If $F, G$ are vector fields, and $F(G(x,y)), G(F(x,y)): \mathbb{R}^2 \rightarrow \mathbb{R}^2 \rightarrow \mathbb{R}^2$ are the identity $(x,y) \mapsto (x,y)$, prove that F is incompressible $\iff$ G is incompressible.
I know you have to suppose F incompressible to get G incompressible and vice versa, but I'm having some trouble. I'm not sure how the composite functions link in with the functions F and G. I can see F(G(x,y)),G(F(x,y)) have divergence 2 also. Just not sure how to tackle it. Hints would be welcome.
We need to show that the divergence of $F$ is zero iff the divergence of $G$ is zero. Denote the two components of $F=(u,v)$. Since you asked for hints, I will give two hints and not elaborate the details.
(1) Try the simplest case first, where $F$ is a linear vector field $F(x,y) = (ax+by,cx+dy)$. The divergence is $a+d$. If you know about matrices then you know that the inverse $G(u,v) = (du-bv,-cu+av)$ divided by the determinant, so the divergence of $G$ is a multiple of the divergence of $F$.
(2) To use the multivariable chain rule, write the derivative of the composition $G\circ F=$ identity as the matrix product $DG\,DF = I$. Use that to relate the divergence of $F$ to that of $G$. (It is true that the divergence of the identity function $(x,y)$ is 2 but that doesn't help.)