TITLE: Prove that there exists a smooth map $g\colon R\to R$ such that $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$ and satisfying $g(2π)=g(0)+2\pi q$.
QUESTION [6 upvotes]: Let $f\colon S^1\to S^1$ be any smooth map. Prove that there exists a smooth map $g\colon\mathbb{R}\to\mathbb{R}$ such that $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$ and satisfying $g(2π)=g(0)+2\pi q$.
The book told me to show that $g(2\pi)=g(0)+2\pi q$, then extend $g(t+2\pi) =g(t)+2\pi q$.
But I don't understand what they mean. How can I get function $g$ from $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$, and what is $g$?

REPLY [3 votes]: $$f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))=(\cos(g(t)+2\pi q),\sin(g(t)+2\pi q))$$
On the other hand 
$$f(\cos(t),\sin(t))=f(\cos(t+2\pi q),\sin(t+2\pi q))=(\cos(g(t+2\pi q)),\sin(g(t+2\pi q)))$$
From these two results you get 
$$g(t)+2\pi q=g(t+2\pi q)$$
where $q\in\mathbb{Z}$. $f$ can be thought as a map taking points from the unit circle and mapping them to another unit circle through the rule 
$$f(\cos(t))=\cos(g(t))$$
and 
$$f(\sin(t))=\sin(g(t))$$
For example of $g(t)=t$ is such a function and $f$ is the identity map.