Index of a plane curve is zero There is a result that seems intuitive to me, but I don't know how to prove it.
Let $\gamma: [0, 1] \longrightarrow \mathbb{R}^{2}$ be a differentiable regular $\left(\gamma'(t) \neq 0, \forall t \in [0, 1]\right)$ plane curve such that
$$\gamma(s) = \gamma(t) \iff s = t,$$ 
and 
$$\pi_{y}(\gamma(0)) \leq \pi_{y}(\gamma(t)) \leq \pi_{y}(\gamma(1)), \ \forall t \in [0, 1],$$
where $\pi_{y}$ is the canonical projection on the $y$-axis. Then the rotation index is zero, that is, the vector
$$\frac{\gamma'(t)}{|\gamma'(t)|}$$
does not complete any loop around the unitary circle $\mathbb{S}^{1}$.
Appreciate.
 A: The crucial fact here is that the mapping $\gamma$ is one-to-one. Here is a suggestion for an approach. See pages 26-28 of my differential geometry text for more details.
Let $\Delta = \{(s,t): 0\le s\le t\le 1\}$. Define the chordal map
$h\colon \Delta\to \Bbb S^1$ by 
$$h(s,t) = \begin{cases} \dfrac{\gamma(t)-\gamma(s)}{\|\gamma(t)-\gamma(s)\|}, & s\ne t \\
\frac{\gamma'(s)}{\|\gamma'(s)\|}, & s=t
\end{cases}.$$
Then $h$ is everywhere defined and continuous (the latter requires a check). 
With a bit of work one can show there is a continuous function $\theta\colon \Delta\to\Bbb R$ so that $h(s,t) = (\cos(\theta(s,t)),\sin(\theta(s,t))$ for all $(s,t)\in\Delta$. You want to see that $|\theta(1,1) - \theta(0,0)|<2\pi$. (For a closed curve, the rotation index would be precisely $(\theta(1,1)-\theta(0,0))/2\pi$.)
Because of the hypothesis on the $y$-coordinate of $\gamma$, we know that for every $t$, $h(0,t)$ lies in the upper semicircle, and the same is true for $h(s,1)$ for every $s$. This means, in particular, that $|\theta(0,1)-\theta(0,0)|<\pi$ and $|\theta(1,1)-\theta(0,1)|<\pi$, so 
$$|\theta(1,1) - \theta(0,0)| \le |\theta(0,1)-\theta(0,0)|+|\theta(1,1)-\theta(0,1)|<2\pi,$$
as required. 
