Inverse image of a subset of the codomain with elements without corresponding elements in domain

functions

I consider the function $f(x) = x^2 : \mathbb{R} \rightarrow \mathbb{R}$ whose image is $[0, + \infty)$. For the sake of simplicity: domain $D = \mathbb{R}$, codomain $C = \mathbb{R}$.

If I consider $A = [-25, 25]$ subset of the codomain $C$, this subset contains elements $[-25, 0)$ which don't have a corresponding element in the domain $D$. In this case is it possible to evaluate the inverse image of $A$? I tried to do it in this way.

According to the definition of the inverse image:

$$
f^{-1}(A) = \lbrace x \in D : f(x) \in A \rbrace
$$

$$
f^{-1}(A) = f^{-1}([0, 25]) = \lbrace x \in [-5, 5] \rbrace
$$

Is it correct?

EDIT: the Mathematica software gave me the same result I wrote.

Best Answer

$A=[-25,25]$ is not a subset of the $f$'s range $[0,\infty),$ so $A$ does not have a preimage under $f.$

$f$'s range (let's call it $R$) is a subset of $f$'s codomain, and $R$'s preimage is $f$'s domain.

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[Math] Typo: in the definition of inverse image $E$ should be replaced by $H$

Inverse image preserves unions, intersections and complements. In topology and analysis we often work with families of subsets of the space, "open sets" or "Borel sets", and so on. These families are often closed under unions, or intersections - or some variation thereof.

Using the fact that inverse images preserves set operations means that we can use it to characterize "good" functions. Continuous functions are those that the preimage of an open set is open; Borel functions are those that the preimage of a Borel set is Borel; and so on.

For this reason we are usually more interested in the inverse image than in the direct image; however that too has its uses (but often coupled with some condition associated with inverse images). For example a bijection which is both open and continuous is in fact a homeomorphism.

[Math] Codomain of a function

What is the point of having values in the codomain that can not be output by the function, how does that aid in describing the function?

Here are a few reasons why we allow some functions to not be surjective.

As Lubin mentioned, the range of a function can be difficult to determine. For example, determining the range of a polynomial of high even degree (such as $P(x) = x^6 - 3x^2 + 6x$) amounts to finding the zeroes of a high-degree polynomial (such as $P'(x) = 6x^5 - 6x + 6$, whose roots are not expressible as radicals), a difficult task in general. We could get around this by defining the codomain of every function $f$ to be $\operatorname{im} f$ (that is, $\{y\,:\,f(x)=y\text{ for some }x\in X \}$), but that doesn't really add any information.
It's nice to separate surjective functions from other functions because surjective functions are dual to injective functions. When I say "dual" I'm referring to, for example, the following fact: a function $f:A\to B$ is injective if and only if there is a function $g:B\to A$ such that $g\circ f=1_A$ (by $1_A$ I mean the identity function on $A$); a function $f:A\to B$ is surjective if and only if there is a function $g:B\to A$ such that $f\circ g=1_B$. When you study the branch of mathematics known as category theory, you'll see that it's very natural to have dual properties like this.

Does this also mean that the domain can include numbers that the are not inputs to the function?

As others have remarked, the domain of a function can include other objects than numbers. For example, you could define a function which takes as input a person and returns his age. In any case, a function must be defined on all possible input values. The answer to your second question is no.

And is it also then true that is a function is "onto" the codomain is the same as the image? So surely any function can be "onto" if you just change the what the codomain is?

That's exactly right. You can make any function onto by changing the codomain. But as I remarked earlier, in general we don't know what the image of a function is and so it doesn't add any information to restrict the codomain.

What I'm really trying to ask I guess is the range/image of a function is defined by the function, what defines the codomain?

The codomain usually arises naturally in the definition of the function. For example, whenever you have a function which returns a number, the natural choice of codomain is $\mathbb R$. Of course, if by "number" you mean "complex number" then the codomain could be $\mathbb C$; if by "number" you mean "quaternion" then the codomain could be $\mathbb H$.

On the other hand, owing to the set-theoretic fact that "there is no set containing everything," it's not possible to pick a single universal codomain for functions.

When I wrote up this answer I realized that I used to ask the same questions as you, but I stopped once I had learned enough mathematics. I can't give you a single profound reason why we don't make all functions surjective besides a pragmatic one: surjectivity is a useful notion, and getting rid of it would be unprofitable.

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[Math] Typo: in the definition of inverse image $E$ should be replaced by $H$

[Math] Codomain of a function

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