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

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 $M = f^{-1}(1,2)$ ?

• $f$ is going from $\mathbb{R}^3$ to $\mathbb{R}^2$ and then can't be invertible (more generally, you can't inverse a differentiable function from $\mathbb{R}^n$ to $\mathbb{R}^m$ if $n>m$). If you want a more explicit argument, you can see that for any $y$, you have $f(0,y,0)=0$, so $f$ is not invertible at all. Jun 19, 2015 at 13:10
• maybe you got to have the implicit function theorem Jun 19, 2015 at 13:11
• This function is non-injective. Consider $(0,y,0)$ and $(0,y',0)$ such that $y \neq y'$. Note $\forall c \in \mathbb{R}$ we have $f(0,c,0)=(0,0)$.
– anak
Jun 19, 2015 at 13:12
• "how does he know that $M=f^{−1}(1,2)$?": do you know the definition of $f^{-1}(A)$ where $A \subset Y$ and $f : X \to Y$ is a map...? Jun 19, 2015 at 13:35

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?
Function $f$ it's not invertible, because for every $y_1,z_1,z_2\in\mathbb{R}$ we have
$$f(0,y_1,z_1)=f(z_1-z_2,-z_1-z_2,z_2).$$
• This is incorrect. There are bijections between $\mathbb{R}^3$ and $\mathbb{R}^2$, just not the one given in OP. Jun 19, 2015 at 13:15