Let $\mathfrak p = \Ker(A \to k)$ be the prime corresponding to $u_0$. \medskip\noindent Assume that $x$ versal at $u_0$ and that $u_0$ is a closed point of $U$. If $H^{-1}(\xi \otimes_A^{\mathbf{L}} k)$ is not surjective, then let $A' \to A$ be an extension with kernel $I$ as in Lemma \ref{lemma-construct-essential-surjection}. Because $u_0$ is a closed point, we see that $I$ is a finite $A$-module, hence that $A'$ is a finite type $\Lambda$-algebra (this fails if $u_0$ is not closed). In particular $A'$ is Noetherian. By property (c) for $A'$ and (i) for $\xi$ we see that $x$ lifts to an object $x'$ over $A'$. Let $\mathfrak p' \subset A'$ be kernel of the surjective map to $k$. By Artin-Rees (Algebra, Lemma \ref{algebra-lemma-Artin-Rees}) there exists an $n > 1$ such that $(\mathfrak p')^n \cap I = 0$. Then we see that $$ B' = A'/(\mathfrak p')^n \longrightarrow A/\mathfrak p^n = B $$ is a small, essential extension of local Artinian rings, see Formal Deformation Theory, Lemma \ref{formal-defos-lemma-essential-surjection}. On the other hand, as $x$ is versal at $u_0$ and as $x'|_{\Spec(B')}$ is a lift of $x|_{\Spec(B)}$, there exists an integer $m \geq n$ and a map $q : A/\mathfrak p^m \to B'$ such that the composition $A/\mathfrak p^m \to B' \to B$ is the quotient map. Since the maximal ideal of $B'$ has $n$th power equal to zero, this $q$ factors through $B$ which contradicts the fact that $B' \to B$ is an essential surjection. This contradiction shows that $H^{-1}(\xi \otimes_A^{\mathbf{L}} k)$ is surjective. \medskip\noindent Assume that $x$ versal at $u_0$. By Lemma \ref{lemma-compute-ext-into-field} the map $H^0(\xi \otimes_A^{\mathbf{L}} k)$ is dual to the map $\Ext^0_A(\NL_{A/\Lambda}, k) \to \text{Ext}^0_A(E, k)$. Note that $$ \Ext^0_A(\NL_{A/\Lambda}, k) = \text{Der}_\Lambda(A, k) \quad\text{and}\quad T_x(k) = \Ext^0_A(E, k) $$ Condition (ii) assures us the map $\Ext^0_A(\NL_{A/\Lambda}, k) \to \text{Ext}^0_A(E, k)$ sends a tangent vector $\theta$ to $U$ at $u_0$ to the corresponding infinitesimal deformation of $x_0$, see Remark \ref{remark-canonical-element}. Hence if $x$ is versal, then this map is surjective, see Formal Deformation Theory, Lemma \ref{formal-defos-lemma-versal-criterion}. Hence $H^0(\xi \otimes_A^{\mathbf{L}} k)$ is injective. This finishes the proof of (1) $\Rightarrow$ (2) in case $u_0$ is a closed point. \medskip\noindent For the rest of the proof assume $H^{-1}(E \otimes_A^\mathbf{L} k) \to H^{-1}(\NL_{A/\Lambda} \otimes_A^\mathbf{L} k)$ is surjective and $H^0(E \otimes_A^\mathbf{L} k) \to H^0(\NL_{A/\Lambda} \otimes_A^\mathbf{L} k)$ injective. Set $R = A_\mathfrak p^\wedge$ and let $\eta$ be the formal object over $R$ associated to $x|_{\Spec(R)}$. The map $d\underline{\eta}$ on tangent spaces is surjective because it is identified with the dual of the injective map $H^0(E \otimes_A^{\mathbf{L}} k) \to H^0(\NL_{A/\Lambda} \otimes_A^{\mathbf{L}} k)$ (see previous paragraph). According to Formal Deformation Theory, Lemma \ref{formal-defos-lemma-versal-criterion} it suffices to prove the following: Let $C' \to C$ be a small extension of finite type Artinian local $\Lambda$-algebras with residue field $k$. Let $R \to C$ be a $\Lambda$-algebra map compatible with identifications of residue fields. Let $y = x|_{\Spec(C)}$ and let $y'$ be a lift of $y$ to $C'$. To show: we can lift the $\Lambda$-algebra map $R \to C$ to $R \to C'$. \medskip\noindent Observe that it suffices to lift the $\Lambda$-algebra map $A \to C$. Let $I = \Ker(C' \to C)$. Note that $I$ is a $1$-dimensional $k$-vector space. The obstruction $ob$ to lifting $A \to C$ is an element of $\Ext^1_A(\NL_{A/\Lambda}, I)$, see Example \ref{example-key}. By Lemma \ref{lemma-compute-ext-into-field} and our assumption the map $\xi$ induces an injection $$ \Ext^1_A(\NL_{A/\Lambda}, I) \longrightarrow \Ext^1_A(E, I) $$ By the construction of $ob$ and (i) the image of $ob$ in $\Ext^1_A(E, I)$ is the obstruction to lifting $x$ to $A \times_C C'$. By (RS*) the fact that $y/C$ lifts to $y'/C'$ implies that $x$ lifts to $A \times_C C'$. Hence $ob = 0$ and we are done.