I need help in solving the following exercise.
Consider a $\mathcal C^1$ function $f : U \subseteq \mathbb R^n \to \mathbb R$ with $U$ an open subset.
In the next exercise, it will be proved that $f$ cannot be injective. Let $\mathbf p \in U$ be a point for which $\nabla f(\mathbf p) \neq \mathbf 0.$ (If this point does not exist, then there is nothing to prove.) (Why?)
Let's say that $f_{x_1}(\mathbf p)\neq 0.$ Consider the function $F:U\subseteq \mathbb R^n \to \mathbb R^n$ defined by $F(\mathbf x)=F(x_1,x_2.....,x_n)=(f(\mathbf x),x_2, \dots, x_n).$ Show that this function has an inverse function $F^{-1}$ defined in some open ball of $F(\mathbf p) \in \mathbb R^n.$ Conclude from here that there exist infinitely many points $\mathbf x \in \mathbb R^n$ such that $f(\mathbf x)=c$ for some $c \in f \ \operatorname{rank}.$ This is to say, conclude that $f(\mathbf x)$ is not injective.
I was successful in proving that there exists some $F^{-1}$ defined in some open ball of $F(\mathbf p) \in \mathbb R^n.$ Therefore, there does exist $x_i = \phi_i(f(\mathbf x),x_2,x_3....x_n)$ for all $i.$ But I don't know how to continue from here to prove that $f(\mathbf x)$ is not injective. Please help. Thanks in advance.
For the WHY point notice that if the gradient vanishes in all points the function has to be constant. For the rest notice that in a neighborhood of that point you can construct a hypersurface, all the points are mapped to the same value if the gradient doesn’t vanish.