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Let $f\colon \mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ defined by: $$f(x,y)=(e^x\cos(y), e^x\sin(y)).$$

Obviously $f$ is a $C^{\infty }$ function. Also: $Jf(x,y)=\det(Df(x,y))=e^{2x}\neq 0$ for all $x$ in $\mathbb{R}^{2}$. So for me, $f$ is a $C^{\infty }$ diffeomorphism of $\mathbb{R}^{2}$. However, in the problem's solution it says that: $f(x,y+2\pi k)=f(x,y)$ where $k\in \mathbb{Z}$ and that $f\colon\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}\setminus\{\left ( 0,0 \right )\}$, then $f$ is not a diffeomorphism.

Can anyone explain the solution of the problem to me? and also what condition is not satisfied so that the function $f$ is not a diffeomorphism?

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$f$ is not injective. – Alex Becker Mar 7 '12 at 19:43
Nor surjective. – Jason DeVito Mar 7 '12 at 19:48
@Alex Becker: Can you please prove that it is not injective? (also, can you please give me the definition of injectivity for a function of two variables?) – M.Krov Mar 7 '12 at 19:51
@Jason DeVito: Can you please prove the non-surjectivity of the function? (also, would you, please, give me the definition of surjectivity of a function of two variables?) – M.Krov Mar 7 '12 at 19:52
@S.k.J Sure. $f(x,y+2\pi k)=f(x,y)$, yet $y+2\pi k\neq y$. Q.E.D. – Alex Becker Mar 7 '12 at 20:54
up vote 2 down vote accepted

A function is a diffeomorphism if the function is differentiable with a differentiable inverse. That is $ f:\mathbb{R}^{n}\rightarrow \mathbb{R}^{m} $ is a diffeomorphism if f is differentiable, $ f^{-1} $ exists, and $ f^{-1} $ is differentiable.

$ f(x,y)=(e^{x}cos(y),e^{x}sin(y)) $ is not invertible on its domain, $ \mathbb{R}^{2} $. You have the proof in your question: $ f(x,y+2\pi k)=f(x,y) $.

Let me be try to be precise...

A function is invertible if it is injective and surjective.
A function is injective if '$ f(a)=f(b) $ implies $ a=b $' or equivalently '$ a\neq b $ implies $f(a)\neq f(b)$'. [Here $a,b$ are points in $\mathbb{R}^{n}$ or the Domain, and $f(a),f(b)$ are points in $\mathbb{R}^{m}$ or the Range - in this case both $\mathbb{R}^{2}$]

As an example, consider the following two points in $\mathbb{R}^{2}$ : $a=(0,0)$ and $b=(0,2\pi )$. We can see that $a\neq b$, but $f(a)=f(0,0)=(1,0)=f(0,2\pi )=f(b)$. This violates the definition of injectivity. It follows that $f$ is not invertible.

Since it is required that our function be invertible in order to be a diffeomorphism, the function above (although differentiable) is not a diffeomorphism.

Here are some links on invertible functions. I hope this helps.

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A diffeomorphism $f:\ {\mathbb R}^2\to{\mathbb R}^2$ is a map $f:\ {\mathbb R}^2\to{\mathbb R}^2$ which is (1) $C^1$ (or $C^\infty$), (2) bijective, and which (3) has a $C^1$ (or $C^\infty$) inverse. The condition that guarantees (3) in the presence of (1) and (2) is the nonvanishing of the Jacobian $Jf(x,y)$ on ${\mathbb R}^2$.

The map $f$ in your example is $C^\infty$ and has $Jf(x,y)>0$ everywhere; but it is neither injective $\bigl($since $f(x,y)=f(x,y+2k\pi)\bigr)$ nor surjective $\bigl($as $(0,0)$ is not in the image$\bigr)$. Therefeore it cannot be a diffeomorphism $f:\ {\mathbb R}^2\to{\mathbb R}^2$. The condition $Jf(x,y)>0$ only guarantees that $f$ is a local diffeomorphism: For all $(x_0,y_0)\in{\mathbb R}^2$ a small disk with center $(x_0,y_0)$ is mapped diffeomorphically onto a disklike domain around the point $f(x_0,y_0)$.

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