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Is there a concept of inverting a non-invertible function by returning a set of the possible solutions?

For example:

$g(x) = x^2$

Would it be possible to create an inverse function $f(y)$ where, for example:

$f(4) = -2,2$

(I'm pretty sure I'm going about this wrong, but I'm still learning so I don't know *how* I'm wrong)

This is a multivalued function (see especially the first example!), or multifunction, or set-valued function. A set-valued map, taking elements of $X$ and producing subsets of $Y$, is often denoted $f : X rightrightarrows Y$.

It can also be denoted more literally by $f : X to 2^Y$, as such maps can be thought of as (ordinary, single-valued) functions from $X$ to the power set of $Y$.

Finally, one could also view them simply as relations with a full domain.

It is common to use the notation $f^-1(A)$, where $A$ is a subset of the value range of $f$, as a shorthand to describe the set $x : f(x) in A $. Furthermore, if $A$ is a set with only one value $x$ it is also somewhat common to just write $f^-1(x)$ instead of $f^-1(x)$. So if in your case the context is clear, it is fine to write $g^-1(4) = 2, -2$.

Also there are functions that are multivalued by default like the complex logarithm for example.

It is possible to have a set-valued function just like your $f$. The only 'but' is that it is not a function $f: mathbb Rto mathbb R$ but actually $f:mathbb Rtomathcal P(mathbb R)$, i.e. its target is the set of parts (or power set) of $mathbb R$.

What you are actually describing by your $f$ is the level sets of $g$. Any more I would add is contained in @LionCoder's comment, so I invite you to continue reading his input instead of repeating what they have said.

To give a function is the same to give the sets where the function is constant: $cmapsto f^-1(c)$. This would be a map $F:mathbbRto power(mathbbR)$. The function $F$ is almost injective (since $f^-1(c)cap f^-1(c')=emptyset$ for $cneq c'$). It might not be injective if $f^-1(c)=f^-1(c')=emptyset$, that is, if $c$ and $c'$ are not in the range of $f$. The result is:

$$F: Image(f)to power(mathbbR)$$

is injective.

This map can never be surjective (by cardinality), so this is the best we can say. If you restrict an injective function to its image, then it becomes bijective.

We usually want functions to only take one value in value space.

If there are several different possible ones, we call them branches of a function.

For example $f^-1(t) = -sqrtt$ is one such branch that is inverse to $f(t) = t^2$