The map is given by completion: given $0 \to N \to C \to B \to 0$ in $\text{Exal}_k(B, N)$ we send it to the completion $C^\wedge$ of $C$ with respect to the inverse image of $J$. Compare with the proof of Lemma \ref{lemma-completion}. \medskip\noindent Since $k \to B$ is of finite presentation the complex $\NL_{B/k}$ can be represented by a complex $N^{-1} \to N^0$ where $N^i$ is a finite $B$-module, see Algebra, Section \ref{algebra-section-netherlander} and in particular Algebra, Lemma \ref{algebra-lemma-NL-homotopy}. As $B$ is Noetherian, this means that $\NL_{B/k}$ is pseudo-coherent. For $g \in J$ the $k$-algebra $B_g$ is smooth and hence $(\NL_{B/k})_g = \NL_{B_g/k}$ is quasi-isomorphic to a finite projective $B$-module sitting in degree $0$. Thus $\text{Ext}^i_B(\NL_{B/k}, N)_g = 0$ for $i \geq 1$ and any $B$-module $N$. By More on Algebra, Lemma \ref{more-algebra-lemma-ext-annihilated-into} we conclude that $$ \text{Ext}^1_B(\NL_{B/k}, N) \longrightarrow \lim_n \text{Ext}^1_B(\NL_{B/k}, N/J^n N) $$ is an isomorphism for any finite $B$-module $N$. \medskip\noindent Injectivity of the map. Suppose that $0 \to N \to C \to B \to 0$ is in $\text{Exal}_k(B, N)$ and maps to zero in $\text{Exal}_k(B^\wedge, N^\wedge)$. Choose a splitting $C^\wedge = B^\wedge \oplus N^\wedge$. Then the induced map $C \to C^\wedge \to N^\wedge$ gives maps $C \to N/J^nN$ for all $n$. Hence we see that our element is in the kernel of the maps $$ \text{Ext}^1_B(\NL_{B/k}, N) \to \text{Ext}^1_B(\NL_{B/k}, N/J^n N) $$ for all $n$. By the previous paragraph we conclude that our element is zero. \medskip\noindent Surjectivity of the map. Let $0 \to N^\wedge \to C' \to B^\wedge \to 0$ be an element of $\text{Exal}_k(B^\wedge, N^\wedge)$. Pulling back by $B \to B^\wedge$ we get an element $0 \to N^\wedge \to C'' \to B \to 0$ in $\text{Exal}_k(B, N^\wedge)$. we have $$ \text{Ext}^1_B(\NL_{B/k}, N^\wedge) = \text{Ext}^1_B(\NL_{B/k}, N) \otimes_B B^\wedge = \text{Ext}^1_B(\NL_{B/k}, N) $$ The first equality as $N^\wedge = N \otimes_B B^\wedge$ (Algebra, Lemma \ref{algebra-lemma-completion-tensor}) and More on Algebra, Lemma \ref{more-algebra-lemma-pseudo-coherence-and-ext}. The second equality because $\text{Ext}^1_B(\NL_{B/k}, N)$ is $J$-power torsion (see above), $B \to B^\wedge$ is flat and induces an isomorphism $B/J \to B^\wedge/JB^\wedge$, and More on Algebra, Lemma \ref{more-algebra-lemma-neighbourhood-equivalence}. Thus we can find a $C \in \text{Exal}_k(B, N)$ mapping to $C''$ in $\text{Exal}_k(B, N^\wedge)$. Thus $$ 0 \to N^\wedge \to C' \to B^\wedge \to 0 \quad\text{and}\quad 0 \to N^\wedge \to C^\wedge \to B^\wedge \to 0 $$ are two elements of $\text{Exal}_k(B^\wedge, N^\wedge)$ mapping to the same element of $\text{Exal}_k(B, N^\wedge)$. Taking the difference we get an element $0 \to N^\wedge \to C' \to B^\wedge \to 0$ of $\text{Exal}_k(B^\wedge, N^\wedge)$ whose image in $\text{Exal}_k(B, N^\wedge)$ is zero. This means there exists $$ \xymatrix{ 0 \ar[r] & N^\wedge \ar[r] & C' \ar[r] & B^\wedge \ar[r] & 0 \\ & & B \ar[u]^\sigma \ar[ru] } $$ Let $J' \subset C'$ be the inverse image of $JB^\wedge \subset B^\wedge$. To finish the proof it suffices to note that $\sigma$ is continuous for the $J$-adic topology on $B$ and the $J'$-adic topology on $C'$ and that $C'$ is $J'$-adically complete by Algebra, Lemma \ref{algebra-lemma-complete-modulo-nilpotent} (here we also use that $C'$ is Noetherian; small detail omitted). Namely, this means that $\sigma$ factors through the completion $B^\wedge$ and $C' = 0$ in $\text{Exal}_k(B^\wedge, N^\wedge)$.