This is a seemingly easy exercise. Yet I am not sure if I am missing any finer details here as this is listed as one of the challenging problems on Dr. Epstein's (Upenn) course site for real analysis. http://www.math.upenn.edu/~cle/amcs608/ps0.pdf
I am doing Problem 7 on this webpage as I am learning the material on my own.
Define a map from $\Bbb R^{2}$ to itself by setting $$F(x, y) = (\sin x \cos y + \sin y \cos x, \cos x \cos y − \sin x \sin y).$$ Does there exist a point $(x_{0}, y_{0})$ such that $F$ is locally invertible in a neighborhood of $F(x_{0}, y_{0})$. You must prove your answer.
What I did:
$$DF(x,y)= \begin{bmatrix} \cos x \cos y-\sin x \sin y & -\sin x \sin y+\cos x \cos y \\ -\sin x \cos y-\cos x \sin y & -\sin y\cos x-\sin x\cos y \\ \end{bmatrix}. $$
$$\det {DF(x,y)}= \cos (x+y) \sin (x+y)- \sin (x+y) \cos (x+y)=0. $$
It seems $F$ is not invertible anywhere, is it correct or have I missed some finer details?
Another point of view:
Identify $\mathbb{R}^2$ with $\mathbb{C}$, then after flipping coordinates $$ F(x,y) = \exp(i(x+y))=\exp(ix)\exp(iy)=(\cos x + i \sin x)(\cos y + i \sin y) $$ so it cannot be locally invertible because for all $x,y$ and all $h$ $$ F(x+h,y-h)=F(x,y) $$