Prove the set of continuous functions on $S^1$ is isomorphic to the set of continuous periodic functions with period $2\pi$

Question

Let $S^1 = \{ (x,y)\in \mathbb{R}^2 | x^2+y^2=1$. Let $z:\mathbb{R} \to S^1$ given by $z=(cos(\theta),sin(\theta))$. Define the map $z^* : C^0 (S^1) \to C^0 (\mathbb{R})$ by $z^* (f)= f \circ z$.

I'm asked to prove that the map $z^*$ is injective onto the subset of continuous periodic functions with period $2\pi$ (i.e. those f which satisfy $f(a+2\pi)=f(a)$ for all $a\in \mathbb{R}$.

Here $C^0 (S^1)$ represents all continuous functions from $S^1$ to $\mathbb{R}$ and $C^0 (\mathbb{R})$ is the set of all continuous functions from $\mathbb{R}$ to itself. I've already proven that $z^*$ is injective and it's image is a subset of the periodic functions with period of $2\pi$. But I'm having trouble with surjectivity.

Answer 1

Suppose $g$ is a $2\pi$-periodic continuous function on $\mathbb{R}$. Then $g=z^*(f)$, where $f(\cos\theta,\sin\theta)=g(\theta)$. Indeed,if $x=\theta+2k\pi$ for $\theta\in[0,2\pi]$, $$ z^*(f)(x)=f\circ z(x)=f(cos(\theta+2k\pi),\sin(\theta+2k\pi))=f(\cos\theta,\sin\theta)=g(\theta)=g(\theta+2k\pi)=g(x) $$ by the $2\pi$-periodicity of $g$.