-compare-schlessinger-H4} we see $\overline{\mathcal{F}}$ satisfies (RS) if and only if the map $$ \overline{\mathcal{F}}(A_1 \times_A A_2) \to \overline{\mathcal F}(A_1) \times_{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A_2) $$ is bijective for any such $f_1, f_2$. This map is at least surjective since that is the condition of (S1) and $\overline{\mathcal{F}}$ satisfies (S1) by Lemmas \ref{lemma-RS-implies-S1-S2} and \ref{lemma-S1-S2-associated-functor}. Moreover, this map factors as $$ \overline{\mathcal{F}}(A_1 \times_A A_2) \longrightarrow \overline{\mathcal{F}(A_1) \times_{\mathcal{F}(A)} \mathcal{F}(A_2)} \longrightarrow \overline{\mathcal{F}}(A_1) \times_{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A_2), $$ where the first map is a bijection since $$ \mathcal{F}(A_1 \times_A A_2) \longrightarrow \mathcal{F}(A_1) \times_{\mathcal{F}(A)} \mathcal{F}(A_2) $$ is an equivalence by (RS) for $\mathcal{F}$. Hence (1) is equivalent to (2). \medskip\noindent Assume (2) holds. Let $x' \to x$ be a morphism in $\mathcal{F}$ lying over a surjective ring map $f: A' \to A$. Let $a \in \text{Aut}_A(x)$. The objects $$ (x', x', a : x \to x), \ (x', x', \text{id} : x \to x) $$ of $\mathcal{F}(A') \times_{\mathcal{F}(A)} \mathcal{F}(A')$ have the same image in $\overline{\mathcal{F}}(A') \times_{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A')$. By (2) there exists maps $b_1, b_2 : x' \to x'$ such that $$ \xymatrix{ x \ar[r]_a \ar[d]_{f_*b_1} & x \ar[d]^{f_*b_2} \\ x \ar[r]^{\text{id}} & x } $$ commutes. Hence $b_2^{-1} \circ b_1 \in \text{Aut}_{A'}(x')$ has image $a \in \text{Aut}_A(x)$. Hence (3) holds. \medskip\noindent Assume (3) holds. Suppose $$ (x_1, x_2, a : (f_1)_*x_1 \to (f_2)_*x_2), \ (x'_1, x'_2, a' : (f_1)_*x'_1 \to (f_2)_*x'_2) $$ are objects of $\mathcal{F}(A_1) \times_{\mathcal{F}(A)} \mathcal{F}(A_2)$ with the same image in $\overline{\mathcal{F}}(A_1) \times_{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A_2)$. Then there are morphisms $b_1: x_1 \to x'_1$ in $\mathcal{F}(A_1)$ and $b_2: x_2 \to x'_2$ in $\mathcal F(A_2)$. By (3) we can modify $b_2$ by an automorphism of $x_2$ over $A_2$ so that the diagram $$ \xymatrix{ (f_1)_*x_1 \ar[r]_a \ar[d]_{(f_1)_*b_1} & (f_2)_*x_2 \ar[d]^{(f_2)_*b_2} \\ (f_1)_*x'_1 \ar[r]^{a'} & (f_2)_*x'_2. } $$ commutes. This proves $(x_1, x_2, a) \cong (x'_1, x'_2, a')$ in $\overline{\mathcal{F}(A_1) \times_{\mathcal{F}(A)} \mathcal{F}(A_2)}$. Hence (2) holds.
