Orbit of vector field crosses transverse section in the same direction

Question

Let $X\in\mathbf{C}^1(U,\mathbb{R}^2)$ a vector field on the open set $U\subset\mathbb{R}^2$. Let $D\subset\mathbb{R}$ open and $f:D\rightarrow U$ be a $\mathbf{C}^1$ map such that $\{f'(x),X_{f(x)}\}$ spans $\mathbb{R}^2,\;\forall x\in D$. Assume $f:D\rightarrow T=f(D)$ is a homeomorphism. $T$ is called a transverse section of the vector field $X$. Given a point $x_0\in U$, we take the positive orbit through it, $O_{x_0}=\{\Phi_t(x_0):t\ge 0\}$ and we take for granted that the set of moments in time when this trajectory intersects $T$ is a discrete set, namely $\tau=\{0<\tau_1<\ldots<\tau_n\}$. What we want to prove is the following:

The orbit $O_{x_0}$ crosses $T$ in the same direction at each time of intersection. In particular, the case from the picture below is impossible.

Attempt I have tried to take a unit orthogonal vector to $T$, say $v$ and if $x(t)=\Phi_t(x_0)$ then I should obtain that both scalar products $x'(0)\cdot v$ and $x'(\tau_1)\cdot v$ have the same sign, so the tangent vectors to the curves at the point when crossing occur are oriented in the same way with $v$.

Answer 1

I'm assuming you mean $f'(x)$ in the span assumption. Similarly, I'll assume $D$ is connected. (I believe I have counterexamples for both?)

Take the function \begin{align*} g:D &\to \mathbb R \\ s &\mapsto \left(f'(s) \times X_{f(s)}\right)\cdot \langle 0,0,1\rangle = \text{ the $z$ component of }\left(f'(s) \times X_{f(s)}\right) \end{align*} By assumption this is nonzero for each $x\in D$ since the two vectors must span the plane at each point. Since everything is $C^0$ and $D$ is connected, $g$ has the same sign for all $x\in D$.