My today's question is about a proof of this book. More precisely we are talking about the proof of Prop. 2.24 on page 52. The book says that we have $\tilde{\mathscr{F}_x}=\mathscr{F}_x$ for all $x\in X$. I tried to verify that but unfolding the definition of elements of $\tilde{\mathscr{F}_x}$ is very technical. So I got stuck. Furthermore, it can't be meant literally because elements of $\tilde{\mathscr{F}_x}$ and $\mathscr{F}_x$ aren't of the same type so to speak. So what is meant by that equality?
Here my approach to unfold the definition of elements of $\tilde{\mathscr{F}_x}$:
An element of $\tilde{\mathscr{F}_x}$ is the equivalence class of a pair $(U,(s_y)_{y\in U})$ with $x\in U$. Each pair $(U',(s'_y)_{y\in U'})$ is equivalent to $(U,(s_y)_{y\in U})$ iff there is a $V\subseteq U\cap U'$ with $x\in V$ and $(s_y)_{y\in U}|_V=(s'_y)_{y\in U'}|_V$. By definition of the restriction that means for all $y\in V$ we have $s_y=s'_y$. Now the equality $s_y=s'_y$ could be itself unfold in a similar way since $s_y$ is again an equivalence class of pairs.
Is it correct what I did so far and does it help me to show $\tilde{\mathscr{F}_x}=\mathscr{F}_x$?
I am not sure if this is what you are getting at but if you have $$s_x\in \mathcal{F}_x=\lim\mathcal{F}(U)$$ then there is some $U$ and $t\in \mathcal{F}(U)$ such that $t$ restricts to $s_x$. So the equivalalence class of the germ $U, (t_y)$ about the point $x$ will be $s_x$.