# Local smooth extensions of a vector field. [closed]

Let $\psi \colon M \to N$ be $C^\infty$. A smooth vector field X along $\psi$ (that is, $X \in C^\infty(M,T(N))$ and $\pi \circ X = \psi$) has local $C^\infty$ extensions in $N$ if given $m \in M$ there exists a neighborhood $U$ of $m$ and a neighborhood $V$ of $\psi(m)$ such that $\psi(U) \subset V$ and there is a $C^\infty$ vector field $Y$ on $V$ such that $Y \circ \psi|U = X|U$.

Does a $C^\infty$ vector field $X$ along an immersion $\psi \colon M \to N$ always have local $C^\infty$ extensions in $N$?

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## closed as no longer relevant by Qiaochu Yuan, Jorge Vitório Pereira, S. Carnahan♦Jun 12 '11 at 8:47

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Why do you bother asking this (for the second time) if you don't read the answers? – Steven Landsburg May 10 '11 at 4:35
@Steven Landsburg it's not exactly the same question. The answer is yes, because for every point $m\in M$ there are local charts $\Phi$ and $\Psi$ of $M$ and $N$ respectively, around $m$ and $\psi (m)$ respectively, such that $\psi (\mathrm{Domain~of~}\Phi)\subset\mathrm{Domain~of~}\Psi$ and such that for all $(x_1,\dots,x_d)\in\mathbb{R}^d$, where $d$ is the dimension of $M$, $\Psi\psi\Phi^{-1}(x_1,\dots,x_d)=(x_1,\dots,x_d,0,\dots,0)$. It's a corollary to the local inversion theorem. In this chart you can extend your vector field along $\psi$ by letting it be constant on the "horizontal". – Olivier Bégassat May 10 '11 at 4:51
This looks like the third time a question like this has been asked. The answer is no for an immersion, because the map $\psi$ does not need to be one-to-one. In particular, if there are distinct points $p \ne q \in M$ such that $\psi(p) = \psi(q)$ but $X(p) \ne X(q)$, then there is no way to define a vector field $Y$ on any neighborhood of $\psi(p) = \psi(q)$. – Deane Yang May 12 '11 at 2:18
Dear Ralph, if you want to delete a question, please use the "delete" button instead of converting your question into a sequence of periods. – S. Carnahan May 25 '11 at 3:38
IMO the question should be closed as the question-asker appears to be a non-participant. – Ryan Budney Jun 11 '11 at 19:18