If $A$ is an object of $\mathcal{C}_\Lambda$, then an object of the fiber category of $\mathcal{H} \times_\mathcal{F} \mathcal{G}$ over $A$ is a triple $(u, v, a)$ where $u \in \mathcal{H}(A)$, $v \in \mathcal{G}(A)$, and $a : f(u) \to g(v)$ is a morphism in $\mathcal{F}(A)$. Consider a diagram in $\mathcal{H} \times_\mathcal{F} \mathcal{G}$ $$ \vcenter{ \xymatrix{            & (u_2, v_2, a_2) \ar[d] \\ (u_1, v_1, a_1) \ar[r] & (u, v, a) } } \quad\text{lying over}\quad \vcenter{ \xymatrix{            & A_2 \ar[d] \\ A_1 \ar[r] & A } } $$ in $\mathcal{C}_\Lambda$ with $A_2 \to A$ surjective. Since $\mathcal{H}$ and $\mathcal{G}$ satisfy (RS), there are fiber products $u_1 \times_u u_2$ and $v_1 \times_v v_2$ lying over $A_1 \times_A A_2$. Since $\mathcal{F}$ satisfies (RS), Lemma \ref{lemma-RS-fiber-square} shows $$ \vcenter{ \xymatrix{ f(u_1 \times_u u_2) \ar[r] \ar[d] & f(u_2) \ar[d] \\ f(u_1) \ar[r] & f(u) } } \quad\text{and}\quad \vcenter{ \xymatrix{ g(v_1 \times_v v_2) \ar[r] \ar[d] & g(v_2) \ar[d] \\ g(v_1) \ar[r] & g(v) } } $$ are both fiber squares in $\mathcal{F}$. Thus we can view $a_1 \times_a a_2$ as a morphism from $f(u_1 \times_u u_2)$ to $g(v_1 \times_v v_2)$ over $A_1 \times_A A_2$. It follows that $$ \xymatrix{  (u_1 \times_u u_2, v_1 \times_v v_2, a_{1} \times_a a_2) \ar[d] \ar[r] & (u_2, v_2, a_2) \ar[d] \\ (u_1, v_1, a_1) \ar[r] & (u, v, a) } $$ is a fiber square in $\mathcal{H} \times_\mathcal{F} \mathcal{G}$ as desired.
Consider the ring map $g : A' \times_A A' \to k[I]$ defined by the rule $g(a_1, a_2) = \overline{a_1} \oplus a_2 - a_1$ (compare with Lemma \ref{lemma-lifting-along-small-extension}). There is an isomorphism $$ A' \times_A A' \xrightarrow{\sim} A' \times_k k[I] $$ 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 of $A' \times_A A'$ and $A' \times_k k[I]$ to $A$. Thus there is a bijection \begin{equation} \label{equation-one} \text{Lift}(x, A' \times_A A') \longrightarrow \text{Lift}(x, A' \times_k k[I]) \end{equation} By Remark \ref{remark-lift-bijections} there is a bijection \begin{equation} \label{equation-two} \text{Lift}(x, A') \times \text{Lift}(x, A') \longrightarrow \text{Lift}(x, A' \times_A A') \end{equation} There is a commutative diagram $$ \xymatrix{ A' \times_k k[I] \ar[r] \ar[d] & A \times_k k[I] \ar[r] \ar[d] & k[I] \ar[d] \\ A' \ar[r] & A \ar[r] & k. } $$ Thus if we choose a pushforward $x \to x_0$ of $x$ along $A \to k$, we obtain by the end of Remark \ref{remark-lift-bijections} a bijection \begin{equation} \label{equation-three} \text{Lift}(x, A' \times_k k[I]) \longrightarrow \text{Lift}(x, A') \times \text{Lift}(x_0, k[I]) \end{equation} Composing (\ref{equation-two}), (\ref{equation-one}), and (\ref{equation-three}) we get a bijection $$ \Phi : \text{Lift}(x, A') \times \text{Lift}(x, A') \longrightarrow \text{Lift}(x, A') \times \text{Lift}(x_0, k[I]). $$ This bijection commutes with the projections on the first factors. By Remark \ref{remark-tangent-space-lifting} we see that $\text{Lift}(x_0, k[I]) = T\mathcal{F} \otimes_k I$. If $\text{pr}_2$ is the second projection of $\text{Lift}(x, A') \times \text{Lift}(x, A')$, then we get a map $$ a = \text{pr}_2 \circ \Phi^{-1} : \text{Lift}(x, A') \times (T\mathcal{F} \otimes_k I) \longrightarrow \text{Lift}(x, A'). $$ Unwinding all the above we see that $a(x' \to x, \theta)$ is the unique lift $x'' \to x$ such that $g_*(x', x'') = \theta$ in $\text{Lift}(x_0, k[I]) = T\mathcal{F} \otimes_k I$. To see this is an action of $T\mathcal{F} \otimes_k I$ on $\text{Lift}(x, A')$ we have to show the following: if $x', x'', x'''$ are lifts of $x$ and $g_*(x', x'') = \theta$, $g_*(x'', x''') = \theta'$, then $g_*(x', x''') = \theta + \theta'$. This follows from the commutative diagram $$ \xymatrix{ A' \times_A A' \times_A A' \ar[rrrrr]_-{(a_1, a_2, a_3) \mapsto (g(a_1, a_2), g(a_2, a_3))} \ar[rrrrrd]_{(a_1, a_2, a_3) \mapsto g(a_1, a_3)} & & & & & k[I] \times_k k[I] = k[I \times I] \ar[d]^{+} \\ & & & & & k[I] } $$ The action is free and transitive because $\Phi$ is bijective.
