hcju/stacksqa huggingface dataset:
The statement makes sense as $B^\wedge$ is an object of (\ref{equation-C-prime}) by Lemma \ref{lemma-topologically-finite-type-Noetherian}. Let $J = \Ker(\alpha)$. The functor of taking $I$-adic completion is exact on finite modules over $A[x_1, \ldots, x_n]$ and agrees with the functor $M \mapsto M \otimes_{A[x_1, \ldots, x_n]} A[x_1, \ldots, x_n]^\wedge$, see Algebra, Lemmas \ref{algebra-lemma-completion-tensor} and \ref{algebra-lemma-completion-flat}. Moreover, the ring maps $A[x_1, \ldots, x_n] \to A[x_1, \ldots, x_n]^\wedge$ and $B \to B^\wedge$ are flat. Hence $B^\wedge = A[x_1, \ldots, x_n]^\wedge / J^\wedge$ and $$ (J/J^2) \otimes_B B^\wedge = (J/J^2)^\wedge = J^\wedge/(J^\wedge)^2 $$ Since $\NL(\alpha) = (J/J^2 \to \bigoplus B\text{d}x_i)$, see Algebra, Section \ref{algebra-section-netherlander}, we conclude the complex $\NL_{B^\wedge/A}^\wedge$ is equal to $\NL(\alpha) \otimes_B B^\wedge$. The final statement follows as $\NL_{B/A}$ is homotopy equivalent to $\NL(\alpha)$ and because the ring map $B \to B^\wedge$ is flat (so derived base change along $B \to B^\wedge$ is just base change).
The statement means the following: for every $n$ we have a well defined complex $\NL_{B_n/A_n}$ of $B_n$-modules and we have transition maps $\NL_{B_{n + 1}/A_{n + 1}} \to \NL_{B_n/A_n}$. See Algebra, Section \ref{algebra-section-netherlander}. Thus we can consider $$ \ldots \to \NL_{B_3/A_3} \to \NL_{B_2/A_2} \to \NL_{B_1/A_1} $$ as an inverse system of complexes of $B$-modules and a fortiori as an inverse system in $D(B)$. Furthermore $R\lim \NL_{B_n/A_n}$ is a homotopy limit of this inverse system, see Derived Categories, Section \ref{derived-section-derived-limit}. \medskip\noindent Choose a presentation $B = A[x_1, \ldots, x_r]^\wedge / J$. This defines presentations $$ B_n = B/I^nB = A_n[x_1, \ldots, x_r]/J_n $$ where $$ J_n = JA_n[x_1, \ldots, x_r] = J/(J \cap I^nA[x_1, \ldots, x_r]^\wedge) $$ The two term complex $J_n/J_n^2 \longrightarrow \bigoplus B_n \text{d}x_i$ represents $\NL_{B_n/A_n}$, see Algebra, Section \ref{algebra-section-netherlander}. By Artin-Rees (Algebra, Lemma \ref{algebra-lemma-Artin-Rees}) in the Noetherian ring $A[x_1, \ldots, x_r]^\wedge$ (Lemma \ref{lemma-topologically-finite-type-Noetherian}) we find a $c \geq 0$ such that we have canonical surjections $$ J/I^nJ \to J_n \to J/I^{n - c}J \to J_{n - c},\quad n \geq c $$ for all $n \geq c$. A moment's thought shows that these maps are compatible with differentials and we obtain maps of complexes $$ \NL_{B/A}^\wedge \otimes_B B/I^nB \to \NL_{B_n/A_n} \to \NL_{B/A}^\wedge \otimes_B B/I^{n - c}B \to \NL_{B_{n - c}/A_{n - c}} $$ compatible with the transition maps of the inverse systems $\{\NL_{B/A}^\wedge \otimes_B B/I^nB\}$ and $\{\NL_{B_n/A_n}\}$. This proves part (1) of the lemma. \medskip\noindent By part (1) and since pro-isomorphic systems have the same $R\lim$ in order to prove (2) it suffices to show that $\NL_{B/A}^\wedge$ is equal to $R\lim \NL_{B/A}^\wedge \otimes_B B/I^nB$. However, $\NL_{B/A}^\wedge$ is a two term complex $M^\bullet$ of finite $B$-modules which are $I$-adically complete for example by Algebra, Lemma \ref{algebra-lemma-completion-tensor}. Hence $M^\bullet = \lim M^\bullet/I^nM^\bullet = R\lim M^\bullet/I^n M^\bullet$, see More on Algebra, Lemma \ref{more-algebra-lemma-compute-Rlim-modules} and Remark \ref{more-algebra-remark-how-unique}.
