Show that the inverse of $g(x)= (\cos(2x) , \sin(2x))$ is not continuous over a specific domain. I am struggling with the following proof and any help would be appreciated.

Let $T=[0,\pi)$ and $g:T \to \mathbb{R}^2$ be defined as $g(x)= (\cos(2x) , \sin(2x))$. Prove that $g$ is continuous on $T$ but that $g^{-1}:g[T] \to T$ is not continuous on $g[T]$.

My attempt is as follows: First $g(x)$ is continuous on $T$ as it's components $\cos(2x) , \sin(2x)$ are continuous on all of $\mathbb{R}$. Next the inverse of $\cos(2x)$ is $\frac{\pm 1}{2}\arccos(x)$, and the inverse of $\sin(2x)$ is $\frac{1}{2}\arcsin(x)$. What I do not understand is why $g^{-1}$ is not continuous on $[-1,1]$. As both arccos and arcsin are both defined on $[-1,1]$.
 A: $g(T)$ is the unit circle centered at the origin. $g$ wraps the interval $[0,\pi)$ around the unit circle counterclockwise, starting at $(1,0)$. The inverse function $g^{-1}$ "unwraps" the circle, but this involves tearing it at the point $(1,0)$. So we should not expect $g^{-1}$ to be continuous at $(1,0)$.
To prove that $g^{-1}$ is not continuous, you need to show that there is an open subset $U$ of $T$ such that $(g^{-1})^{-1}(U)$ is not open in $g(T)$, i.e., $g(U)$ is not open in $g(T)$. Alternatively, $g^{-1}$ is not continuous at $(1,0)$ if there is a sequence $(a_n,b_n)$ of points in $g(T)$ such that $(a_n,b_n) \to (1,0)$, but $g^{-1}(a_n,b_n) \not\to 0=g^{-1}(1,0)$. Do you see how to show either of these?
A: Pick a sequence $(x_n)$ in $[0,\pi)$ such that $x_n\to \pi$.
Denote for each $n\in\mathbb N$, $u_n=g(x_n)$ and notice that $$u_n=g(x_n)\to (1,0).$$
Now consider $g^{-1}:S^1\to [0,\pi)$ and notice that $$g^{-1}(u_n)=g^{-1}(g(x_n))=x_n\to \pi.$$
Since $g^{-1}((1,0))=0\neq\pi$, the limit above shows that $g^{-1}$ is not continuous at $(1,0)$
A: Hint: If $g^{-1}$ is continuous, then $g$ must be an open mapping. Note that $[0,\pi/2)$ is open in $T$. But what is 
$
g([0,\pi/2))?
$
