Give an example of a diffeomorphism of the class $C^1$ that takes the set $\left\{ (x, y) \in \mathbb R^2: 1 < x^2 +y^2 < 2, x < y < 2x \right\}$ on square $\left\{ (x, y) \in \mathbb R^2: |x| < 1, |y| < 1 \right\}$.
I did this task but I need to make sure it's correct and if it is a sufficient explanation of the problem:
$K=\{(x,y)\in \mathbb R^2: 1 <x^2+y^2<2, x<y<2x \}$
$L=\{(x, y) \in \mathbb R^2: |x| < 1, |y| < 1\}$
$\Phi(r, \alpha) = (r\cos \alpha, r \sin \alpha)$ - diffeomorphism of the class $C^1$
$\Phi \left( (1,\sqrt{2}) \times ( \frac{\pi}{4}, \arctan 2) \right) =K$
$\Phi^{-1}$ is also diffeomorphism of the class $C^1$
$f(x,y)=(x-1,y-\frac\pi4)$
$g(x,y)=\left( (2\sqrt{2}+2)x-1, \frac{2}{\arctan 2 - \frac{\pi}{4}}y-1 \right)$
$\Phi^{-1}(K)=(1, \sqrt{2}) \times ( \frac{\pi}{4}, \arctan 2)$
$f \left( (1, \sqrt{2}) \times (\frac{\pi}{4}, \arctan 2) \right) = (0, \sqrt{2}-1) \times (0, \arctan 2 - \frac{\pi}{4})$ - injective, class $C^1$
$g \left( (0, \sqrt{2}-1) \times (0, \arctan 2 - \frac{\pi}{4}) \right) =L$ - injective, class $C^1$
So $g \circ f \circ \Phi^{-1}$ is searched diffeomorphism of the class $C^1$
You did the task correctly. $g \circ f \circ \Phi^{-1}:K\to L$ is a wanted diffeomorphism.
You wrote your solution in such a clear way that it is self-explanatory to anyone who knows the basics of diffeomorphism, polar coordinates, affine transformation of $\Bbb R^2$ and composition of maps or diffeomorphisms. If affine transformation sounds unfamiliar, replace it with translation and invertible linear transformation.
Here are some minor drawbacks
"$\Phi(r, \alpha) = (r\cos \alpha, r \sin \alpha)$ - diffeomorphism of the class $C^1$".
While this can be understood as correct implicitly, it could be considered wrong as well, since $\Phi$ is not a diffeomorphism from $\Bbb R^2$, the default domain to $\Bbb R^2$, the default codomain. It is not even a homeomorphism because of the origin. To be correct explicitly, you should have defined $\Phi$ as "$\Phi:(1,\sqrt2)\times$ $(\frac\pi4,\arctan2)\to K$, $\Phi(r, \alpha) = (r\cos \alpha, r \sin \alpha)$".
"$f(x,y)=(x-1,y-\frac\pi4)$
$f \left( (1, \sqrt{2}) \times (\frac{\pi}{4}, \arctan 2) \right) = (0, \sqrt{2}-1) \times (0, \arctan 2 - \frac{\pi}{4})$"
It looks you define $f:\Bbb R^2\to\Bbb R^2$ first. Then you are using the restriction of $f$ to $\left( (1, \sqrt{2}) \times (\frac{\pi}{4}, \arctan 2) \right)$ domain-wise and $(0, \sqrt{2}-1) \times (0, \arctan 2 - \frac{\pi}{4})$ codomain-wise.
It should be clearer to present the default domain $\Bbb R^2$ and the default codmain $\Bbb R^2$ of $f$ explicitly. Then we can apply the proposition, "If $\psi$ is a diffeomorphism of some $C^*$ class, then its restriction to an open subset $M$ onto $\psi(M)$ is also a diffeomorphism of the same $C^*$ class."
Or you can just write "$f:\left( (1, \sqrt{2}) \times (\frac{\pi}{4}, \arctan 2) \right)$ $\to (0, \sqrt{2}-1) \times (0, \arctan 2 - \frac{\pi}{4})$, $f(x,y)=(x-1,y-\frac\pi4)$".
"$f$ … - injective, class $C^1$".
While this is not wrong, it is very misleading. Since a diffeomorphism is wanted, we should explain that so is $f$ so that we can later apply "a composition of diffeomorphisms is also a diffeomorphism". However, a map that is injective and $C^1$ is not necessarily a $C^1$-diffeomorphism. For example, $h:\Bbb R\to\Bbb R, h(x)=x^3$ is injective and $C^1$. $h$ is not a diffeomorphism since $h^{-1}$ is not differentiable at $0$.
What you should have written is "$f$ … - a diffeomorphism of class $C^1$". In case more logic is expected, append "since $f$ is a translation."
"$g(x,y)=\left( (2\sqrt{2}+2)x-1, \frac{2}{\arctan 2 - \frac{\pi}{4}}y-1 \right)$
$g \left( (0, \sqrt{2}-1) \times (0, \arctan 2 - \frac{\pi}{4}) \right) =L$ - injective, class $C^1$"
Here are the same defects as before.
By the way, $g \circ f \circ \Phi^{-1}:K\to L$ is, in fact, a $C^\infty$- diffeomorphism.