Observe that taking the tensor product $\NL_{B/A}^\wedge \otimes_B C$ makes sense as $\NL_{B/A}^\wedge$ is well defined up to homotopy by Lemma \ref{lemma-NL-up-to-homotopy}. Also, $(B, IB)$ is pair where $B$ is a Noetherian ring (Lemma \ref{lemma-topologically-finite-type-Noetherian}) and $C$ is in the corresponding category (\ref{equation-C-prime}). Thus all the terms in the $6$-term sequence are (well) defined. \medskip\noindent Choose a presentation $B = A[x_1, \ldots, x_r]^\wedge/J$. Choose a presentation $C = B[y_1, \ldots, y_s]^\wedge/J'$. Combinging these presentations gives a presentation $$ C = A[x_1, \ldots, x_r, y_1, \ldots, y_s]^\wedge/K $$ Then the reader verifies that we obtain a commutative diagram $$ \xymatrix{ 0 \ar[r] & \bigoplus C \text{d}x_i \ar[r] & \bigoplus C \text{d}x_i \oplus \bigoplus C \text{d}y_j \ar[r] & \bigoplus C \text{d}y_j \ar[r] & 0 \\ & J/J^2 \otimes_B C \ar[r] \ar[u] & K/K^2 \ar[r] \ar[u] & J'/(J')^2 \ar[r] \ar[u] & 0 } $$ with exact rows. Note that the vertical arrow on the left hand side is the tensor product of the arrow defining $\NL_{B/A}^\wedge$ with $\text{id}_C$. The lemma follows by applying the snake lemma (Algebra, Lemma \ref{algebra-lemma-snake}).
For each $n \geq 1$ we set $A_n = A/I^n$, $B_n = B/I^nB$, and $C_n = C/I^nC$. We have \begin{align*} H^{-1}(\NL_{B/A}^\wedge \otimes_B C) & = \lim H^{-1}(\NL_{B/A}^\wedge \otimes_B C_n) \\ & = \lim H^{-1}(\NL_{B/A}^\wedge \otimes_B B_n \otimes_{B_n} C_n) \\ & = \lim H^{-1}(\NL_{B_n/A_n} \otimes_{B_n} C_n) \end{align*} The first equality follows from More on Algebra, Lemma \ref{more-algebra-lemma-consequence-Artin-Rees} and the fact that $H^{-1}(\NL_{B/A}^\wedge \otimes_B C)$ is a finite $C$-module and hence $I$-adically complete for example by Algebra, Lemma \ref{algebra-lemma-completion-tensor}. The second equality is trivial. The third holds by Lemma \ref{lemma-NL-is-limit}. The maps $H^{-1}(\NL_{B_n/A_n} \otimes_{B_n} C_n) \to H^{-1}(\NL_{C_n/A_n})$ are injective by More on Algebra, Lemma \ref{more-algebra-lemma-transitive-lci-at-end}. The proof is finished because we also have $H^{-1}(\NL_{C/A}^\wedge) = \lim H^{-1}(\NL_{C_n/A_n})$ similarly to the above.
