Smooth lifting criteria of smooth vector fields given smooth surjective submersion whose fibers are connected. I am working on an exercise:
Suppose $F : M \to N$ is a smooth submersion, where $M$ and $N$ are positive-dimensional smooth manifolds. Given $X \in \mathfrak{X}(M)$ and $Y \in \mathfrak{X}(N)$, we say that $X$ is a lift of $Y$ if $X$ and $Y$ are $F$-related. A vector field $V \in \mathfrak{X}(M)$ is said to be vertical if $V$ is everywhere tangent to the fibers of $F$ (or, equivalently, if $V$ is $F$-related to the zero vector field on $N$).
Assume in addition that $F$ is surjective with connected fibers. Show thata vector field $X \in \mathfrak{X}(M)$ is a lift of a smooth vector field on $N$ if and only if $[V, X]$ is vertical whenever $V \in \mathfrak{X}(M)$ is vertical.
I have previously shown one direction, and also have shown that if $F$ is surjective, $X$ is a lift of some smooth vector field on $N$ if and only if $d F_p(X_p) = d F_q(X_q)$ if $p, q$ are in the same fiber.
To show that $X \in \mathfrak{X}(M)$ is a lift of a smooth vector field on $N$ if $[V, X]$ is vertical whenever $V \in \mathfrak{X}(M)$ is vertical, I considered below approach.
Since each fiber is connected by the assumption, I thought maybe I can show $\{ q \in F^{-1}(F(p)) :  d F_p(X_p) = d F_q(X_q)\}$ is both open and closed, and conclude that $\{ q \in F^{-1}(F(p)) :  d F_p(X_p) = d F_q(X_q)\}$ is $F^{-1}(F(p))$ for each $p \in M$, but I am not sure how  to show that $\{ q \in F^{-1}(F(p)) :  d F_p(X_p) = d F_q(X_q)\}$ is open.
 A: For one direction, since $X$  is a lift of smooth vector field on $N$,there exist a vector field on $N$ ,$F$-related to $X$, so by natruality of Lie braket $$dF([V,X]) = [dF(V),dF(X)] = [0,dF(X)] = 0$$
For the other direction ,the point is to show that the value $dF(X)$ is locally constant on the fiber as you have noticed.Intuitively inside the small neiborhood center at $p$ of the firber,there exist a ray connected any point $p \to q$ inside the neiborhood.the vector field $V$ corresponding to this integral curve in the fiber is verticle.So we know $[V,X] $ is verticle.
Using the definition of Lie derivative we know $$[V,X] = \frac{d}{dt}(d\theta_{-t})X$$ is verticle ,where $V$ is the vector field correspond to the flow on the fiber.
i.e. $dF(\frac{d}{dt}(d\theta_{-t})X) = 0$,what we want to show is that $$\frac{d}{dt}dF(X) = 0$$ along the integral curve.
To do this $$dF(\lim_t \frac{X_{\theta_t(p)} - (d\theta_t)_p(X)}{t}) = \lim_t(\frac{dF(X_{\theta_t(p)}) - dF\circ d\theta (X_p)}{t}) =0$$
Observe that $F\circ \theta = F$ since the flow is along the verticle direction,so ,we know
$$\lim_t \frac{dF_{\theta_t(p)(X)} - dF(X_p) }{t} = 0$$
that's exactly the locally constant of $dF(X)$!
A: Probably, this is too late but I would like to know if my proof has a mistake.
First, let $S_c$ denote a fiber s.t. $c\in N$ and $S_c=F^{-1}(c)$, thus $F$ smooth submersion implies $S_c$ is a properly embedded submanifold of $M$, whose codimension is equal to the dimension of $N$($\text{dim }N=n$). In addition, there exist $(V,(y^1,y^{2},\cdots,y^{n})=\psi)$ a chart about $c$, this chart allows us to define $b^{s}:F^{-1}(V)\to\mathbb{R}$ from the following equation,
$$dF_r(X_r)=\sum_{s=1}^{n}b^{s}(r)\left.\frac{\partial}{\partial y^{s}}\right\vert_{F(r)}$$
, i.e. $b^{s}(r)=X_r(y^s\circ F)$. Our goal is to prove $b^s(r)$ is constant on $S_c$.
Therefore, if $[W,X]=0$ as $W$ is a vertical vector field.
$$[W,X](y^{s}\circ F)=0\Rightarrow W_r(X(y^{s}\circ F))=X_r\left(W(y^{s}\circ F)\right)=0$$
Here, $X(y^{s}\circ F):F^{-1}(V)\to\mathbb{R}\Rightarrow X_r(y^{s}\circ F)=dF_r(X_r)y^{s}=b^s(r)$, then $W_r b^{s}=0$ as $r\in F^{-1}(V)$.
Let $p\in S_c$ and let's define $H_c=\left\lbrace q\in S_c, dF_p(X_p)=dF_q(X_q)\right\rbrace$, by definition $p\in H_c$.
Claim: $H_c$ is open in $S_c$, analogously $S_c-H_c$ is open in $S_c$.
Let's suppose $q\in H_c$ then there exists $(U,(x^1,x^2,\cdots,x^m)=\varphi)$ a slice chart about $q$ s.t. $U\cap S_c=\left\lbrace q\in U,  x^{k+1}(q)=x^{k+2}(q)=\cdots=x^m(q)=0\right\rbrace$ , $U\subseteq F^{-1}(V)$ and $\varphi(U)$ is a ball, thus if $r\in U\cap S_c\rightarrow W_r\in\text{span}\left\lbrace \partial^x_1\vert_r,\cdots,\partial^x_k\vert_r\right\rbrace$
\begin{align*}
W_r b^s=0&\Rightarrow 0=\left.\frac{\partial}{\partial x^i}\right\vert_r b^s\\
\text{If, }\hat{b}^s(a^1,a^2,\cdots,a^m)&=b^s\circ\varphi^{-1}(a^1,a^2,\cdots,a^m)\\
\hat{b}^s(a^1,a^2,\cdots,a^m)&=\hat{b}^s(a^{k+1},a^{k+2},\cdots,a^m)
\end{align*}
Conclusion, $b^s$ is constant on $U\cap S_c$ for all $c\in V$ then $dF_r(X_r)=dF_q(X_q)=dF_p(X_p)$ for all $r\in U\cap S_c$, thus $H_c$ is open in $S_c$. On the other hand, if $q\not\in H_c$ then $dF_r(X_r)=dF_q(X_q)\not=dF_p(X_p)$ for all $r\in U\cap S_c$, i.e. $H_c$. complement is open in $S_c$.
Then, the connectedness of the fibers implies $dF_p(X_p)=dF_q(X_q)$ as $F(p)=F(q)$ and you probably know this is a sufficient condition to assure the existence of $Y$ smooth vector field on $N$, $Y_{F(p)}=dF_p(X_p)$, in fact this the unique vector field which is F-related to $X$.
