Example 4.21 in Baby Rudin: How is the map $f^{-1}$ not continuous at the point $(1,0) = f(0)$? Let $f \colon [0,2\pi ) \to \{ (x,y) \in \mathbb{R}^2 \mid x^2 + y^2 = 1\}$ be defined as 
$$f(t) = ( \cos t , \sin t) \ \ \mbox{ for all } \ t \in [0, 2\pi).$$ 
Then the map $f$ is bijective and continuous, as asserted by Rudin. But how do we rigorously show that the inverse map is not continuous at the point $(1,0) = f(0)$? 
Intuitively, it is obvious since if we approach the point $(1,0)$ along that part of the unit circle which lies in the fourth quadrant, we can make the  distance in $\mathbb{R}^2$ between $f(t)$ and $f(0)$ as small as we please, but the distance in $\mathbb{R}$ between $t= f^{-1}\left(f(t)\right)$ and $0= f^{-1}\left(f(0)\right) = f^{-1}\left( (1,0) \right)$ approaches $2\pi$. 
But how to formulate this into a rigorous, $\epsilon$-$\delta$ argument? 
 A: Approach the point $(1,0)$ from above and below (in the circle). The limits of $f^{-1}$ are $0$ and $2\pi$ respectively.
Take $\epsilon=1$ then for every $\delta>0$ take a point $a=(x,\sqrt{1-x^2})$  and $b=(x,-\sqrt{1-x^2})$ such that $d(a,(1,0))<\delta$ and $d(b,(1,0))<\delta$.
But $f^{-1}(a)=\arctan(\frac{\sqrt{1-x^2}}{x})$ and $f^{-1}(b)=2\pi-\arctan(-\frac{\sqrt{1-x^2}}{x})=2\pi+\arctan{\frac{\sqrt{1-x^2}}{x}}$. Then $|f^{-1}(b)-f^{-1}(a)|=2\pi>\epsilon$.
A: The set $[0,\varepsilon)$, for small $\varepsilon>0$, is an open subset of the image of $f^{-1}$ whose inverse-image under $f^{-1}$ (which is the same as its direct image under $f$) is not an open subset of the domain of $f^{-1}$.  That proves $f^{-1}$ is not everywhere continuous.
But to use $\varepsilon$-$\delta$ methods to show that $f^{-1}$ fails to be continuous at the one point $(1,0)$, suppose we have a small $\varepsilon>0$.  We have $f^{-1}(1,0)=0$.  No matter how small we make $\delta>0$, some points in the $\delta$-neighborhood of $(1,0)$ get mapped to points near $2\pi$ and thus not in the $\varepsilon$-neighborhood of $0$.
A: Assume that $f^{-1}:\>S^1\to[0,2\pi[\ $ is continuous at $z_0:=(1,0)$. As $f^{-1}(z_0)=0$ there is a  $\delta>0$ such that
$$|f^{-1}(z)-f^{-1}(z_0)|<1$$
for all $z\in S^1$ satisfying $|z-z_0|<\delta$. Let $\delta':=\min\bigl\{{\delta\over2},{\pi\over4}\bigr\}$. Then the point $$z_1:=(\cos\delta',-\sin\delta')$$ satisfies $z_1\in S^1$ and $|z_1-z_0|<\delta$, but 
$$f^{-1}(z_1)=2\pi-\delta'\geq{7\pi\over4}>1$$
– a contradiction.
