# Inverse image of regular value is a regular curve

Define a regular curve in analogy with a regular surface. Prove that the inverse image of a regular value of a differentiable function $$f:U \subset \mathbb{R}^2 \to \mathbb{R}$$ is a regular plane curve. Give an example of such a curve which is not connected.

My attempt:

Let $a$ be a regular value and $p \in f^{-1}(a)$. Since $a$ is a regular value, we know that either $f_x(p) \neq 0$ or $f_y(p) \neq 0$ (or both). WLOG, assume $f_x(p) \neq 0$. Now, consider the function $$F:U \subset \mathbb{R}^2 \to \mathbb{R},$$ where $(x,y) \mapsto (f(x,y),y)$. Then it is easy to see that $J_F|_p$, the Jacobian matrix of $F$, is invertible. We invoke the inverse function theorem, which ensures that there exist an open neighborhood $W$ of $p$ and an open neighborhood $V$ of $f(p)$ such that $$F^{-1}: V \to W$$ where $F^{-1}(u,v)=(g(u,v),v)$ for some differentiable function $g$.

I think I am near the end. There is a similar proof in DoCarmo's book. But I simply find the last part subtle. I hope that you can give me some more details to move on.

• I think it should be $F$ : $U$ $\rightarrow$ $\mathbb{R}^2$ at the 4th line of your argument, – glimpser Jan 22 at 4:56

Let $I = \{y \in \mathbb{R} \, | (a,y) \in W \}$. Then $I$ is an open subset of $\mathbb{R}$. Consider the parametrization $\mathbf{x} \colon I \rightarrow \mathbb{R}^2$ given by $\mathbf{x}(y) = F^{-1}(a,y)$. The map $\mathbf{x}$ is smooth, one-to-one, and onto $f^{-1}(a) \cap U$. It is also open (as a map $\mathbf{x} \colon I \rightarrow f^{-1}(a) \cap U$) since if $V \subseteq I$ is open , then $\mathbf{x}(V) = F^{-1}(W \cap (\mathbb{R} \times V)) \cap f^{-1}(a)$ and $W \cap (\mathbb{R} \times V)$ is open in $W$ and $F^{-1}$ is a diffeomorphism. Thus, $\mathbf{x}$ is a homeomorphism onto $f^{-1}(a) \cap U$.
Finally, $\frac{d\mathbf{x}}{dy}|_{y_0} = \frac{\partial F^{-1}}{\partial v}|_{(a,y_0)} \neq 0$ for all $y_0 \in I$ because $dF^{-1}|_{(a,y_0)}$ must be invertible.