Inverse of a continuos function

Question

If $f:I\subseteq \Bbb R \rightarrow \Bbb R$ is injective and everywhere continuous in its domain then $f^{-1}$ is also continuous everywhere. Here $I$ cannot be discrete set and $f$ is single variable function The above statement is always correct or not? I thought it is always correct. Am I correct or not?

Answer 1

Let $I=(0,1]\cup(2,3]$ and consider $$ f(x) = \begin{cases} x & x\in(0,1] \\ x-1 & x \in (2,3] \end{cases} $$

Your claim is true if $I$ is an interval, but "not discrete" is not enough.