Continously differentiable function is not injective

Question

I learnt about the implicit function theorem and had to prove the following:

Let $F \in C^1(\mathbb{R}^2, \mathbb{R}).$ Show with the implicit function theorem that $F$ is not injective.

Proof:

Assume $F(x,y) \in C^1(\mathbb{R}^2, \mathbb{R})$ in an open set $U$ whicht contains $(x_0,y_0)$ such that:

1. $F(x_0,y_0) = 0$

2. $\frac{\partial F}{\partial y}(x_0,y_0) \neq 0$

According to the implicit function theorem:

$\exists!$ function $y=f(x) \in C^1$ in the open set $U$ which contains $x_0$ s.t.

1. $f(x_0)=y_0$
2. $F(x,f(x))=0, $ $\forall x \in U$

$\Rightarrow$ There is more than one solution for $F(x,f(x))=0 $. This equation is true for all $x \in U.$

$\Rightarrow f$ is not injective.

Was my understanding of this theorem correct and is this proof valid?

Answer 1

You've only given a partial proof of the claim. Note that there are 3 cases to be considered.

1. There is a point $(x_0, y_0) \in \Bbb{R}^2$ such that $\dfrac{\partial F}{\partial y}(x_0,y_0) \neq 0$.
2. There is a point $(x_0, y_0) \in \Bbb{R}^2$ such that $\dfrac{\partial F}{\partial x}(x_0,y_0) \neq 0$.
3. For every point $(x_0, y_0) \in \Bbb{R}^2$, we have $\dfrac{\partial F}{\partial x}(x_0,y_0) = \dfrac{\partial F}{\partial y}(x_0,y_0) = 0$.

You've given a correct proof for case (1) using the implicit function theorem. Case (2) is very similar. Case 3 is easy enough because it implies that $F$ is actually a constant function on $\Bbb{R}^2$, so definitely not injective. To see that $F$ is constant, use the identity (use the fundamental theorem of calculus to prove this): \begin{align} F(x,y) &= F(0,0) + \int_0^x \dfrac{\partial F}{\partial x}(s,0) \, ds + \int_0^y \dfrac{\partial F}{\partial y}(x,s) \, ds. \end{align}