I have learned the inverse function theorem which ensures that a regular mapping (which has its inverse) is a (local) diffeomorphism. But I wonder whether a diffeomorphism is regular. I guess the answer would be 'yes', but I have no idea.

  • $\begingroup$ what do you mean by a regular mapping? $\endgroup$ – weierstrash Jun 12 at 8:01
  • $\begingroup$ its tangent map is 1-1. $\endgroup$ – Oh JoonSuk Jun 12 at 8:02
  • $\begingroup$ I'm not sure if I understand what you're trying to say. I think it has something to do with the rank of a map $\endgroup$ – weierstrash Jun 12 at 8:03
  • $\begingroup$ right. regularity means Jacobian matrix of a map has a full rank. $\endgroup$ – Oh JoonSuk Jun 12 at 8:06

If $M$ and $N$ are smooth manifolds and if $F: M \to N$ is a diffeomorphism, then by definition, there is a smooth map $g: N \to M$, such that $f \circ g = I_M$ and $g \circ f = I_N$. Now via chain rule, we get that for every $p \in M$

\begin{equation} I_{M_{*}} = f_{*} \circ g_{*}: T_{f(p)}(N) \to T_{f(p)}(N)\\ I_{N_{*}} = g_{*} \circ f_{*}: T_p(M) \to T_p(M) \end{equation}

Where the the '*' denotes the derivative. So $f_*$ is an isomorphism from $T_p(M) $ to $T_{f(p)}(N)$. Since this is true for every $p$, $f_*$ is regular. S

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  • $\begingroup$ thank you for help! $\endgroup$ – Oh JoonSuk Jun 17 at 5:21

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