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Proof. This proof is the same as the proof of Lemma 93.12.4 but easier. We suggest the reader to skip the proof. The map is given by henselization: given $0 \to N \to C \to B \to 0$ in $\text{Exal}_ k(B, N)$ we send it to the henselization $C^ h$ of $C$ with respect to the inverse image $J_ C \subset C$ of $J$. Compare with the proof of Lemma 93.8.8.

Since $k \to B$ is of finite presentation the complex $\mathop{N\! L}\nolimits _{B/k}$ can be represented by a complex $N^{-1} \to N^0$ where $N^ i$ is a finite $B$-module, see Algebra, Section 10.134 and in particular Algebra, Lemma 10.134.2. As $B$ is Noetherian, this means that $\mathop{N\! L}\nolimits _{B/k}$ is pseudo-coherent. For $g \in J$ the $k$-algebra $B_ g$ is smooth and hence $(\mathop{N\! L}\nolimits _{B/k})_ g = \mathop{N\! L}\nolimits _{B_ g/k}$ is quasi-isomorphic to a finite projective $B$-module sitting in degree $0$. Thus $\text{Ext}^ i_ B(\mathop{N\! L}\nolimits _{B/k}, N)_ g = 0$ for $i \geq 1$ and any $B$-module $N$. Finally, we have

The first equality by More on Algebra, Lemma 15.33.8 (or rather its analogue for henselizations of pairs). The second by More on Algebra, Lemma 15.99.2. The third because $\text{Ext}^1_ B(\mathop{N\! L}\nolimits _{B/k}, N)$ is $J$-power torsion, the map $B \to B^ h$ is flat and induces an isomorphism $B/J \to B^ h/JB^ h$ (More on Algebra, Lemma 15.12.2), and More on Algebra, Lemma 15.89.3. This concludes the proof by the description of $\text{Exal}_ A(B, N)$ as $\text{Ext}^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ given just above Lemma 93.12.4. $\square$