Given the function $f:[0,2\pi)\to S^1$, $\varphi\mapsto (\cos(\varphi), \sin(\varphi))^t$. Show that $f$ is continuous and a bijection, but not a homeomorphism.
That $f$ is continuous is clear, since every component is continuous.
Furthermore it is continuous differentiable.
When I want to show, that $f$ is a bijection it is easy to see, that $f$ is injective, since for
$f(x)=f(y)\Leftrightarrow (\cos(x), \sin(x))=(\cos(y),\sin(y))\Leftrightarrow \cos(x)=\cos(y)\wedge\sin(x)=\sin(y)\stackrel{x,y\in [0,2\pi)}{\Leftrightarrow} x=y$
But how can I show, that $f$ is a surjection?
To show, that $f$ is not a homeomorphism, I have to verify, that $f^{-1}$ is not continuous.
Can I use the inverse function theorem?
I get:
$Df(\varphi)=\begin{pmatrix}-\sin(\varphi)&0\\0&\cos(\varphi)\end{pmatrix}$
With determinant $\operatorname{det}Df(\varphi)=-\sin(\varphi)\cos(\varphi)$
Where $Df(\varphi)$ is not invertible for $\varphi=0$.
Thanks in advance for hints and comments.
Best Answer
The function $f$ is surjective becasue if $(x,y)\in S^1$, there is a $\theta\in[0,2\pi)$ such that $(x,y)=(\cos\theta,\sin\theta)$; just take $\theta=\arccos x$ if $y\geqslant0$ and $\theta=2\pi-\arccos x$ otherwise.
And $f^{-1}$ is discontinuous because $\lim_{n\to\infty}\left(\cos\left(2\pi-\frac1n\right),\sin\left(2\pi-\frac1n\right)\right)=(1,0)=f(0)$, but $\lim_{n\to\infty}f^{-1}\left(\cos\left(2\pi-\frac1n\right),\sin\left(2\pi-\frac1n\right)\right)$ doesn't exist (in $[0,2\pi)$).