I am studying Differential Geometry, I have some books, but I am really unable to find out some exercises (solved, or with hints) as to initiate me. I found some exercises without explanations and I am trying to solve them. I have the following exercise:

Let $f:\mathbb{R}^2\rightarrow\mathbb{R}, \ f(x,y) = y^2 $

Check if $0$ is a regular value of $f$.

I know, by the definition, that $\gamma$ is a regular value of $f$, if $d_xf$ is surjective for any $x$ from $f^{-1}(\gamma)\neq\emptyset$. I read somewhere that I have to compute the Jacobian matrix of f, and compute the rank. And after that? Can you, please, help me?

The Jacobian matrix is $\begin{bmatrix}0 \\2y \end{bmatrix}$, which has rank 1, right? What does that means? Is 0 a regular value?

EDIT: I think I said nonsense things. The rank of the Jacobian matrix should be maximal (1 in this case) when subsituting $y$ with 0. But in this case the rank is $0$. So $0$ is not a regular value. Am I right?

Thank you!

  • 1
    $\begingroup$ Yes, $0$ is not a regular value. $\endgroup$ – Jerry Jun 25 '18 at 8:11
  • $\begingroup$ You also should double-check rows and columns. For me, the Jacobian matrix is the transpose of yours. $\endgroup$ – Ted Shifrin Jun 25 '18 at 14:24

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