Let $X$ be a topological space, and $U$ an open subset. Denote $j:U\to X$ the inclusion.

Let $\mathcal F$ be a sheaf on $U$. We define $j_!\mathcal F$ to be the sheaf associated to the presheaf $$\mathcal V\mapsto\left\{ \begin{array}{ll} \mathcal F(V) & \mbox{if } V\subseteq U \\ 0 & \mbox{otherwise} \end{array} \right.$$

I know that the stalk $(j_!\mathcal F)_x$ is equal to $\mathcal F_x$ for $x\in U$, and $0$ for $x\notin U$, and that $j_!\mathcal F$ is the only sheaf with this property.

Is it true that $j_!\mathcal F(V)=0$ whenever $V\nsubseteq U$? Is this true under suitable conditions on $X$?

I'm particularly interested in the case that $X=\mathbb P^1_\mathbb C$ and $\mathcal F=\mathcal O_X$.


No, it is not true. Here is a counterexample:

Take $X=\mathbb C$, $U=\mathbb D(0;1)$ (the open disk of radius $1$ centered at zero) and $\mathcal F=\mathcal C^\infty _U$, the sheaf of smooth functions on $U$ .
Let now $f\in \mathcal C^\infty (\mathbb C)$ be a bump function with compact support $supp(f)\subset U$, such that $f(0)=1$.
This function can be seen as a non-zero section $0\neq f\in \Gamma(\mathbb C,j_!(\mathcal C^\infty_U))$ and you have your counterexample with $V=\mathbb C\nsubseteq U$

Edit: a general result
Let $M$ be $\mathcal C^k$ manifold and $U\subset M$ a relatively compact open subset .
Then the restriction morphism $$\operatorname {res}:\Gamma(M, j_!(\mathcal C^k_U))\stackrel {\cong}{\to}\mathcal C^k_c(U):\phi\mapsto \phi|U$$ between the global sections of $j_!( \mathcal C^k_U)$ and the $\mathcal C^k$ functions defined only on $\text U$ and with compact support is an isomorphism .
This result explains the above example and may help build an intuition for the functor $j_!$.

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  • $\begingroup$ Dear @Benja, I already gave an answer here $\endgroup$ – Georges Elencwajg Jan 9 '14 at 15:54

Let $j_{!}^{\operatorname{pre}}\mathcal{F}$ denote the presheaf on $X$ sending $V \mapsto \mathcal{F}(V)$ if $V \subseteq U$ and $V \mapsto 0$ if $V \not\subseteq U$. Here is a sufficient condition for $j_{!}^{\operatorname{pre}}\mathcal{F}$ to be a sheaf.

Claim: Suppose $X$ is irreducible and, for any inclusion $V_{2} \subseteq V_{1}$ of nonempty open subsets of $U$, the restriction map $\mathcal{F}(V_{1}) \to \mathcal{F}(V_{2})$ is injective. Then $j_{!}^{\operatorname{pre}}\mathcal{F}$ is a sheaf.

(Note that the hypotheses are satisfied if $X$ is irreducible and $\mathcal{F}$ is a constant sheaf, or if $X$ is an integral scheme and $\mathcal{F} = \mathcal{O}_{U}$.)

Proof: Let $V$ be a nonempty open subset of $X$, and let $V = \bigcup_{\alpha} V_{\alpha}$ be an open cover of $V$. Suppose $s \in j_{!}^{\operatorname{pre}}\mathcal{F}(V)$ such that $s|_{V_{\alpha}} = 0$ for all $\alpha$. If $V \subset U$, then we have $s = 0$ since $\mathcal{F}$ is a sheaf. If $V \not\subseteq U$, then $s = 0$ by definition of $j_{!}^{\operatorname{pre}}\mathcal{F}(V)$. Thus $j_{!}^{\operatorname{pre}}\mathcal{F}$ is a separated presheaf.

Let $V$ be a nonempty open subset of $X$, and let $V = \bigcup_{\alpha \in I} V_{\alpha} \cup \bigcup_{\beta \in J}V_{\beta}'$ be an open cover of $V$ such that $V_{\alpha} \subseteq U$ for all $\alpha$ and $V_{\beta}' \not\subseteq U$ for all $\beta$. We have that $V \subseteq U$ if and only if $J = \emptyset$; if either $I = \emptyset$ or $J = \emptyset$, then the sheaf condition is satisfied; suppose $I \ne \emptyset$ and $J \ne \emptyset$. Let $\{s_{\alpha}\}_{\alpha \in I} \cup \{s_{\beta}'\}_{\beta \in J}$, with $s_{\alpha} \in j_{!}^{\operatorname{pre}}\mathcal{F}(V_{\alpha})$ and $s_{\beta}' \in j_{!}^{\operatorname{pre}}\mathcal{F}(V_{\beta}')$, be a compatible collection of sections (i.e. which agree on intersections); here $s_{\beta}' = 0$ for all $\beta$ since $j_{!}^{\operatorname{pre}}\mathcal{F}(V_{\beta}') = 0$. Since $X$ is irreducible, we have $V_{\alpha} \cap V_{\beta}' \ne \emptyset$ for every $\alpha$; then $s_{\alpha}|_{V_{\alpha} \cap V_{\beta}'} = s_{\beta}'|_{V_{\alpha} \cap V_{\beta}'} = 0|_{V_{\alpha} \cap V_{\beta}'} = 0$ in $j_{!}^{\operatorname{pre}}\mathcal{F}(V_{\alpha} \cap V_{\beta}') = \mathcal{F}(V_{\alpha} \cap V_{\beta}')$. Since $\mathcal{F}(V_{\alpha}) \to \mathcal{F}(V_{\alpha} \cap V_{\beta}')$ is injective by assumption, we have that $s_{\alpha} = 0$. Then the compatible collection consists of zero sections, so $s=0 \in j_{!}^{\operatorname{pre}}\mathcal{F}(V)$ restricts to all $s_{\alpha}$ and $s_{\beta}'$.

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