If P(f)(A) is defined as f[A], then P is covariant. If P(f)(A) is defined as , then it is contravariant.
I have read that the powerset constuction when viewed as an endofunctor is contravariant but I cant find any explanation of why it is contravariant and not covariant. Any clues?
ie: if f is an arrow A->B why is the arrow P(f) an arrow PB->PA
I did wonder if the author was saying if you view it as contravariant as opposed to covariant then various other things follow. So is it necessarily contravariant?
Thanks,
Chad.
the covariant power-set functor arrow P(f) is not a lattice isomorphism (considering meet as intersection, and join as union) from P(A) to P(B), whereas the contravariant P(f) is from P(B) to P(A). naively, you can think of this as: "functions may lose information" (they may identify distinct elements), whereas "pre-images preserve information" (more precisely, they respect the partition of A induced by f).
this is why it is more natural to consider pre-images of topologies, than images of topologies (topologies are lattices, defined by bases...if we wish say that continuous functions are "nearness-morphisms" we have to account for the fact that two points "not near" in A may indeed be "near" in the image set f(A), for example: constant functions).
monomorphisms (or as functors: faithful functors) are "nicer" than regular morphisms. often these have special names in any given category (injection, embedding, sub-thingy, full rank) to indicate their distinguished status.
the contravariant power-set functor is an example of a Hom-functor: since we can regard P(A) as 2^{A} = Hom(A,2) (where "2" is any two-element set, typically
{ {},{{}} }, or its more "usual" name {0,1}) and Hom is contravariant in the first argument.
some people use opposite (or dual) categories to make everything "covariant". for example, when creating a partial order, we get to choose what "up" ("bigger", "join") means, usually we do so to aid our intuition, or following convention (in a complemented lattice, you can see how arbitrary this choice actually is: for example, "smaller open set" means "bigger closed complement", and vice-versa).