Properties of local ring of restriction sheaf Let $X$ be a scheme and $\mathcal{F}$ a sheaf on $X$. I am trying to figure out if it is true that for any open affine $U \subset X$ such that $x \in U$, we have that $$\mathcal{F}_x = (\mathcal{F} \vert_U)_x$$
To show this I would need to show that every equivalence class of $\mathcal{F}_x$ contains $(W, s)$ where $W \subset U$. 
I am having a hard time showing this, maybe because it isn't even a true statement. 
I was thinking that maybe it is possible to show the existence of an affine open so that we have the above equality.
Are either of these ideas correct? 
 A: Let $\mathcal F$ be a sheaf on a topological space $X$, and $x \in X$.  The set $\mathcal S$ of open neighborhoods of $x$ is directed: one says that $U \leq V$ if and only if $U$ contains $V$.  This gives us for each pair of elements $U \leq V$ in $\mathcal S$ a morphism $$\rho_{UV}: \mathcal F(U) \rightarrow \mathcal F(V)$$ with the property that $\rho_{UU} = 1_{\mathcal F(U)}$ and $\rho_{UV} \circ \rho_{VW} = \rho_{UW}$ for $U \leq V \leq W$.  Thus $\mathcal S$, the objects $\mathcal F(U) : U \in \mathcal S$ and the morphisms $\rho_{UV} : U \leq V$ form a directed system, and the stalk $\mathcal F_x$ is by definition the direct limit $$\varinjlim \mathcal F(U)$$ It is an object, $\mathcal F_x$, together with morphisms $r_U: \mathcal F(U) \rightarrow \mathcal F_x : U \in \mathcal S$, satisfying the following property:
$\ast$:If $A$ is any object, and $s_U: \mathcal F(U) \rightarrow A, U \in \mathcal S$ is a collection of morphisms with the property that $s_U = s_V \circ \rho_{UV}$ for $U \leq V$, then there is a unique morphism $\eta: \mathcal F_x \rightarrow A$ such that $r \circ s_U = r_U$ for all $U \in \mathcal S$.
The object $\mathcal F_x$ together with the morphisms $r_U$ is unique: if $\mathcal F'_x$ is another object, together with morphisms $r'_U: \mathcal F(U) \rightarrow \mathcal F'_x : U \in \mathcal S$ satisfying a similar property $\ast$, then there is a unique isomorphism $\eta: \mathcal F_x \rightarrow \mathcal F'_x$  such that $\eta \circ r_U = r'_U$ for all $U \in \mathcal S$.
Now, fix an open neighborhood $W$ of $x$, and let $\mathcal S' \subseteq \mathcal S$ be the set of open neighborhoods of $x$ which are contained in $W$.  Then $(\mathcal F|W)_x$ is defined analogously as a direct limit of $\mathcal F(U) : U \in \mathcal S'$.  It comes with morphisms $\mathfrak r_U: \mathcal F(U) \rightarrow (\mathcal F|W)_x$ satisfying a similar universal property to $\ast$.
You want to show that $\mathcal F_x$ and $(\mathcal F|W)_x$ are isomorphic.  This can be done by showing that $\mathcal F_x$ is still the direct limit of the $\mathcal F(U)$, except now you only consider those $U$ in the smaller indexing set $\mathcal S'$.  This will imply that $\mathcal F_x$ satisfies the same universal property as $(\mathcal F|W)_x$, hence there is a unique isomorphism $(\mathcal F|W)_x \rightarrow \mathcal F_x$ satisfying some nice compatability properties.
