sheaf defined on the support of a sheaf Let $X$ be a topological space and $\mathcal{F}$ a sheaf on $X$. Let $Y$ be the support of $\mathcal{F}$ (Hartshorne, exercise 1.14), i.e., $Y = \left\{ P \in X: \, F_P \neq 0\right\}$. Is it true that $F$ induces a sheaf $G$ on $Y$? What is the precise definition? Do we need $Y$ to be closed for such sheaf $G$ to exist? In what cases is this sheaf isomorphic to $F$ (see comment by Georges)?
Motivation: Let $(X,\mathcal{O}_X)$ be a scheme and $\mathcal{I}$ a quasi-coherent ideal sheaf on $X$ with support $Y$. Then Hartshorne casually says that $(Y,\mathcal{O}_X/\mathcal{I})$ is a scheme (proof of Proposition II.5.9); yet some work is required to show that $\mathcal{O}_X/\mathcal{I}$ is a sheaf on $Y$.
Remark: Intuitively, it would seem that we could define $G$ as follows: For any open $U'$ inside $Y$ let $T_{U'}$ be the set of all open sets $U$ of $X$ such that $U \cap Y = U'$. Then define $G(U') = \lim_{U \in T_{U'}} F(U)$.
 A: 1) Yes, the sheaf  $\mathcal F$ induces a sheaf on its support $Y$, independently of whether $Y$ is closed or not.   
2) Actually, given a completely arbitrary subset $S\subset X$, there exists a restricted sheaf $\mathcal F\vert S$ on $S$, whose stalks satisfy $(\mathcal F\vert S)_s=\mathcal F_s$ for all $s\in S$.
Beware that it doesn't even make sense to inquire about  $(\mathcal F\vert S)_x$ for $x\in X\setminus S$ because $\mathcal F\vert S$ is only defined on $S$ and has no stalk at a point $x$ not in $S$ !   
3) Even more generally, given a continuous map of topological spaces $f: Z\to X$ and a sheaf  $\mathcal F$ on $X$, there is an associated sheaf $f^{-1}\mathcal F$  on $Z$ satisfying $(f^{-1}\mathcal F)_z=(\mathcal F)_{f(z)}$ for all $z\in Z$, and the special case of an inclusion $j:S\subset X$ yields the restriction sheaf $\mathcal F\vert S=j^{-1}(\mathcal F)$.  
4) About your last question : in the case of a subset inclusion  $j: Y \hookrightarrow  X$ there is a canonical morphism $\mathcal F\to j_\ast (\mathcal F\vert Y)$ of sheaves on $X$.
If $Y$ is closed and if $\mathcal F$ has support in $Y$, then that canonical morphism is an isomorphism of sheaves on $X$: $$\mathcal F\stackrel {\cong}{\to} j_\ast (\mathcal F\vert Y)=j_\ast (j^{-1}\mathcal F)$$
