The Stacks project

Proof. 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 93.8.10.

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 $N$. By More on Algebra, Lemma 15.102.1 we conclude that

is an isomorphism for any finite $B$-module $N$.

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

for all $n$. By the previous paragraph we conclude that our element is zero.

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

The first equality as $N^\wedge = N \otimes _ B B^\wedge$ (Algebra, Lemma 10.97.1) and More on Algebra, Lemma 15.65.3. The second equality because $\text{Ext}^1_ B(\mathop{N\! L}\nolimits _{B/k}, N)$ 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 15.89.3. Thus we can find a $C \in \text{Exal}_ k(B, N)$ mapping to $C''$ in $\text{Exal}_ k(B, N^\wedge )$. Thus

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

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 10.97.10 (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 $\text{Exal}_ k(B^\wedge , N^\wedge )$. $\square$