We will use Lemma \ref{lemma-equivalent-with-artin-smooth} without further mention. \medskip\noindent Assume (1). Recall that formation of $\NL_{B/A}$ commutes with localization, see Algebra, Lemma \ref{algebra-lemma-localize-NL}. Hence by the very definition of smooth ring maps (in terms of the naive cotangent complex being quasi-isomorphic to a finite projective module placed in degree $0$), we see that $\NL_{B/A}$ satisfies the fourth equivalent condition of More on Algebra, Lemma \ref{more-algebra-lemma-ext-1-annihilated} with respect to the ideal $IB$ (small detail omitted). Since $\NL_{B^\wedge/A}^\wedge = \NL_{B/A} \otimes_B B^\wedge$ by Lemma \ref{lemma-NL-is-completion} we conclude (2) holds by More on Algebra, Lemma \ref{more-algebra-lemma-base-change-property-ext-1-annihilated}. \medskip\noindent Assume (2). Choose a presentation $B = A[x_1, \ldots, x_n]/J$, set $N = J/J^2$, and consider the element $\xi \in \Ext^1_B(\NL_{B/A}, J/J^2)$ determined by the identity map on $J/J^2$. Using again that $\NL_{B^\wedge/A}^\wedge = \NL_{B/A} \otimes_B B^\wedge$ we find that our assumption implies the image $$ \xi \otimes 1 \in \Ext^1_{B^\wedge}(\NL_{B/A} \otimes_B B^\wedge, N \otimes_B B^\wedge) = \Ext^1_{B^\wedge}(\NL_{B/A}, N) \otimes_B B^\wedge $$ is annihilated by $I^c$ for some integer $c \geq 0$. The equality holds for example by More on Algebra, Lemma \ref{more-algebra-lemma-base-change-RHom} (but can also easily be deduced from the much simpler More on Algebra, Lemma \ref{more-algebra-lemma-pseudo-coherence-and-base-change-ext}). Thus $M = I^cB\xi \subset \Ext^1_B(\NL_{B/A}, N)$ is a finite submodule which maps to zero in $\Ext^1_B(\NL_{B/A}, N) \otimes_B B^\wedge$. Since $B \to B^\wedge$ is flat this means that $M \otimes_B B^\wedge$ is zero. By  Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) this means that $M = I^cB\xi$ is annihilated by an element of the form $g = 1 + x$ with $x \in IB$. This implies that for every $b \in I^cB$ there is a $B$-linear dotted arrow making the diagram commute $$ \xymatrix{ J/J^2 \ar[r] \ar[d]^b & \bigoplus B\text{d}x_i \ar@{..>}[d]^h \\ J/J^2 \ar[r] & (J/J^2)_g } $$ Thus $(\NL_{B/A})_{gb}$ is quasi-isomorphic to a finite projective module; small detail omitted. Since $(\NL_{B/A})_{gb} = \NL_{B_{gb}/A}$ in $D(B_{gb})$ this shows that $B_{gb}$ is smooth over $\Spec(A)$. As this holds for all $b \in I^cB$ we conclude that $\Spec(B_g) \to \Spec(A)$ is smooth over $\Spec(A) \setminus V(I)$ as desired.
By Lemma \ref{lemma-NL-base-change} there is a map $$ \NL_{B_1/A_1} \otimes_{B_2} B_1 \to \NL_{B_2/A_2} $$ which induces and isomorphism on $H^0$ and a surjection on $H^{-1}$. Thus the result by More on Algebra, Lemmas \ref{more-algebra-lemma-two-term-base-change}, \ref{more-algebra-lemma-base-change-property-ext-1-annihilated}, and \ref{more-algebra-lemma-surjection-property-ext-1-annihilated} the last two applied with the principal ideals $(f_1) \subset B_1$ and $(f_2) \subset B_2$.
