Let $X$ be any topological space, $S$ - any category (e.g. of sets). Consider a new category $C$: its objects are only open subsets of $X$ and a set of morphisms from $U$ to $V$ is nonempty if and only if $U\subset V$. In this case we define the only one morphism $\phi_{UV}:U\rightarrow V$, that includes naturally $U$ into $V$. Thus we can define a presheaf of sets on $X$ as a contravariant functor $\Phi$ from $C$ to $S$ (that functor turns inclusion $U\subset V$ into restriction $res_{VU}$ - a morphism in $S$).
Perhaps, the following question is too popular, but i cannot get the answer myself.
Is it possible to define a sheaf in the same way? I mean some simple categorical terms.
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1$\begingroup$ Would sheafification be the sort of thing you're looking for? en.wikipedia.org/wiki/Gluing_axiom#Sheafification $\endgroup$– Dan RustMay 7, 2013 at 12:09
1 Answer
Yes, it is. If the target category is complete, the gluing axiom translates into $F(U) \rightarrow \prod_i F(U_i) \rightrightarrows \prod_{i,j} F(U_i \cap U_j)$ being a difference kernel diagram. This is actually a very common definition of sheafs, especially since it is no longer dependend on the starting category being the category of open sets of a topological space. On can (and does) define (pre)sheaves on any category in this way. This is especially useful when working with Grothendieck topologies etc.
Greetings.
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1$\begingroup$ Little more detail can be found on Stacks Project -- Tag 0071. $\endgroup$– PeceMay 7, 2013 at 12:18
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4$\begingroup$ "On can (and does) define (pre)sheaves on any category in this way." On sites, not arbitrary categories. Or are you refering to the canonical topology on a category? $\endgroup$ May 7, 2013 at 13:24
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1$\begingroup$ @CSA: "the", not "a" canonical topology. See en.wikipedia.org/wiki/… $\endgroup$ Apr 21, 2015 at 18:12