Let $S^1=\{(x,y):x^2+y^2=1\}$ be the unit circle. Prove that $G:[0,1)\to S^1$, $G(t)=(\cos(2\pi t),\sin(2\pi t))$ is not a homeomorphism 
Let $S^1=\{(x,y):x^2+y^2=1\}$ be the unit circle with the subspace topology induced from $\mathbb{R}^2$. Prove that $G:[0,1)\to S^1$, $G(t)=(\cos(2\pi t),\sin(2\pi t))$ is not a homeomorphism

I think $G^{-1}(x,y)$ will not be continuous but I'm not sure how to prove this.
I believe the issue is that $[0,a)$ is open in the subspace topology on $[0,1)$ but because a circle is a loop the image will never be open in the circle? So the inverse won't be continuous as the preimage of this open set, will not be open in the circle.
I'm not sure how to compute the inverse of this map however, so I'm not sure how I could show that this won't be a homeomorphism.
 A: There are two ways to going about it: show that $[0,1)$ and $S^1$ are not homeomorphic  is the first. If you’ve covered compactness you can say $S^1$ is compact and the first space is not ( not closed in $\Bbb R$ or give a cover etc. ) or if you’ve covered connectedness you can say that we can remove any point of the circle and are left with a connected set, while for $[0,1)$ we can only remove $0$ to have that property. Etc. 
You can also directly show that $G^{-1}$ is not continuous. It’s not really needed to give a formula, just look at how $G$ works: a number $0 \le t < 1$ is scaled to an angle $0 \le 2\pi$ (in radians) and sent to the unique point in the complex plane of radius $1$ and argument (angle with the positive real axis) equal to that angle. So the inverse for a point on $S^1$ we can compute by taking the argument, which lies in $[0,2\pi)$ and scaling it back to $[0,1)$. Now look at the sequence on the circle that converges to $(1,0)$ coming from below the positive real axis. So $(x_n,y_n)= (\cos(2\pi(-\frac1n)),\sin(2\pi(-\frac1n))$, which has argument $2\pi ( 1-\frac1n)$ so $G^{-1}(x_n,y_n)=1-\frac1n \not \to 0=G^{-1}(1,0)$. So $G^{-1}$ is not sequentially continuous. 
A: The answer in the comments is the way to go. But if you want to show directly that 
$G^{-1}$ is not continuous, note that 
$(\cos 2\pi/n,\sin 2\pi/n)\to (1,0)$ and $G^{-1}(\cos 2\pi/n,\cos 2\pi/n)\to 0$ 
whereas 
$(\cos (2\pi(1-1/n)),\sin (2\pi(1-1/n)))\to (1,0)$ and $G^{-1}(\cos (2\pi(1-1/n)),\sin (2\pi(1-1/n)))\to 1$
