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no need for \phi when there is \emptyset
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darij grinberg
darij grinberg
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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(\phi)$$F(\emptyset)$ is a set consisting of exactly one point. $\phi$$\emptyset$ is the empty set.

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 $F(\phi)$ is a set consisting of exactly one point. $\phi$ is the empty set.

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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Rex
Rex
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A simple(possibly trivial) question about Grothendieck Topologies

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 $F(\phi)$ is a set consisting of exactly one point. $\phi$ is the empty set.