Given:
$$f:\mathbb{R}_{>0} \times \mathbb{R} \rightarrow \mathbb{C} – \{0\}; \:\: f(r, \varphi) = re^{2\pi i \varphi}$$
Proof that f is not a diffeomorphism.
I don't know how to do this. My approach was:
I think we have proven that the above function is bijective in Analysis 1. So it's missing that it and the inverse function are (totally) differentiable and the derivations are continuous.
So I calculated the Jacobi matrix:
$$J_f = \begin{bmatrix}
\cos \varphi & -r \sin \varphi \\
\sin \varphi & r \cos \varphi
\end{bmatrix}$$
As $\sin, \cos$ are continuous, it follows that the derivative is too. Now I can explicitly write down the inverse matrix and the matrix values will consist of $\sin, \cos$ and multiplications, divisions, etc. of it. So it is also continuous.
So something is very wrong with my approach…
Best Answer
You're probably off by some factors of $2\pi$. but anyway, what you've shown is that the derivative of the polar coordinate transformation is invertible at every point. But this is not the same as being a diffeomorphism. Being a diffeomorphism requires 3 things
In your case, $f$ is not even injective (it is surjective though), so it's not bijective, and thus is not a diffeomorphism. Why is $f$ not injective? Think about the periodic properties of the trig functions.
As an aside, the inverse function theorem allows you to conclude that because the derivative of $f$ is invertible at every point, $f$ is a local diffeomorphism (i.e for any point $p$ of the domain there is an open set $U$ of the domain, an open set $V$ in the target about $f(p)$ such that the restriction $f:U\to V$ is a diffeomorphism... anyway, in this special case, there's no need to invoke the IFT, one can write down explicit formulas for the local inverses). But this is not the same as $f$ itself being a diffeomorphism, because $f$ can fail to be injective (as shown by your example for $f$).