Co-domain & Image

Question

I understand much so that the image is a subset of the co-domain of a function. It is also my understanding that the co-domain of a function is arbitrary, which would by extension mean whether or not a function is surjective is also arbitrary (For $f:X \rightarrow Y$, if $im(f)=Y$, we say the function $f$ is surjective). As an author, this frustrates me. I wish to know the benefit of distinguishing between the image and co-domain of a function. It seems it would be far more clear to speak of a function as so: $$f(\mathbb{N})=A \text{ where } im(f)=A \text{ and } f:\mathbb{N} \rightarrow A$$ If one wished to speak of the value of a function at a particular value $n$ then you would see the traditional $f(n)=k \text{ where } k \in A \text{ and } n\in \mathbb{N}$.

Feelings, comments, suggestions?

Thanks,

Jordan

Answer 1

One practical reason for making the distinction between range and codomain is that one doesn't always know the range of a function in advance of defining it. For example, what's the range of $f(x) = {\rm sin}(x) + {\rm cos}^2(x) + e^{-x^2} + \pi$? Off hand, I have no idea, but I certainly know it's a subset of $\mathbb{R}$.

I tend to think of the codomain as the "natural target" of the function, and the range as the set of realizable outputs within that target.

Answer 2

The best answer is that you should spend a couple of hours reading the first 20 pages or so of Sets for Mathematics. This should clarify everything enormously, and I can only sketch the story here. But anyway, that story goes something like this:

There's two different paradigms at work here. In one paradigm (called "material set theory"), functions do not have unique codomains, because codomains are weird and artificial in that framework. In the other paradigm (called "structural set theory"), functions do have unique codomains, because codomains are built very deeply into the foundations of that framework and there's simply no way to get rid of them.

Let me elaborate a bit.

From the perspective of material set theory, sets can overlap in all sorts of complicated ways, e.g. we have $\mathbb{C} \supseteq\mathbb{R}$. This means, for example, that whether we have $\cos_{\mathbb{R}} : \mathbb{R} \leftarrow \mathbb{R}$ or $\cos_{\mathbb{R}} : \mathbb{C} \leftarrow \mathbb{R}$ is (forgive the pun) immaterial and irrelevant, and we can safely ignore the distinction. So in this framework, the whole concept of a codomain can be safely ignored.

However, from the perspective of structural set theory and/or category theory, sets "float free" in the universe, and they don't have apriori inclusions; so for example, its not really correct to say "it holds that $\mathbb{C} \supseteq \mathbb{R}$," because $\mathbb{C} \supseteq \mathbb{R}$ is not a boolean (truthvalue). What's really going on, under this view, is that there's a unique homomorphism of $\mathbb{R}$-algebras of type $\mathbb{C} \leftarrow \mathbb{R},$ call it $\mathbb{C} \supseteq \mathbb{R}.$ Hence, we have:

$$\cos_{\mathbb{R}} : \mathbb{R} \leftarrow \mathbb{R}, \qquad (\mathbb{C} \supseteq \mathbb{R}) \circ \mathrm{cos}_\mathbb{R} : \mathbb{C} \leftarrow \mathbb{R}$$

By an abuse of notation, we sometimes omit the $(\mathbb{C} \supseteq \mathbb{R})$ and wind up with a situation where $\mathrm{cos}_\mathbb{R}$ appears to have two different codomains, but strictly speaking, these are different functions.

Anyway, like I said, you really have to run off and read the first 20 pages or so of Sets for Mathematics (or a similar book) for this to all make sense.

Answer 3

A function is not defined just by the rule of assignment. You must also specify the domain and the codomain (also commonly called range). Neither of these is arbitrary. In particular, you must have in mind a specific codomain before you can tell if a function is surjective.

So "$f(x)=x$ for $x\in[0,1]$" does not completely define a function. one must also say, for example, that $f:[0,1]\to [0,1]$, or (say) that $f:[0,1]\to\mathbb R$; the first function would be surjective, the second not. But be assured that they are different functions. The codomain is just as important as the domain in defining a function.

Answer 4

There are some situations in which a function $f:X\to Y$ has certain properties, but the function $g:X\to Y'$ (where $Y\subset Y'$) for which $f(x)=g(x)$ for all $x\in X$ has different properties. Examples: (1) Suppose a $S\subset T$. Then $i_x:S\to S$ is the identity function and the "same" function with specified codomain $T$ is the inclusion function. If you allow the codomain to be unspecified, then the identity function and the inclusion function are the "same function". It makes more sense to say they are different functions and to point out that these two functions have the same $graph$. (2) The function $t\mapsto \sin^2 t+cos^2 t:R\to R\times R$ has the unit circle as image. The complement of the image is disconnected. If you specify the codomain to be the complex numbers, the complement of the image is connected. (3) Addition of real numbers is a binary operation on the reals. It is not a binary operation on the complex numbers.

Now, it must be admitted that in a great many situations in math, it doesn't make any difference what the codomain is, for example in a first-year calculus text. In that case, you can talk about a function without saying what its codomain is and nothing bad will happen. Then the distinction can remain in the background.