Let $\phi : \mathbb{R}^{2} \rightarrow \mathbb{R}$ be a continuously differentiable function and define the mapping $\mathbf{F} : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ by
$$\mathbf{F}(x, y) = (\phi(x, y), -\phi(x, y)).$$
Why can't $\mathbf{F}$ be bijective? Now suppose $\psi : \mathbb{R}^{2} \rightarrow \mathbb{R}$ is a different function for which there was a point $(x_0, y_0)$ such that $\psi(x, y) \geq \psi(x_0, y_0)$ for all $(x, y)$ in $\mathbb{R}^{2}$. Why can't the mapping defined by
$$\mathbf{G}(x, y) = (\phi(x, y), \psi(x,y)) $$
be bijective?
I'm not sure if this is true or not (I would appreciate it if someoen can confirm), but I think that a mapping is bijective on $\mathbb{R}^{n}$ if and only if each of its components are bijective on $\mathbb{R}$.
If my above claim is true, then $\mathbf{G}$ wouldn't be bijective because the second component function won't hit everything in $\mathbb{R}$ (that is, it wouldn't hit anything less than $\psi(x_0, y_0)$.
Now what about $\mathbf{F}$? I don't have a good reason for this one. Can someone please help me?