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
Comments (0)
There are also: