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Let $M$ and $N$ be $n$-dimensional smooth manifolds. A map $F: M \to N$ is smooth if for each $p \in M$ there exists smooth charts $(U, \varphi)$ containing $p$ and $(V, \psi)$ containing $F(p)$ such that $F(U) \subset V$ and the composite map $\psi \circ F \circ \varphi^{-1} : \varphi(U) \to \psi(V)$ is smooth.

My confusion is: Does this definition hold for all smooth charts containing $p$?

In other words, if $(W, \sigma)$ is any chart for containing $p$, in any smooth atlas for $M$, will there be a corresponding chart $(V_2, \psi_2)$ containing $F(p)$ such that $F(W) \subset V_2$ and $\psi_2 \circ F \circ \sigma^{-1} : \sigma(W) \to \psi_2(V_2)$ is smooth.?

Or is the definition saying that if $F$ is smooth then there exists at least one chart containing $p$ satisfying the conditions.

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The answer is yes.
Suppose $(\phi_1, U_1)$, $(\phi_2, U_2)$ and $(\psi_1, V_1)$, $(\psi_2, V_2)$ are pairs of charts around $p$, $F(p)$ in $M$, $N$ respectively, satisfying $F(U_i) \subset V_i$ for $i=1,2$ and $\psi_1 \circ F \circ \phi_1^{-1}$ is smooth on $\phi(U)$. Then $W=U_1 \cap U_2$ is an open neighborhood of $p$ and (restricted to $\phi_2 (W)$) we have: $\psi_2 \circ F \circ \phi_2^{-1} = (\psi_2 \circ \psi_1^{-1})\circ (\psi_1 \circ F \circ \phi_1^{-1}) \circ (\phi_1 \circ \psi_1^{-1})$ (always be careful with the domains of charts). Now the three paranthesis are smooth, the one in the middle by assumption, the other two because you have the compatibility condition on $M$, $N$.
Remark: The condition $F(U_i) \subset V_i$ needs to be given in the definition of smoothness, i.e. the "counterpart" chart needs to be provided. This is a difference between Analysis on Manifolds and Real Analysis. In $\mathbb{R}^n$ the differentiability condition implies continuity, not so here, for any neighborhood $V$ of $F(p)$ in $N$ you can't garantee there will be a "counterpart" $U$ in $M$ with $F(U) \subset V$ so that the coresponding local map between Euclidean spaces makes sense. You have to assume it.

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  • $\begingroup$ The only issue here is that you have shown $F$ to be smooth using the charts $(W, \phi_2 \mid_W)$ and $(V_2, \psi_2)$, but not on the charts $(U_2, \phi_2)$ and $(V_2, \psi_2)$. $\endgroup$ – Al Jebr Sep 12 '18 at 22:29
  • $\begingroup$ Also, how can we intersect $U$ and $V$ if $U\subset M$ and $V\subset N$? $\endgroup$ – Al Jebr Sep 12 '18 at 22:34
  • $\begingroup$ That's all it takes, it's a local condition, forget about having to restrict the domain of your charts because first you define the smoothness at a point, then extend it to larger domains. $\endgroup$ – Laz Sep 12 '18 at 22:35
  • $\begingroup$ Who is intersecting $U$ and $V$? $\endgroup$ – Laz Sep 12 '18 at 22:37
  • $\begingroup$ You have $W=U \cap V$ in second sentence $\endgroup$ – Al Jebr Sep 12 '18 at 22:37
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The definition just requires the two charts $(U, \phi)$ containing $p$ and $(V, \psi)$ containing $f(p)$. But if two such charts exist, the definition of a smooth manifold will imply that $\psi' \circ F \circ \phi'$ will be smooth whenever $(U', \phi')$ is a chart in $M$ containing $p$ and $(V', \phi')$ is a chart in $N$ containing $f(p)$.

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  • $\begingroup$ How does the definition imply this? $\endgroup$ – Al Jebr Sep 12 '18 at 22:31
  • $\begingroup$ The definition says that the transition maps between a neighbourhood of $p$ in $U$ and a neighbourhood of $p$ in $U'$ are smooth. Similarly for $V$ and $V'$. Wiring the various smooth maps together gives you what you need. $\endgroup$ – Rob Arthan Sep 12 '18 at 23:07

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