Let $X$ be a topological space and $\mathcal{F}$ a sheaf on $X$ such that $$\mathcal{F}_x\neq 0$$ $$\mathcal{F}_y=0\,\forall\,y\in X\setminus\{x\}$$ $$\overline{\{x\}}=x$$
I'm trying to figure out if the following is true:
For all open neighbourhood $U$ of $x$, we have $\mathcal{F}(U)\simeq \mathcal{F}_x$.
I know there is an injective morphism $\mathcal{F}(U)\hookrightarrow \prod_{p\in U}\mathcal{F}_p\simeq \mathcal{F}_x$. To prove surjectivity, we must take an arbitrary $f\in\mathcal{F}(V)$ for some $V$ containing $x$ and find $g\in\mathcal{F}(U)$ such that $f\big|_W=g\big|_W$ for some $W\subset U\cap V$ containing $x$.
Is it possible to do?
By the condition that $\mathcal{F}_y=0$ for all $y\neq x$ plus the injective morphism $\mathcal{F}(U)\to \prod_{p\in U} \mathcal{F}_p$, we see that if $U$ is an open subset of $X$ not containing $x$, then $\mathcal{F}(U)=0$. Now let $V\subset W$ be two open neighborhoods of $x$, and write $W=V\cup (W\setminus \{x\})$. The sheaf $\mathcal{F}$ takes the value zero on the latter set in this decomposition, because $W\setminus \{x\}$ is an open subset of $X$ not containing $x$. We then get that $\mathcal{F}(V)=\mathcal{F}(W)$ by the sheaf condition. As any two neighborhoods of $x$ have a common refinement, we see that the object $\mathcal{F}(V)$ for any open subset $V\subset X$ containing $x$ must be the same, and thus the stalk which is the limit over all open neighborhoods of $x$ must be exactly this object. So we're done.