$p\mapsto c_p$ is continuous where $p\in S^1$ and $c\in \mathbb{R}$. Let $f: S^1\to S^1$ be a diffeomorphism. Consider the derivative map $$ Df_p:T_p S^1\to T_{f(p)}S^1,\ (p,ip)\mapsto (f(p),ic_pf(p)). $$ we say $f$ is orientation preserving if given $p\in S^1,\ c_p>0$. Now I want to prove $p\mapsto c_p$ is a continuous function. Can anyone give some hint?
Thanks. 
 A: Note that $TS^1\cong S^1\times\mathbb{R}$ as is stated in the coments. The diffeomorphism takes the points in $TS^1$ to $S^1$ and the tangent space at each point to the $\mathbb{R}$ componant. You "turn the tangents" to get the cylinder. Then projection onto the second coordinate is continuous. At this point you can divide by $|f(p)|$, which you know is never zero.
A: You can also show this is true locally.
Let $(U,\phi)$ be a smooth chart for $p$ and $(V,\psi)$ a smooth chart for $f(p)$. Since continuity is a local condition, it is enough to show your map is continuous in $U$.
We can write $F=\psi\circ f\circ\phi^{-1}$ for the coordinate representation, which is a diffeomorphism from $\phi(U)\subset\mathbb{R}$ to $\psi(V)\subset\mathbb{R}$. Then for $\phi(q)\in\phi(U)$, your $c_q$ is just $\frac{dF}{dx}(\phi(q))$. This is the composition of continuous maps, and so is continuous.
Note this works because the tangent space $T_{f(q)}S^1$ is one-dimensional, so $c_p$ is just 
$$ \frac{\lvert df_q(v)\rvert}{\lvert v\rvert}$$
for any nonzero $v\in T_qS^1$.
