Homeomorphism: can $f:X\mapsto X'$ be continuous, bijective without $f^{-1}:X'\mapsto X$ being continuous?

Question

In our definition of homeomorphic topological spaces there must exist a bijective function $f:X\mapsto X'$ such that $f$ is continuous on $X$ and $f^{-1}$ is continuous on $X'$.

Is it necessary to assure that $f^{-1}$ is continuous on $X'$?

Answer 1

You have to assume this, it is not automatic:

If $X=\{1,2\}$ in the discrete topology and $Y=X$ in the indiscrete/trivial topology, then $f(x)=x$ is continuous from $X$ to $Y$ but its inverse (the same identity) is not continuous from $Y$ to $X$.

Answer 2

The requirement that $f^{-1}$ is continuous, is essential. Consider for instance the function $f:[0,2 \pi)\rightarrow S_1$ (the unit circle in $\mathbb{R}^2$ defined by $f(\phi )=(\cos \phi ,\sin \phi ))$.

This function is bijective and continuous, but not a homeomorphism ($S^{1}$ is compact but $[0,2\pi )$ is not).

The function $f^{-1}$ is not continuous at the point $(1,0)$, because although $f^{-1}$ maps $(1,0)$ to $0$, any neighbourhood of this point also includes points that the function maps close to $2\pi $, but the points it maps to numbers in between lie outside the neighbourhood.