Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be a smooth function and let $g:\mathbb{R}^n \rightarrow \mathbb{R}$ be defined by $g(x_1,...,x_n)=x_1^5+...+x_n^5$. Suppose $g\circ f\equiv 0$. Show that $\det Df\equiv 0$.
I was going to start the solution from $n=2$ and then use induction on $n$.
So, for $n=2$, $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a smooth function and $g:\mathbb{R}^2 \rightarrow \mathbb{R}$ where $g(x_1,x_2)=x_1^5+x_2^5$. Then $Df=\begin{pmatrix} f_{x_1x_1} & f_{x_1x_2} \\ f_{x_2x_1} & f_{x_2x_2} \end{pmatrix}$
But now I am stuck in proving the result. Am I in the correct track? Can anyone plese give me a hint to continue?
Ok so since $g\circ f$ is $0$, the image of $f$ is the line $x_1+x_2=0$ which has no min, max or saddle.
If det$Df(x_0)>0$ and $f_{x_1x_1}(x_0)>0$ OR det$Df(x_0)>0$ and $f_{x_1x_1}(x_0)<0$, we have contradiction.
But if det$Df(x_0)>0$ and $f_{x_1x_1}(x_0)=0$, then det$Df(x_0)=0$ or, det$Df(x_0)<0$(Since $f_{x_1x_2}=f_{x_2x_1}$). So this is impossible.
If det$Df(x_0)<0$ we have saddle and it is impossible.
So finally we conclude that det$Df(x)=0$ for all $x$. Am I correct?
We know $g \circ f \equiv 0$. Therefore, by applying the chain rule we get $$Dg(f(x)) Df(x) \equiv 0.$$ Let us reason with the facts that $\operatorname{rank}(Dg(f(x))) \leq 1$ and $\operatorname{rank}(Df(x)) \leq n$.
We can't have $\operatorname{rank}(Dg(f(x))) = 1$ and $\operatorname{rank}(Df(x)) = n$ because this would mean the product is not zero.
If $\operatorname{rank}(Dg(f(x))) = 0$ then the whole $Dg(f(x))$ is zero. Designating the component functions of $f$ as $f_i$ we conclude that $f_i \equiv 0$ and $f \equiv 0$, leading to $\det(Df) \equiv 0$ because $f$ is the zero function.
If $\operatorname{rank}(Dg(f(x))) = 1$ then this means that $\operatorname{rank}(Df(x)) < n$. From this it follows that $\det(Df) \equiv 0$.