I am a beginner in Topology. I was going through Munkres book where I came across this example.
The mapping $[0,1)\to S^1$ (unit circle) is bijective and continuous, but $f^{-1}$ is not continuous.
The function $f(t)=(\cos 2\pi t, \sin 2\pi t)$ and $S^1$ is a subspace of the plane $\mathbb{R}^2$. I don't understand how the inverse is not continuous. Can someone please explain this simple thing to me?
Thanks a lot
Best Answer
We need $(f^{-1})^{-1}(U) \subset S^1$ to be open for any open $U \subset [0,1)$, i.e. $f(U)$ needs to be open. Any interval of the form $[0,\epsilon)$ for $0<\epsilon<1$ is open in $[0,1)$, but $f([0,\epsilon))$ is not open in $S^1$.