What is the stalk $F_0$ of the presheaf $F$ at $0 \in X = \{0, 1\}$ given the discrete topology? If you can look at example 1.5 here, it is shown that $F_0= F(\{0\})$.
I was wondering if anyone here can explain in some more detail the example given.
I am confused by how we know that there is some $b \in F(\{0\})$ such that $a\sim  b$? How does it follow that $F_0 \subseteq F(\{0\})$ and why do we have $a = b$? How do we get $F(\{0\})\subseteq F_0$?
 A: The most important thing to understand is the equivalence relation. In general $s\sim t$ with $s\in\mathcal{F}(U)$ and $t\in\mathcal{F}(V)$ iff there exists $W\subset U\cap V$ such that $s_{|W}=t_{|W}$. In particular $s$ is always equivalent to its restriction ! In other words : $s\simeq s_{|W}$ for any $W\subset U$.
So let us see what is going on with a sheaf $\mathcal{F}$ on $X=\{0,1\}$. The stalk at $0$ is given by $\mathcal{F}(\{0\})\sqcup\mathcal{F}(\{0,1\})/\sim$
Claim 1 : Every element in the stalk is equivalent to some element in $\mathcal{F}(\{0\})$. Indeed, it $a\in\mathcal{F}_0$, then either $a\in\mathcal{F}(\{0\})$ and we are done, or $a\in\mathcal{F}(\{0,1\})$. But then $a\sim a_{|\{0\}}$ and $a_{|\{0\}}\in\mathcal{F}(\{0\})$. This proves the claim.
We just showed that the map $\mathcal{F}(\{0\})\rightarrow\mathcal{F}_0$ is onto.
Claim 2 : it is also into. Indeed, if $a,b\in\mathcal{F}(\{0\})$ have the same image, then $a\sim b$, but this mean that there is $W\subset\{0\}$ such that $a_{|W}=b_{|W}$. Since $0\in W\subset\{0\}$, we see that $W=\{0\}$. Hence $a=a_{|W}$, same for $b$, and thus $a=b$. This prove the claim.
Conclusion : $\mathcal{F}(\{0\})\rightarrow\mathcal{F}_0$ is bijective. We may write it as an equality of sets.
(Note however that I don't like the claim in the linked paper that $\mathcal{F}_0\subset\mathcal{F}(\{0\})$. Strictly speaking, this is not an inclusion. Moreover, and most importantly, the claims was meant to say that any element of the stalk is equivalent to some element of $\mathcal{F}(\{0\})$, but it does not say that this element was unique (this follows from the second part of the proof).)
A: The index set for the direct limit forming the stalk, in this case, has a largest element $\{ 0 \}$.  (Since the ordering is by reverse inclusion, the largest element with respect to this ordering is actually the smallest open neighborhood of 0 with respect to regular inclusion.)  Therefore, the direct limit is canonically isomorphic to the final object in the direct limit, $F(\{ 0 \})$.
A: You may find it easier to use the universal property of direct limits instead of working with that glueing construction, which can become rather unwieldy.
Let $I$ be a directed set (partially ordered, and for any $i, j \in I$ there exists a $k$ with $i, j$ both $\leq k$), and $R_i : i \in I$ a directed system of rings.  Suppose that $I$ has a unique maximal element $i_0$.  Then $R_{i_0}$ can easily be shown to be the direct limit of the $R_i$.
An example where you can see what I'm saying is obvious is the following: let $R_1 \subseteq \cdots \subseteq R_n$ be rings, directed by inclusion.  Then $R_n$ is the direct limit of the $R_i$.
Now if $\mathscr F$ is a presheaf of rings on a discrete topological space $X$, then for any $x \in X$, the stalk $\mathscr F_x$ is the direct limit of the rings $\mathscr F(U)$ for $U$ an open set containing $x$.  Our direct set is the set of open sets of $X$ containing $x$, ordered by reverse inclusion, and this directed set has a unique maximal element, which is the singleton set containing $x$.
A: $\newcommand{\F}{\mathscr F}
\newcommand{\dirlim}[1]{\varinjlim\limits_{#1}}
$
Consider the stalk of $\F$ at point $x$ :
$$\F_x := \dirlim{  \substack{U \ni x \\ U \subset X \text{ open}}} \F(U)
 \quad=\quad  \dfrac{ \bigsqcup\limits_{\substack{U \ni x \\ U \subset X \text{ open}}} \F(U) }{ \sim }$$
In our case, we have
$$\F_0 = \dfrac{\F(\{0\})  \sqcup  \F(X)}{ \sim }$$
where $X = \{0,1\}$ has the discrete topology.
Here is how the equivalence relation $\sim$ works. Basically, we identify sections (i.e. elements of $\F(U)$ for some open $U \subset X$) whenever they coincide on some open neighborhood of $x$ (contained in the intersection of their domains of definition).


