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I need to show that

$$\phi:(x,y)\to(\sin\frac{y}{2}-x, \sin\frac{x}{2}-y)$$

Is a $C^1$-diffeomorphism. So, I need to show it's injective. How can I do this? Just explicitly setting $\phi(x, y)=\phi(x', y')$ just leads to a non-linear system that I can't even begin to solve. Is there some trick for showing that a multivariable function is injective?

Jack M
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2 Answers2

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Solving $\phi(x,y) = \phi(x',y')$ is easier than it looks. You get the pair of equations $$ \sin \frac{y'}2 - x' = \sin \frac y2 - x, \quad \sin \frac{x'}2 - y' = \sin \frac x2 - y $$ Look at the first equation. It leads to $$ |x - x'| = \left| \sin \frac y2 - \sin \frac{y'}2 \right| $$ Since the sine function is Lipschitz with constant $1$ you obtain $|x - x'| \le \dfrac 12 |y - y'|$.

Use the same argument with the other equation, and conclude $x = x'$ and $y = y'$.

Umberto P.
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Decompose your function into $$ \phi' : (x,y) \to (x-\sin\frac{y}{2}, y-\sin\frac{x}{2}) $$ and the reflection through origin.

Now $\phi'$ is the identity with a small periodic perturbation, so you only need to check that the perturbation is not enough to break injectivity (hint: partial derivatives of the perturbation are almost everywhere less than 1 in absolute value).

rewritten
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