Consider the obstruction class $$ o(\psi_n) \in \Ext^1_B(\NL_{B/A}^\wedge, I^nC/I^{2n}C) $$ of Remark \ref{remark-improve-homomorphism}. For any $C/I^nC$-module $N$ we have \begin{align*} \Ext^1_B(\NL_{B/A}^\wedge, N) & = \Ext^1_{C/I^nC}(\NL_{B/A}^\wedge \otimes_B^\mathbf{L} C/I^nC, N) \\ & = \Ext^1_{C/I^nC}(\NL_{B/A}^\wedge \otimes_B C/I^nC, N) \end{align*} The first equality by More on Algebra, Lemma \ref{more-algebra-lemma-upgrade-adjoint-tensor-RHom} and the second one by More on Algebra, Lemma \ref{more-algebra-lemma-two-term-base-change}. In particular, we see that $\Ext^1_{C/I^nC}(\NL_{B/A}^\wedge \otimes_B C/I^nC, N)$ is annihilated by $I^cC$ for all $C/I^nC$-modules $N$. It follows that we may apply Local Cohomology, Lemma \ref{local-cohomology-lemma-bound-two-term-complex} to see that $o(\psi_n)$ maps to zero in $$ \Ext^1_{C/I^nC}(\NL_{B/A}^\wedge \otimes_B C/I^nC, I^{n'}C/I^{2n'}C) = \Ext^1_B(\NL_{B/A}^\wedge, I^{n'}C/I^{2n'}C) = $$ where $n' = n - (d + 1)c$. By the discussion in Remark \ref{remark-improve-homomorphism} we obtain a map $$ \psi'_{2n'} : B \to C/I^{2n'}C $$ which agrees with $\psi_n$ modulo $I^{n'}$. Observe that $2n' > n$ because $n \geq 2(d + 1)c + 1$. \medskip\noindent We may repeat this procedure. Starting with $n_0 = n$ and $\psi^0 = \psi_n$ we end up getting a strictly increasing sequence of integers $$ n_0 < n_1 < n_2 < \ldots $$ and $A$-algebra homorphisms $\psi^i : B \to C/I^{n_i}C$ such that $\psi^{i + 1}$ and $\psi^i$ agree modulo $I^{n_i - tc}$. Since $C$ is $I$-adically complete we can take $\varphi$ to be the limit of the maps $\psi^i \bmod I^{n_i - (d + 1)c} : B \to C/I^{n_i - (d + 1)c}C$ and the lemma follows.
Consider the obstruction class $$ o(\psi_n) \in \Ext^1_B(\NL_{B/A}^\wedge, a^nC/a^{2n}C) $$ of Remark \ref{remark-improve-homomorphism}. Since $a$ is a nonzerodivisor on $C$ the map $a^c : a^nC/a^{2n}C \to a^nC/a^{2n}C$ is isomorphic to the map $a^nC/a^{2n}C \to a^{n - c}C/a^{2n - c}C$ in the category of $C$-modules. Hence by our assumption on $\NL_{B/A}^\wedge$ we conclude that the class $o(\psi_n)$ maps to zero in $$ \Ext^1_B(\NL_{B/A}^\wedge, a^{n - c}C/a^{2n - c}C) $$ and a fortiori in $$ \Ext^1_B(\NL_{B/A}^\wedge, a^{n - c}C/a^{2n - 2c}C) $$ By the discussion in Remark \ref{remark-improve-homomorphism} we obtain a map $$ \psi_{2n - 2c} : B \to C/a^{2n - 2c}C $$ which agrees with $\psi_n$ modulo $a^{n - c}C$. Observe that $2n - 2c > n$ because $n > 2c$. \medskip\noindent We may repeat this procedure. Starting with $n_0 = n$ and $\psi^0 = \psi_n$ we end up getting a strictly increasing sequence of integers $$ n_0 < n_1 < n_2 < \ldots $$ and $A$-algebra homorphisms $\psi^i : B \to C/a^{n_i}C$ such that $\psi^{i + 1}$ and $\psi^i$ agree modulo $a^{n_i - c}C$. Since $C$ is $I$-adically complete we can take $\varphi$ to be the limit of the maps $\psi^i \bmod a^{n_i - c}C : B \to C/a^{n_i - c}C$ and the lemma follows.
Let $C \to C'$ be the quotient of $C$ by $C[a^\infty]$. The $A$-algebra $C'$ is $I$-adically complete by Algebra, Lemma \ref{algebra-lemma-quotient-complete} and the fact that $\bigcap (C[a^\infty] + a^nC) = C[a^\infty]$ because for $n \geq d$ the sum $C[a^\infty] + a^nC$ is direct. For $m \geq d$ the diagram $$ \xymatrix{ 0 \ar[r] & C[a^\infty] \ar[r] \ar[d] & C \ar[r] \ar[d] & C' \ar[r] \ar[d] & 0 \\ 0 \ar[r] & C[a^\infty] \ar[r] & C/a^m C \ar[r] & C'/a^m C' \ar[r] & 0 } $$ has exact rows. Thus $C$ is the fibre product of $C'$ and $C/a^mC$ over $C'/a^mC'$ for all $m \geq d$. By Lemma \ref{lemma-get-morphism-nonzerodivisor} we can choose a homomorphism $\varphi' : B \to C'$ such that $\varphi'$ and $\psi_n$ agree as maps into $C'/a^{n - c}C'$. We obtain a homomorphism $(\varphi', \psi_n \bmod a^{n - c}C) : B \to C' \times_{C'/a^{n - c}C'} C/a^{n - c}C$. Since $n - c \geq d$ this is the same thing as a homomorphism $\varphi : B \to C$. This finishes the proof.
