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Is this claim correct?

If so, is it because identity is the diffeomorphism?

WishingFish
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1 Answers1

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Yes, the identity map is a diffeomorphism. The reason is that it is smooth and its inverse (the identity map again!) is smooth. Finally, its bijective for obvious reasons.

The following exercise provides more interesting examples of diffeomorphisms from $\mathbb{R}^{k}$ to $\mathbb{R}^k$.

Exercise 1: Let $T:\mathbb{R}^{k}\to \mathbb{R}^{k}$ be the linear operator defined by the invertible $k\times k$ matrix $A$ by the rule $T(x)=Ax$ for all $x\in \mathbb{R}^{k}$. Prove that:

(a) $T$ is bijective

(b) $T$ is smooth and its derivative at all points of $\mathbb{R}^{k}$ is $A$

(c) $T^{-1}$ is smooth and its derivative at all points of $\mathbb{R}^{k}$ is $A^{-1}$

Therefore, $T:\mathbb{R}^{k}\to \mathbb{R}^{k}$ is a diffeomorphism!

Exercise 2: Prove or give a counterexample: every diffeomorphism from $\mathbb{R}^{k}$ to itself is of the form given by Exercise 1.

Exercise 3: Prove that if $m\neq n$, then $\mathbb{R}^{m}$ is not diffeomorphic to $\mathbb{R}^{n}$. (Hint: use the implicit function theorem.)

I hope this helps!

Amitesh Datta
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    I would say: it is smooth and bijective for obvious reasons. Finally its inverse (the identity map again!) is smooth. (I mean you are talking about the inverse before saying it is bijective.) – wildildildlife Jul 06 '13 at 14:01
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    @wildildildlife Yes, you're right. In fact, that sentence has more flaws than just mathematical ones. E.g., I wrote "finally, its bijective ..."; the "its" should be "it's". I don't know what I was thinking at the time ... – Amitesh Datta Jul 06 '13 at 14:06