Is inverse image of a point a set?

Question

This is purely a matter of notation. Suppose $f(x) = x^2$. We say that $f(2) = 4$. We also say that $f(\{2\}) = \{4\}$.

We can also say that $f^{-1}(0) = 0$ without much grief because there exists a single solution for that case. However, does it make sense to ask what $f^{-1}(2)$ when it's an inverse image, or do inverse images only apply to sets, that is $f^{-1}(\{2\})$?

Answer 1

Inverse image applied to elements only makes sense if the function is injective, in which case we can restrict the range to the image and get a bijective map for which $f^{-1}$ is a well-defined map. Otherwise we reserve the notation $f^{-1}$ for the map given by $f^{-1} : \mathcal P(Y) \to \mathcal P(X)$ which sends a subset of $Y$ to a subset of $X$.

However, it is common to write $f^{-1}(a)$ for $f^{-1}(\{a\})$, for purposes of brevity. It is up to the context to determine if the function was bijective or not to begin with. One can write $f^{-1}(a) = \{b\}$ to make this clear, if it would so happen that the inverse image was a singleton.

Hope that helps,

Answer 2

Note that $f^{-1}(2)=\{\sqrt{2},-\sqrt{2}\}$, and in general for $x>0$ we have $$f^{-1}(x)=\{\sqrt{x},-\sqrt{x}\}.$$