Openness and differentiation

Question

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.

Answer 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 $f(x)\in V\subset f(A)$.

Answer 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/