Prove the function $f(x, y, z) = (x^3, y^2-z^2, yz)$ is an open map from $\mathbb{R^3}$ to $\mathbb{R^3}$ (i.e for every open set $U$ of $\mathbb{R^3}$, $f(U)$ is open).
I know, as an application of the inverse function theorem, that if the determinant of the Jacobian is non-zero in every point of the domain, then $f$ is an open map. However, the determinant of the Jacobian of $f$ is:
$3x(2y^2+2z^2) = 0 \iff x = 0 $ or $(y, z) = (0, 0)$
I've also tried the following. If $g(x, y, z) = (x^3, y, z)$ and $h(x, y, z) = (x, y^2 - z^2, yz)$, then $f = h \circ g$. Since the composition of open maps is an open map, and $g$ is clearly open since it is bijective, then all that is there left to prove is that $h$ is an open map.
Again, the only way I know of proving a map is open is either by definition, or using the inverse function theorem. Are there any other useful and simple results that provide sufficient conditions for a map to be open?
For a map $f\colon X → Y$, being open is a local property. It suffices to check that for any $x ∈ X$ there’s an open neighbourhood $U_x$ of $x$ such that $f|_{U_x}$ is open. Therefore, you can investigate those points at which you can’t argue with the inverse function theorem separately.
You can also use that holomorphic functions are always open in this case:
The map $x^3 \colon ℝ → ℝ$ is open (check $x = 0$ separately). The map $ℝ^2 → ℝ^2,~(y,z) ↦ (y^2 - z^2, 2yz)$ is open because it corresponds to $ℂ → ℂ,~w ↦ w^2$ split into real and imaginary parts and holomorphic maps are open. (Or you use the inverse function theorem directly and check $(y,z) = 0$ separately). It only differs from the map you’re intersted in by the map $ℝ^2 → ℝ^2,~(y,z) ↦ (y,2z)$, which is a homeomorphism.
Then you can use the stability of being open under products: If $f \colon X → Y$ and $g \colon X' → Y'$ are open, then so is $f×g \colon X×X' → Y×Y'$.