# Given $F: M \to N$ a smooth submersion, every smooth vector field on $N$ has a lift in $M$

I am currently stuck on Problem 8-18 (b) in Lee's Introduction to Smooth Manifolds.

We have a smooth submersion $$F: M \to N$$ and a smooth vector field $$Y$$ on $$N$$. The problem in part (b) asks to show that if $$\dim M \neq \dim N$$, then $$Y$$ has a lift, but the lift is not unique. I am struggling to show the existence of a lift globally.

If we assume that $$\dim M = \dim N$$ instead (this is part (a) of the problem) then it is clear that $$dF_p: T_pM \to T_pN$$ is an isomorphism for all $$p \in M$$ so we can define the lift of $$Y$$ by $$X_p = dF_p^{-1}\left(Y_{F(p)}\right)$$ for all $$p$$ in $$M$$. By the rank theorem, there are smooth charts $$(U, \varphi)$$ for $$M$$ containing $$p$$ and $$(V, \psi)$$ for $$N$$ such that $$F(U) \subseteq V$$ and $$\psi \circ F \circ \phi^{-1} = \text{id}$$. Then working in the coordinate basis $$\frac{\partial}{\partial x^i}$$ associated to $$(U, \varphi)$$ and $$\frac{\partial}{\partial y^i}$$ associated to $$(V, \psi)$$ and using the Jacobian of $$dF_p$$ which is just $$I_n$$ in this case, we see that $$X^i = Y^i \circ F$$ so the components of $$X$$ are smooth, and so $$X$$ is a smooth lift of $$Y$$.

This argument fails when $$\dim M \gt \dim N$$ because there is no canonical way to choose a lift of $$Y_{F(p)}$$. We now have more than one choice for picking an element in $$dF_p^{-1}\left(Y_{F(p)}\right)$$ since $$dF_p$$ is not injective. And there is no clear way to make that arbitrary choice behave smoothly over all $$p \in M$$.

One idea I had was that by the local section theorem, for each $$p \in M$$, we can choose a smooth local section $$\sigma_p: V_p \to M$$ of $$F$$ defined on a neighborhood $$V_p$$ of $$F(p)$$ in $$N$$ whose image contains $$p$$, and define $$X_p = d\left(\sigma_p\right)_{F(p)}\left(Y_{F(p)}\right)$$. However it is not clear that this defines a smooth vector field because if we want to apply the gluing lemma to stitch together the smooth vector fields $$d\sigma_p \circ Y$$ defined on $$V_p$$, we need to know that $$d\sigma_p \circ Y$$ agrees with $$d\sigma_{p'} \circ Y$$ on their overlaps $$V_p \cap V_{p'}$$.

Another approach is to work in coordinates as in part (a) and try to define $$X$$ explicitly by $$X = \sum\limits_{i=1}^n (Y^i \circ F) \frac{\partial}{\partial x^i}$$ on each chart. But again it is not clear how to show that definitions agree on overlaps between different charts. For example, you could have a situation where $$X_{p'} = \sum\limits_{i=1}^n (Y^i \circ F) \frac{\partial}{\partial x^i}\bigg\vert_{p'} + \sum\limits_{i=n+1}^mc_i\frac{\partial}{\partial x^i}\bigg\vert_{p'}$$ for some $$p'$$ arbitrarily close to $$p$$, where the nonzero terms $$c_i$$ for $$i \gt n$$ arise from the change of coordinates when writing $$\frac{\partial}{\partial {x'}^i}\bigg\vert_{p'}$$ in terms of $$\frac{\partial}{\partial x^i}\bigg\vert_{p'}$$.

• There's a hint that might be useful to you in the paragraph just below Exercise 2.24 on page 44 of my book. Commented Dec 21, 2022 at 22:28
• @JackLee Ah I think I see it now, I got too focused on the first approach to constructing smooth functions and forgot about partitions of unity. So it looks like we can construct smooth vector fields $X^{(p)} = \sum_{i=1}^n (Y^i \circ F)\frac{\partial}{\partial x^i}$ on charts containing $p$, and then glue them together as $X = \sum_{p \in M} \psi_p X^{(p)}$ where $\left\{\psi_p\right\}$ is a smooth partition of unity subordinate to the coordinate cover of $M$. Each $X^{(p)}$ is $F$-related to $Y$ by construction, and so $X$ also is by linearity. Commented Dec 22, 2022 at 5:00

Here is my solution. Suppose $$m=\dim M \neq \dim N=n$$ and $$Y \in \mathfrak{X}(N)$$. For any p $$p \in M$$ let $$(U_p,x^i)$$ be a chart centered at $$p$$ and $$(V_{F(p)},y^j)$$ centered at $$F(p)$$ so the representation of $$F: M \to N$$ is $$\hat{F}(x^1,\dots,x^m) = (x^1,\dots,x^n).$$ If there is a smooth vector field $$X$$ that is $$F$$-related to $$Y$$, then $$\forall x \in U$$ we have $$Y_{F(x)} = Y^j(F(x)) \partial_{y^j}\big|_{F(x)} = dF_x(X_x) = X^i(x) \frac{\partial F^j}{\partial x^i}(x) \partial_{y^j} \big|_{F(x)} = X^i(x) \, \delta^j_i \partial_{y^j}\big|_{F(x)}.$$ So the first $$n$$ components of $$X$$ at $$(U_p,x^i)$$ must satisfy $$X^i = Y^i \circ F|_{U_p}$$. The rest of the components of $$X$$ can be chosen arbitrarily. So define local vector $$X_p : U_p \to TM$$ $$F$$-related to $$Y$$ as $$X_p = X_p^i \partial/\partial x^i$$ with $$X_p^i = Y^i \circ F|_{U_p}$$ for $$i=1,\dots,n$$. This construction can be done for every point in $$M$$, so by partition of unity we can blend them to get a global vector field $$X$$, that we hope still $$F$$-related to $$Y$$.
Suppose $$(\psi_p)_{p \in M}$$ is a partition of unity on $$\{U_p\}_{p \in M}$$, define a smooth vector field $$X =\sum_p \psi_p X_p$$, where $$\psi_p X_p$$ interpreted as local extension of $$\psi_p|_{U_p} X_p$$ to $$M$$ that zero outside $$\text{supp }\psi_p \subseteq U_p$$. This vector field $$F$$-related to $$Y$$ by the following computation \begin{align*} dF_x(X|_x) &= dF_x \Big( \sum_{p} (\psi_p X_p)(x) \Big) \\ &= dF_x\Big( \sum_{i=1}^N \psi_{p_i}(x) X_{p_i}|_x \Big) \\ &= \sum_{i=1}^N \psi_{p_i}(x)\, dF_x(X_{p_i}|_x) \\ &= \sum_{i=1}^N \psi_{p_i}(x)\, Y_{F(x)} \\ &= Y_{F(x)}. \end{align*} So $$X$$ is a lift of $$Y$$. It is not unique because the construction depends on the choice of partition of unity local components $$X_p = X^i_p \partial/\partial x^i$$.