Lemma 52.24.1. In Situation 52.16.1 assume

$A$ has a dualizing complex and is $I$-adically complete,

$I = (f)$ generated by a single element,

$A$ is local with maximal ideal $\mathfrak a = \mathfrak m$,

one of the following is true

$A_ f$ is $(S_2)$ and for $\mathfrak p \subset A$, $f \not\in \mathfrak p$ minimal we have $\dim (A/\mathfrak p) \geq 4$, or

if $\mathfrak p \not\in V(f)$ and $V(\mathfrak p) \cap V(f) \not= \{ \mathfrak m\} $, then $\text{depth}(A_\mathfrak p) + \dim (A/\mathfrak p) > 3$.

Then with $U_0 = U \cap V(f)$ the completion functor

\[ \mathop{\mathrm{colim}}\nolimits _{U_0 \subset U' \subset U\text{ open}} \textit{Coh}(\mathcal{O}_{U'}) \longrightarrow \textit{Coh}(U, f\mathcal{O}_ U) \]

is an equivalence on the full subcategories of finite locally free objects.

**Proof.**
It follows from Lemma 52.15.8 that the functor is fully faithful (details omitted). Let us prove essential surjectivity. Let $(\mathcal{F}_ n)$ be a finite locally free object of $\textit{Coh}(U, f\mathcal{O}_ U)$. By either Lemma 52.20.4 or Proposition 52.22.2 there exists a coherent $\mathcal{O}_ U$-module $\mathcal{F}$ such that $(\mathcal{F}_ n)$ is the completion of $\mathcal{F}$. Namely, for the application of either result the only thing to check is that $(\mathcal{F}_ n)$ satisfies the $(2, 3)$-inequalities. This is done in Lemma 52.20.6. If $y \in U_0$, then the $f$-adic completion of the stalk $\mathcal{F}_ y$ is isomorphic to a finite free module over the $f$-adic completion of $\mathcal{O}_{U, y}$. Hence $\mathcal{F}$ is finite locally free in an open neighbourhood $U'$ of $U_0$. This finishes the proof.
$\square$

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