I am not completely sure if this a direct consequence of the inverse function theorem.

Assume that we have a function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ that we can write in terms of coordinates $x,y.$

Does the fact that $D_2f \neq 0$ mean then that we can also write $y$ as a function of $x,f$?

I feel as if my question is not completely rigorous, as $f$ is again a function depending on $x,y$ so there is somehow a circular argument here, but the question is: Assuming that I know what $x$ and $f(x,y)$ are. Does $D_2f \neq 0$ mean that I can reconstruct what $y$ was?


1 Answer 1


The inverse function theorem and implicit function theorem are "cousins" of each other. You can prove one and then deduce the other. Your intuition is guiding you from the inverse function theorem towards the implicit function theorem.

Your description is slightly inaccurate (but easily fixable) in the sense that a function $f: \mathbb{R}^2 \to \mathbb{R}$ itself does not define $y$ as a function of $x$, but, the relation $f(x,y) = 0$ together with the condition $D_2(f) \neq 0$ indeed determines locally $y$ as a function of $x$. You can look up for example the book by Boothby on differentiable manifolds, or, actually, many books on differential geometry contain this.

  • $\begingroup$ No, actually I think I figured out what I want $G_x:=F(x,.):\mathbb{R} \rightarrow \mathbb{R}$ such that $y \mapsto F(x,y).$ Now by the inverse function theorem, the map $G_x^{-1}$ exists iff $D_2F(x,y) \neq 0$ for all $y.$ Hence, $G_x^{-1}$ reconstructs $y$, if $f(x,y)$ is known. Additionally, we need to know $x$ in order to know which $G_x^{-1}$ we have to use. Thus, $y$ is fully determined in terms of $x$ and $f(x,y)$. But okay, I think I will accept your answer anyway, because my question was poorly stated $\endgroup$
    – wewasss
    Jul 2, 2015 at 16:28
  • $\begingroup$ I understand better your question. Roughly speaking, you have $z = F(x,y)$, and you want to understand under which conditions you can express $y$ as a function of $x$ and $z$. You can also cast this problem in a form where you can apply the implicit function theorem. Indeed, let $H(x,y,z) = z - F(x,y)$. You are interested in the set determined by $H(x,y,z) = 0$, and would like to express $y$ as a function of $x$ and $z$. A sufficient condition for this is $H_y \neq 0$, or equivalently $\partial_y(F) \neq 0$. $\endgroup$
    – Malkoun
    Jul 3, 2015 at 18:42

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