Homeomorphism and diffeomorphism

[rayman123]

Homework Statement

I have some problems with understanding these two things.

Homeomoprhism is a function f $$f: M\rightarrow N$$ is a homeomorphism if if is bijective and invertible and if both $$f, f^{-1}$$ are continuous.
Here comes an example, let's take function
$$f(x) = x^{3}$$ it is clear that f(x) is continuous and bijective and at the same $$f^{-1}(x)\rightarrow x^{\frac{1}{3}}$$ is also continuous which indicates that f is a homeomorphism. No strange things here

Diffeomorphism- given two manifolds M, N and a bijective map f is called a diffeomorphism if both
$$f: M\Rightarrow N$$
$$f^{-1}: N\Rightarrow M$$ are of class $$C^{\infty}$$ (if these functions have derivatives of all orders) f is called diffeomorphism


Going back to my example.

f is clearly a homeomorphism but :

why is $$f(x)=x^{3}$$ of a class $$C^{\infty}$$?? what do they mean by 'all orders'
$$\frac{df}{dx}=3x^{2}$$
$$\frac{d^2f}{dx^2}=6x$$
$$\frac{d^3f}{dx^3}=6$$
$$\frac{d^4f}{dx^4}=0...$$


and its inversion
$$f^{-1}=x^{\frac{1}{3}}$$ is not of a $$C^{\infty}$$...because
$$\frac{df^{-1}}{dx}=\frac{1}{3}x^{\frac{-2}{3}}$$
it is not definied at x=0

 

[Dick]

So f is a homeomorphism but not a diffeomorphism. What's wrong with that?

 

[rayman123]

nothing is wrong with that, you did not read my message carefully and overlooked what I was asking for. The problem was that I did not exactly understand the concept of a function which has derivatives of all orders, what do they mean by that?
Apparently f has 4 derivatives until we get 0 but why its inversion is not a smooth function (because it is not definied at x=0?)

We could differentiate the inversion function as many times as we wanted to...

 

[Dick]

nothing is wrong with that, you did not read my message carefully. The problem was that I did not exactly understand the concept of a function which has derivatives of all orders, what do they mean by that?
Apparently f has 4 derivatives until we get 0 but why its inversion is not a smooth function (because it is not definied at x=0?)

We could differentiate the inversion function as many times as we wanted to...

f is C^infinity. It has derivatives of all orders. f^(-1) isn't. It isn't even differentiable at 0. On the other hand if you define them on a domain that doesn't include 0, like [1,infinity) then they are both C^infinity and it's also a diffeomorphism.