Preimage of $x^2$ Let $f(x)=x^2$ and let $B=[-1,1]$. My book then says that the preimage $f^{-1} (B)=[-1,1]$
But this seems unresonable to me. $f$ does does not map the domain to negative values, i.e. $ f: A  \rightarrow B$ always means that $B\geq 0$. Is there an explanation for this perhaps? Thank you.
 A: Note that while the notation for preimages looks like an inverse function, the analogy is somewhat loose. In particular it is not expected that $f(f^{-1}(B))$ will necessarily equal $B$.
By definition $f^{-1}(B)$ is the set of all $x$ such that $x^2\in B$. This set is exactly $[-1,1]$, because the squares of all numbers between $-1$ and $1$ happen to be between $-1$ and $1$ and the squares of a number outside this interval is always larger than $1$ and therefore not in $B$.
In particular, for example, $-\frac 12$ needs to be in $f^{-1}(B)$ becasue $(-\frac 12)^2 = \frac 14 \in B$.
$[-1,1]$ is also the preimage of some other sets, such as $[0,1]$ or $(-2,1]$ or $[-23\pi+\log 3, 1]$ or $(-\infty,1]$. That is not problematic.
A: Even though $x^2$ does not map to $[-1,0)$, the pre-image of $x^2$ to $[-1,1]$ must include all $x$ that map to anywhere in the interval. Since $(-1)^2 = 1$, it is necessary to include $-1$ in the pre-image.
In other words, $B$ is "larger" than it needs to be, The answer would be the same for $B=[0,1]$.
