Let $M$ be a differentiable manifold, $U\subseteq M$ an open neighborhood of $p\in M$ and $f:U\to\mathbb{R}$ differentiable. Then there exists $F:M\to\mathbb{R}$ differentiable such that $F=f$ in a neighborhood of $p$.

I found this problem in a book, and it's solved in the case $M=\mathbb{R}^n$, but it says the general case is easy by using charts, but I don't understand it yet.

This is what I have tried. Suppose $\dim M=n$ and take a chart $(V,\phi)$ with $p\in V$. Then $$f\circ\phi^{-1}:\phi (U\cap V)\to \mathbb{R}$$

is differentiable, where $\phi(U\cap V)$ is a neighborhood of $\phi(p)$ in $\mathbb{R}^n$.

In $\mathbb{R}^n$, we can now find $F:\mathbb{R}^n\to\mathbb{R}$ such that $F=f\circ\phi^{-1}$ in some neighborhood of $\phi (p)$, say $V_0\subseteq\phi(U\cap V)$.

From here I can't see how to obtain a differentiable function $M\to\mathbb{R}$. First I thougt something like $F\circ\phi$, but it's not defined in the whole manifold $M$.

Any hint? Thank you.


Hint multiply f by a cut off function g such that the support of g is contained in the domain of a chart which contains $p$ and whose value at $p$ is 1.

  • $\begingroup$ these are often called "bump" functions. $\endgroup$ – James S. Cook Feb 29 '16 at 3:04
  • $\begingroup$ We can take the cutt off function to be constant equal to 1 in a neighborhood of $p$ so that the resulting function $F$ is equal to $f$ in this neighborhood. $\endgroup$ – Eduardo Longa Feb 29 '16 at 3:06

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