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Def 1. Let $f:I\to J$. The function $f(x)$ is $C^r$-diffeomorphism if $f(x)$ is a $C^r$-homeomorphism such that $f^{-1}(x)$ is also $C^r$.

Def 2. Let $f:I\to J$. The function $f(x)$ is homeomorphism if $f(x)$ is one-to-one, onto, and continuous, and $f^{-1}(x)$ is also continuous.

From these two definitions, can I say that $f(x)=\frac{1}{x}$ is a diffeomorphism (in its domain of definition)? I think yes, because: let $f:I\to J$, and $I=J=R-\{0\}$, then $f^{-1}(x)$ is continuous in $J=R-\{0\}$. What do you think?

Thanks

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    $f=f^{-1}$; but to conclude that it is a *diffeomorphism* you need to say more than just that $f$ and $f^{-1}$ are continuous; that only makes it a *homeo* morphism.2012-07-19
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    Then in this case the function is a diffeomorphism in its domain of definition (because it is a homeomorphism )? So?2012-07-20
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    No: homeomorphism is **weaker** than "diffeomorphism". A homeomorphism only requires the function and its inverse to both be continuous. A diffeomorphism requires **more** than mere continuity (*hint*: are the function that are continuous but not differentiable?) The point is that your argument only shows that $f$ is a homeomorphism, and you are jumping from that to claiming it is a diffeomorphism. I'm not saying your conclusion is wrong, I'm saying your justification is lacking.2012-07-20
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    To try to explain what I mean, say you want to prove that a certain number is a multiple of 20. You look at the last digit and note that it is a $0$, and you say, "Aha! That shows that the number can be expressed as $10k$ for some $k$, so this shows it is a multiple of $20$." I'm saying "Well, you only showed that it is a multiple of $10$; that's not enough to conclude it is a multiple of $20$, though." Your argument only establishes that $f,f^{-1}\in\mathcal{C}^0$; but in order to be a diffeomorphism you need more than that (though this is *necessary*, it is not sufficient).2012-07-20

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Posting CW just to have an answer recorded as such: yes, $f(x)=1/x$ is a $C^\infty$-diffeomorphism of $\mathbb R\setminus\{0\}$ onto itself. Indeed, it has derivatives of all orders, and so does its inverse, because the inverse is $f$ itself.