Periodic Function as Mapping from Unit Circle

Theorem
Let $\SS$ denote the unit circle whose center is at the origin of the Cartesian plane $\R^2$.

Let $p: \R \to \SS$ be the mapping defined as:
 * $\forall x \in \R: \map p x = \tuple {\cos x, \sin x}$

Let $f: \R \to \R$ be a periodic real function whose period is $2 \pi$.

Then there exists a well-defined real-valued function $f': \SS \to \R$ such that:
 * $f = f' \circ p$

where $f' \circ p$ denotes the composition of $f'$ with $p$.

Proof
Let $f': \SS \to \R$ be defined as:


 * $\forall \tuple {x, y} \in \SS: \map {f'} {x, y} = \map f x$

Consider the inverse $p^{-1}: \SS \to \R$ of $p$:
 * $\forall \tuple {x', y'} \in \SS: p^{-1} \sqbrk {x', y'} = \set {x \in \R: \cos x = x', \sin x = y'}$

Let $\RR$ be the equivalence relation on $\R$ induced by $p$:


 * $\forall \tuple {x, y} \in \R \times \R: \tuple {x, y} \in \RR \iff \map p x = \map p y$

That is:

Let $f'$ be defined as:


 * $f' = f \circ p^{-1}$

Then by the Quotient Theorem for Sets:


 * $f'$ is well-defined $f$ is periodic with period $2 \pi$.

It follows from Conditions for Commutative Diagram on Quotient Mappings between Mappings that $f$ and $f'$ are related by the commutative diagram:


 * $\begin{xy} \xymatrix@L+2mu@+1em{

\R \ar[r]^*{p} \ar@{-->}[rd]_*{f := f' \circ p} & \SS \ar[d]^*{f'} \\ & \R }\end{xy}$

Also see

 * Mapping from Unit Circle defines Periodic Function
 * Periodic Function is Continuous iff Mapping from Unit Circle is Continuous