5
$\begingroup$

Given that $A$ is an open set in $\mathbb R^n$ and $f:A \to \mathbb R^n$ is differentiable, and its derivative is non-singular at every point in $A$, prove that $f(A)$ is open in $\mathbb R^n$

Note $f$ is differentiable, not continuously differentiable.

  • 5
    Note that the tag "open-problem" is for problems for which no solution is known, *not* for problems concerning open sets.2012-03-21

2 Answers 2

2

The answer is given in this excellent post by Terence Tao

https://terrytao.wordpress.com/2011/09/12/the-inverse-function-theorem-for-everywhere-differentiable-maps/

1

By the inverse function theorem for each $x\in A$ there exists open sets $x\in U$ and $f(x)\in V$ so that $f|_U:U\to V$ is a diffeomorphism. So in particular $f(U)=V$ hence $f(x)\in V\subset f(A)$.

  • 3
    But $f$ was *not* assumed to be continuously differentiable which is needed in the usual proof of the inverse function theorem via the Banach fixed point theorem.2012-03-21