In this setting, why is the inverse function not continuous? Definition for continuity used here:   
$f: A \rightarrow B$ is continuous if given any $x$ in $A$ and any neighborhood $N$ of the point $f(x)$ in $B$, the set $f^{-1}(N)$ is a neighborhood of $x$ in $A$.     

I am trying to understand why the inverse function $g = f^-1$ is not continuous.
OK... $f$ maps $0$ to $1 + i \cdot 0$ So $g$ maps  $1 + i \cdot 0$ to $0$.  
So I think it must be because of the following. If we take a neighborhood $N$ (on the unit circle) of the complex number $z = 1 + i \cdot 0$ then we need to show that the set $g(N)$ is not a neighborhood of the point $0$ ( from the real interval $[0, 1)$ ). 
But why is it not a neighborhood, it seems like a neighborhood to me?! I think $g(N)$ has the form something like $[0, 0+\epsilon_1) \cup (1-\epsilon_2, 1)$ which seems like a neighborhood of $0$ given that our whole space/sub-space of the reals is $[0, 1)$ 
Or... do I need to pick another $z$ to show that $g$ is not continuous?! I doubt it.    
 A: Prove that the continuous image of a compact set is compact.
Now if the inverse map were continuous, then [0,1) would
be the continuous image of a compact set, hence compact.
As [0,1) is not compact, continuity would contradict the above theorem. 
A: We have that    
$f : [0, 1) \rightarrow \mathbb{C}, f(x) = e^{2 \pi i x}$
Obviously $f$ is a bijection from  $[0, 1)$ to the unit circle $B$ from $\mathbb{C}$. 
We want to prove that the inverse function  
$g: B \rightarrow [0, 1)$ 
is not continuous.   
Let's take the point $z_0 = 1 + 0 \cdot i \in B$ and $x_0=0 \in [0,1)$.  
Clearly $f(x_0) = z_0$ 
$\Rightarrow g(z_0) = x_0$ 
Let's pick a small $\epsilon$ neighborhood (in the real interval $[0, 1)$) of the point $g(z_0) = x_0 = 0$ and denote this neighborhood by $N$.   
$\Rightarrow N = [0, \epsilon)$ for some small $\epsilon \gt 0$.    
Now by definition the set $g^{-1}(N) $ is the set $ \{z\ |\ g(z) \in N\} = M$ 
i.e. it is the set of those points from the unit cirlce which $g$ sends into $N$.  
$\Rightarrow$ $M = \{ z = e^{2 \pi i x}\ |\ x \in [0, \epsilon)\}$ which is a tiny piece of the unit circle which lies in 1st quadrant. Since M does not contain the respective complex points from 4th quadrant ( I mean the points $\{ z = e^{2 \pi i (-x)}\ |\ x \in (0, \epsilon)\}$ ), this set $M$ is not a neighborhood of the point $z_0 = 1$ from the unit circle $B$.  
$\Rightarrow$ the set $g^{-1}(N)$ is not a neighborhood of the point $z_0 = 1$ in the unit circle $B$ 
$\Rightarrow$ $g$ is not continuous at the point $z_0$ 
