Let $C$ be the category of sets. Define coverings {$U_i\to U$} to be jointly surjective maps, i.e. $U$ is the union of the images of $U_i$. Then if $F$ is a sheaf of sets on $C$, is it clear that $F(\emptyset)$ is a set consisting of exactly one point. $\emptyset$ is the empty set.
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2$\begingroup$ I think the definition of a "sheaf of sets" includes the axiom that for every covering $\left\{U_i\right\}$ of an open set $U$, the map $F\left(U\right)\to\prod\limits_i F\left(U_i\right)$ is injective, and (this is something I am not 100% sure about, but I can't believe that Grothendieck would define otherwise) this includes the case when $U=0$ and the covering consists of zero sets. In this case, of course, $\prod\limits_i F\left(U_i\right)=\text{ empty product }=\text{ 1-element set}$, so $F\left(\emptyset\right)\to\text{ 1-element set}$ must be injective, so $F\left(\emptyset\right)$ ... $\endgroup$– darij grinbergCommented Feb 7, 2011 at 21:56
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2$\begingroup$ ... can't have more than one element. $\endgroup$– darij grinbergCommented Feb 7, 2011 at 21:56
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$\begingroup$ Read $\emptyset$ for $0$, sorry. $\endgroup$– darij grinbergCommented Feb 7, 2011 at 21:56
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$\begingroup$ Ah!, I didn't know that the empty product is taken to be the 1-element set. I am reading some notes by Artin and there he says that every such functor is representable and that $F(U)=Hom(U,F(e))$. Thus, $F(\phi)$=1-element set. So how to show that $F(\phi)$ cannot be empty? Thanks. $\endgroup$– RexCommented Feb 7, 2011 at 22:04
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3$\begingroup$ Right, the key is that the empty product of sets is the 1-element set. Intuitively, this is for the same reasons that the empty product of numbers is 1, or the empty sum of numbers is 0. Rigorously, the categorical definition of product implies that an empty product is precisely a terminal object; and the terminal object of the category of sets is the 1-element set. $\endgroup$– Tom LeinsterCommented Feb 8, 2011 at 1:39
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