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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!

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    $\begingroup$ $f$ 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. $\endgroup$
    – Sylvain L.
    Jun 19, 2015 at 13:10
  • $\begingroup$ maybe you got to have the implicit function theorem $\endgroup$
    – janmarqz
    Jun 19, 2015 at 13:11
  • $\begingroup$ 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 $f(0,c,0)=(0,0)$. $\endgroup$
    – anak
    Jun 19, 2015 at 13:12
  • $\begingroup$ "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...? $\endgroup$ Jun 19, 2015 at 13:35

3 Answers 3

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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?

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

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

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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... ;)

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  • $\begingroup$ This is incorrect. There are bijections between $\mathbb{R}^3$ and $\mathbb{R}^2$, just not the one given in OP. $\endgroup$
    – vadim123
    Jun 19, 2015 at 13:15
  • $\begingroup$ See, for example, this question and its answers. $\endgroup$
    – vadim123
    Jun 19, 2015 at 13:22

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