Continuous function on the circle and real line 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?

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 $\phi = f \circ \tau$.
