Winding number of $f$ is equal to $\frac{1}{2\pi}\int_{S^1}f^*(d\theta).$

Question

Let $f: S^1 \to \mathbb{R}^2 - \{0\}$ be a smooth map. Define the winding number of $f$ about 0 and prove that it equals $\frac{1}{2\pi}\int_{S^1}f^*(d\theta).$

I have no clue, except for the winding number of $f$ around 0 is just the degree of $u$, where $$u(x) = \frac{f(x)}{|f(x)|}.$$

Degree. $Y$ is connected and has the same dimension as $X$. We define the degree of an arbitrary smooth map $f: X \to Y$ to be the intersection number of $f$ with any point, $\deg(f) = I(f,\{y\}).$

Intersection number. If $f: X \to Z$ is transversal to $Z$, $f, Z$ are appropriate for intersection theory, then $f^{-1}(Z)$ is a finite number of points, each with an orientation number $\pm 1$ provided by the preimage orientation.

Thank you~

Answer 1

I am sorry for having no time to write a complete answer. But I suggest you to read the definition of degree in more detail (what does it tell you about the behavior of $u$?). The wikipedia article on winding number and degree may be a good start.

Answer 2

Hints: (1) The degree theorem tells you that if $f\colon M \to N$ is a smooth map between compact oriented $n$-manifolds and $\omega$ is an $n$-form on $N$, then $$\int_M f^*\omega = \text{deg}(f)\int_N\omega\,.$$ (2) Show that $\displaystyle\int_{S^1} f^*d\theta = \displaystyle\int_{S^1} \Big(\frac f{|f|}\Big)^*d\theta $.