Currently I'm studying differentiable manifolds using the books of Boothby and Lee. I encounter the following problem: Suppose $M$ and $N$ are smooth manifolds, $U$ an open subset of $M$, and $F:U\rightarrow N$ a smooth map. Then, for every $p \in U$, there exist an open neighborhood $V \subset U$ of $p$, such that the restriction $F_{|V}$ can be extended to a smooth mapping $F^∗:M\rightarrow N$. Here smooth means $C^\infty$. I try to use the extension lemma or gluing lemma from Lee's book but in both cases the range space is $\mathbb{R}$ or $\mathbb{R^n}$. Of course we can use charts to make the range space Euclidean space but I can't proceed.


closed as off-topic by Stefan Waldmann, Deane Yang, YCor, Stefan Kohl, Pace Nielsen Mar 26 '18 at 4:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Stefan Waldmann, Deane Yang, YCor, Pace Nielsen
If this question can be reworded to fit the rules in the help center, please edit the question.


This is a well-known basic fact, so it might be not suitable for MO, but let me sketch an answer. You are looking for a local extension, and everything can be arranged to happen in a local chart. So, take a local chart $(P, \psi)$ at $F(p)$ in $N$, with $\psi : P \to R^n$ a diffeomorphism, and a local chart $(W, \phi)$ at $p$, such that $W \subset U$, $F(W) \subset P$ and $\phi : W \to D^m \subset R^m$ a diffeomorphism. Then, without loss of generality, we can consider $M = R^m$, $N = R^n$, $U = D^m$ (open unit disk), $p = 0 \in R^m$ and $F : D^m \to R^n$ with $F(0) = 0$.

Let $\lambda : R \to R$ be a $C^\infty$ function such that $\lambda(-\infty, 1/3] = 1$ and $\lambda [2/3, +\infty) = 0$. Let $F^*(x) = \lambda(\|x\|^2) F(x)$ for $x \in D^m$ and $F^*(x) = 0$ for $x \in R^m - D^m$. This concludes the proof with $V = D^m_{1/3}$ (open disk of radius $1/3$).


Not the answer you're looking for? Browse other questions tagged or ask your own question.