Lemma 89.7.4. In the situation above, $\widehat{\mathcal{F}}_\mathcal {I}(R)$ is equivalent to the category $\widehat{\mathcal{F}}(R)$.

Proof. An equivalence $\widehat{\mathcal{F}}_\mathcal {I}(R) \to \widehat{\mathcal{F}}(R)$ can be defined as follows. For each $n$, let $m(n)$ be the least $m$ that $I_ m \subset \mathfrak m_ R^ n$. Given an object $(\xi _ n, f_ n)$ of $\widehat{\mathcal{F}}_\mathcal {I}(R)$, let $\eta _ n$ be the pushforward of $\xi _{m(n)}$ along $R/I_{m(n)} \to R/\mathfrak m_ R^ n$. Let $g_ n : \eta _{n + 1} \to \eta _ n$ be the unique morphism of $\mathcal{F}$ lying over $R/\mathfrak m_ R^{n + 1} \to R/\mathfrak m_ R^ n$ such that

$\xymatrix{ \xi _{m(n + 1)} \ar[rrr]_{f_{m(n)} \circ \ldots \circ f_{m(n + 1) - 1}} \ar[d] & & & \xi _{m(n)} \ar[d] \\ \eta _{n + 1} \ar[rrr]^{g_ n} & & & \eta _ n }$

commutes (existence and uniqueness is guaranteed by the axioms of a cofibred category). The functor $\widehat{\mathcal{F}}_\mathcal {I}(R) \to \widehat{\mathcal{F}}(R)$ sends $(\xi _ n, f_ n)$ to $(R, \eta _ n, g_ n)$. We omit the verification that this is indeed an equivalence of categories. $\square$

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