Talk:Codomain

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[edit] Errors in the Article

The first point I would like to make is that there is no sense in writing something like "let f be a function from R to R, where f(x) = x^(1/2)." This is because there is no such function. Suppose to the contrary that there is such a function f. Because it is a function, we know that for all a in R (its domain), there exists b in R (its codomain) such that b = f(a) = a^(1/2). However, it is easy to see that for a = -1, no b in R satisfies the statement b = a^(1/2) = (-1)^(1/2). This leads to a contradiction. Therefore, we conclude that there is no such function. Another interpretation would be that the "function" f is not well-defined, although this terminology is misleading as it suggests that f is in fact a function, which we have just disproved.

Of course, there is nothing wrong with supposing that such a function exists. It is just that doing so serves no purpose, since any statement follows from a false supposition. At any rate, the article should be changed to fix what is clearly a mistake.

The second point I would like to make is that even if the functions f and g were properly defined, it is trivial to show that f and g are in fact the same. By treating them as the sets they really are, proving that f = g is a simple task of proving set equality. This disproves the article's current claim that the functions are not the same.

-Your Friendly Anonymous Mathematician

[edit] Example

[edit] Copied from article

Let the function f be a function on the real numbers:

f\colon \mathbb{R}\rightarrow\mathbb{R}

defined by

f\colon\,x\mapsto x^2

The codomain of f is R, but clearly f(x) never takes negative values, and thus the range is in fact the set R+—non-negative reals, i.e. the interval [0,∞):

0\leq f(x)<\infty

One could have defined the function g thus:

g\colon\mathbb{R}\rightarrow\mathbb{R}^+ <== does not include zero!!
g\colon\,x\mapsto x^2

While f and g have the same effect on a given number, they are not, in the modern view, the same function since they have different codomains.

The codomain can affect whether or not the function is a surjection; in our example, g is a surjection while f is not.

[edit] Discussion

Why should f and g be considered different functions? No doubt this is very confusing for those who are meeting functions for the first time; also I know of no use for the surjection concept (although I still have to read that article). Brianjd

I have a math degree and I was never taught any such thing. I was taught that the range is the set of all outputs, and that if someone says x->x^2 has a range of R, then they were just wrong or imprecise.
Secondly the use of the words possible and actualized is totally confusing
Thirdly I can understand how x->x^2 is different if you are talking about R and C, but when both map R->R+ I don't see how misstating the proper range changes the function at all. This seems to be some kind of semantic nuance that is only important when abusive language is used with classes of functions.
I just don't get it. I'm gonna check my analysis books when I get home.

It's just a case of mathematicians wanting to complicate things and make their profession seems advanced.

I too agree that it's stupid to consider a "mapped to" set to be anything different than the range of the function doing the mapping. But mathematicians allow for the mapped to set to have some useless elements to which the function will never map to (i.e., the codomain of f(x) = x^2 being R has no use, why not limit it to R+?).

It is important in many areas of advanced mathematics, particularly algebraic topology to distinguish between the range of the function and the codomain. The reason is that when you apply a functor like homology or fundamental group to a function, the resulting homorphisms can be entirely different. For example proofs of the fundamental theorem of algebra and its generalizations to quaternions and octonions using algebraic topology techniques rely on clever use of this distinction. Fiedorow 20:09, 7 December 2005 (UTC)


[edit] CODOMAIN - "set of all possible values" ?

Assume f: A ->B, If the ran(f) is a proper subset of B, then how are the values in the set B-ran(f) "possible values" of the function f defined on A?

The set of all positive values is ran(f) not B. B is a set which contains the set of all positive values (a set being mapped "into" by f defined on A)

To describe ran(f) as the set of "actual values" implies that the function is actually used for all members of A. This is true if we're graphing y=f(x) or some other mechanism that actually feeds all members of A into f. But a function isn't like this, it's just a machine or rule that "can" take any value in A and map it to it's corresponding value in ran(f).

ran(f) is the set of all "possible" values!

From the point of view adopted in this context, the domain A and the codomain B are part of the data associated to the function f. Admittedly this doesn't make much sense when one considers one function in isolation. However it is very helpful when one is considering large collections of functions. One wants to say things like "consider all continuous functions with domain A and codomain B". Fiedorow 15:29, 19 December 2005 (UTC)
Yes but to explain the codomain as the set of possible values and range as the set of actual values is vague and makes even less sense in light of your comments above (and even less sense when going back and reading the explanation again).
The range is the "set of all possible output values from the function." The codomain is "the set that contains the set of possible output values of the function, and thus is a superset of the range."

[edit] On the composition example

The article gives an example where two functions with the same graph are given, but with different stated codomains, and claims that in one caase a certain composition is possible and in another case it is not. This could be expanded into a longer explanation of why it is sometimes necessary to track the codomain explicitly; it is not about the ability of the functions qua functions to be composed, but about the ability of the functions qua morphisms in a category to be composed. In a noncategorical context, all that is needed in order to compose f and g into f \circ g is that the range of g is contained in the domain of f, and this is a property of the graphs alone. CMummert 13:06, 14 October 2006 (UTC)

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