31
$\begingroup$

So as far as I understand, a function $f\colon A \to B$ is surjective if and only if for every $b\in B$ there exists $a\in A$ such that $f(a) = b$.

My question is when is this actually relevant? Couldn't you arbitrarily define the set $B$ so that any elements never "used" are removed from the set, leaving you with a surjective function?

$\endgroup$
13
  • 22
    $\begingroup$ It's not the same function then. $\endgroup$
    – Bernard
    Jun 16 '17 at 0:03
  • 7
    $\begingroup$ @Bernard It's not the same arrow in the category of sets, say, but it is indeed exactly the same set of ordered pairs. $\endgroup$ Jun 16 '17 at 5:20
  • 2
    $\begingroup$ This, by the way, is a very good question, showing the right level of skepticism that is necessary for a deeper understanding of math. It's not enough to accept what your book or teacher tells you - you have to fight with it a bit to really appreciate it. $\endgroup$ Jun 16 '17 at 7:09
  • 2
    $\begingroup$ @Bernard Changing the codomain of a function may or may not result in a different function - it depends on whose definition you're using. If a function is just a set of ordered pairs then a codomain need not be specified. $\endgroup$ Jun 16 '17 at 7:15
  • 5
    $\begingroup$ @JairTaylor: I stick to Bourbaki's definition: it's a triple $(E, F,G)$ satisfying $G\subset E\times F$ and a uniqueness condition. $\endgroup$
    – Bernard
    Jun 16 '17 at 8:17
40
$\begingroup$

Yes, we can arbitrarily only look at the range of functions, but this often misses the point. When we study a function, $f: A \rightarrow B$, often we're interested in the properties of $A$ and $B$ just as much as we are the properties of $f$. So, if we want to learn about $B$, and we know that we can do this somehow using surjective function $f$ from $A$ to $B$, just looking at the range of $f$ means we've given up looking at $B$, which is what we wanted to learn about in the first place.

Edit: Since this seems like we're just talking about very basic stuff, here's a very basic property. Let's say we want to know if $A$ and $B$ have the same cardinality. How do we know that? That there exists a bijective function $f: A \rightarrow B$. If we look at the range of $f$ instead of the codomain, we're no longer thinking about the cardinality of $B$, we're thinking about something else all together.

$\endgroup$
1
  • 2
    $\begingroup$ Exactly, as my example in the comments illustrates, we note that a topological embedding $f:A \rightarrow f(A) \subseteq B$ is indeed surjective, but the point is that an isomorphic copy of $A$ exists as a subset of $B$. A similar thing can be / is done in every category. $\endgroup$
    – Kaj Hansen
    Jun 16 '17 at 0:15
31
$\begingroup$

The other answers are good, but I'd like to add one thing. Suppose a function $f:\mathbb{R}\to\mathbb{R}$ is given. Is it possible to solve the equation $f(x)=b$, for some particular $b$?

If we know that $f$ is surjective, then we can be sure that a solution exists for any choice of $b$. If not, then we need to worry about whether $b$ is in the range of $f$ or not.

In linear algebra, this comes up a lot. The range of a linear function, given by a matrix $A$, so $f(x)=Ax$, is called the column space of $A$. Sometimes, the column space is the entire codomain, and sometimes it is a subspace. Whether or not a function like that is surjective becomes an interesting question, not only for solving equations, but for answering other questions about the structure of the function.

For example, what if the domain is $\mathbb{R}^4$, and the codomain is $\mathbb{R}^3$. Then what kind of subset of the domain solves the equation $f(x)=(0,0,0)$? If we know that $f$ is surjective, then we can answer that the set mapping to zero is a one-dimensional subspace. If $f$ is not surjective, then the set mapping to zero will have greater dimension.

$\endgroup$
2
  • $\begingroup$ +1, I used to think like OP, and this is the sort of things that made the concept of surjectivity interesting. $\endgroup$
    – Arnaud D.
    Jun 16 '17 at 8:25
  • $\begingroup$ As I recall, this is fundamental in group theory and provides a basis for equivalence relations in A. $\endgroup$
    – rrogers
    Jun 21 '17 at 13:47
7
$\begingroup$

$\newcommand{\Reals}{\mathbf{R}}\newcommand{Ratls}{\mathbf{Q}}$Mathematical problems often come in the form of,

"Some value $y$ depends deterministically on data $x$; is every prospective value $y$ an actual value?"

Two common formulations are:

  • Let $Y$ be a set of prospective values (specified in advance by the context of an external question), $X$ the set of allowable inputs, and $f:X \to Y$ a mapping representing the dependence $y = f(x)$.

    The question above means Is $f$ surjective?

  • Let $Y \subset Z$ be a set of prospective values, $X$ the set of allowable inputs, and $f:X \to Z$ a mapping representing the dependence $y = f(x)$.

    The question above means Is $Y \subset f(X)$?

Here's a selection of five examples, four of them kind of the same:

  1. Is every real number the square of some real number?

    That is, if $f:\Reals \to \Reals$ is defined by $f(x) = x^{2}$, is $f$ surjective? (Answer: No. For instance, $-1$ is not in the image.)

  2. Is every non-negative real number the square of some real number?

    That is, if $f:\Reals \to \Reals$ is defined by $f(x) = x^{2}$, is $[0,\infty)$ contained in the image of $f$? (Answer: Yes, though proving this "existence of real square roots" requires non-trivial use of the completeness axiom for the real numbers, even though the result is usually introduced into the curriculum many years prior to a careful analysis course.)

  3. Is every positive rational number the square of some rational number?

    That is, if $f:\Ratls \to \Ratls$ is defined by $f(x) = x^{2}$, is $[0,\infty) \cap \Ratls$ contained in the image of $f$? (Answer: No. For instance, $2$ is not in the image.)

