0
$\begingroup$

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?

  • 0
    @Zhe$n$ Lin: but obviously that cannot be Jr.'s definition.2011-03-17

2 Answers 2

3

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

0

Because presheaf deals with restriction. Please read 1st chapter of Red book by Mumford for more details. It is explained very well there.