This is a generalization of the question given at Definition of Function.

I think it fairly clear that the Bourbaki and the ordered pair definition of functions are equivalent. A deeper question is whether or not the conception of functions as arrows-as in category theory without specification-is equivalent.

Let me be a little more specific what I mean. Currently,there are roughly 2 kinds of foundations proposed for mathematics: "internal" and "external".

An internal foundation for mathematics is one based on a primitive building block from which the functions are ultimately constructed explicitly from the building blocks. Examples are,of course,set theories-particularly Zermelo-Frankel set theory-where sets are defined as abstract collections that satisfy a finite number of axioms (union,extensionality,etc).

We then define ordered pairs explicitly from sets (to be precise,the set of ordered pairs ={{x},{x,y}} of elements is a subset of the power set of the power set of the union of sets where the elements {x} and {x,y} are taken). A function is then defined as a subset of this set where no 2 different ordered pairs have the same first member.

An extensionalist foundation-such as Paul Taylor and to a lesser extent, William Lawvere,have proposed-takes the FUNCTIONS to be the primitive elements as arrows that are specified in some manner. There are then 2 ways sets can be obtained from functions:
1) the sets can simply be defined as the domain and ranges of maps without specifying thier composition explicitly. I think this is what Taylor does,more or less-but I had a lot of difficulty understanding exactly what he's proposing since most of it is couched in heavy logical language.
2) A primitive function-called the membership function-is proposed which assigns single elements to collections and they are built up as the ranges of collections of membership maps. The rest of set-theoretic constructions are built up in a similar manner using specialized functions. This is what William Lawvere and Robert Rosebraugh do in thier fascinating book,*Sets For Mathematics*.

The extensionalist approach,of course,is favored by those who want to replace set theoretic models with categorical ones for mathematics. In some respects,it is simpler-arrows are easier to deal with then axiomatic sets. It also allows us to sidestep the sticky cardinality issues that come up in category theory,when dealing with collections that are too large to be sets,that started all this debating in the first place. Unfortunately,what makes set theory difficult is also what gives it it's tremendous power as a foundation for mathematics: It allows us to develop all objects in mathematics explicitly and unambiguously.There's absolutely no doubt 2 sets are equal in axiomatic set theory since in principle,you can see that they indeed have the same elements.

Personally,I favor a foundation that allows us to do both.Thinking we need to jettison one for the other is tantamount to saying we can't accept quantum theory without rejecting general relativity.

My question: Can these 2 conceptions of function-internal and external-be shown to be logically equivalent?Of course,the question's probably not that simple-you'd probably need to specify which "theory" you're talking about-for example,Taylor or Lawvere/Rosebaugh's.

So would they be equivelent in:

a) Paul Taylor's vrs. classical ZFC models? b) Lawvere/Rosebaugh's vrs. classical ZFC models?

It seems to me the model proposed in Lawvere/Rosebaugh should be equivalent since the membership function more or less is the same as the membership relation in axiomatic set theory. In Taylor's though-it's not so clear.

don'tlet us sidestep issues of cardinality/size, but on the other hand, theydo(many logicians would argue) let us develop mathematics quite as "explicitly and unambiguously" as traditional set theory does. The difference is not so much one of content or expressive power (most foundational theories on either side are equivalent to something on the other side), but rather of language and emphasis. $\endgroup$