Let $x \in \mathcal{U}$ be a point which its trajectory for $x' = F(x)$ is not injective.
a) Show that this trajectory is defined for all $t \in \mathbb{R}$;
b) Show that $\{\tau \in \mathbb{R}; f^{\tau} (x) = x\}$ is a closed set and a subgroup of $(\mathbb{R}, +)$;
c) Conclude that either $x$ is a stacionary point or there exists $T > 0$ such that $f^T(x) = x$ and the restriction of the trajectory to $\left[0, T\right)$ is injective.
Attempt: I was able to show both a) and b), but I don't know how to proceed in order to prove $c)$. Suppose that $x$ is not a stationary point. Therefore, there exists $T_1 > 0$ such that $f^{T_1}(x) \neq x$. Now, consider the set $\left[0, T_1\right)$. My initial idea was to consider the smallest $T \in (0, T_1)$ such that $f^T(x) = x$, but I can't guarantee the existence of such $T$ in $(0, T_1)$. Can someone give me a hint?
Any help would be appreciated!
You have either $F(x)=0$ and get a stationary solution, or in the nontrivial case $F(x)\ne 0$. This means by some Taylor argument that $f^t(x)=x+tF(x)+O(t^2)\ne x$ for some small interval $t\in(-\delta,\delta)\setminus\{0\}$, which implies that indeed a smallest positive $T$ exists.