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Let $f:\mathbb R^n\to \mathbb R^m $ is class $C^1$

Also $f^{-1}(B)$ is bounded whenever $B$ is bounded

and $\nabla f_i(x)$ are linearly independent for each $x$.

Then $f$ is onto.

Why? I have no idea to explain this.

Maddy
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1 Answers1

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First note that by hypotesis $f$ is a subersion, since the jacobian matrix \begin{equation} J_f=\left[ \begin{array}{c} \nabla f_1 \\ .\\ .\\ .\\ \nabla f_n \end{array} \right] \end{equation} has maximum rank at each point, by local subersion theorem this is an open map. If we conclude that it's also a closed map then it's onto, by connectedness. Since $\{y_1,...,y_n,...\}$ is bounded, we have $f^{-1}(\{y_1,...,y_n,...\})$ is bounded. Then, $\{x_1,...,x_n,...\}$ is bounded, so by Bolzano-Weierstrass theorem there is a convergent subsequence, $(x_{n_j})_{j\geq1}$ that converges to a point $x_0$, since $f$ is continuous, $lim f(x_{n_j})=f(x_0)$ by uniqueness of limit we have $f(x_0)=y$ and $f(\mathbb{R}^n)$ is closed, since it is a nonempty set and $\mathbb{R}^m$ is connected, we have that $f$ is onto.

math_man
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  • note that I used an important fact that n=m, because the hypotesis just give me $n\leq m$ since the $df:\mathbb{R}^n\to \mathbb{R}^m$ is injective by hypotesis, remember that the matrix that represents $df$ is $J_f$. So, if the result says that f is onto, than it is not possible $n<m$, because in this case Sard's theorem says that $f(\mathbb{R}^n)$ has measure zero. So, $m=n$. – math_man Mar 19 '14 at 20:01
  • I think you are using circular logic for $n=m$. I think $n \leq m$ is sufficient to state your answer. – Maddy Mar 20 '14 at 07:22