2
$\begingroup$

Given a continuous map $f:M_{1}\rightarrow M_{2}$ between differentiable manifolds, a map is smooth if for all $p\in M_{1}$ with there exist charts $\varphi_{1}:U_{1}\rightarrow V_{1}$ and $\varphi_{2}:U_{2}\rightarrow V_{2}$ in $M_{1},M_{2}$ respectively (with $p\in U_{1},f(p)\in U_{2}$) such that the map $\varphi_{2}\circ f\circ\varphi_{1}^{-1}$ is a smooth map between Euclidean spaces.

Why is the smoothness of this map independent of the charts? This is clear to me if we change either of the $\phi_{i}$ or $V_{i}$ (immediate from compatibility), or if we make $U_{1}$ smaller, but I can't see why it should be true in general.

  • 5
    There's a condition on the charts. Do you see why that might help?2012-04-08
  • 0
    Whoops, I was being stupid. Replacing $U_{1},V_{1},\varphi_{1}$ with $U'_{1},V'_{1},\varphi'_{1}$, then looking at the transition map between the image of $U_{1}\cap U'_{1}$ under $\varphi_{1}$ to that under $\varphi'_{1}$ gives that $\varphi_{2}\circ f\circ \varphi_{1}^{-1}$ as a map on $U_{1}\cap U_{1}^{'}$ is smooth at $p$ by an obvious diagram chase. I had forgotten that smoothness of maps between Euclidean spaces is also a local property (so I was worried about points in $U_{1}'$ but not $U_{1}$)...2012-04-08
  • 0
    @Carl Since you've figured it out yourself, it's encouraged to _answer_ your own question. A further benefit of this is that the question will be removed from the [Unanswered queue](http://math.stackexchange.com/unanswered) (which is already amply flooded as-is).2013-05-22

1 Answers 1