The equivalence of the definitions of stalks The definition of direct limits is:

I'm trying to see how this definition works in the stalks:
The index $I$ is the open sets containing $x$ under the inclusion and the restrictions $\rho_{UV}:\mathcal F(U)\to \mathcal F(V)$ are the $f_{ik}$.
The $f_i$ are the maps where $s\in \mathcal F(U)$, $s\mapsto s_x\in \mathcal F_x$
I hope I'm right by now, I continue...
Then, In order prove the definition of direct limit is equivalent with the definition of $\mathcal F_x$, I have to demonstrate the diagram commutes using the definition of $\mathcal F_x$ using equivalence classes:

I'm struggling to prove the commutativity of the inner triangle $(f_j\circ f_{ij}=f_i)$, I need help in this part.
Thanks a lot
 A: Ok, I see: you need to verify that, actually, definition 1.3 of these notes gives you a direct limit. Particularly, remarks 1.4 (2). Isn't it?
Well, I would say that this is by definition again.   :-)
You want to prove that $\tau_U = \tau_V \circ \rho_{VU}$, where $V \subset U$.
So, take any $s \in \mathcal{F}(U)$. Then, on one hand 
$$
\tau_U(s) = [(s,U)] \ .
$$
(I'm omitting this subindex $\sim$: we all understand it.) On the other hand,
$$
(\tau_V \circ \rho_{VU})(s) = [(\rho_{VU}(s) = s_{\vert V}, V)] \ .
$$
By definition, two classes in $\mathcal{F}_x$ are the same if there is some $W \subset U\cap V$ such that, when restricted to this $W$ they coincide, right?
Ok, in our case, just take $W = V$. Of course, $s_{\vert V} = s_{\vert V}$.  :-)
A: Use the explicit construction of directed colimits (unfortunately often called direct limits, although they are not limits), which I have mentioned for example in your previous question. An element of $\mathrm{colim}_{x \in U \text{ open}} F(U)$ is an equivalence class $[s]$, where $s \in F(U)$ for some $x \in U$ open, and we have $[s]=[t]$ for $s \in F(U), t \in F(V)$ iff there are maps $W \to U$ and $W \to V$ in the index category, i.e. inclusions $W \subseteq U$, $W \subseteq V$, such that $s,t$ have the same image in $F(W)$, i.e. $s|_W = t|_W$.
