How to prove that for a sheaf functor F, F(empty set) = terminal object? The empty set is covered by an empty family of sets. Therefore any two sections of F(empty set) are vacuously equal, so the F(empty set) can be at most a singleton. Now, why does this mean that it is in fact the terminal object, as Vakil claims in his Algebraic geometry notes? In Sets, this is true, but why in general? In particular, what happens if we are talking about a sheaf of rings with unity. Then I think that we actually have a problem because rings with unity have at least two elements.
 A: For any index set $I$ we can consider the covering of $U=\emptyset$ by the sets $U_i=\emptyset$ for $i\in I$. This gives us that $F(U)$ in
$$ \tag1F(U)\to\prod_{i\in I}F(U_i){{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}} \prod_{i,j\in I}F(U_i\cap U_j)$$
is an equalizer. For $U=\emptyset$ we may in fact pick $I=\emptyset$ and this still works. But the empty product of whatever is terminal, so $(1)$ becomes
$$ F(\emptyset)\to T{{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}}T$$
where $T$ is terminal and the two arrows on the right are $\operatorname{id}_T$.
The equalizer condition says that for any object $X$ together with a morphism $X\to T$ such that in $X\to T{{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}}T$ both paths to the right $T$ give the same morphism, there exists a unique morphism $X\to F(\emptyset)$  that makes
$$\tag2 \begin{matrix}F(\emptyset)&\longrightarrow&T\\\uparrow&\nearrow\\X\end{matrix}$$
commute. But the condition about the morphism $X\to T$ is not really a condition: There exists a unique morphism from $X$ to $T$ any way and the equality of the two paths is always given (again by uniqueness of morphism to $T$). On the other hand, any morphism $X\to F(\emptyset)$ make $(2)$ commutative. Therefore $F(\emptyset)$ is simply an object such for any $X$ there exists a unique morphism $X\to F(\emptyset)$. - In other words, $F(\emptyset)$ is terminal.
Now that $(1)$ can be used to define the notion of sheaf only for a category with products including the empty product, hence also having terminal objects. The category Ring of rings with unity has products only for the nonempty case, so $(1)$ does not apply. More precisely, we can use $(1)$ for Ring valued sheaves only if $I\ne \emptyset$. Still, the formulation of locality with sections also postulates that $F(\emptyset)$ has only one element, which is impossible. Everything makes sense again if we augment Ring by adding the zero ring artificially as terminal object.
A: If $F$ is a sheaf valued in a category $C$ with terminal object $t$, then the empty covering of $\emptyset$ shows that $F(\emptyset) \to t \rightrightarrows t$ is an equalizer. Since $t$ is terminal, the two morphisms $t \rightrightarrows t$ agree, hence $F(\emptyset) \to t$ is an isomorphism.
