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If $\varphi$ is a continuous function defined on the unit circle $T =\{z \in \mathbb{C} : |z| = 1 \}$, then $f (t) = \varphi(e^{it})$ is a continuous periodic function with period $2\pi$. Conversely, if $f$ is a continuous function on the real line of period $2\pi$, there exists a unique continuous function $\varphi$ defined on the unit circle $T$ such that $f (t) = \varphi(e^{it})$. Is my guess correct? Can anyone lead me a proof for this?

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    @user10805: you are right. Surely there is a unique function $\phi$ such that $f(t)=\phi(e^{it})$, so you want to know why $\phi$ is continuous. The map $\mathbb{R}\to T$, $t\mapsto e^{it}$, has locally a homeomorphism, so $\phi$ is continuous iff $f$ is.2011-05-15

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Let $X$ be a topological space and $f\colon \mathbb R \to X$ be continuous and $2\pi$-periodic. Since $f$ is $2\pi$-periodic $\phi\colon T \to X$ is well-defined by $\phi(e^{it}) = f(t).$ On the other hand with the condition $\forall t\in \mathbb R\colon f(t) = \phi(e^{it}),$ there is only this possible definition, showing the uniqueness. Finally let us prove the continuity of $\phi$. Let $\tau\colon T \to [0,2\pi)$ be defined by $\tau(e^{it}) = t$ for an adequate $t\in [0,2\pi)$. Then $\tau$ is certainly continuous (but not homeomorphic). Continuity now follows from $\phi = f \circ \tau$.