Inverse function of $f(x,y,z) = (xy-z^2, x+z)$?

Question

How do you determine the inverse function $f^{-1}: \mathbb{R}^2 \to \mathbb{R}^3$ of

$f: \mathbb{R}^3 \to \mathbb{R}^2 , f(x,y,z) = (xy-z^2, x+z) $ ?

Or to put it into a bigger context:

I have to show that $M := \{(x,y,z) \in \mathbb{R}^3 | xy-z^2 = 1 \text{ and } x+z = 2 \} $ is a submanifold of $\mathbb{R}^3$. The professor's approach is to show that $(1,2)$ is a regular value of f (from above). Because $M = f^{-1}(1,2) $ and because of the Submersion Theorem, $M$ is a submanifold.

Now, how does he know that $M = f^{-1}(1,2) $ ?

Thanks in advance for your help!

Answer 1

In order to be invertible, the function must be one-to-one. However $f$ sends both $(2,2,-2)$ and $(3,3,-3)$ to $(0,0)$. Thus it is not one-to-one, and not invertible. What would $f^{-1}(0,0)$ be?

Answer 2

At the risk of a down vote, I do not think it is possible. The reason being is the first formula is inherently lossy. Information is lost as you go from 3 elements to 2. But I look forward to being wrong.

Edit: Refreshing my memory, the OP looks like a set of simultaneous equations to solve. One could treat one of the variables as free and define the others accordingly except for the issue found by the better answer... ;)

Answer 3

Function $f$ it's not invertible, because for every $y_1,z_1,z_2\in\mathbb{R}$ we have

$$f(0,y_1,z_1)=f(z_1-z_2,-z_1-z_2,z_2).$$