Why write codomain instead of image when defining a function.

Question

$X \xrightarrow{\quad f \quad } Y $

Given some function like the one above, why must the function have as the object on the left the domain which it must cover but have as the right hand object the codomain which $f$ need not be onto?

I have looked online and don't understand why we don't restrict $Y$ to be the image of which $f$ is onto.

I would assume that much like the distinction between partial functions and relations and functions, there is some edifice of theorems and axioms that may fall apart if we don't allow $Y$ to be an overset of the image of $f$.

However, I would like some concrete motivating examples of why we have chosen this as our criteria and or notation.

Thank you in advance for your help, I really appreciate this!

Hi! Great question. You are absolutely right that domain and codomain don't play a symmetric role at all and that changing the domain would be a big change, whereas changing the codomain is a much smaller change. However, there are reasons why we don't want to force ourselves to always use the image instead of the codomain. Sometimes the domain is hard to calculate. Sometimes we have two functions f:A->B and g:A->B and f and g don't have the same image, but we want them to have the same domain and codomain. Sometimes we want to study "all functions A->B" including those whose image is not B. — Stef, Nov 27 '23 at 09:21
Related: Codomain as part of the definition of a function?; Why is it important to have a discrepancy between image and codomain?; If there is a need to distinguish between the image and codomain, why not do this on the input side of a function? — Stef, Nov 27 '23 at 09:22
One piece of utility from allowing functions which are just injections, is due to cardinality.if there exists an injection from A to B then |A| ≤ |B|. So, even if all functions were surjective it would be usefull to define injective Relations — Michael Carey, Nov 27 '23 at 19:37
another example, cardinal exponentiation, relies on defining $κ^λ$ = | { f | f: λ $\to$ κ } |, so this notion of function is deeply connected with basic operations on cardinals, a special case being $2^n$, which is discussed in an excellent answer below — Michael Carey, Nov 27 '23 at 19:48
Often, when we are studying functions from $X$ to $Y$, we are just as interested in $X$ and $Y$ as the functions themselves. If, say, $X$ and $Y$ are vector spaces, then we might want to consider all of the linear maps between them. In such a situation, it would be quite unnatural to restrict each function to its image. — Joe, Nov 28 '23 at 22:50

score 4 · Answer 1 · answered Nov 27 '23 at 17:50

Here is an easy example.

If $A$ is a set, then $\mathcal P(A)$ is the power set of $A$, i.e. $\{B\mid B\subseteq A\}$. In many contexts, we want to identify $\mathcal P(A)$ with $2^A$, that is the set of all functions $A\to 2$ by identifying $B$ with $$1_B(a)=\begin{cases}1 & a\in B\\0 & a\notin B.\end{cases}$$

If we followed your suggestion that a function is always surjective, then $A$ and $\varnothing$ cannot be written as functions $A\to 2$, since $\varnothing$ is $A\to\{0\}$ and $A$ is $A\to\{1\}$. (This is, of course, under the assumption that $A$ has at least two elements, but most sets do anyway.)

So in that case, $\mathcal P(A)$ would be $2^A\cup\{\{0\}^A,\{1\}^A\}$, which is fairly horrendous if we're being honest. And this can be much more annoying when you're studying $L^p$ spaces and other function spaces. So, instead, we just have a codomain and we separate it from the image/range of the function.

score 2 · Answer 2 · answered Nov 27 '23 at 11:41

I am not 100% sure if I am going to provide the example you wanted, but let me try.

A couple of days ago, I was reminded of Cantor's diagonal argument. It's a well-known proof of the fact that there is no bijection between $ \mathbb{N} $ and $ \mathbb{R} $. In order to even state an idea of the proof, we need to assume that there is a bijection $ f $ between $ \mathbb{N} $ and $ \mathbb{R} $. At the end, we will show that $ f $ can't be a surjection. And surjection is explicitly defined in terms of image being equal to codomain.

I am quite sure that there should be a workaround (it may or may not involve some categorical magic), but I think this example shows why it's quite natural to talk about codomains.

Why write codomain instead of image when defining a function.

2 Answers2