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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 $X_p = X(p)$.

C.F.G
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Falcon
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  • Can you do this locally, using the structure of a submersion? – Ted Shifrin Nov 22 '20 at 23:05
  • Could you be a bit more precise ? – Falcon Nov 22 '20 at 23:09
  • There are local coordinates $(x,y)$ on $M$ and $y$ on $N$ so that $\pi(x,y)=y$. – Ted Shifrin Nov 22 '20 at 23:16
  • Hint: Take charts $(\phi, p\in U)$ and $(\psi, f(p)\in V)$ and consider the Jacobian of $F:=\psi\circ f\circ \phi^{-1}$, try to modify it so as to apply the Inverse Function Theorem in some way. – Matematleta Nov 22 '20 at 23:35
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    If you know some Riemamnian geometry, picking a Riemannian metric on $M$ gives a smooth choice of complementary subspace to $\ker \pi_\ast$. Then $\pi_\ast$ gives an isomorphism between these complementary subspaces amd tangent spaces in $N$, and this can used to pull vectorfields back. – Jason DeVito - on hiatus Nov 23 '20 at 00:22
  • You can do that locally as in here : https://math.stackexchange.com/questions/3701046/show-that-there-is-a-pi-i-related-smooth-vector-field-for-each-smooth-vector/3701108#3701108 – Kelvin Lois Nov 23 '20 at 03:29

1 Answers1

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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 $d\pi(X_p)=\sum_\alpha \tau_\alpha Y_{\pi(p)}=Y_{\pi(p)}$.