Differential Geometry - Does there exists curve such that $f(\gamma'(t))\ge 0$ $f(x,y)=x$? I tried to define reparametrization of $\gamma$ such as $\tilde{\gamma}$ and tried to use Hopf's Umlaufsatz theorem but could not do it. Can you help?
 A: There does not exist a (differentiable) simple closed curve as required. Write $\gamma(t) = (\gamma_1(t), \gamma_2(t))$. Then the $\gamma_i : I \to \mathbb R$ are differentiable and we must have $\gamma'_1(t) \ge 0$. This means that $\gamma_1$ is increasing. We have $\gamma_i(0) = \gamma_i(1) = \xi_i$ because $\gamma$ is closed. Thus $\gamma_1(t) = \xi_1$ for all $t$.  The set $\gamma_2(I)$ is a compact connected subset of $\mathbb R$, thus $\gamma_2(I) = [a,b]$ for suitable $a \le b$. Now there are two arguments showing the non-existence of $\gamma$.


*

*argument based on the Jordan curve theorem: $\gamma$ is a Jordan curve and therefore $\gamma(I) = \{ \xi_1 \} \times [a,b]$ should separate the plane into a bounded "interior" component an unbounded "exterior" component. But $\mathbb R^2 \setminus \gamma(I)$ is connected, thus there is no bounded interior component.

*argument based on the IVT: It is impossible that $a = b$ because in that case $\gamma$ would be constant (and not a simple closed curve). Hence $a < b$ and we have $\xi_1 > a$ or $\xi_1 < b$. Let us assume that $\xi_1 > a$; the other case is treated similarly. We have $a = \gamma_2(\tau)$ for some $\tau \in (0,1)$ (recall $\gamma_2(0) = \gamma_2(1) = \xi_1$). Choose $a' \in (a,\xi_1)$. By the IVT there exist $\tau'$ between $0$ and $\tau$ such that $\gamma_2(\tau') = a'$ and $\tau''$ between and $\tau$ and $1$ such that $\gamma_2(\tau'') = a'$. Thus $\tau' < \tau''$ and $\gamma(\tau') = \gamma(\tau'')$ which is impossible for a simple closed curve.