(A) By Definition \ref{definition-rig-smooth-homomorphism} we can choose a $c \geq 0$ such that $I^c$ annihilates $\Ext^1_B(\NL_{B/A}^\wedge, N)$ for all $B$-modules $N$. \medskip\noindent (B) By More on Algebra, Lemmas \ref{more-algebra-lemma-approximate-complex} and \ref{more-algebra-lemma-approximate-complex-graded} there exists a constant $c_1 = c(g_1, \ldots, g_m, K)$ such that for $n \geq c_1 + 1$ the complex $$ P^{\oplus t} \xrightarrow{K'} P^{\oplus m} \xrightarrow{g'_1, \ldots, g'_m} P \to B' \to 0 $$ is exact and $\text{Gr}_I(B) \cong \text{Gr}_I(B')$. \medskip\noindent (C) Let $d_0 = d(\text{Gr}_I(B))$ and $q_0 = q(\text{Gr}_I(B))$ be the integers found in Local Cohomology, Section \ref{local-cohomology-section-uniform}. \medskip\noindent We claim that $n = \max(c_1 + 1, q_0 + (d_0 + 1)c, 2(d_0 + 1)c + 1)$ works where $c$ is as in (A), $c_1$ is as in (B), and $q_0, d_0$ are as in (C). \medskip\noindent Let $g'_1, \ldots, g'_m$ and $K'$ be as in the lemma. Since $g_i = g'_i \in I^nP$ we obtain a canonical $A$-algebra homomorphism $$ \psi_n : B \longrightarrow B'/I^nB' $$ which induces an isomorphism $B/I^nB \to B'/I^nB'$. Since $\text{Gr}_I(B) \cong \text{Gr}_I(B')$ we have $d_0 = d(\text{Gr}_I(B'))$ and $q_0 = q(\text{Gr}_I(B'))$ and since $n \geq \max(q_0 + (1 + d_0)c, 2(d_0 + 1)c + 1)$ we may apply Lemma \ref{lemma-get-morphism-general-better} to find an $A$-algebra homomorphism $$ \varphi : B \longrightarrow B' $$ such that $\varphi \bmod I^{n - (d_0 + 1)c}B' = \psi_n \bmod I^{n - (d_0 + 1)c}B'$. Since $n - (d_0 + 1)c > 0$ we see that $\varphi$ is an $A$-algebra homomorphism which modulo $I$ induces the isomorphism $B/IB \to B'/IB'$ we found above. The rest of the proof shows that these facts force $\varphi$ to be an isomorphism; we suggest the reader find their own proof of this. \medskip\noindent Namely, it follows that $\varphi$ is surjective for example by applying Algebra, Lemma \ref{algebra-lemma-completion-generalities} part (1) using the fact that $B$ and $B'$ are complete. Thus $\varphi$ induces a surjection $\text{Gr}_I(B) \to \text{Gr}_I(B')$ which has to be an isomorphism because the source and target are isomorphic Noetherian rings, see Algebra, Lemma \ref{algebra-lemma-surjective-endo-noetherian-ring-is-iso} (of course you can show $\varphi$ induces the isomorphism we found above but that would need a tiny argument). Thus $\varphi$ induces injective maps $I^eB/I^{e + 1}B \to I^eB'/I^{e + 1}B'$ for all $e \geq 0$. This implies $\varphi$ is injective since for any $b \in B$ there exists an $e \geq 0$ such that $b \in I^eB$, $b \not \in I^{e + 1}B$ by Krull's intersection theorem (Algebra, Lemma \ref{algebra-lemma-intersect-powers-ideal-module-zero}). This finishes the proof.