We prove $F(A) \to G(A)$ is a bijection for all $A \in \Ob(\mathcal{C}_\Lambda)$ by induction on $\text{length}_A(A)$.  For $A = k$ the statement follows from the assumption that $F$ and $G$ are deformation functors. Suppose that the statement holds for rings of length less than $n$ and let $A'$ be a ring of length $n$. Choose a small extension $f : A' \to A$.  We have a commutative diagram $$ \xymatrix{ F(A') \ar[r] \ar[d]_{F(f)} & G(A') \ar[d]^{G(f)} \\ F(A) \ar[r]^{\sim} & G(A) } $$ where the map $F(A) \to G(A)$ is a bijection.  By smoothness of $F \to G$, $F(A') \to G(A')$ is surjective (Lemma \ref{lemma-smooth-morphism-essentially-surjective}).  Thus we can check bijectivity by checking it on fibers $F(f)^{-1}(x) \to G(f)^{-1}(\varphi(x))$ for $x \in F(A)$ such that $F(f)^{-1}(x)$ is nonempty. These fibers are precisely $\text{Lift}(x, A')$ and $\text{Lift}(\varphi(x), A')$ and by assumption we have an isomorphism $d\varphi \otimes \text{id} : TF \otimes_k \Ker(f)  \to TG \otimes_k \Ker(f)$. Thus, by Lemma \ref{lemma-free-transitive-action} and Remark \ref{remark-free-transitive-action-functorial}, for $x \in F(A)$ such that $F(f)^{-1}(x)$ is nonempty the map $F(f)^{-1}(x) \to G(f)^{-1}(\varphi(x))$ is a map of sets commuting with free transitive actions by $TF \otimes_k \Ker(f)$. Hence it is bijective.
It follows that $\mathit{Aut}(x)$ satisfies (RS) from the fully faithfulness of the functor $\mathcal{F}(A_1 \times_A A_2) \to \mathcal{F}(A_1) \times_{\mathcal{F}(A)} \mathcal{F}(A_2)$ in Lemma \ref{lemma-RS-2-categorical}.
We apply Remark \ref{remark-localize-cofibered-groupoid} to the functor $\mathit{Aut}(x) : \mathcal{C}_A \to \textit{Sets}$ and the element $\text{id}_{x_0} \in \mathit{Aut}(x)(k)$ to get a predeformation functor $F = \mathit{Aut}(x)_{\text{id}_{x_0}}$. By Lemmas \ref{lemma-Aut-functor-RS} and \ref{lemma-localize-RS} $F$ is a deformation functor. By definition $T_{\text{id}_{x_0}} \mathit{Aut}(x) = TF = F(k[\epsilon])$ which has a natural $k$-vector space structure specified by Lemma \ref{lemma-tangent-space-functor}. \medskip\noindent Addition is defined as the composition $$ F(k[\epsilon]) \times F(k[\epsilon]) \longrightarrow F(k[\epsilon] \times_k k[\epsilon]) \longrightarrow F(k[\epsilon]) $$ where the first map is the inverse of the bijection guaranteed by (RS) and the second is induced by the $k$-algebra map $k[\epsilon] \times_k k[\epsilon] \to k[\epsilon]$ which maps $(\epsilon, 0)$ and $(0, \epsilon)$ to $\epsilon$. If $A \to B$ is a ring map in $\mathcal{C}_\Lambda$, then $F(A) \to F(B)$ is a homomorphism where $F(A) = \mathit{Aut}(x)_{\text{id}_{x_0}}(A)$ and $F(B) = \mathit{Aut}(x)_{\text{id}_{x_0}}(B)$ are groups under composition. We conclude that $+ : F(k[\epsilon]) \times F(k[\epsilon])\to F(k[\epsilon])$ is a homomorphism where $F(k[\epsilon])$ is regarded as a group under composition. With $\text{id} \in F(k[\epsilon])$ the unit element we see that $+(v, \text{id}) = +(\text{id}, v) = v$ for any $v \in F(k[\epsilon])$ because $(\text{id}, v)$ is the pushforward of $v$ along the ring map $k[\epsilon] \to k[\epsilon] \times_k k[\epsilon]$ with $\epsilon \mapsto (\epsilon, 0)$. In general, given a group $G$ with multiplication $\circ$ and $+ : G \times G \to G$ is a homomorphism such that $+(g, 1) = +(1, g) = g$, where $1$ is the identity of $G$, then $+ = \circ$. This shows addition in the $k$-vector space structure on $F(k[\epsilon])$ agrees with composition. \medskip\noindent Finally, (2) is a matter of unwinding the definitions. Namely $T_{\text{id}_{x_0}} \mathit{Aut}(x)$ is the set of automorphisms $\alpha$ of the pushforward of $x$ along $A \to k \to k[\epsilon]$ which are trivial modulo $\epsilon$. On the other hand $T_{\text{id}_{x_0}} \mathit{Aut}(x_0)$ is the set of automorphisms of the pushforward of $x_0$ along $k \to k[\epsilon]$ which are trivial modulo $\epsilon$. Since $x_0$ is the pushforward of $x$ along $A \to k$ the result is clear.
This is just the analogue of Lemma \ref{lemma-free-transitive-action} in the setting of automorphism sheaves. To be precise, we apply Remark \ref{remark-localize-cofibered-groupoid} to the functor $\mathit{Aut}(x') : \mathcal{C}_{A'} \to \textit{Sets}$ and the element $\text{id}_{x_0} \in \mathit{Aut}(x)(k)$ to get a predeformation functor $F = \mathit{Aut}(x')_{\text{id}_{x_0}}$. By Lemmas \ref{lemma-Aut-functor-RS} and \ref{lemma-localize-RS} $F$ is a deformation functor. Hence Lemma \ref{lemma-free-transitive-action} gives a free and transitive action of $TF \otimes_k I$ on $\text{Lift}(\text{id}_x, A')$, because as $\text{Lift}(\text{id}_x, A')$ is a group it is always nonempty. Note that we have equalities of vector spaces $$ TF = T_{\text{id}_{x_0}} \mathit{Aut}(x') \otimes_k I = \text{Inf}_{x_0}(\mathcal{F}) \otimes_k I $$ by Lemma \ref{lemma-Aut-functor-tangent-space}. The equality $\text{Inf}(x'/x) = \text{Lift}(\text{id}_x, A')$ of Remark \ref{remark-infaut-lifting-equalities} finishes the proof.
The morphism $\mathcal{F}_{x_0} \to \overline{\mathcal{F}_{x_0}}$ is an equivalence if and only if $\mathcal{F}_{x_0}$ is fibered in setoids, cf.\ Categories, Section \ref{categories-section-fibred-in-setoids} (a setoid is by definition a groupoid in which the only automorphism of any object is the identity). We prove that $\text{Inf}_{x_0}(\mathcal{F}) = 0$ if and only if this condition holds for $\mathcal{F}_{x_0}$.  Obviously if $\mathcal{F}_{x_0}$ is fibered in setoids then $\text{Inf}_{x_0}(\mathcal{F}) = 0$.  Conversely assume $\text{Inf}_{x_0}(\mathcal{F}) = 0$.  Let $A$ be an object of $\mathcal{C}_\Lambda$. Then by Lemma \ref{lemma-infaut-trivial}, $\text{Inf}(x/x_0) = 0$ for any object $x \to x_0$ of $\mathcal{F}_{x_0}(A)$. Since by definition $\text{Inf}(x/x_0)$ equals the group of automorphisms of $x \to x_0$ in $\mathcal{F}_{x_0}(A)$, this proves $\mathcal{F}_{x_0}(A)$ is a setoid.
