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 $[0,1]$?
 A: Consider the sequence $x_n = f(1 - \frac{1}{n})$. On the circle, $x_n = e^{2\pi i - \frac{2 \pi i}{n}} \to e^{2\pi i} = 1 + 0i$ (which is identified with the point $(1,0)$ on the plane). However, $f^{-1}(x_n) = 1 - \frac{1}{n}$ has no limit in $[0,1)$ and in particular we don't have $\lim_{n \to \infty} f^{-1}(x_n) = f^{-1}(1) = 0$ as the limit doesn't even exist!
The circle $S^1$ is not homeomorphic to $[0,1]$. You can see that by noticing that if you remove any point from $S^1$, you are left with a connected space while if you remove any point that is not equal to $0,1$ from $[0,1]$, you get a disconnected space. Thus, $S^1$ and $[0,1]$ cannot be homeomorphic. If this is the first time you see such an argument, I really recommend writing down explicitly all the details (assume that there exists a homeomorphism $f \colon S^1 \rightarrow [0,1]$, show that $f|_{S^1 \setminus \{ f^{-1}(1/2) \}}$ must be a homeomorphism between ${S^1 \setminus \{ f^{-1}(1/2)}$ and $[0,1/2) \cup (1/2,1]$ and show that this is not possible). 
A: 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 $U=[0, \epsilon[\cup ]1-\delta, 1[$.  
