I am trying to prove that if $\gamma(t)=(x(t),y(t))$ ,a function from the closed interval $[0,1]$ to $\mathbb{R^2}$ is a simple closed unit speed curve such that $\gamma '(0)=\gamma '(1)$. Then the tangent vector rotates exactly by $2 \pi$.
My Try: I think I have a proof I am not sure though.I am trying to use the idea of covering spaces to prove this. So first I observe that considering the tangent vector to be $x'(t)+iy'(t)$, we get a map from $[0,1]$ to $S^1$ and by our assumption it is a closed loop.
Now I use $\mathbb{R} $ to cover $S^1$ and the covering map is $t \rightarrow (\cos (2 \pi t), \sin (2 \pi t))$. Now I lift the path $\gamma '(t)$ to $\widetilde{\gamma '(t)}$.
Now we observe that a circle $\alpha(t)$ which passes through the point $\gamma (0)$ has the property that the lift $\widetilde{\alpha '(t)}$ has its endpoint at $1$.
As suggested by Andrey Ryabichev we observe that one of the two pieces obtained because of simple closed regular curve is diffeomorphic to the disc.Now we know that $\gamma(t)$ and $\alpha (t)$ are homotopic by a differentiable homotopy $H(s,t)$. Now I differentiate this map to get a homotopy from $\widetilde{\alpha '(t)}$ to $\widetilde{\gamma '(t)}$ . This means they both should have the same endpoint. Therefore we are done. Is this proof correct?
Here is my proof, that doesn't use covering circle by line at all.
By the Jordan–Schönflies theorem, there exists an isotopy $F_t:\mathbb R^2\to \mathbb R^2, t\in [0,1]$ of $F_0=\mathrm{Id}_{\mathbb R^2}$, such that $F_1(\gamma)$ is the unit circle (because $\gamma$ is simple). This isotopy doesn't change the fact $\gamma'\ne0$, so it preserves a rotation number.