# The Stacks Project

## Tag 00A8

Lemma 6.31.9. Let $X$ be a topological space. Let $j : U \to X$ be the inclusion of an open subset. The functor $$j_! : \mathop{\textit{Sh}}\nolimits(U) \longrightarrow \mathop{\textit{Sh}}\nolimits(X)$$ 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. $\square$

The code snippet corresponding to this tag is a part of the file sheaves.tex and is located in lines 4751–4762 (see updates for more information).

\begin{lemma}
\label{lemma-equivalence-categories-open}
Let $X$ be a topological space.
Let $j : U \to X$ be the inclusion of an open subset.
The functor
$$j_! : \Sh(U) \longrightarrow \Sh(X)$$
is fully faithful. Its essential image consists exactly
of those sheaves $\mathcal{G}$ such that
$\mathcal{G}_x = \emptyset$ for all $x \in X \setminus U$.
\end{lemma}

\begin{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.
\end{proof}

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