Homeomorphism to circle.

Question

I want to show that $f:(0,2\pi]\to S^1$ defined by $t\to (\sin t, \cos t)$ is not a homeomorphism. I will do this by showing that its inverse is not continuous.

The inverse is defined by $(\sin t, \cos t)\to t$. So at $(0,1)$ we want to have $\lvert 2(1-\cos t)\rvert <\delta \Rightarrow \lvert t-2\pi\rvert<\epsilon$ for any $\epsilon$. If we take $\epsilon=\pi/2$, no matter how close $\cos t$ and $1$ are there will always be $t$'s close to $0$. So $f^{-1}$ is not continuous.

Can somebody explain this step of $\lvert 2(1-\cos t)\rvert <\delta \Rightarrow \lvert t-2\pi\rvert<\epsilon$ for any $\epsilon$ how this came?

Answer 1

The distance from $(0,1)$ to $(\sin t,\cos t)$ is$$\sqrt{\sin^2t+\bigl(\cos(t)-1)\bigr)^2}=\sqrt{2-2\cos t}.$$This is where the $2(1-\cos t)$ comes from. If $f^{-1}$ was continuous at $(0,1)$, then, when this number is small, the distance from $t$ to $2\pi$ ought to be small too.

Answer 2

Note that the distance between $(\sin(t),\cos (t))$ and the point $(0,1)$ is $$\sqrt {(1-\cos (t))^2+\sin^2(t)}= \sqrt {1-2\cos (t)+1}= |2(1-\cos (t)|$$

So if $|2(1-\cos (t)|<\delta$ you need to have $|t-2\pi | < \epsilon$

Note that $2\pi$ is the preimage of $(0,1)$

Answer 3

Although you want to show that $f^{-1}$ is not continuous, it is much easier to show that $f$ is not a closed map. To see this consider $C = (0,\pi]$ which is closed in $(0,2\pi]$. Then $(0,1) = f(2\pi) \notin f(C)$, but clearly $(0,1) \in \overline{f(C)}$.

Answer 4

As an alternative you can show that $f$ is not a homeomorphism by showing that $(0,2\pi]$ cannot be homeomorphic to $S^1$. For that purpuse there can be several strategies depending on you knowledge. One elementary way is noticing that erasing a point from $S^1$ doesn't disconnect the space, while any point in $(0,2\pi)$ disconnects $(0,2\pi]$. If you know about algebraic topology, the interval is contractible while $S^1$ is not.