I want to know if I unwrapped the definition of ideal sheaves correctly.
Let $\mathscr{O}$ be a sheaf of rings on topological space $X$. The definition of an ideal sheaf is as follows: The ideal sheaf $\mathscr{I}$ is a subsheaf of $\mathscr{O}$ as sheaves of abelian groups, such that for each open set $U\subset X$, $\mathscr{I}(U)$ is an ideal of $\mathscr{O}(U)$.
On the other hand by definition a subsheaf $\mathscr{F}'$ of a sheaf $\mathscr{F}$ is such that for every open subset $U$, the group $\mathscr{F}'(U)$ is a subgroup of $\mathscr{F}(U)$ and the restriction maps are $\rho_{UV}^{\mathscr{F}'}=\rho_{UV}^{\mathscr{F}}|_{\mathscr{F}'(U)}$.
Combining the two definition if $U\subset V$, and $\rho^\mathscr{O}_{UV}: \mathscr{O}(V)\to \mathscr{O}(U)$, then $\rho_{UV}^{\mathscr{I}}:\mathscr{I}(V)\to \mathscr{I}(U)$ only makes sense if $\rho^{\mathscr{O}}_{UV}(\mathscr{I}(V))\subset \mathscr{I}(U)\subset \mathscr{O}(U)$. The left hand side is the image of an ideal, which is not necessarily an ideal of $\mathscr{O}(U)$ unless $\rho^{\mathscr{O}}_{UV}$ is surjective.
So this suggests that $\mathscr{I}(V)$ is actually $\mathscr{I}(U)^e=\mathscr{I}(U)\cdot\mathscr{O}(V)$ the the extension of $\mathscr{I}(U)$. So is this true?