Implicit Inverse function Theorems on $\mathbb R^2$ and injective functions

Question

I have the following question on the applications of Implicit Inverse function Theorems on $\mathbb R^2$ and injective functions:

Let $U\subset\mathbb{R}^{2}$ be an open set containing $(0,0)$ and let $f:U\to\mathbb{R},$ $f\in C^{1}(U)$ such that $f(0,0)=0,$ and $f_{x}(0,0)\ne 0$.

Prove that there exists a $C^{1}$ function $\psi$ defined on a small neighbourhood of $(0,0)$ so that $\psi(0,0)=0$ and $f(\psi(x,y),y)=x$.

Further show that $f$ is not injective in any neighbourhood of $(0,0)$.

I attempted a solution for the first part but could not write a clear solution for the second part.

Define $T:U\to\mathbb{R}$ such that:

$$T(x,y,z)=f(z,y)-x$$ $$\frac{\partial T}{\partial z}|_{(0,0,0)}=\frac{\partial f}{\partial z}|_{(0,0)}\ne 0$$ $$\text{and } \space T(0,0,0)=0$$ Those two imply that $T(x,y,\psi(x,y))=0$ in a small neighbourhood of $(0,0)$, by implicit function theorem.

Therefore, $f(\psi(x,y),y)=x$ on a small neighbourhood of $(0,0)$.For the second part I could not construct a clear argument. I know any neighbourhood contains an open ball from point set topology.

Yes, that second part is obvious. But for the first part, I do not understand your definition of $T$ nor how you use it. Here is a clean solution, using the local inversion theorem rather that the implicit function theorem. — Anne Bauval, Jun 21 '23 at 11:45
In this case there exists no Jacobian determinant just the vector $(1,0)$ comes from partial derivatives. — maths and chess, Jun 21 '23 at 15:21

score 1 · Answer 1 · answered Jun 21 '23 at 12:41

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Your definition of $T$ is a great idea but as written it doesn't quite work; you treat $U$ as a subset of $\Bbb R^3$ (three variables $x,y,z$) but $U\subseteq\Bbb R^2$. However it is easily fixed. You can define $T:\Bbb R\times U\to\Bbb R$ through $(x,y,z)\mapsto f(z,y)-x$ and the rest follows as in your attempt.

In comments it's been shown that $f$ cannot be injective, using the solution $\psi$. However, this is not special to $C^1$ functions or to the implicit function theorem: it's generally true that a continuous $f:U\to\Bbb R^m$ is never injective, where $U\subseteq\Bbb R^n$ is open and $n>m$. This follows from the Borsak-Ulam theorem... but in general, this theorem is difficult. An easy special case is for $m=1$.

Fix $x_0\in U$; there is a ball $B(x_0,\delta)\subseteq U$ for some $\delta>0$. So $U$ contains a copy of the unit circle, e.g. the boundary $\partial B(x_0,\delta/2)$. Overall there is a composite: $$g:S^1\hookrightarrow U\overset{f}{\longrightarrow}\Bbb R$$

If $f$ were injective then $g$ would be injective. But this is not possible!

$g(S^1)$ is compact and connected, so it is an interval $[a,b]$ for some $a\le b$. Without loss of generality, $a<b$, since otherwise $g$ obviously is not injective (it is constant).

Pick $z_1\in g^{-1}(a)$ and $z_2\in g^{-1}(b)$. There are two arcs $z_1\to z_2$ on the unit circle - call them $\gamma_1,\gamma_2:[0,1]\to S^1$. $g\circ\gamma_1:[0,1]\to[a,b]$ is, by the intermediate value theorem, surjective. In particular there is $t\in[0,1]$ with $g(\gamma_1(t))=g(\gamma_2(1/2))$ and necessarily $0<t<1$. But then $g$ is not injective, because $\gamma_1(t)$ and $\gamma_2(1/2)$ are distinct elements of $S^1$.

It follows $f$ cannot be injective.

answered Jun 21 '23 at 12:41

FShrike

40,125

Thank you that was my absent-mindedness. – maths and chess Jun 21 '23 at 14:57
I really could not understand why $g(S^{1})$ is an interval.The problem is $g(S^{1})$ is in $\mathbb{R}^{2}$ not in $\mathbb{R}$. – maths and chess Jun 21 '23 at 15:05
Do you mean up to homoeomorphism? – maths and chess Jun 21 '23 at 15:05
@tutkudoruk No, $g$'s codomain is $\Bbb R$ because $f$'s codomain is $\Bbb R$ – FShrike Jun 21 '23 at 17:41
Thank you,I did not understand an important point. How did you use intermediate value theorem to say the surjectivity of the composite function? – maths and chess Jun 21 '23 at 18:40
@tutkudoruk Because: (1) it is continuous and (2) it hits both $a$ and $b$ (the endpoints) so by the IVT it must hit every single point in between $a$ and $b$. This is extremely similar to the argument that $g(S^1)$ is an interval by connectivity (the IVT is a very quick corollary of the classification of connected subsets of $\Bbb R$) – FShrike Jun 21 '23 at 18:41
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Okay,thank you I understand now. – maths and chess Jun 21 '23 at 18:43
The final point is they are distinct because they are at distinct arcs. – maths and chess Jun 21 '23 at 18:45

Implicit Inverse function Theorems on $\mathbb R^2$ and injective functions

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