Find a continuous bijective function $f: [0,1) \to S^1$ such that $f^{-1}$ is not continuous

Question

Find a continuous bijective function $f: [0,1) \to S^1$ such that $f^{-1}$ is not continuous, where $S^1=\{(x,y)\in \Bbb{R}^2 : x^2+y^2=1\}$

Please any help or direction

Answer 1

$t \to (\cos\, 2\pi t, \sin 2\pi t)$ is such a function. The inverse is no continuous at $(1,0)$. [Take a sequence $(a_n,b_n) \to (1,0)$ with $b_n >0$ for $n$ even and $b_n <0$ for $n$ odd. You will see that the image of this under $f^{-1}$ oscillates between $0$ and $1$].

Answer 2

Consider $g=f^{-1}: S^1 \rightarrow [0,1)$ where $$f:[0,1) \ni \theta \mapsto (\cos 2 \pi \theta, \sin 2\pi \theta)\in S^1$$

Note that every element of $S^1$ can be written as $$(\cos \theta , \sin \theta);0 \leq \theta <2 \pi$$

So $g$ is of the form $$g((\cos \theta , \sin \theta))=\frac{\theta}{2 \pi}$$ Let $[0,1/2)$ be an open subset of $[0,1)$. Now, $$g^{-1}\Big([0,1/2)\Big)=\Big\{(\cos \theta , \sin \theta): 0 \leq \theta < \pi \Big\}$$ That is, $g^{-1}\Big([0,1/2)\Big)$ is the $$\text{set of points in $S^1$ with positive y coordinate together with $(1,0)$}$$ This set is not open, since any open ball around $(1,0)$ would include some poi.nts with negative $y$ coordinate. Thus $g=f^{-1}$ is not continuous at $(1,0)$

Answer 3

Correct if wrong.

$f(t):= (\cos2πt,\sin 2πt)$ , $t \in [0,1)$

is bijective, and continuos . The inverse

$f^{-1}$ is not continuous at $P(1,0) \in S^2$.

Consider

$a_n=$

$\small{ (\cos (2π(1-1/n),\sin (2π(1-1/n)}$ $\in S^2$,

$n$ positive integer.

$\lim_{n \rightarrow \infty} a_n= (1,0)= f(0).$

However the sequence

$f^{-1}(a_n) = (1/(2π))(2π(1-1/n))$,

does not converge to $0$.