S is a surface in $\mathbb{R}^{3}$ parameterized by a function $f:S\rightarrow(a,b)^{2}\subset\mathbb{R}^{2}$
$F$ is the function defined by:
$F:T^{1}S\rightarrow(a,b)^{2}\times S^{1}$ ($T^{1}S$ is the unit tangent bundle of $S,$ and $S^{1}$ is the unit circle)
$F(x,s)=(f(x),\dfrac{Tf(x).s}{\mid Tf(x).s\mid}$) ($Tf(x)$ is the tangent map)
Is $F$ a bijection ?
If yes, how to compute $F^{-1}$ ?
Assuming $f$ is a bijection, $f$ and $f^{-1}$ are differentiable, and $\det(Tf(x)) \neq 0$ for any $x$, then $F$ is a bijection. The inverse formula would be
$F^{-1} (y, w) = (f^{-1}(y), \frac{Tf^{-1}(y).w}{|Tf^{-1}(y).w|})$
Applying $F$ to this directly gives back the identity. The main point is that
$Tf(x)^{-1} = Tf^{-1}(f(x))$