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On p.221 of Voisin's book on Hodge theory, there are two claims:

a) Let $B$ be a contractible smooth manifold. There exists a vector field $\chi$ on $B$ whose flow $\Phi_t$ is global and, given any neighborhood $U$ of the point we contract to, we can find $T$ such that, for all $t \geq T$, $Im_t$ is contained in $U$.

b) Let $B$ and $X$ be smooth manifolds, $\chi$ a global vector field on $B$, $\phi: X \rightarrow B$ a submersion [WAS MISSING, NOW ADDED] which is smooth and proper. Then $\chi$ lifts to a global vector field $\chi'$ on $X$. [as now stated see Need help understanding a lift of a vector field ] How do we prove these?

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Nick Stevenson
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    Statement (b) is false. For example, let $B = X = \mathbb R$, and let $\phi\colon X\to B$ be $\phi(x) = x^3$. Then $\phi$ is smooth, and it's proper because it's a homeomorphism. But the vector field $\partial/\partial x$ has no lift to $X$. – Jack Lee Aug 06 '14 at 20:57
  • @JackLee It would be great to have that as an answer instead of a comment. Deserves more visibility than it has right now. – Mark Fantini Aug 06 '14 at 20:58
  • @Mark, I think it would be better to give the OP an opportunity to modify the question. I think he's missing a hypothesis in (b). (And besides, I don't have any idea how to answer question (a).) – Jack Lee Aug 06 '14 at 21:47
  • I think in (b) should be a smooth proper submersion. But in the sense of algebraic geometry, a smooth morphism would also be a submersion, so…. – user40276 Aug 06 '14 at 23:53
  • OK, now (b) makes sense. The hint given in the last comment to the question you linked to should help: Locally you can find coordinates in which $\phi$ has the form $(x^1,\dots,x^k,x^{k+1},\dots,x^n)\mapsto (x^1,\dots,x^k)$, and in these coordinates, finding a smooth local lift is easy. Then patch together with a partition of unity. But I still don't know how to do part (a). – Jack Lee Aug 24 '14 at 15:32

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