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Tag 00A8

Chapter 6: Sheaves on Spaces > Section 6.31: Open immersions and (pre)sheaves

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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