Lemma 58.17.1. Let $X$ be a Noetherian scheme and let $Y \subset X$ be a closed subscheme with ideal sheaf $\mathcal{I} \subset \mathcal{O}_ X$. Assume the completion functor

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

is fully faithful on the full subcategory of finite locally free objects (see above). Then the restriction functor $\textit{FÉt}_ X \to \textit{FÉt}_ Y$ is fully faithful.

Proof. 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 $X$ and a morphism $t : Y \to U$ over $X$ there exists a unique section $s : X \to U$ such that $t = s|_ Y$. Picture

$\xymatrix{ & U \ar[d]^ f \\ Y \ar[r] \ar[ru] & X \ar@{..>}@/^1em/[u] }$

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

$s^\sharp : f_*\mathcal{O}_ U \longrightarrow \mathcal{O}_ X$

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 $X$. Since $f_*\mathcal{O}_ U$ is a finite locally free $\mathcal{O}_ X$-module, we conclude that we get a unique $\mathcal{O}_ X$-module map $\sigma : f_*\mathcal{O}_ U \to \mathcal{O}_ X$ 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

$\xymatrix{ f_*\mathcal{O}_ U \otimes _{\mathcal{O}_ X} f_*\mathcal{O}_ U \ar[r] \ar[d]_{\sigma \otimes \sigma } & f_*\mathcal{O}_ U \ar[d]^\sigma \\ \mathcal{O}_ X \otimes _{\mathcal{O}_ X} \mathcal{O}_ X \ar[r] & \mathcal{O}_ X }$

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 conclude. $\square$

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