TITLE: How to prove that for a sheaf functor F, F(empty set) = terminal object?
QUESTION [17 upvotes]: 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.

REPLY [8 votes]: 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.