By More on Algebra, Lemma \ref{more-algebra-lemma-henselization-Noetherian-pair} the ring $C^h$ is Noetherian. Say $J = (g_1, \ldots, g_m)$. The ring $C^h$ is a filtered colimit of \'etale $C$ algebras $C'$ such that $C/IC \to C'/IC'$ is an isomorphism (see proof of More on Algebra, Lemma \ref{more-algebra-lemma-henselization}). Pick an $C'$ such that $g_1, \ldots, g_m$ are the images of $g'_1, \ldots, g'_m \in C'$. Setting $B = C'/(g'_1, \ldots, g'_m)$ we get a finite type $A$-algebra. Of course $(C, IC)$ and $C', IC')$ have the same henselizations and the same completions. It follows easily from this that $B^\wedge = (C^h/J)^\wedge$.
The statement of the lemma makes sense as $C \to C'$ is flat (Algebra, Lemma \ref{algebra-lemma-etale}) hence induces an isomorphism $C/I^nC \to C'/I^nC'$ for all $n$ (More on Algebra, Lemma \ref{more-algebra-lemma-neighbourhood-isomorphism}) and hence an isomorphism on completions. Let $C^h$ be the henselization of the pair $(C, IC)$, see More on Algebra, Lemma \ref{more-algebra-lemma-henselization}. Then $C^h$ is the filtered colimit of the algebras $C'$ and the maps $C \to C' \to C^h$ induce isomorphism on completions (More on Algebra, Lemma \ref{more-algebra-lemma-henselization-Noetherian-pair}). Thus it suffices to prove there exists an $A$-algebra map $B \to C^h$ which is congruent to $\psi$ modulo $I^N$. Write $B = A[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$. The ring map $\psi$ corresponds to elements $\hat c_1, \ldots, \hat c_n \in C^\wedge$ with $f_j(\hat c_1, \ldots, \hat c_n) = 0$ for $j = 1, \ldots, m$. Namely, as $A$ is a Noetherian G-ring, so is $C$, see More on Algebra, Proposition \ref{more-algebra-proposition-finite-type-over-G-ring}. Thus Smoothing Ring Maps, Lemma \ref{smoothing-lemma-henselian-pair} applies to give elements $c_1, \ldots, c_n \in C^h$ such that $f_j(c_1, \ldots, c_n) = 0$ for $j = 1, \ldots, m$ and such that $\hat c_i - c_i \in I^NC^h$. This determines the map $B \to C^h$ as desired.
Let $T$ be the $m \times m$ matrix with entries $\partial f_j/\partial x_i$ for $1 \leq i, j \leq n$. Let $K \in D(B)$ be represented by the complex $T : B^{\oplus m} \to B^{\oplus m}$ with terms sitting in degrees $-1$ and $0$. By More on Algebra, Lemmas \ref{more-algebra-lemma-silly} we have $g : K \to K$ is zero in $D(B)$. Set $J = (f_1, \ldots, f_m)$. Recall that $\NL_{B/A}$ is homotopy equivalent to $J/J^2 \to \bigoplus_{i = 1, \ldots, n} B\text{d}x_i$, see Algebra, Section \ref{algebra-section-netherlander}. Denote $L$ the complex $J/J^2 \to \bigoplus_{i = 1, \ldots, m} B\text{d}x_i$ to that we have the quotient map $\NL_{B/A} \to L$. We also have a surjective map of complexes $K \to L$ by sending the $j$th basis element in the term $B^{\oplus m}$ in degree $-1$ to the class of $f_j$ in $J/J^2$. Picture $$ \NL_{B/A} \to L \leftarrow K $$ From More on Algebra, Lemma \ref{more-algebra-lemma-two-term-surjection-map-zero} we conclude that multiplication by $g$ on $L$ is $0$ in $D(B)$. On the other hand, the distinguished triangle $B^{\oplus n - m}[0] \to \NL_{B/A} \to L$ shows that $\Ext^1_B(L, N) \to \Ext^1_B(\NL_{B/A}, N)$ is surjective for every $B$-module $N$ and hence annihilated by $g$. This proves part (1). If $n = m$ then $\NL_{B/A} = L$ and we see that (2) holds.
