Let $f: A \to B$ be given by $f(x) = |x|$, such that $A = [-1,0] \cup [1,2]$ and $B = [0,2]$.
I understand that $f$ does not have an inverse, since $f(-1) = f(1)$.
However, albeit my textbook is saying $f$ does not have an inverse $f^{-1}: B \to A$, it is also saying $f^{-1}$ is not continuous on $B$. How can that second sentence be made since $f^{-1}$ does not exist?