If $f$ takes $[-1,1]$ onto $[-1,1]$ then $f^{-1}(\{f(0)\})=\{0\}$

Question

Consider the statement:

If $f$ takes $[-1,1]$ onto $[-1,1]$ then $f^{-1}(\{f(0)\})=\{0\}$.

My book tells me this is suppose to be false, but I don't understand why.

We know:

If $f:X\to Y$ has an inverse function, then $f^{-1}(f(x))=x$ and $f(f^{-1}(y))=y$ for all $x\in X$ and $y\in Y$. From this definition, it would appear that our original statement is true.

If this is indeed false, do I need to find a function $f$ such that $[-1,1]$ onto $[-1,1]$ can't be inverted?

Answer 1

Counter example: $f(x)=\sin(\pi x)$. Then, $f:[-1,1]\to[-1,1]$ is onto

but $f^{-1}(f(0))=f^{-1}(0)=\{-1,0,1\}.$

Answer 2

$f$ may not necessarily have an inverse because it may not necessarily be one-to-one. Consider functions $g:[-1,0]\to[-1,1]$ and $h:(0,1]\to[-1,1]$ where $g,h$ are onto, and let $f=g\cup h$