Lemma 58.17.3. Let $X$ be a Noetherian scheme and let $Y \subset X$ be a closed subscheme with ideal sheaf $\mathcal{I} \subset \mathcal{O}_ X$. Let $\mathcal{V}$ be the set of open subschemes $V \subset X$ containing $Y$ ordered by reverse inclusion. Assume the completion functor

$\mathop{\mathrm{colim}}\nolimits _\mathcal {V} \textit{Coh}(\mathcal{O}_ V) \longrightarrow \textit{Coh}(X, \mathcal{I}), \quad \mathcal{F} \longmapsto \mathcal{F}^\wedge$

defines is fully faithful on the full subcategory of finite locally free objects (see above). Then the restriction functor $\mathop{\mathrm{colim}}\nolimits _\mathcal {V} \textit{FÉt}_ V \to \textit{FÉt}_ Y$ is fully faithful.

Proof. Observe that $\mathcal{V}$ is a directed set, so the colimits are as in Categories, Section 4.19. The rest of the argument is almost exactly the same as the argument in the proof of Lemma 58.17.1; we urge the reader to skip it.

Since the category of finite étale coverings has an internal hom (Lemma 58.5.4) it suffices to prove the following: Given $U$ finite étale over $V \in \mathcal{V}$ and a morphism $t : Y \to U$ over $V$ there exists a $V' \geq V$ and a morphism $s : V' \to U$ over $V$ such that $t = s|_ Y$. Picture

$\xymatrix{ & & U \ar[d]^ f \\ Y \ar[r] \ar[rru] & V' \ar@{..>}[ru] \ar[r] & V }$

Finding the dotted arrow $s$ is the same thing as finding an $\mathcal{O}_{V'}$-algebra map

$s^\sharp : f_*\mathcal{O}_ U|_{V'} \longrightarrow \mathcal{O}_{V'}$

which reduces modulo the ideal sheaf of $Y$ to the given algebra map $t^\sharp : f_*\mathcal{O}_ U \to \mathcal{O}_ Y$. By Lemma 58.8.3 we can lift $t$ uniquely to a compatible system of maps $t_ n : Y_ n \to U$ and hence a map

$\mathop{\mathrm{lim}}\nolimits t_ n^\sharp : f_*\mathcal{O}_ U \longrightarrow \mathop{\mathrm{lim}}\nolimits \mathcal{O}_{Y_ n}$

of sheaves of algebras on $V$. Observe that $f_*\mathcal{O}_ U$ is a finite locally free $\mathcal{O}_ V$-module. Hence we get a $V' \geq V$ a map $\sigma : f_*\mathcal{O}_ U|_{V'} \to \mathcal{O}_{V'}$ whose completion is $\mathop{\mathrm{lim}}\nolimits t_ n^\sharp$. To see that $\sigma$ is an algebra homomorphism, we need to check that the diagram

$\xymatrix{ (f_*\mathcal{O}_ U \otimes _{\mathcal{O}_ V} f_*\mathcal{O}_ U)|_{V'} \ar[r] \ar[d]_{\sigma \otimes \sigma } & f_*\mathcal{O}_ U|_{V'} \ar[d]^\sigma \\ \mathcal{O}_{V'} \otimes _{\mathcal{O}_{V'}} \mathcal{O}_{V'} \ar[r] & \mathcal{O}_{V'} }$

commutes. For every $n$ we know this diagram commutes after restricting to $Y_ n$, i.e., the diagram commutes after applying the completion functor. Hence by faithfulness of the completion functor we deduce that there exists a $V'' \geq V'$ such that $\sigma |_{V''}$ is an algebra homomorphism as desired. $\square$

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