For a non-surjective function, $f : \Bbb{R} \to \Bbb{R}^+$ is better than $f : \Bbb{R} \to \Bbb{R}$?

Question

Example:

From this we can tell no negative real number can be the image of any element of the domain. Thus not surjective because the range is not equal to the codomain, which means the function associates a any real number to a positive real number only. Is it better to write it this way $f : \Bbb{R} \to \Bbb{R}^+$ then?

Answer 1

It depends on the context, but you're right that writing $f : \Bbb{R} \to \Bbb{R}^+$ contains superior information about the rule $x \mapsto |x|$. There are a few reasons why you might choose to write a more general codomain than what is called for, for example:

• Sometimes a definition will tie you to a codomain. For example, the definition of function composition takes a function $f : A \to B$ and $g : B \to C$ and produces a function $g \circ f : A \to C$, defined by $(g \circ f)(x) := g(f(x))$. Now, if we take, for example, $g : \Bbb{R} \to \Bbb{R} : x \mapsto x^3 - x$, then we can only really compose with $f$ it is has codomain $\Bbb{R}$, to match the domain of $g$. Of course, it's a simple thing to do to extend a codomain (or even limit the domain of $g$ to $\Bbb{R}^+$), but one or the other should be done in order to make the composition kosher.
• Sometimes the codomain is easy, but the range is unknown. As per Hagen von Eitzen's suggestion, we can take any object conjectured to exist (but whose existence is unknown) and turn it into a function to $\{0, 1\}$ with unknown range. For example: $$f : \Bbb{N} \to \{0, 1\} : n \mapsto \begin{cases} 1 & \text{if $n$ is an odd perfect number} \\ 0 & \text{otherwise.} \end{cases}$$ Nobody knows the range of this function, but we do know the codomain is $\{0, 1\}$.
• Sometimes, when you're writing mathematics, you choose a more general codomain, because writing the range as the codomain becomes an unproven assertion that demands proof, and writing a proof will interrupt the flow of the writing.
• And sometimes there are pedagogical reasons. You might want to show your students that not every function needs to be surjective, to give the term "surjective" some meaning!

Answer 2

It depends on how you define what a function is (i.e., whether you take the set of ordered-pairs alone a la Bourbaki, or use something like the source-target predicate).

If you use the Bourbaki definition (i.e., $f\colon A\to B$ is $\{(a,f(a))\mid a\in A\}\subseteq A\times B$ and the Kuratowski ordered pair $(a,b)=\{\{a\},\{a,b\}\}$), then the codomain $B$ cannot be recovered from $f$.

On the other hand, if you build the codomain into the function as in source-target predicate, then you get the absolute value $\mathbb{R}\to\mathbb{R}$ and $\mathbb{R}\to\mathbb{R}^+$ are two different functions (which are related by the inclusion $\mathbb{R}^+\to\mathbb{R}$), which may or may not be what you want.