Hellow, I need some ideas for this problem:
Let $f:\mathbb{R}^m\to \mathbb{R}^n$ a continuous function and let $g_1,\dots, g_n:\mathbb{R}^n\to \mathbb{R}$ fuctions in $C^1(\mathbb{R}^n)$ such that:
- $g_j\circ f: \mathbb{R}^m\to \mathbb{R}$ are in $C^1(\mathbb{R}^m)$.
- The vectors $\{\nabla g_1(f(x)),\dots,\nabla g_n(f(x))\}$ are linear independent for all $x\in \mathbb{R}^m$.
Show that $f$ is in $C^1(\mathbb{R}^m,\mathbb{R}^n)$.
My attempt:
I considered the function $G:\mathbb{R}^n\to \mathbb{R}^n$ defined by, $$G(x)=(g_1(x),\dots,g_n(x))$$ and I applied the inverse function theorem to $G$ and I tried to use the relation $f=g_j^{-1}\circ(g_j\circ f)$ for some $j\in \{1,\dots,n\}$, but I could not prove that $g_j^{-1}$ there exists and It is in $C^1(\mathbb{R})$. So, I need help for this problem, maybe a different approach.
Thanks.
Your approach is the right way to do this: Define $G \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ via $G(y) := (g_1(y),\ldots,g_n(y))$. Fix $y = f(x)$, then $D_{y} G$ is invertible and thus we can use the local inverse theorem, that is a consequence of the implict function theorem, that there is open neighborhoods $U$ of $y$ and $V$ of $G(y)$ such that $G$ restricted to $U$ is a diffeomorphism from $U$ onto $V$.
Since $f$ is continious, we may take a neighborhood $U'$ of $x$ such that $f(U') \subset U$. Then $G \circ f$ (restriced to U') is continuously differentiable in $x$ and has image in $V$. Thus we can use the inverse function $G^{-1} \colon V \rightarrow U$ to get that $G^{-1} \circ G \circ f = f$ is continuously differentiable in $U'$.
Since this holds for any $x \in \mathbb{R}^m$, we get the conlusion.