Consider the category of all open sets of a given topological space where the morphism are inclusions,why one can see a Presheaf as a contravariant functor?
Presheaf on a topological space
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category-theory
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0@Zhe$n$ Lin: but obviously that cannot be Jr.'s definition. – 2011-03-17
2 Answers
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Suppose your presheaf $F$ has values in some category, say rings. Then to each open $U $ there corresponds a ring $F(U)$, and to the inclusion map $U\to V$ where $U\subseteq V$ there corresponds map $V\to U$ in the opposite direction, sometimes called the restriction.
Then you should verify that the axioms of a functor hold. The domain of your functor is just the poset category where the objects are open sets and the morphisms are inclusion maps
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Because presheaf deals with restriction. Please read 1st chapter of Red book by Mumford for more details. It is explained very well there.