Show that $f^{-1}: S^{1} \to [0,2\pi)$ is not continuous

Question

I have $f^{-1}: S^{1} \to [0,2\pi)$ where $(cos(t), sin(t)) \mapsto t$. I want to show that this function is not continuous by using the epsilon-delta defintion

$\forall\epsilon >0 \, \exists\delta>0 \,\, \forall t\in S^{1}: |t-t_{0}|<\delta \Rightarrow |f^{-1}(t)-f^{-1}(t_0)|< \epsilon.$

Now, if I look at the point $(1,0)$ I have that:

$|t-t_{0}| = |(cos(t), sin(t) ) - (1,0)| = \sqrt{2(1-cos(t))}< \delta$ and

$|f^{-1}(x)-f^{-1}(x_0)| = |t-0|=t < \epsilon$.

So:

$\sqrt{2(1-cos(t))}< \delta \Rightarrow t <\epsilon$.

I think I made a mistake, since I don't see how I could choose $\epsilon$ to satisfy these conditions. Should I consider another point than $(1,0)?$

Answer 1

Asserting that $f^{-1}$ is discontinuous means at $(1,0)$ means that there is some $\varepsilon>0$ such that$$(\forall\delta>0)\bigl(\exists(x,y)\in S^1\bigr):\bigl\|(x,y)-(1,0)\bigr\|<\delta\text{ and }\bigl|f^{-1}(x,y)-f^{-1}(1,0)\bigr|\geqslant\varepsilon.$$Take $\varepsilon=1$. For any $\delta>0$, take $t\in[1,2\pi)$ so close to $2\pi$ that $\bigl\|(\cos t,\sin t)-(1,0)\bigr\|<\delta$. Then$$\bigl|f^{-1}(\cos t,\sin t)-f^{-1}(1,0)\bigr|=|t|\geqslant1.$$

Answer 2

Alternatively, if $f^{-1}$ were continuous, we'd have a homeomorphism. But this is impossible, because only one space is compact.