*

*If $a, b \in \F(\{0\})$, then $a \sim b$ iff $a=b$ (iff $a\vert_{ \{0\} } = b\vert_{ \{0\} }$).

*If $a, b \in \F(X)$, then $a \sim b$ iff [$a=b$ or $a\vert_{ \{0\} } = b\vert_{ \{0\} }$]

*If $a \in \F(\{0\}), b \in \F(X)$, then $a \sim b$ iff $a\vert_{ \{0\} } = b\vert_{ \{0\} }$

*If $b \in \F(\{0\}), a \in \F(X)$, then $a \sim b$ iff $a\vert_{ \{0\} } = b\vert_{ \{0\} }$.
You notice that in all the cases, expect the second case, the sections are identified if and only if their restrictions to the open set $\{0\}$ are equal. 
This leads to the following guess : $\F_0 \cong \F( \{0\} )$ as commutative rings.

1)
We still need to study the second case, namely when we have a global section $a \in \F(X)$.
Notice that $a \sim a \vert_{  \{0\} }$ (this is the section $b$ you were looking for : just take $b=a \vert_{  \{0\} }$ — notice that we only have $a \sim b$, not $a=b$). Indeed, we always have
$s:=a\vert_U \sim t:=a\vert_V$ for any open $U, V \subset X$, since $s\vert_{U \cap V} = a\vert_{U \cap V} = t\vert_{U \cap V}$.
Therefore, the equivalence class of $a$ equals  the equivalence class of $a\vert_{ \{0\} }$ in the stalk $\F_0$.
2)
As for the claim "$\F_0 \subset \F(\{0\})$", we construct a ring morphism
$$f : \F_0 \to \F(\{0\}),  \quad [s]_{\sim} \mapsto s\vert_{ \{0\} }.$$


*

*It is well-defined because, from the 4 cases above, we have 
$$a \sim b \implies a\vert_{ \{0\} } = b\vert_{ \{0\} }.$$
(Notice that, in the second case, $a=b \in \F(X)$ always implies $a\vert_{ \{0\} } = b\vert_{ \{0\} }$).

*It is also injective. 
It is clear that if $f(s) = f(t)$ for some $s, t \in \F(\{0\})$, then $s=t$ (because $s=s\vert_{ \{0\} }$ in that case), so that $[s]_{\sim}=[t]_{\sim}$.
When $s, t \in \F(X)$ are such that $f(s)=f(t)$, then we we have 
$$s \sim s\vert_{ \{0\} } = t\vert_{ \{0\} } \sim t$$
so that we again find $[s]_{\sim}=[t]_{\sim}$.
Thus, we can identify $\F_0$ with a subring of $\F( \{0\} )$, since $f$ is an injective ring morphism.
Finally, $f$ is also surjective : if $t \in \F( \{0\} )$, then $[t]_{\sim}$ is an element of $\F_0$ which is mapped to $t$ under $f$.

Personally, I would have rather worked with the ring morphism
$$g : \F(\{0\}) \to \F_0,  \quad s  \mapsto [s]_{\sim}$$
You can try to show that this is an isomorphism, whose inverse if the morphism $f$ from above.
More generally, you can think of an element of the stalk $\F_x$ as a "germ" $[(U, s)]_{\sim}$ where $s \in \F(U)$ and $U \subset X$ is an open neighboorhood of $x$.
