Lemma 86.5.5. Let $I = (a)$ be a principal ideal of a Noetherian ring $A$. Let $B$ be an object of (86.2.0.2). Assume given an integer $c \geq 0$ such that $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , N)$ is annihilated by $a^ c$ for all $B$-modules $N$. Let $C$ be an $I$-adically complete $A$-algebra. Assume given an integer $d \geq 0$ such that $C[a^\infty ] \cap a^ dC = 0$. Let $n > \max (2c, c + d)$. For any $A$-algebra map $\psi _ n : B \to C/a^ nC$ there exists an $A$-algebra map $\varphi : B \to C$ such that $\psi _ n \bmod a^{n - c} = \varphi \bmod a^{n - c}$.

**Proof.**
Let $C \to C'$ be the quotient of $C$ by $C[a^\infty ]$. The $A$-algebra $C'$ is $I$-adically complete by Algebra, Lemma 10.96.10 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

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 86.5.4 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. $\square$

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