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?

The answer is no: take $f^{-1}(x) = e^{ix}$, defined from $[0,2\pi)$ to $\mathbb{S}^1$ (the unit sphere in the plane)
This function is clearly continuous.
Unfortunately, its inverse cannot be continuous since otherwise $[0,2\pi)$ would be compact being the image of the compact set $\mathbb{S}^1$ under a continuous function.
I hope it helps, let me know if the details are clear enough :)
EDIT: something easier to check: the identity map from the reals with the lower limit topology ($\mathbb{R}_l$) to the real with the standard topology ($\mathbb{R}$) is another counterexample!
This is immediate to check since the inverse of the identity map is of course the identity map itself which now goes from $\mathbb{R}$ to $\mathbb{R}_l$. Clearly the pre image of the open set $[a,b)$ is not open!
