Inverse of continuous function If we have continuous function $f$ between two topological spaces, such that it is one to one and onto, is it true that we conclude $f^\rm{-1}$  is continuous?

No. As a counterexample, take
$
f:[0,1) \to S^1
$
given by
$$
f(x) = (\cos(2\pi x),\sin(2\pi x))
$$
or, depending on your preferred definition of $S^1$,
$$
f(x) = e^{2 \pi i x}
$$
However, it is useful to note that if $f:X \to Y$ is one to one and onto with $X$ compact and $Y$ Hausdorff, then $f^{-1}$ must be continuous since any continuous map between these spaces takes closed sets to closed sets.
