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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.

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Note that the tag "open-problem" is for problems for which no solution is known, not for problems concerning open sets. –  Alex Becker Mar 21 '12 at 2:37

2 Answers 2

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

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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. –  t.b. Mar 21 '12 at 3:00

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