Is there exist a smooth function $f:\mathbb{R}^{m}\to M$ such that $f|_{U′}=\phi^{-1}$?

Question

Let $M$ topological smooth manifold and $(U,\phi)$ chart fixed with $\phi(U)=U′$ open in $\mathbb{R}^{m}$. Is there exist a smooth function $f:\mathbb{R}^{m}\to M$ such that $f|_{U′}=\phi^{-1}$? I stuck in this. Thanks for some help

Att

Answer 1

Let $M = U = \mathbb R$ and consider the chart $\arctan : U \to \mathbb R$. Any $f$ satisfying your requirements would be a smooth map $\mathbb R \to \mathbb R$ satisfying $f|_{(-\pi/2, \pi/2)} = \tan$; but this is clearly impossible because $\lim_{x \to \pi/2} \tan (x)= \infty$, so $f$ cannot even be continuous.