How is this example not a homeomorphism? 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
 A: 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 $S^1$.
A: What is the image of $[0,1/3)$?  It's open in $[0,1)$ but its image is not open in $S_1$.  Let $g=f^{-1}$.  If $g$ were continuous then $g^{-1}(\text{open})=\text{open}$.  But $g^{-1}([0,1/3))$ is not open.  So $g$ is not continuous.
A: Name $P$ the point $f(0)$. What are the inverse images of points close to $P$ under $f$?
A: The intuition is that a homeomorphism captures the idea of two spaces being the same.  Since an interval and a circle do not have the same shape, you should expect this map to not be a homeomorphism. 
Formally, you should investigate an open neighborhood of 0 in $[0,1)$.  If $f^{-1}:S^1 \to [0,1)$ were continuous, then the preimage of this open neighborhood will be an open set in $S^1$.
A: Intuitively the inverse of f is ripping open the circle.  So consider a pair of points on the circle that are on opposite sides of (1,0).  Make them very close.  Now consider the inverse of f.  It sends each of the pair close to a different end of [0,1).  How can this be a continuous map?
