TITLE: On the $j_!$ of a sheaf
QUESTION [10 upvotes]: 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 $\mathcal F=\mathcal O_X$.

REPLY [2 votes]: 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 $s_{\beta}'$.