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Let $M$ and $N$ be two smooth manifolds and $f:M\longrightarrow N$ a smooth map. A vector field $X\in\mathfrak{X}(M)$ is projectable along $f$ whenever there is a vector field $Y\in\mathfrak{X}(N)$ such that $$df\circ X=Y\circ f.$$ Is it true that, under these conditions, the flow of $X$ is constante on the fibers of $f$?
Thanks.
The answer to your question is no. To clearly see why, let us consider $M=\mathbb{R}^{2}$, $N=\mathbb{R}$, $f\colon (x,y)\;\mapsto\;q=f(x,y)=x$, $X=\frac{\partial}{\partial x} + \frac{\partial}{\partial y}$ and $Y=\frac{\partial}{\partial q}$. A direct computation shows that $Tf\circ X = Y\circ f$, and it is clear that the flow of $X$ is not constant along the fibre of $f$. What is important to note, though, is that the flow of $X$ decouples into a part which is purely "horizontal" with respect to $f$ (the flow of $\frac{\partial}{\partial x}$) plus a part which is purely "vertical" with respect to $f$ (the flow of $\frac{\partial}{\partial y}$), and these two flows commute (the decomposition obeys the so-called "principle of composition of independent motions") because $[\frac{\partial }{\partial x},\,\frac{\partial}{\partial y}]=0$.
