Consider a control system of the form $$ \dot x(t) = X(x(t)) + u(t)\, Y(x(t)) \qquad \qquad (*) $$ where $X,Y$ are two smooth vector fields, and $u$ is the (bounded measurable) control function.
Take a trajectory $x(\cdot)$ and a control $u(\cdot)$ solution of $(*)$ for $t\in [0,T]$ and such that $Y(x(t))=0$ on $[0,T]$.
It is stated that if $x(\cdot)$ is analytic and injective, then we can assume that $x(\cdot)$ is associated to $u\equiv 0$.
Why do we need the injectivity or analycity of the trajectory ? I would have said: any control function can be candidate since $Y(x(t))=0$.
EDIT: In fact we are interested in a system of the form : $$ \dot z = H_X(z(t)) + u(t)\, H_Y(z(t)) $$ where $z(t)=(x(t),p(t))$ satisfies $H_Y(z(t))=0$.
The remaining question: Why to we need the injectivity of $x(\cdot)$ to state that we can consider the zero control to be associated to $x(\cdot)$.
here is the solution, I see why now it's not obvious.