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In literature on logic and set theory, there seem to be two different definitions of functions, one more general than the other. First of all, a function $f\colon X\to Y$ consists of three element $(X, Y, f)$, where $f\subseteq X\times Y$ satisfies $$\forall x\in X\quad\exists!y\in Y:(x,y)\in f. $$

On the other hand, a wff $P(x,y)$ with free variables $x,y$ gives rise to a function $x\mapsto y$ if $$ \forall x\quad\exists!y:P(x,y). $$ (Of course, $P$ could have other free variables, but I don't notate them.) The last definition, which allows our functions to be defined on our entire universe of sets, is the one used, for instance, in the Axiom Schema of Replacement. In many cases, I also see that people use the notation $y = f(x)$ even though $f$ cannot be defined as an object in our set universe.

I think that, even though we can always guess from the context which definition we use, the two definitions have a significant difference, first of all in their general applicability. My question is: Do I understand everything correctly? And in this case, is there no terminology to distinguish between functions in the two senses described? Isn't it somewhat sloppy to use the same terminology? Would it be totally wrong to call the first one a function and the second one a map, for instance?

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    $\begingroup$ The second notion, modulo the evident notion of equivalence, is properly called a function class. $\endgroup$ – Zhen Lin Jul 21 '14 at 9:30
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    $\begingroup$ There have been some wonderful threads that dealt with the pros and cons of using the definition of a function as a triplet, and as a set of ordered pairs with a certain properties; as well as threads about how both end up describing the notion of a function pretty well. Have you read all of them, or at least some of them? $\endgroup$ – Asaf Karagila Jul 21 '14 at 9:30
  • $\begingroup$ Yes, I would say I have. This question is mostly about terminology. $\endgroup$ – Gaussler Jul 21 '14 at 9:35
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    $\begingroup$ Following on the comment by Zhen Lin, I call the second kind, whose domain is a proper class, a "class function". $\endgroup$ – Carl Mummert Jul 21 '14 at 11:47
  • $\begingroup$ Now people cannot decide if it is "class function" or "function class" ;-) . But I guess both of them make sense. The function is defined on a class, hence it is a class function. On the other hand, its restriction to any set gives a function in the usual sense; hence it is a class of functions, i.e. a function class. $\endgroup$ – Gaussler Jul 21 '14 at 15:29

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