* Re: Freyd Categories
2010-09-23 11:01 Freyd Categories David Leduc
2010-09-25 19:27 ` Paul Levy
@ 2010-09-25 21:48 ` Toby Bartels
1 sibling, 0 replies; 3+ messages in thread
From: Toby Bartels @ 2010-09-25 21:48 UTC (permalink / raw)
To: categories; +Cc: David Leduc
David Leduc wrote:
>A Freyd category is essentially a functor.
>So why is it called a category?!
As I understand it, one says a Freyd category on C,
where C is the source/domain of the functor.
So a Freyd category on C is a category K (with certain structure),
together with a functor C -> K (with certain properties).
More simply, a Freyd category on C is a category K
together with certain extra stuff.
Calling that whole business (a category together with ...) a "category"
is an abuse of language akin to terms like "partially ordered set";
a partially ordered set is really a set together with certain structure.
There may be a better reason why Power and Thielecke used this terminology,
but if so, they don't explain it in the papers that I have found
(but I have not been able to read their first paper on the subject,
Environments, Continuation Semantics and Indexed Categories,
so maybe there is an explanation there).
>More generally, is there a way to see a functor as being a category?
More generally, a functor is a pair of categories equipped with extra stuff,
which is trivial, but I don't think that there's anything deeper than that.
You could do something like the trick that encodes a function as a set
(as everything must be) in the foundation of material set theory:
a function f: X -> Y is the set {(a,b) in X x Y | f(a) = b}.
But this set is isomorphic, in the cateory of sets, to X itself,
so really we need this set together with maps from it to X and Y.
Similarly, think of a functor F: X -> Y as the category
with {(a,b) in Ob X x Ob Y | F(a) = b} as set of objects
and Hom((a,b), (c,d)) := {(f,g) in X(a,c) x Y(b,d) | F(f) = g},
with the obvious composition; but again, this is isomorphic to X,
so we really need it together with functors from it to X and Y.
So I don't think that this really helps anything.
--Toby
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