# Application of inverse function theorem?

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?

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.
• 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 Jul 2, 2015 at 16:28
• 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 $\partial_y(F) \neq 0$. Jul 3, 2015 at 18:42