When is an inverse a multifunction?

Question

Define $f:\mathbb{R} \to \mathbb{R}$ given by $f(x)=x^2$. If $f^{-1}(x)$ is interpreted as a function, it is undefined, since $f$ is not injective. If $f^{-1}$ is interpreted as a multifunction, it returns $\{-\sqrt{x},\sqrt{x}\}$. When is each of these interpretations normally made?

Answer 1

This is why we include domains and codomains in function definitions. Given an invertible mapping $f:A\to B$, the inverse mapping is $f^{-1}:B\to A$. Given any mapping $f:A\to B$, the preimage mapping is $f^{-1}:\mathcal P(B)\to\mathcal P(A)$, where $\mathcal P(X)$ refers to the power set of $X$.

Therefore, $f^{-1}(2)$ would refer to the inverse function, and $f^{-1}(\{2\})$ would refer to the pre-image function.

In practice, few people work with domains where there is any risk of not understanding which function is meant. I will note that my foundations professor who had noble but futile dreams of fixing mathematical notation, thought that $f_>$ and $f^<$ were distinct but still clear notations indicating the image and pre-image functions.

Answer 2

We define inverse of an element as other element of the same kind such that their product is identity.In other words notion of inverse is understood as defined in case of groups. So this means inverse of a single valued function is expected to be a single valued function only unless otherwise stated. If obviously one is not restricting to single valued function then multiple valued functions are also acceptable as inverses and suitably the inverse is defined in that case.