  4. If $y:\Reals \to \Reals$ is a continuous function, does there exist a differentiable function $x:\Reals \to \Reals$ such that $x' = y$?

    That is, if $X$ is the set of differentiable, real-valued functions on $\Reals$, and $Y$ is the space of continuous functions, and $Z$ the space of all functions, and if $f(x) = x'$, is $Y$ contained in the image of $X$? (Yes: One of the fundamental theorems of calculus guarantees that every continuous function on $\Reals$ is the derivative of some differentiable function.)

  5. Let $(M, g_{0})$ be a compact Kähler manifold. If $\rho$ is a smooth $(1, 1)$-form in the cohomology class $2\pi\, c_{1}(M)$, does there exist a Kähler metric $g$ whose Kähler form is cohomologous to the Kähler form of $g_{0}$ and whose Ricci form is $\rho$?

    Analogously to the preceding example, you can imagine that there is a partial differential equation of the abstract form $\rho = f(g)$, and the question amounts to surjectivity of the Ricci curvature operator $f$. The answer turns out to be "yes"; largely for this resolution of the Calabi conjecture, S. T. Yau was awarded the Fields Medal in 1982.

    The take-away is, not only is surjectivity interesting, but proving that a specific mapping is surjective can constitute a major work in a distinguished mathematical career.

$\endgroup$
5
$\begingroup$

To fully specify a function requires three things.

  • A domain $X$, a set of allowable inputs
  • A co-domain $Y$, a set of allowable outputs
  • A rule $f$ that, for each $x\in X$ specifies some $f(x) \in Y$.

Failure to include all three means you have failed to properly define a function. For a function to be onto, every element $y\in Y$ must have some $x\in X$ so that $f(x) = y$. A function is not just a rule; it is these three items.

$\endgroup$
3
  • 1
    $\begingroup$ But this raises the OP's question: Why do we care if $f(X) \subsetneq Y$? Why specify $Y$ as the co-domain when we could just restrict the the codomain only to the image of $f$? $\endgroup$
    – fleablood
    Jun 16 '17 at 0:22
  • $\begingroup$ Think about reflexivity of Banach Spaces. That is about the surjectivity of the evaluation map $j: V \rightarrow V^{**}$. $\endgroup$ Jun 16 '17 at 0:32
  • $\begingroup$ This is not illuminating. Of course, what you are presenting here is Bourbaki's definition of a function, but this is by no means the standard definition used in mathematical literature. Most mathematics texts identify a function with its Bourbabki graph. So your answer begs the question: why should we use Bourbaki's definition and not the standard definition? $\endgroup$
    – Angel
    Nov 4 '21 at 13:52
3
$\begingroup$

The definition of a function is not just the rule $f$. Instead, a function is defined by a domain $A$ and a codomain $B$ together with a rule $f$ that takes every $x\in A$ and returns a unique element of $B$. So, if we were to change the set $B$ as you suggest, we are actually changing the function itself, so although the new function will indeed be surjective, it will not be the same function that you started with.

For example, $f(x)=x^2$, with $f:\mathbb{R}\rightarrow \mathbb{R}$ is not surjective, but $f(x)=x^2$, with $f:\mathbb{R}\rightarrow [0,\infty) $ is surjective. The rule and domain are the same for both functions, but the codomains differ, so the two functions are not the same.

As for why the concept of surjectivity is important, one example is that if a function is both surjective and injective (i.e. both 1-1 and onto), then the function is called bijective, and showing that a function is bijective is one of the most common tools in analysis. For example, one way to show that two sets have the same cardinality is to construct a bijection from one set to the other. As another example, a function is invertible if and only if it is bijective.

$\endgroup$
10
  • $\begingroup$ In response to your first paragraph, define $f: \mathbb R\to \mathbb R$ given by $f(x)=x^2$, and $g: \mathbb R\to [0,\infty)$ given by $g(x)=x^2$. How are the sets $f$ and $g$ different? Both are equal to $\{(x,y)\in\mathbb R^2\,\vert\, y=x^2\}$. $\endgroup$
    – florence
    Jun 16 '17 at 0:18
  • $\begingroup$ The difference, I think, is that while the first is equal to the set you describe, the second is equal to $\{(x,y)\in \mathbb{R} \times [0,\infty) | \ y=x^2 \}$ $\endgroup$
    – M_B
    Jun 16 '17 at 0:25
  • $\begingroup$ It's easy to show that $\{(x,y)\in \mathbb R^2 \, \vert \, y=x^2\}=\{(x,y)\in \mathbb R\times [0,\infty) \, \vert \, y=x^2\}$. In fact, $f=g$. A function $h$ is a set of ordered pairs with the property that $(x,y), (x,y)\in h \implies x=z$; it doesn't contain any information as to what its codomain is intended to be. $\endgroup$
    – florence
    Jun 16 '17 at 0:28
  • 2
    $\begingroup$ I see my mistake, thanks Florence, that is a nice explanation. I need to run out for a bit, but I will update my post tonight. $\endgroup$
    – M_B
    Jun 16 '17 at 0:44
  • 1
    $\begingroup$ Re: any debate about the definition of a function, the definition I'm referring to is the standard one. Second, functions (according to the definition I gave) are already relations. In general, a relation does not specify a "codomain". A relation has a range, and you can consider any superset of the range to be the codomain. As for surjectivity, strictly speaking you can't just stay a function $f$ is onto, rather, you need to specify that it is onto $X$, for some possible codomain $X$. But usually the $X$ you're talking about is implicit, and so this generally isn't an issue. $\endgroup$
    – florence
    Jun 18 '17 at 1:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.