Lemma 6.31.9. Let $X$ be a topological space. Let $j : U \to X$ be the inclusion of an open subset. The functor

$j_! : \mathop{\mathit{Sh}}\nolimits (U) \longrightarrow \mathop{\mathit{Sh}}\nolimits (X)$

is fully faithful. Its essential image consists exactly of those sheaves $\mathcal{G}$ such that $\mathcal{G}_ x = \emptyset$ for all $x \in X \setminus U$.

Proof. Fully faithfulness follows formally from $j^{-1} j_! = \text{id}$. We have seen that any sheaf in the image of the functor has the property on the stalks mentioned in the lemma. Conversely, suppose that $\mathcal{G}$ has the indicated property. Then it is easy to check that

$j_! j^{-1} \mathcal{G} \to \mathcal{G}$

is an isomorphism on all stalks and hence an isomorphism. $\square$

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