Denote $p_i : \mathcal{H}_i \to \mathcal{G}$ and $q : \mathcal{G} \to \mathcal{F}$ be the given maps. Let $A' \to A$ be a small extension in $\mathcal{C}_\Lambda$. An object of $\mathcal{H}_1 \times_\mathcal{F} \mathcal{H}_2$ over $A'$ is a triple $(x'_1, x'_2, a')$ where $x'_i$ is an object of $\mathcal{H}_i$ over $A'$ and $a' : q(p_1(x'_1)) \to q(p_2(x'_2))$ is a morphism of the fibre category of $\mathcal{F}$ over $A'$. By pushforward along $A' \to A$ we get $(x_1, x_2, a)$. Lifting this to an object of $\mathcal{H}_1 \times_\mathcal{G} \mathcal{H}_2$ over $A$ means finding a morphism $b : p_1(x_1) \to p_2(x_2)$ over $A$ with $q(b) = a$. Thus we have to show that we can lift $b$ to a morphism $b' : p_1(x'_1) \to p_2(x'_2)$ whose image under $q$ is $a'$. \medskip\noindent Observe that we can think of $$ p_1(x'_1) \to p_1(x_1) \xrightarrow{b} p_2(x_2) \quad\text{and}\quad p_2(x'_2) \to p_2(x_2) $$ as two objects of $\textit{Lift}(p_2(x_2), A' \to A)$. The functor $q$ sends these objects to the two objects $$ q(p_1(x'_1)) \to q(p_1(x_1)) \xrightarrow{b} q(p_2(x_2)) \quad\text{and}\quad q(p_2(x'_2)) \to q(p_2(x_2)) $$ of $\textit{Lift}(q(p_2(x_2)), A' \to A)$ which are isomorphic using the map $a' : q(p_1(x'_1)) \to q(p_2(x'_2))$. On the other hand, the functor $$ q : \textit{Lift}(p_2(x_2), A' \to A) \to \textit{Lift}(q(p_2(x_2)), A' \to A) $$ defines a injection on isomorphism classes by Lemma \ref{lemma-free-transitive-action} and our assumption on tangent spaces. Thus we see that there is a morphism $b' : p_1(x_1') \to p_2(x'_2)$ whose pushforward to $A$ is $b$. However, we may need to adjust our choice of $b'$ to achieve $q(b') = a'$. For this it suffices to see that $q : \text{Inf}(p_2(x'_2)/p_2(x_2)) \to \text{Inf}(q(p_2(x'_2))/q(p_2(x_2)))$ is surjective. This follows from our assumption on infinitesimal automorphisms and Lemma \ref{lemma-lifted-automorphisms-torsor}.
Let $A' \to A$ be a small extension. Let $x \in \mathcal{F}(A)$. Let $y' \to f(x)$ be a morphism in $\mathcal{G}$ over $A' \to A$. Consider the functor $\text{Lift}(A', x) \to \text{Lift}(A', f(x))$ induced by $f$. We have to show that there exists an object $x' \to x$ of $\text{Lift}(A', x)$ mapping to $y' \to f(x)$, see Lemma \ref{lemma-smoothness-small-extensions}. By condition (2) we know that $\text{Lift}(A', x)$ is not the empty category. By condition (2) and Lemma \ref{lemma-free-transitive-action} we conlude that the map on isomorphism classes is surjective as desired.
Let $A' \to A$ be a small extension in $\mathcal{C}_\Lambda$ and let $x \in \mathcal{F}(A)$ with image $y \in \mathcal{G}(A)$. Assume there is a lift $y' \in \mathcal{G}(A')$. According to Lemma \ref{lemma-easy-check-smooth} all we have to do is check that $x$ has a lift too. Take the image $z' \in \mathcal{H}(A')$ of $y'$. Since $\mathcal{F} \to \mathcal{H}$ is smooth, there is an $x' \in \mathcal{F}(A')$ mapping to both $x \in \mathcal{F}(A)$ and $z' \in \mathcal{H}(A')$, see Definition \ref{definition-smooth-morphism}. This finishes the proof.
