On Mon, 28 May 2012 13:43:33 -0400 (EDT)
Jay Sulzberger
No. The point is that, by definition, a category, call it C, is a struct with two sets, Obj(C) and Mor(C), and further operations:
1. head: Mor(C) -> Obj(C)
2. tail: Mor(C) -> Obj(C)
3. id: Obj(C) -> Mor(C)
4. *: Mor(C) x Mor(C) -> Mor(C)
where head and tail and id are everywhere defined single valued maps. They are all maps of sets. *, read "composition of morphisms" is a map of sets, with signature as displayed, but is not usually everywhere defined. We have then several "equational" axioms, which C is required to satisfy to be a category.
(set theoretical note: We have, partly implicitly, ruled out categories which are not "small". See standard texts for this locus of difficulty.)
By the axioms, any object b of C must have defined its associated identity morphism id[b]. For many categories, b will always be an actual set, and id[b] will be the unique map of sets defined by
(id[b])(x) = x , for all x in b
where (id[b])(x) is read "the result of applying id[b] to the element x of b".
But, as explained, many categories have objects which are not sets. Indeed, often, no object is a set.
The definition of category never mentions whether or not the objects are sets. And, as we have seen, there are many categories whose objects are not sets. (Perhaps categorically better: many categories are not directly presented as having objects which are sets.)
to repeat: The concept "category" is larger in extension than the concept "category whose objects are sets and whose morphisms are maps of sets".
ad representations of categories:
http://en.wikipedia.org/wiki/Yoneda_Lemma [page was last modified on 1 April 2012 at 05:17]
I guess that this made me think of idA as idA(x) = x for each x of A. Later when I saw other (more general) definitions I did not read carefully to realize the difference.
Thanks a lot for making this clear to me.
-- Manfred
I will let stand my restatement of what you already know ;)
oo--JS.
Thanks a lot for the detailed example and explanations. I will study your post thoroughly. -- Manfred