Function with continuous inverse is continuous? If function $\textbf{F}^{-1}(x)$ is an inverse of function $\textbf{F}$ and $\textbf{F}^{-1}(x)$ is continuous. Is it true that $\textbf{F}(x)$ is continuous too?

I think  the map $f:[0,1) \rightarrow S^1$ given by $x \rightarrow (\sin 2\pi x, \cos 2\pi x)$ is a counter.
For a somewhat-trivial example , take $id: (X,\text{discrete} )\rightarrow (X, \text{indiscrete})$ , for any set $X$ .
As a general result, if $f A \rightarrow B$ is a continuous bijection; $A$ is compact and $B$ is Hausdorff, then $f^{-1}$ is necessarily continuous.
For counters from a space to itself with the same topology, see:
https://mathoverflow.net/questions/30661/non-homeomorphic-spaces-that-have-continuous-bijections-between-them
