Non-bijective inverse?

Question

I'm at the beginning of J. Munkres's Topology and in the section introducing functions he gives this particularly weird example (attached below). The function $f(x) = 3x^2 + 2$ is neither injective, nor surjective, so its inverse doesn't make any mathematical sense. Moreover, even if we for a second speculatively accept its existence, why is $f^{-1}(2) = -1$, for example, when it should be $0$? I'm really confused by this, so thanks in advance for any clarification.

Answer 1

The notation $f^{-1}(S)$ where $S$ is a set is the preimage. So for your example we have $f(x) = 3x^2+2$ and are considering the preimage of $[2,5]$. This is the following \begin{equation} f^{-1}([2,5]) = \{x\in\mathbb{R}\,:\, f(x)\in[2,5]\}. \end{equation}

Answer 2

In general, $f^{-1}([a, b])\neq [f^{-1}(a), f^{-1}(b)]$, even for continuous functions (and even for bijective functions).

In this case, we're first looking at $f([0, 1]) = [2, 5]$. Then we apply $f^{-1}$ to that interval. By definition, $f^{-1}([2, 5])$ is the collection of all $x$ so that $f(x)\in [2, 5]$. And it so happens that this collection of $x$-values is the interval $[-1, 1]$.

Answer 3

Note that $f([0,1])=\{f (x)\mid x\in[0,1]\} $ is a set. And for any set $A\subseteq \mathbb R $, $f^{-1}(A):=\{y\in\mathbb R\mid f (y)\in A\} $. Even without injectivity, this definition makes sense.

Answer 4

$f^{-1}(2)\ne1,$ but $2\le3x^2+2\le5\implies 0\le3x^2\le3\implies 0\le x^2\le 1\implies-1\le x\le 1,$

so $f^{-1}([2,5])=[-1,1]$.