Restriction of a sheaf of modules Let $X$ be a scheme and $Y$ be a closed subscheme. For $\mathcal{F}$ a sheaf of modules on $X$ to be the pushforward of a sheaf of modules on $Y$ via the inclusion $i: Y \rightarrow X$ is necessary and sufficient that $\mathcal{I}\mathcal{F} = 0$, where $\mathcal{I}$ is the ideal defining $Y$. When we consider sheaves on a topological space $Z$ we know that a sheaf $\mathcal{F}$ such that $\mathcal{F} \vert_U = 0$, $U \subset Z$ open, is isomorphic to $j_{\ast} j^{-1} \mathcal{F}$, where $j : Z - U \rightarrow X$. I think that the same is not true for schemes, as the condition above seems stronger to me as $\mathcal{I}\mathcal{F}$ need not to be zero on $Y$ if it is on $X-Y$, but I am not able to find a counterexample. Any help?
 A: Working from the assumption that your question is: 

Are the conditions $\mathcal{I}\mathcal{F}=0$ and $\mathcal{F}|_U=0$ equivalent for a sheaf of $\mathcal{O}_X$-modules?

Edit:
I was careless before. The correct answer is that they are not equivalent. $\newcommand\calF{\mathcal{F}}\newcommand\calI{\mathcal{I}}\newcommand\calO{\mathcal{O}}\newcommand\pp{\mathfrak{p}}\newcommand\Spec{\operatorname{Spec}}$
Counterexample:
Take the sky scraper sheaf $k[x]_{(x)}$ on $\Bbb{A}^1$ at $(x)$. This is a sheaf of $\calO_X$-modules in the obvious way, but $(x)k[x]_{(x)}$ is certainly nontrivial. (Note that this is not a quasicoherent sheaf though.) For a quasicoherent sheaf, take the module $k[x]/(x^2)$ over $k[x]$, which has support precisely $(x)$, but $(x)k[x]/(x^2)\ne 0$. 
However $\calI\calF=0$ does imply $\calF|_U=0$, so you are correct, $\calI\calF$ is stronger.
Proof.
If $\calI\calF=0$, but $\calF|_U\ne 0$, then choose a point $x\in U$ with $\calF_x\ne 0$. Now restrict to an affine open nhood of $x$, $\Spec R$ with $x=\pp$, and $\calI$ and $\calF$ given by an ideal $I\subseteq R$ and $R$-module $M$ respectively on this open subset. Then $IM=0$. Since $x=\pp\not\in Z$, $I\not\subseteq \pp$, but then localizing at $\pp$, we have $I_\pp = R_\pp$, but then we get
$$0=(IM)_\pp=I_\pp M_\pp = R_\pp M_\pp = M_\pp \ne 0,$$
contradiction. $\blacksquare$
Another way to see this is the fact that $\calI\calF = 0$ implies that $\calI$ is a subsheaf of the annihilator sheaf of $\calF$, so $Y$ contains the support of $M$ (the closed subscheme cut out by the annihilator sheaf). Thus $\calF|_U=0$.
