Is the codomain defined by us when we define a function?

Question

For example if I define a function such as

$f(x) = x^2$

then set the domain to be {$x|x\in\mathbb{Z}$}

it follows then, that the range is {$f(x)|f(x)\in\mathbb{Z^+}$}

from my understanding of the codomain, it is defined by us when we create a function, therefore in this case would the codomain of the function be undefined, as we haven't set it or would it be $\mathbb{Z}$ or $\mathbb{Z^+}$

If you haven't defined the codomain, then you haven't defined the function. — Randall, Jul 25 '23 at 13:00
${x \mid x \in \mathbb{Z}}$ is just the same as $\mathbb{Z}$, of course. ${f(x) \mid f(x) \in \mathbb{Z}^+}$ is an unusual notation since $f(x)$ is not a variable. — aschepler, Jul 25 '23 at 13:01
See the post Codomain of a function as well as Isn't codomain of a function ambiguous? and Codomain as part of the definition of a function — Mauro ALLEGRANZA, Jul 25 '23 at 13:01
@Randall So then the codomain can never be interpreted based on the function definition itself? It must also be defined alongside the function. — Dan Lupu, Jul 25 '23 at 13:05
Yes, correct. The definition of a function $f: A \to B$ includes a reference to $B$ already. — Randall, Jul 25 '23 at 13:07
@aschepler I was taught to write the range as {$y | y\in$ (Whatever set)} so I just wrote that with f(x) instead, is this wrong? If it is what would be the correct way to express the range? — Dan Lupu, Jul 25 '23 at 13:07
You could write ${f(x) \mid x \in A}$ for the range of $f: A \to B$, but that is not what you've written. — Randall, Jul 25 '23 at 13:08
@Randall What about a transformation for example, if an nxn matrix has a rank such that the rank < n then space is "squished" onto a lower dimension, in this case the codomain of the function would be $R^n$ where n = the rank of the matrix, wouldn't the codomain be defined in this case just based on the function itself, as a transformation is basically just a function. — Dan Lupu, Jul 25 '23 at 13:10
@Randall oh okay that makes more sense, and looks much neater. — Dan Lupu, Jul 25 '23 at 13:10
A function which maps different matrices of the same size to matrices of different sizes would be unusual, but it could be defined. A simple codomain could be the union of several matrix spaces, $\bigcup_{d=1}^n \mathbb{R}^{(d \times d)}$. — aschepler, Jul 25 '23 at 13:13

score 0 · Accepted Answer · answered Jul 25 '23 at 13:09

If we want to be precisely technical, we should always specify the codomain when defining a function. A common way of writing this is

Define function $f: \mathbb{Z} \to \mathbb{Z}$ by $f(x) = x^2$.

Though in less strict contexts, we sometimes use an implicit codomain, especially when the exact codomain set doesn't matter. If you say "Define function $f$ on $\mathbb{Z}$ by $f(x) = x^2$" without mentioning a codomain, I'd probably assume the codomain is also $\mathbb{Z}$. If it makes sense for the formula, a codomain equal to the domain often makes sense. Or often if you don't specify the codomain, it's because it doesn't actually matter: everything we later say about $f$ is true no matter whether the codomain of $f$ is $\mathbb{Z}^+$ (defined to include $0$!) or $\mathbb{Z}$ or $\mathbb{R}$.

Is the codomain defined by us when we define a function?

1 Answers1