# 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?

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How continuous is "continuous"? – J. M. May 15 '11 at 6:40
To detail a little bit J.M.'s question, because I wonder too : when you say $\varphi$ is a continuous function defined on the circle, you don't specify its image... what are the domains and codomains of your functions? After that it'll be easier to tell. – Patrick Da Silva May 15 '11 at 7:16
Well in Artin's Algebra there is a Lemma: The continuous homomorphisms $\varphi: \mathbb{R}^{+} \to U_{1}$ are of the form $\varphi(x)=e^{icx}$ for some $c \in \mathbb{R}$. – user9413 May 15 '11 at 7:17
@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. – user8268 May 15 '11 at 7:49

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 $\phi = f \circ \tau$.