Isolating $f$ from $h(x,y,z)=f(g(x,y,z))$

Question

I know that give the system $h(x)=f(g(x))$, $f$ can be isolated by composijg both sides with $g^{-1}(x)$ resulting in $h(g^{-1}(x))=f(g(g^{-1}(x)))$ which simplifies to $f(x)=h(g^{-1}(x))$. Thats simple enough, but can be the same be done for multivariable functions, such a the omes in the title? I tried the previously stated strategy but it quickly stopped making sense. You start off with $h(x,y,z)=f(g(x,y,z))$, composing with the inverse of $g$ gives: $h(g^{-1}(x,y,z))=f$. But I ended having more questions like 'If $g$ has three inputs and one output normally, does its inverse have 1 input and 3 outputs', 'What would the inverse of $g$ take as an input, some other variable $w$?', and most importantly 'What is the input for $f$? The mysterious variable $w$?'. Any help would be greatly appreciated, thank you.

Answer 1

I think that it would be rare for a function $g: \mathbb{R}^3 \to \mathbb{R}$ to be invertible or one-to-one. There are bizarre space-filling curves so I'm not counting anything out. But basic results in smooth topology would say that this is either rare or impossible, and I bet it's also rare for continuous maps. So, I guess I agree with your intuition: this is not likely to be a usable technique. I'd love to be proven wrong.