# Proving that inverse of the unit circle parametrization is not continuous. [duplicate]

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.

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.

• The proof is correct. Also you can note that $[\pi,2\pi)$ is open but $f([\pi,2\pi))$ is not open. And further, note that $[0,2\pi)$ and $S^1$ have some fundamental differences for example if we remove one point $x\ne 0$ from $[0,2\pi)$ then we get a disconnected subspace, which is not the case for $S^1$. Commented Apr 23, 2023 at 10:29

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.
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.
• Their proof uses the principle: $g$ is continuous if and only if $g(\overline{A})\subseteq\overline{g(A)}$ for all sets $A$ Commented Apr 23, 2023 at 9:51
• Yes. Broken down to points, I used the condition of continuity $p \in cl_{\underline X}(A) \Rightarrow g(p) \in cl_{\underline Y}(g[A])$ to show it doesn't hold for $p := (1,0)$.