**Proof.**
Consider the obstruction class

\[ o(\psi _ n) \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , I^ nC/I^{2n}C) \]

of Remark 88.5.2. For any $C/I^ nC$-module $N$ we have

\begin{align*} \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , N) & = \mathop{\mathrm{Ext}}\nolimits ^1_{C/I^ nC}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B^\mathbf {L} C/I^ nC, N) \\ & = \mathop{\mathrm{Ext}}\nolimits ^1_{C/I^ nC}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C/I^ nC, N) \end{align*}

The first equality by More on Algebra, Lemma 15.99.1 and the second one by More on Algebra, Lemma 15.84.6. In particular, we see that $\mathop{\mathrm{Ext}}\nolimits ^1_{C/I^ nC}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C/I^ nC, N)$ is annihilated by $I^ cC$ for all $C/I^ nC$-modules $N$. It follows that we may apply Local Cohomology, Lemma 51.22.7 to see that $o(\psi _ n)$ maps to zero in

\[ \mathop{\mathrm{Ext}}\nolimits ^1_{C/I^ nC}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C/I^ nC, I^{n'}C/I^{2n'}C) = \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , I^{n'}C/I^{2n'}C) = \]

where $n' = n - (d + 1)c$. By the discussion in Remark 88.5.2 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 $n \geq 2(d + 1)c + 1$.

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.
$\square$

## Comments (0)