Is there something wrong with this problem statement? [Analysis Preliminaries]

Question

The problem seems to be asking the pre-image of $f(B)=[-1,1]$ where $f(x)=x^2$. The solutions manual provided that the pre=image is $f^{-1}(B)=[-1,1]$ which does not seem right to me.

Answer 1

The pre-image is $[-1, 1]$. The numbers from that set, squared, will be in $B$, and any number outside of that set, squared, will fall outside $B$.

I am not sure what is confusing you. Perhaps the fact that not every element of $B$ will be an image of something in $[-1,1]$? That is true but not relevant for deciding what the pre-image is. Incidentally, what is true is that $f(f^{-1}(B))\subset B$ where the inclusion is strict in our case (the LHS is $[0,1]$).

Answer 2

Let $B = [-1,1]$, and let $f(x) = x^2$.

Claim $f^{-1}(B) = [-1,1]$.

Proof:

First suppose $x \in [-1,1]$. \begin{align*} \text{Then}\;&x \in [-1,1]\\[4pt] \implies\;&|x| \le 1\\[4pt] \implies\;&0 \le x^2 \le 1&&\text{[since $x^2 \ge 0$]}\\[4pt] \implies\;&-1 \le x^2 \le 1\\[4pt] \implies\;&x^2 \in [-1,1]\\[4pt] \implies\;&f(x) \in B\\[4pt] \implies\;&x \in f^{-1}(B)\\[4pt] \end{align*} It follows that $[-1,1] \subseteq f^{-1}(B)$.

Next, suppose $x \in f^{-1}(B)$. \begin{align*} \text{Then}\;&x \in f^{-1}(B)\\[4pt] \implies\;&f(x) \in B\\[4pt] \implies\;&x^2 \in [-1,1]\\[4pt] \implies\;&x^2 \in [0,1]&&\text{[since $x^2 \ge 0$]}\\[4pt] \implies\;&0 \le x^2 \le 1\\[4pt] \implies\;&|x| \le 1\\[4pt] \implies\;&-1 \le x \le 1\\[4pt] \implies\;&x \in [-1,1]\\[4pt] \end{align*} It follows that $f^{-1}(B) \subseteq [-1,1]$.

Thus, we have both inclusions, hence $f^{-1}(B)=[-1,1]$.