Part (1) follows from Remark \ref{remark-groupoid-in-functors-complete-restrict}. For (2), the ``only if'' direction is clear from the definition of a prorepresentable groupoid in functors. Conversely, assume $U$ and $R$ are prorepresentable, say $U \cong \underline{R_0}|_{\mathcal{C}_\Lambda}$ and $R \cong \underline{R_1}|_{\mathcal{C}_\Lambda}$ for objects $R_0$ and $R_1$ of $\widehat{\mathcal{C}}_\Lambda$. Since $\underline{R_0} \cong \widehat{\underline{R_0}|_{\mathcal{C}_\Lambda}}$ and $\underline{R_1} \cong \widehat{\underline{R_1}|_{\mathcal{C}_\Lambda}}$ by Remark \ref{remark-restrict-complete-continuous-functor} we see that the completion $(U, R, s, t, c)^\wedge$ is a groupoid in functors of the form $(\underline{R_0}, \underline{R_1}, \widehat{s}, \widehat{t}, \widehat{c})$. By Lemma \ref{lemma-CLambdahat-pushouts} the pushout $\underline{R_1} \times_{\widehat{s}, \underline{R_1}, \widehat{t}} \underline{R_1}$ exists. Hence $(\underline{R_0}, \underline{R_1}, \widehat{s}, \widehat{t}, \widehat{c})$ is a representable groupoid in functors on $\widehat{\mathcal{C}}_\Lambda$. Finally, the restriction $(\underline{R_0}, \underline{R_1}, s, t, c)|_{\mathcal{C}_\Lambda}$ gives back $(U, R, s, t, c)$ by Remark \ref{remark-groupoid-in-functors-complete-restrict} hence $(U, R, s, t, c)$ is prorepresentable by definition.
Statement (2) is equivalent to (3) since the inverse $i: R \to R$ of $(U, R, s, t, c)$ is an isomorphism and $t = s \circ i$. By definition (1) is equivalent to (2) and (3) together, hence it is equivalent to either of them individually. \medskip\noindent Finally we prove (2) is equivalent to (4). Unwinding the definitions: \begin{enumerate} \item[(2)] Smoothness of $s: R \to U$ amounts to the following condition: If $f: B \to A$ is a surjective ring map in $\mathcal{C}_\Lambda$, $a \in R(A)$, and $y \in U(B)$ such that $s(a) = U(f)(y)$, then there exists $a' \in R(B)$ such that $R(f)(a') = a$ and $s(a') = y$. \item[(4)] Smoothness of $U \to [U/R]$ amounts to the following condition: If $f: B \to A$ is a surjective ring map in $\mathcal{C}_\Lambda$ and $(f, a) : (B, y) \to (A, x)$ is a morphism of $[U/R]$, then there exists $x' \in U(B)$ and $b \in R(B)$ with $s(b) = x'$, $t(b) = y$ such that $c(a, R(f)(b)) = e(x)$.  Here $e : U \to R$ denotes the identity and the notation $(f, a)$ is as in Remarks \ref{remarks-cofibered-groupoids} (\ref{item-construction-associated-cofibered-groupoid}); in particular $a \in R(A)$ with $s(a) = U(f)(y)$ and $t(a) = x$. \end{enumerate} If (4) holds and $f, a, y$ as in (2) are given, let $x = t(a)$ so that we have a morphism $(f, a): (B, y) \to (A, x)$. Then (4) produces $x'$ and $b$, and $a' = i(b)$ satisfies the requirements of (2). Conversely, assume (2) holds and let $(f, a): (B, y) \to (A, x)$ as in (4) be given. Then (2) produces $a' \in R(B)$, and $x' = t(a')$ and $b = i(a')$ satisfy the requirements of (4).