Assume $B^\wedge$ satisfies the equivalent conditions of Lemma \ref{lemma-equivalent-with-artin}. The naive cotangent complex $\NL_{B/A}$ is a complex of finite type $B$-modules and hence $H^{-1}$ and $H^0$ are finite $B$-modules. Completion is an exact functor on finite $B$-modules (Algebra, Lemma \ref{algebra-lemma-completion-flat}) and $\NL_{B^\wedge/A}^\wedge$ is the completion of the complex $\NL_{B/A}$ (this is easy to see by choosing presentations). Hence the assumption implies there exists a $c \geq 0$ such that $H^{-1}/I^nH^{-1}$ and $H^0/I^nH^0$ are annihilated by $I^c$ for all $n$. By Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) this means that $I^cH^{-1}$ and $I^cH^0$ are annihilated by an element of the form $g = 1 + x$ with $x \in IB$. After inverting $g$ (which does not change the quotients $B/I^nB$) we see that $\NL_{B/A}$ has cohomology annihilated by $I^c$. Thus $A \to B$ is \'etale at any prime of $B$ not lying over $V(I)$ by the definition of \'etale ring maps, see Algebra, Definition \ref{algebra-definition-etale}. \medskip\noindent Conversely, assume that $\Spec(B) \to \Spec(A)$ is \'etale over $\Spec(A) \setminus V(I)$. Then for every $a \in I$ there exists a $c \geq 0$ such that multiplication by $a^c$ is zero $\NL_{B/A}$. Since $\NL_{B^\wedge/A}^\wedge$ is the derived completion of $\NL_{B/A}$ (see Lemma \ref{lemma-NL-is-limit}) it follows that $B^\wedge$ satisfies the equivalent conditions of Lemma \ref{lemma-equivalent-with-artin}.
By Lemma \ref{lemma-NL-base-change} there is a map $$ \NL_{B_1/A_1} \otimes_{B_2} B_1 \to \NL_{B_2/A_2} $$ which induces and isomorphism on $H^0$ and a surjection on $H^{-1}$. Thus the result by More on Algebra, Lemma \ref{more-algebra-lemma-two-term-surjection-map-zero}.
Uniqueness follows from the fact that $C^h$ is a subring of $C^\wedge$, see for example More on Algebra, Lemma \ref{more-algebra-lemma-henselization-Noetherian-pair}. The final assertion follows from the fact that $C^h$ is the filtered colimit of these $C$-algebras $C'$, see proof of More on Algebra, Lemma \ref{more-algebra-lemma-henselization}. Having said this we now turn to the proof of existence. \medskip\noindent Let $\varphi : B^\wedge \to C^\wedge$ be the given map. This defines a section $$ \sigma : (B \otimes_A C)^\wedge \longrightarrow C^\wedge $$ of the completion of the map $C \to B \otimes_A C$. We may replace $(A, I, B, C, \varphi)$ by $(C, IC, B \otimes_A C, C, \sigma)$. In this way we see that we may assume that $A = C$. \medskip\noindent Proof of existence in the case $A = C$. In this case the map $\varphi : B^\wedge \to A^\wedge$ is necessarily surjective. By Lemmas \ref{lemma-rig-etale} and \ref{lemma-exact-sequence-NL} we see that the cohomology groups of $\NL_{A^\wedge/\!_\varphi B^\wedge}^\wedge$ are annihilated by a power of $I$. Since $\varphi$ is surjective, this implies that $\Ker(\varphi)/\Ker(\varphi)^2$ is annihilated by a power of $I$. Hence $\varphi : B^\wedge \to A^\wedge$ is the completion of a finite type $B$-algebra $B \to D$, see More on Algebra, Lemma \ref{more-algebra-lemma-quotient-by-idempotent}. Hence $A \to D$ is a finite type algebra map which induces an isomorphism $A^\wedge \to D^\wedge$. By Lemma \ref{lemma-rig-etale} we may replace $D$ by a localization and assume that $A \to D$ is \'etale away from $V(I)$. Since $A^\wedge \to D^\wedge$ is an isomorphism, we see that $\Spec(D) \to \Spec(A)$ is also \'etale in a neighbourhood of $V(ID)$ (for example by More on Morphisms, Lemma \ref{more-morphisms-lemma-check-smoothness-on-infinitesimal-nbhds}). Thus $\Spec(D) \to \Spec(A)$ is \'etale. Therefore $D$ maps to $A^h$ and the lemma is proved.