Why does a function $f(x)$ have to be bijective in order to have a $f^{-1}(x)$ (in Euclidean plane)?

Question

Yes, the title explains itself. I have no take on this one, or insights for that matter. For example, why isn't it enough for a certain function to be injective to have its $f^{-1}(x)$?

Answer 1

It is the matter of the nuances of definitions.

If a function from a domain $A$ to a codomain $B$ is merely a subset of $A\times B$, as in the Wikipedia definition, then the function does not "encode" $B$. You end up with the same set of pairs regardless of whether you view $f$ as a function of $A$ to $B$, or of $A$ to a superset $B'\supset B$, or of $A$ to $f(A)=\{f(a)\mid a\in A\}$. In that sense, you are right, you can invert $f$ as a relation, and the result is a function if and only if $f$ was injective.

On the other hand, it is often argued (and this is how I was taught functions in my youth) that it is necessary for the function to "encode"/specify both its domain and its codomain too. For example (I think this probably comes from Bourbaki), a function is defined as a triplet $(f,A,B)$, where $f$ is a function in the previous sense, and $A, B$ are the domain and the codomain, respectively. Then, the inverse $f^{-1}$ exists, as a function $B\to A$ if and only if both it exists in the previous sense, and $f(A)=B$, i.e. $f$ is bijective.

As it turns out, most of the time, even intuitively, we regard the functions as defined in the second definition above (certainly not the first definition), or something similar that encodes the codomain. We implicitly assume that the function "knows" its codomain somehow, i.e. the function is not just a set of ordered pairs, but it also "knows" which (potentially larger) set the images belong to. Thus the requirement for $f^{-1}$ to exist is $f$ to be bijective, not merely injective.

Answer 2

Surjectivity is necessary to define $f^{-1}$ over $\mathbb R$, and injectivity is necessary to define the map $f^{-1}$ as a function. In fact if $f$ is not surjective then $\exists x\in \mathbb R$ such that $x\notin f(\mathbb R)$ then $f^{-1}(x)=\emptyset$. If $\exists x_1, x_2$ such that $f(x_1)=f(x_2)$ then $x_1, x_2\in f^{-1}(f(x_1))$, then it's not generally a funcion.