2
$\begingroup$

A possible explanation is that some yet unsolved advanced functions have a known codomain, but not a known range.

Got any better ideas?

$\endgroup$
  • 1
    $\begingroup$ Please define what you mean by range; codomain has only one use, but range has many ... $\endgroup$ – kcrisman Oct 8 '16 at 17:45
  • 1
    $\begingroup$ You can tell them they need both in order to compute cokernels :) (en.wikipedia.org/wiki/Cokernel) $\endgroup$ – Dag Oskar Madsen Oct 8 '16 at 22:36
  • $\begingroup$ @kcrisman range = image of a function {the set of all $x$ such that there exists an $x$ in the domain of the function $f$, such that $y=f(x)$} $\endgroup$ – Buffer Over Read Oct 8 '16 at 23:06
  • 1
    $\begingroup$ @Dag: A dangerous gambit, since it's pretty easy to come up with the idea that a cokernel is not a property of a homomorphism, but a property of a group and a homomorphism with range in that group. $\endgroup$ – Hurkyl Oct 8 '16 at 23:43
  • 1
    $\begingroup$ If I don't get around to managing to compose an answer, here are three points. (1) Codomain is a simpler notion when we are paying attention to the types of objects. (2) I think the historical origins comes from binary relations -- if you have a notion of domain for such a relation, the codomain is the "converse domain": the domain of the opposite relation. (3) There are other concepts one can work with here, such as "partial function from the universe to itself". I think a number of people have that in mind when they think of the word "function" -- thus there may be a need to disambiguate. $\endgroup$ – Hurkyl Oct 9 '16 at 7:34
4
$\begingroup$

Codomain is a primary notion in the definition of a function, range is something that can be computed given the whole definition, one cannot possibly replace the other. What bothers me is that you seem to imply that you introduced codomain after range, and I don't see how that could be.

The only possible question left would be "why not always introduce functions that have their range equal to their codomain?", i.e. "why don't we only define onto functions?" which might be your real question. To that, there are several answers:

  • the one you gave: it might be difficult to determine the range. This is a strong one, and you don't need to involve advanced functions. We want to be able to define the function $x\mapsto x^4-x^3+12x-7$ from $\mathbb{R}$ to $\mathbb{R}$ without having to determine its range!

  • in many occasion we want to have a larger codomain, e.g. when we consider composed function: the codomain of one function should match the domain of the other one; when we consider paths in, say, the plane: we don't want the path to take its value in some subset of the plane, but to consider it as taking its values in the plane itslef; imagine you are modeling a physical problem with a function, say the temperature at a point expressed as a function of the time: you would not like to have to know the range before getting started. This last case points to an important point: domain and codomain really are about the type of the input and output of the function (in the physics example, both are real numbers but physically one is a time and the other a temperature, in maths a function could have a integer input and real output, or real number input and point output, etc.)

$\endgroup$
  • $\begingroup$ As the devil's advocate, I'd point out that in my evil convention where there isn't a notion of codomain, I have no trouble defining the function $x \mapsto x^4 - x^3 + 12x - 7$ for $x \in \mathbb{R}$; in particular, there is no need to determine the range in order to define a function! It is worth knowing that the range is a subset of $\mathbb{R}$, but that doesn't need to be built into the definition of the function. Also, note that we used to express such a function $f$ as $f \subseteq \mathbb{R} \times \mathbb{R}$ before the idea of $f : \mathbb{R} \to \mathbb{R}$ took over. $\endgroup$ – Hurkyl Oct 9 '16 at 17:57
  • $\begingroup$ And if I double down on evil, the semantics could be based on composition of relations -- I can compose $f \circ g$ even when the range of $g$ is not a subset of the domain of the function $f$. In fact, we even teach students to compute the domain of such composites -- e.g. questions like "what is the domain of $\sqrt{1-x^2}$"? $\endgroup$ – Hurkyl Oct 9 '16 at 18:01
  • $\begingroup$ @Hurkyl at the danger of digressing wildly from this post... why do you say the idea $f \subseteq \mathbb{R} \times \mathbb{R}$ predates the $f: \mathbb{R} \rightarrow \mathbb{R}$ view of a function? I'm not being argumentative, I'm curious, are you thinking of $f \subseteq \mathbb{R} \times \mathbb{R}$ as interchangeable with thinking of the function as its graph? I'm not sure how 19th century folks really thought, but, I would put the formula first... my apologies in advance if this comment is dumb. $\endgroup$ – James S. Cook Oct 9 '16 at 23:35
  • 1
    $\begingroup$ @JamesS.Cook: I've not seen it myself, but I have been told that's how the actual history of notation goes. My impression is that $f \subseteq \mathbb{R} \times \mathbb{R}$ stemmed from the early work in formal logic (Russel?) viewing a function as a special kind of relation (and identifying a relation with its graph), and the arrow notation came from category theory. $\endgroup$ – Hurkyl Oct 10 '16 at 14:05
1
$\begingroup$

I'll give a brief answer. Basically, $f: A \rightarrow B$ paired with a rule $f(x) \in B$ for each $x \in A$ defines a function. There are three moving parts, the choice of $A$, the choice of $B$ (the codomain) and the rule $f(x)$. These choices do not exist in isolation. We must have $f(x) \in B$ so the function is into and we must have $f(x)$ is a single element in $B$ so the function is single-valued. One very nice aspect of this definition is it allows us to change the domain without changing the formula and it allows us to change the codomain without changing the formula. Why this matters? Certainly good references to problems in higher mathematics which require the codomain have already been mentioned in the comments. I would point out the terminology allows us to take any function and create a corresponding bijection $\iota \circ f \circ s$ where $s: C \rightarrow A$ is a map which selects just one point in each fiber of the domain and $\iota: f(A) \rightarrow B$ is the inclusion map. If we were to exchange the range $f(A)$ for the codomain $B$ then it would be difficult or impossible to have this discussion.

But, I think your answer has the right idea for beginning students. We could say this: $f: A \rightarrow B$ is like planning a trip. The $B$ describes where you may possibly go whereas $f(A)$ is where you actually go. This reflects the view that we may hold a function with partial understanding. In particular, this is a reality for students who when given $f(x)$ do not entirely understand it's content (until they do much analysis of say, $f(x) = \frac{x}{x-2}$ etc.)

$\endgroup$

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.