Let $$ \xymatrix{                            &     (A_2, x_2) \ar[d]^{(f_2, a_2)} \\ (A_1, x_1) \ar[r]^{(f_1, a_1)} &     (A, x) } $$ be a diagram in $[U/R]$ such that $f_2: A_2 \to A$ is surjective. The notation is as in Remarks \ref{remarks-cofibered-groupoids} (\ref{item-construction-associated-cofibered-groupoid}). Hence $f_1: A_1 \to A, f_2: A_2 \to A$ are maps in $\mathcal{C}_\Lambda$, $x \in U(A)$, $x_1 \in U(A_1)$, $x_2 \in U(A_2)$, and $a_1, a_2 \in R(A)$ with $s(a_1) = U(f_1)(x_1)$, $t(a_1) = x$ and $s(a_2) = U(f_2)(x_2)$, $t(a_2) = x$. We construct a fiber product lying over $A_1 \times_A A_2$ for this diagram in $[U/R]$ as follows. \medskip\noindent Let $a = c(i(a_1), a_2)$, where $i: R \to R$ is the inverse morphism. Then $a \in R(A)$, $x_2 \in U(A_2)$ and $s(a) = U(f_2)(x_2)$. Hence an element $(a, x_2) \in R(A) \times_{s, U(A), U(f_2)} U(A_2)$. By smoothness of $s : R \to U$ there is an element $\widetilde{a} \in R(A_2)$ with $R(f_2)(\widetilde{a}) = a$ and $s(\widetilde{a}) = x_2$. In particular $U(f_2)(t(\widetilde{a})) = t(a) = U(f_1)(x_1)$. Thus $x_1$ and $t(\widetilde{a})$ define an element $$ (x_1, t(\widetilde{a})) \in U(A_1) \times_{U(A)} U(A_2). $$ By the assumption that $U$ satisfies (RS), we have an identification $U(A_1) \times_{U(A)} U(A_2) = U(A_1 \times_A A_2)$. Let us denote $x_1 \times t(\widetilde{a}) \in U(A_1 \times_A A_2)$ the element corresponding to $(x_1, t(\widetilde{a})) \in U(A_1) \times_{U(A)} U(A_2)$. Let $p_1, p_2$ be the projections of $A_1 \times_A A_2$.  We claim $$ \xymatrix{ (A_1 \times_A A_2, x_1 \times t(\widetilde{a})) \ar[d]_{(p_1, e(x_1))} \ar[rr]_-{(p_2, i(\widetilde{a}))} & & (A_2, x_2) \ar[d]^{(f_2, a_2)} \\ (A_1, x_1) \ar[rr]^{(f_1, a_1)} & & (A, x) } $$ is a fiber square in $[U/R]$. (Note $e: U \to R$ denotes the identity.) \medskip\noindent The diagram is commutative because $c(a_2, R(f_2)(i(\widetilde{a}))) = c(a_2, i(a)) = a_1$. To check it is a fiber square, let $$ \xymatrix{ (B, z) \ar[d]_{(g_1, b_1)} \ar[rr]_{(g_2, b_2)} & & (A_2, x_2) \ar[d]^{(f_2, a_2)} \\ (A_1, x_1) \ar[rr]^{(f_1, a_1)} & & (A, x) } $$ be a commutative diagram in $[U/R]$. We will show there is a unique morphism $(g, b) : (B, z) \to (A_1 \times_A A_2, x_1 \times t(\widetilde{a}))$ compatible with the morphisms to $(A_1, x_1)$ and $(A_2, x_2)$. We must take $g = (g_1, g_2) : B \to A_1 \times_A A_2$. Since by assumption $R$ satisfies (RS), we have an identification $R(A_1 \times_A A_2) = R(A_1) \times_{R(A)} R(A_2)$. Hence we can write $b = (b'_1, b'_2)$ for some $b'_1 \in R(A_1)$, $b'_2 \in R(A_2)$ which agree in $R(A)$. Then $((g_1, g_2), (b'_1, b'_2)) : (B, z) \to (A_1 \times_A A_2, x_1 \times t(\widetilde{a}))$ will commute with the projections if and only if $b'_1 = b_1$ and $b'_2 = c(\widetilde{a}, b_2)$ proving unicity and existence.
Since $U$ and $R$ are deformation functors $[U/R]$ is a predeformation category. Since (RS) holds for deformation functors by definition we see that (RS) holds for [U/R] by Lemma \ref{lemma-smooth-RS-groupoid-in-functors-quotient}. Hence $[U/R]$ is a deformation category. Statements (2) and (3) follow directly from the definitions.