Why $S^1$ does not homeomorphic to $[0, 1)$? The map $f:[0,1)\to S^1$  given by $f(x)=e^{2\pi ix}$ is a continuous bijection.
However since $S^1$ is compact and  $[0,1)$ is not compact, $f$ can not be a homeomorphism.
For any $z\in S^1,$ we have $f^{-1}(z)=\dfrac{\log(z)}{2\pi i}.$ I suppose $f^{-1}:S^1\to[0,1)$ is not continuous.
But I could not explain it precisely. 
How can this happen? 
Since $S^1$ not homeomorphic to $[0,1),$ is it homeomorphic to $[0,1]$?

With the topology induced by $\Bbb R$, the spaces $\Bbb S^1$ and $[0,1[$ aren't homeomorphic. But if you change the topology on the set $[0,1[$, the function $f(x)=e^{2\pi i x}$ become an homeomorphism. Precisely, endowed $[0,1[$ with a coarser topology $\tau$ whose open set are:
$U=(\epsilon, 1-\delta) \quad \epsilon, \delta >0$, and $U=[0, \epsilon[\cup ]1-\delta, 1[$.  
