# Continuous bijection which is not a homeomorphism.

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 $\varphi=0$.

• You can show it's not a homeomorphism by showing the two spaces are different. What happens when you remove a point from the interval? What about the circle? Commented Sep 1, 2018 at 17:31
• Can you explain what is it $\varphi\mapsto (\cos(\varphi), \sin(\varphi))^t$, please? and $t$? Commented Sep 1, 2018 at 17:31
• @Piquito $t$ notes the transposed vector. It is for esthetics. Instead of writing $\begin{pmatrix}\cos(\varphi)\\\sin(\varphi)\end{pmatrix}$ you write simply $(\cos(\varphi),\sin(\varphi))^t$. Commented Sep 1, 2018 at 17:34
• @SteveD If you remove a point from the interval it is not connected anymore, but the circle would stay connected. I would like the "most elementary" way to solve this. Is my approach wrong? Could it be fixed? How about showing that $f$ is a surjection? Commented Sep 1, 2018 at 17:35
• I'm not sure how anything could be more elementary than removing a point :) Commented Sep 1, 2018 at 17:44

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 $[0,2\pi)$).
• How do you get $\lim_{n\to\infty} \sin\cos(2\pi-\tfrac1n))=0$? Should it not be $\sin(1)\neq 0$? Commented Sep 2, 2018 at 16:17
• I think you mean just $\sin(2\pi-\tfrac1n)$ and it is a typo. Commented Sep 2, 2018 at 16:18
To see that the inverse is not continuous note that there exists a sequence of points $(y_n)$ s.t $y_n\to (1,0)$ but $f^{-1}(y_n)\to 2\pi\ne f^{-1}(1,0)=0.$ Consider the sequence of points given by $$(\cos (2\pi-n^{-1}), \sin (2\pi-n^{-1}) )$$ for example.
• How do you know what $f^{-1}$ is? Commented Sep 1, 2018 at 17:43
• Once you show that the map is surjective (see the other answer) and injective (as you have shown), for $y\in S^{1}$, $f^{-1}(y)$ is the unique value of $x\in [0,2\pi)$ such that $f(x)=y$. There is at least one such value of $x$ since $f$ is surjective and at most one value of $x$ since $f$ is surjective. This is the definition of $f^{-1}$. In particular $f^{-1}(y_n)=2\pi-n^{-1}$ where $y_n=(\cos (2\pi-n^{-1}), \sin (2\pi-n^{-1}) )$. Commented Sep 1, 2018 at 17:49