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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.

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There's a condition on the charts. Do you see why that might help? –  Dylan Moreland Apr 8 '12 at 3:02
    
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 $U_{1}$)... –  LCL Apr 8 '12 at 3:57
    
@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 (which is already amply flooded as-is). –  Lord_Farin May 22 '13 at 21:10

1 Answer 1

The OP answered own question in comments:

Replacing $U_1$, $V_1$, $\phi_1$ with $U_1'$, $V_1'$, $\phi_1'$, then looking at the transition map between the image of $U_1\cap U_1'$ under $\phi_1$ to that under $\phi_1'$ gives that $\phi_2\circ f\circ \phi_1^{-1}$ as a map on $U_1\cap U_1'$ is smooth at $p$ by an obvious diagram chase.

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