If Jacobian of $f$ at $\bar{x}$ is onto, is Jacobian of points near $\bar{x}$ also onto?

Question

The question:

Assumme that $f: \mathbb{R}^n \to \mathbb{R}^m$ is $\mathcal{C}^1$ around $\bar{x}$. Show that if $\nabla f(\bar{x})$ is onto, then $\nabla f(x)$ is also onto for all $x$ near $\bar{x}$.

I have an approach for this question already, as I will sketch it out later. I have two questions related to this problem:

1. Is there a way to just utilize "pure" algebra to tackle with this problem, without using tools from analysis?
2. Can we reduce the assumption of being $\mathcal{C}^1$ around $\bar{x}$ just to only ensuring the existence of $\nabla f(x)$ around $\bar{x}$?

Thank you. Here is my sketch:

1. Prove that the set $\mathscr{L}$ of all onto linear mapping from $\mathbb{R}^n$ to $\mathbb{R}^m$ is open in $\mathbb{R}^{m \times n}$.
2. Since $f$ is $\mathcal{C}^1$ around $\bar{x}$, the mapping $\nabla f$ that turns $x$ into $\nabla f(x)$ is continuous around $\bar{x}$.
3. Let $V$ be an open neigborhood of $\nabla f(\bar{x})$ that sits in $\mathscr{L}$ (thanks to 1.). Then from 2., $\nabla f^{-1} V = U$ is open in $\mathbb{R}^n$. That should be the points "near $\bar{x}$" that we are looking for.

Answer 1

Your proof is right.

1. No, you need analysis (continuity) to link what happens from one point to a nearby point (the very notion of 'nearby' already requires topology).
2. No. Consider first $\phi:\Bbb{R}\to\Bbb{R}$ defined as \begin{align} \phi(x)=\begin{cases}x^2\sin(1/x)&\text{if $x\neq 0$}\\0&\text{if $x=0$}\end{cases}. \end{align} This function is differentiable on $\Bbb{R}$, with \begin{align} \phi'(x)&= \begin{cases} 2x\sin(1/x)-\cos(1/x)&\text{if $x\neq 0$}\\ 0&\text{if $x=0$}, \end{cases} \end{align} so clearly $\phi'$ is not continuous at $0$. Now, consider the function $f(x)=\phi(x)+ax$ where $0<a<1$. Then, $f$ is also differentiable everywhere, but $f'$ not continuous at the origin. We now have that $f'(0)=a\neq 0$. Now, consider sequences of points $\{x_n\}_{n=1}^{\infty}$ and $\{y_n\}_{n=1}^{\infty}$ defined as $x_n=\frac{1}{(2n+1)\pi}$ and $y_n=\frac{1}{2n\pi}$. We have $0<x_n<y_n$ for all $n$, and we have \begin{align} f'(x_n)&=\phi'(x_n)+a =[0-(-1)]+a =1+a >0. \end{align} Also, we have \begin{align} f'(y_n)=\phi'(y_n)=[0-(1)]+a=a-1<0. \end{align} By Darboux's intermediate value theorem for derivatives, there exists a point $\zeta_n\in (x_n,y_n)$ such that $f'(\zeta_n)=0$, so we're done (actually, $f'$ is $C^1$ away from the origin, so if I wanted, I could just apply the usual IVT to $f'$ to deduce the existence of $\zeta_n$). To be more explicit, I gave you a differentiable function $f:\Bbb{R}\to\Bbb{R}$, such that $f'(0)\neq 0$, but such that there exists a sequence of points $\{\zeta_n\}_{n=1}^{\infty}$ such that $f'(\zeta_n)=0$ for all $n$, and such that $\zeta_n\to 0$. Hence, although $f'(0)\neq 0$ (i.e the induced linear mapping $h\mapsto f'(0)h$ is in fact a linear isomorphism), there is no neighborhood of the origin on which $f'$ remains surjective.

Btw, this function $f$ is the classic counterexample to the very intuitively appealing, but unfortunately incorrect statement that

"If $f$ is differentiable at a point $a$ with $f'(a)\neq 0$, then there is some open interval around $a$ on which $f$ is strictly monotone".

(the correct statement requires $f$ to be differentiable on an open interval and $f'$ nowhere vanishing on that interval).