Let f: U $\rightarrow R^{m}$ differentiable in $U \in R^m$. If $|f(x)|$ is constant in $U$, then for all $x \in U$ $, f'(x)$ is not injective. Hint: Derive the function $||f(x)||^{2}$. And verify that the image of $f'(x)$ can not contain the vector $f(x)$, i.e, $f'(x).v \neq f(x)$ for all $v \in R^m$.
I'm trying to solve as follows.
Let $\phi : U \rightarrow R$ defined by $\phi (x)= ||f(x)||^2 = <f(x),f(x)> =||f(x)||.||f(x)||$.
Thus,
$\phi'(x) = 2<f'(x),f(x)> = 2||f(x)||'.||f(x)|| = 0$
$\phi'(x).v = 2<f'(x).v,f(x)> = 0$ for all $v \in R^m$.
Therefore $f'(x).v$ is ortogonal to $f(x)$. Then $f'(x).v \neq f(x)$ for all $v \in R^m$.
My solution is correct? Would have another way to solve it? Thank you.
You should mention also that for linear transformation of the finite dimensional space the notions 'to be injective' and 'to be surjective' are equivalent due to existence of determinant.