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I always defined a function as a set of ordered pairs $(x,y)$ such that some property holds, and now that I think about it, I think it doesn't make sense to talk about surjectiveness with this definition.

When defining a function as a tuple $(f,A,B)$ where $f$ is a functional relation, $A=\text{Dom}f$ and $B\supseteq\text{Im}f$ it does make sense to talk about surjectiveness, because $B$ may be a proper superset of $\text{Im}f$, so we need some differentiation to talk about surjective or non-surjective functions with this definition.

But now, as I see it, if we define a function as set of ordered pairs, it doesn't make sense to talk about surjectiveness, but if we define it as a tuple it does make sense, so, the choice of the definition of a function tells wether this concept of surjectiveness makes sense, which should not happen, as a function is a function, and its properties should not change when changing the definition(if the definitions are equivalent, obviously).

So, what am I missing here, or, which is the "correct" definition of a function? Certainly surjectiveness has helped to tackle some problems, but the choice of definition should not change the properties, I think. Am I correct?

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  • $\begingroup$ Your opening paragraph seems needlessly vague about "a set of ordered pairs... such that some property holds." You should explicitly write down the definition before jumping to the conclusion that "surjectiveness with this" doesn't make sense. In particular I suspect that once you write down the definition that all functions from set $A$ to set $B$ satisfy, you will at once realize that a specialization of this definition gives the surjective functions (from $A$ onto $B$). $\endgroup$ – hardmath Feb 5 '19 at 1:11
  • $\begingroup$ @hardmath I assumed that people would know what the definition of a functional relation is, i just didn't explicitly say it. $\endgroup$ – Garmekain Feb 5 '19 at 13:19
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With every function $f\colon A\to B$, we can associate its graph $\Gamma_f = \{(x,f(x))\in A\times B\mid x\in A\}$. If you identify function $f$ with its graph $\Gamma_f$, then it doesn't make sense to talk about surjectivity anymore, since you lost information about codomain.

On the other hand if you do remember that $\Gamma_f\subseteq A\times B$, then surjectivity makes sense.

The point is, domain and codomain are essential part of any definition of function. It is not enough just to give a set of ordered pairs.

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The abstract definition of a function comes from the definition of a relation. A relation $R$ on two sets $A$ and $B$ is any subset of their Cartesian product, so

$$R\subseteq A\times B$$

If $(x,y)\in R$, then we write $xRy$, read as $x$ is related to $y$.

A relation $f$ is called a function if the following criteria hold:

First, for any $x\in A$, there exists a $y$ such that $xfy$.

Second, if $xfy$ and $xfz$, then $y=z$.

The first criteria says that each element of the domain $A$ must be related to something in the target $B$. The second criteria states that each element in the domain is related to exactly one element in the target. Your definition of a function is intuitively correct, since you describe it as a collection of ordered pairs, which it is by definition of being a relation. The issue is that the function is necessarily a subset of a Cartesian product. This means that we need to define some collection that we call the target set. You could, in theory, be given a function without knowing the target set, and not know if it is surjective or not. For example, if I define the function

$$f(x)=x$$

then this appears to be a surjective function by how it is naturally described. However, if I tell you that the target set is the set of complex numbers, then this is not a surjective function.

Ultimately, a function is generally created to serve some kind of purpose, and so the idea of surjectivity is one we impose to discuss how this purpose is performed. A surjective operation reaches every target we can ostensibly think of (or in any sense care about), while a nonsurjective function does not.

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