$F$-related Vector fields and surjective submersion

Question

I want to show this fact:

Let $M, N$ be two manifolds, $\pi : M \to N$ a surjective submersion and $X$ a vector field over $M$. If $d \pi_q(X_q) = d \pi_p(X_p)$ whenever $\pi(p) = \pi(q)$ then there exists a unique vector field $Y$ over $N$ such that $Y_{\pi(p)} = d \pi_{p}(X_p)$ for every $p \in M$.

I don't know where to start. I could define $Y : N \to TN$ such that $Y_p = d \pi_q (X_q)$ where $q$ is an element of $\pi^{-1}(p)$, but that seems to be too... ugly and not much rigorous, even if it's the right path I can't see how $Y$ can be smooth. Since $\pi$ is a submersion I tried to use its normal local form but following this road is even worse: I can't even see how to build properly $Y$. Can you give me a hint please?

Thanks. English is not my mother tongue, please excuse any errors on my part.

Answer 1

I would argue as follows:

1) First of all, the relation $Y_{\pi(p)}=(d\pi)_{p}(X_{p})$ determines $Y$ uniquely on all of $N$ since $\pi$ is surjective. As a consequence, it suffices to construct $Y$ locally on an arbitrary coordinate chart $U_{\alpha}$. Indeed, on the overlap of such charts $U_{\alpha}$ and $U_{\beta}$, these locally defined vector fields will agree by the uniqueness, hence they patch together to a smooth vector field $Y$.

2) Let us now locally construct the vector field $Y$. Since $\pi$ is a submersion, it is locally given by $$\pi:M\rightarrow N: (x_{1},\ldots,x_{k},y_{1},\ldots,y_{l})\mapsto (x_{1},\ldots,x_{k})$$ Let the vector field $X$ be given by $$ X=\sum_{i=1}^{k}f_{i}(x,y)\partial_{x_{i}}+\sum_{j=1}^{l}g_{j}(x,y)\partial_{y_{j}}. $$ The given condition on $X$ means exactly that the $f_{i}$ don't depend on the $y$-coordinates, i.e. $f_{i}(x,y)=f_{i}(x)$. Consequently, we should define $Y$ locally by $$ Y=\sum_{i=1}^{k}f_{i}(x)\partial_{x_{i}}. $$

Answer 2

You are in the right track. Given $y \in N$, set $Y_y = {\rm d}\pi_x(X_x)$, where $x \in \pi^{-1}(y)$ is any point. This is well defined since ${\rm d}\pi_{x_1}(X_{x_1}) = {\rm d}\pi_{x_2}(X_{x_2})$ whenever $x_1,x_2 \in M$ satisfy $\pi(x_1)=\pi(x_2)$. As a matter of fact, this also ensures the uniqueness of such $Y$, as another $\tilde{Y}$ would have to satisfy $\tilde{Y}_y = {\rm d}\pi_x(X_x)$ for some $x \in \pi^{-1}(y)$, which is precisely how we defined $Y_y$.

So what is left to do is showing that such $Y$ is smooth. This follows from the "universal property" of surjective submersions (Proposition 7.17 in John Lee's Introduction to Smooth Manifolds, 2nd edition): if $P$ is a smooth manifold and $F\colon N \to P$ is a given map, then $F$ is smooth if and only if $F\circ \pi$ is smooth. We take $P = TN$ and $F = Y$. That is to say, smoothness of $Y$ follows from smoothness of $Y\circ \pi = {\rm d}\pi \circ X$.