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In light of answers provided here (Confusion with the definition of a "Set Function" in the context of ZFC) and here (How to define a function when you are unsure if the codomain exists as a set?), which effectively established the difference between class functions and set functions, it dawns on me that I now have no idea how the concept of surjectivity finds its way into ZFC.

In the book that I have been reading (Terrance Tao's Analysis I), there is a section about functions where the author writes:

"Informally, a function $f:X \to Y$ from one set $X$ to another set $Y$ is an operation which assigns to each element $x$ in $X$, a single element $f(x)$ to $Y$."

Having learned that a "set function" is defined as the set of all ordered pairs that satisfy a class function, I think I understand why Tao uses the word "informally" in his description of $f$ (because there is no explicit concept of codomain built into a set function). However, the definition of set function has made me revisit the idea of surjectivity, which, using the basis of the above informal definition of a function, Tao defines as:

"A function $f$ is surjective if every element in $Y$ comes from applying $f$ to some element in $X$."

Clearly, Tao's definition of surjectivity is based on his informal definition of a function. For Tao's informal definition, the "codomain" is baked into the idea of the function. For the more formal definition of a set function, however, there is no codomain (only a domain for which the class function "acts" on). Later on in the book, Tao provides an alternative definition of function:

"One can define a function $f:X \to Y$ to be an ordered triple $(X,Y,G)$ [...], $G=\{(x,f(x)):x \in X\}$ that obeys the vertical line test, [...]."

For this definition of function, it seems like $G$ is actually behaving as the "set function".

So my question is basically as follows: where is the $Y$ coming from in the context of set functions? I can define the range of a set function $f$ as:

$$ \operatorname{ran}(f) = \{y\in \cup\cup f: \exists x \;(x,y)\in f \}.$$

But if I set this equal to $Y$, then my functions are always surjective. At which point, I don't understand the point of even defining surjectivity in the first place. What is the motivation / how does one decide to choose a codomain that is not equal to the range?

Cheers~

S.C.
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  • When you say that $f$ is surjective, you're implicity also stating what is the set it covers. Just like if I say that $G$ is a group, I'm implicitly specifying a group operation. – Asaf Karagila Mar 11 '20 at 15:41
  • @AsafKaragila I'm not sure I understand the point you are making with respect to the question. Perhaps it is because my question was poorly written. What I am really interested in is the final inquiry I pose: What is the motivation / how does one decide to choose a codomain that is not equal to the range? – S.C. Mar 11 '20 at 15:48
  • Context. That's how. – Asaf Karagila Mar 11 '20 at 15:49
  • @AsafKaragila Could you provide an accessible example, please? – S.C. Mar 11 '20 at 15:49
  • The function real-valued function $(x,y)\mapsto x$ is surjective. – Asaf Karagila Mar 11 '20 at 15:50
  • @AsafKaragila no, no. I'm not asking for a surjective function. I'm asking what the motivation is for making NON-surjective functions. – S.C. Mar 11 '20 at 15:51
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    I'm not entirely sure I understand your concern, but if it's literally what you ask in the last sentence, then the answer is mostly by long-standing historical tradition. For well over 150 years in many areas of mathematics, one speaks of real-valued functions, complex-valued functions, etc., and very often (until fairly recent times such as after WW 2 or so) there was no necessity to impose a range/co-domain distinction. For instance, pick any text written before 1940, and see if you can find any place in it where this distinction would make a nontrivial difference in what is dealt with. – Dave L. Renfro Mar 11 '20 at 16:14
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    Constant functions are pretty great motivation when studying functions from one set to another. – Asaf Karagila Mar 11 '20 at 16:48
  • @AsafKaragila the answer provided by Arturo Magidin (answer with 18 votes) from this post (https://math.stackexchange.com/questions/59432/domain-co-domain-range-of-a-function) is actually identifying the issue that I have presented here/above. – S.C. Mar 12 '20 at 00:00
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    the answer provided by Arturo Magidin --- I haven't looked at Tao's notes, but the fact that this is "Analysis I" and not "axiomatic set theory" strongly suggests to me that you might be pushing his explanations beyond their intended axiomatic level. And I suspect the issue you're asking about is dealt with in different ways, depending on the author, in axiomatic set theory texts. I just checked Enderton's text, and there functions are a certain type of relation, and a relation is a subset of a specified Cartesian product, (continued) – Dave L. Renfro Mar 12 '20 at 06:22
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    and he goes on to define the domain and range of a relation (in the standard way), although it is not clear to me how he deals with the notion of being surjective, as this seems to be a concept dependent on the choice of $B$ in the Cartesian product $A \times B$ for which the set of ordered pairs belongs to, but the same set of ordered pairs can clearly be defined by using different choices of $B.$ However, I believe the notion is not strictly needed to define cardinal equivalence. In any event, I recommend looking through some standard axiomatic set theory texts. – Dave L. Renfro Mar 12 '20 at 06:40
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    Incidentally (this occurred to me while I was writing the earlier comments, but I needed to get back to work on my day-job stuff), I think I understand your concern, namely how does ZFC typically take care of the notion of surjective functions when, for example, the function ${(a,a) \in {\mathbb R} \times {\mathbb R}: 3 < a < 5}$ is equal (by set extensionality) to the function ${(a,a) \in [0,6] \times [0,6]: 3 < a < 5}.$ I'm pretty sure one can avoid range/co-codomain issues when defining "has the same cardinality", (continued) – Dave L. Renfro Mar 12 '20 at 14:13
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    and in the middle of p. 43 Enderton discusses the fact that (in formal set theory) onto is not an adjective that applies to the function itself, [what follows is my own interpretation of what Enderton says] but rather onto is a kind of meta-linguistic description about a specific representation of the function, in the same way that numerator is not an adjective that applies to a rational number itself, but rather numerator is an adjective that applies to a certain representation of the rational number. – Dave L. Renfro Mar 12 '20 at 14:14
  • @DaveL.Renfro that is certainly much closer to what I had in mind than what the other answers offered. Thank you very much! – S.C. Mar 12 '20 at 14:51

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