Proving that inverse of the unit circle parametrization is not continuous.

Question

Statement:

Let us have a continuous and bijective unit-circle parametrization map:

$f: [0, 2\pi) \rightarrow S$

$\phi \mapsto cos(\phi) + i \cdot sin(\phi)$

We prove that $f^{-1}$ is not continuous.

Proof:

1. We have $(1,0) \in \overline {f[(\pi, 2\pi)]} \ \ \subset S$

2. But $f^{-1}(1,0) = 0 \notin \overline {f^{-1} \circ f[(\pi, 2\pi)]} = \overline{(\pi, 2\pi)} = [\pi, 2\pi)$.

3. So $f^{-1}$ in point $(1,0) \in S$ is not continuous, hence $f^{-1}$ is not continuous.

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I have not used the compactness criteria and wanted to prove it in a "classical" way.

Not sure it is correct and rigorous enough but the idea is somewhat like that.

Answer 1

Your proof is correct.

Noting $[0,2\pi)$ is not compact while $S^1$ is, continuity of $f^{-1}$ would imply an impossible homeomorphism $[0,2\pi)\cong S^1$. So it cannot be continuous.

By far the most intuitive, for me, proof is this: $\lim_{z\to1}f^{-1}(z)$ doesn’t exist! So it cannot be continuous.

Answer 2

Honestly I cannot follow your proof. I guess you can just take a sequence of points $(\cos \phi_n, \sin \phi_n)$ with $\phi_n\to 2\pi^-$. Its limit on the circle is $(1,0)$. But the sequence of pre-images does not have a limit in $[0,2\pi)$. So $f^{-1}$ is discontinuous at that point.