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Suppose that $S^1$ is the unit circle in $\mathbb{C}$ and suppose that $g\colon [0,2\pi]\rightarrow \mathbb{C}$ is continuous such that $g(2\pi) = g(0)$.

I have to show that $h\colon S^1 \rightarrow \mathbb{C}$, $x \mapsto g(t(x))$ is continuous. where $t(x)$ is the number such that $x = e^{it(x)}$.

I guess that I have to show that if $y \in B(x,\epsilon)$, then $y \in B(t(x),\epsilon)$ (if $\epsilon$ is small enough). This directly implies the result. Can anyone help me to show this?

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1 Answer 1

Presumably $t: S^1 \to [0,2 \pi)$, with $z= e^{i t(z)}$, for $z \in S^1$. Since $t(1) = 0$ and $\lim_{\theta \uparrow 2 \pi} t(e^{i \theta}) = 2 \pi$, it is clear that $t$ is not continuous at $z=1$.

By restricting $t$ to appropriate segments of $S^1$, it is straightforward to write down explicit formulae for $t$ that demonstrate continuity (in fact smoothness) on $S^1 \setminus\{1\}$.

For example, if we let $\Delta = \{z \in S^1 | \operatorname{Re} z > 0 \}$, then $t(z) = \begin{cases} \arcsin (\operatorname{Im} z), & \operatorname{Im} z \ge 0 \\ 2\pi+\arcsin (\operatorname{Im} z), & \operatorname{Im} z < 0 \end{cases}$.

It follows that $t$ is continuous as a map from $S^1 \setminus\{1\}$ to $(0,2 \pi)$.

Hence $g \circ t$ is continuous on $S^1 \setminus\{1\}$, the only point that needs to be checked is $z=1$.

Let $I_\delta = [0, \delta) \cup (2 \pi -\delta, 2 \pi]$. Since $g$ is continuous, with $g(0) = g(2 \pi)$, for any $\epsilon>0$ there exists some $\delta>0$ such that if $\tau \in I_\delta$, then $|g(\tau) - g(0)| < \epsilon$. It should be clear from the formula for $t$ above that for any $\delta>0$, there exists some $\eta$ such that if $|z-1| < \eta$, with $z \in S^1$, then $t(z) \in I_\delta$. Hence $|g(t(z))-g(0)| < \epsilon$, and so $g \circ t$ is continuous at $z=1$.

An alternative approach: Well, this is equivalent, but is tidier in some sense.

Let $X = \mathbb{R}/ 2 \pi \mathbb{Z}$. Define $\tilde{g}: X \to \mathbb{C}$ as $\tilde{g}([x]) = g(x)$ (with $ x\in [0,2 \pi)$), and $\tilde{t}: S^1 \to X $ as $\tilde{t}(z) = [t(z)]$. In a manner similar to above, one shows that $\tilde{g}$ is well defined and continuous, and that $\tilde{t}$ is continuous (with range $X$), hence so is $\tilde{g} \circ \tilde{t}$. Since $g \circ t = \tilde{g} \circ \tilde{t}$, this shows that $g \circ t$ is continuous.

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