# is a diffeomorphism regular?

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.

• what do you mean by a regular mapping? – weierstrash Jun 12 at 8:01
• its tangent map is 1-1. – Oh JoonSuk Jun 12 at 8:02
• 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 – weierstrash Jun 12 at 8:03
• right. regularity means Jacobian matrix of a map has a full rank. – 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 $$p \in M$$

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

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

• thank you for help! – Oh JoonSuk Jun 17 at 5:21