Pullback of a vector field under a surjective submersion. Let $\pi: M \rightarrow N$ be a smooth map. If we consider a vector field $Y$ on $N$, I know that, if $\pi$ is a local diffeomorphism, there exists a unique vector field $X$ on $M$ such that $\pi_*X = Y$. I was wondering if we can weaken this hypothesis: Is there a vector field $X$ with $\pi_*X = Y$ if $\pi$ is only surjective submersion, not a local diffeomorphism? It seems quite intuitive because for each $p$, as $\pi_{*p}$ is surjective, there is at least one $X_p$ such that $\pi_{*p}X_p = Y(p)$, but I have some difficulties to show that we can construct a smooth $X$ with $X_p = X(p)$.
 A: Yes, there's at least one such vector field, provided that $\pi$ is a smooth submersion (not necessarily surjective).
By the rank's theorem, we can always find local coordinates $(U,\varphi),(V,\vartheta)$ such that  $\vartheta\circ\pi\circ\varphi^{-1}(x_1,\dots,x_m)=(x_1,\dots,x_n)$. Locally, $Y=\sum Y^i\partial_{\vartheta^i}$. If we define $X$ as $\sum_{i=1}^n Y^i(\pi(x))\partial_{\varphi^i}$, it is then clear that
$d\pi (X_p)=Y_{\pi(p)}$: the problem is that $X$, defined this way, is not a global vector field.
In order to obtain one, we must use a slighlty different construction: let $(U_\alpha,\varphi)$ be a open cover of $M$ (such that , together with coordinates $(V_\alpha,\vartheta_\alpha)$ of $N$, $\pi$ has the required form) and let $\sum\tau_\alpha$ be the partition of unity subordinated to that cover: we define
$$X=\sum_\alpha\sum_{i=1}^n \left(\tau_\alpha Y^i_\alpha\right)(\pi(x))(\partial_{\varphi_\alpha^i}$$
This is clearly a vector field. To see that it satisfies our requirements, it suffices to observe that
$d\pi(X_p)=\sum_\alpha \tau_\alpha Y_{\pi(p)}=Y_{\pi(p)}$.
