Let $U\subset\mathbb{R}^n$ be an open set and $f:U\to\mathbb{R}^n$ a differentiable function such that $\det \operatorname{J}f(a)\neq 0\; \forall a\in U$. Prove that if $V$ is an open subset of $U$ then $f(V)$ is an open subset of $\mathbb{R}^n$

I'm not sure how could I do this, I tried to sketch a proof as follows:

Let $V$ be an open set as the problem says, I have to prove that if $a\in V$ -therefore $a$ is an interior point- then $f(a)$ is an interior point of $f(V)$.

$\det \operatorname{J}f(a)\neq 0$ implies that $f$ is smooth for every $a\in U$ and by the inverse function theorem exists $f^{-1}$ and $Y\subset\mathbb{R}^n $ in which $f^{-1}$ is continuous. Suppose $f(V)\subset Y$ is not open, therefore for some $y\in f(V)$ we have $B_{\epsilon}(y) \not\subset f(V)$ for every $\epsilon>0\;$ Since $f$ is invertible we can get $a$ with $f^{-1}$ such that $f(a) = y$ and since $V$ is open $\exists\delta>0$ such that $B_{\delta}(a)\subset V$.

The continuity of $f^{-1}$ implies that because of the openness of $B_{\delta}(a)$ the set $f(B_{\delta}(a))$ must be open in $f(V)$ which contradicts that $B_{\epsilon}(y)\not\subset f(V)$ for every $\epsilon>0\;$.

Is this ok?

  • 2
    $\begingroup$ Yep that's the idea! This proof is correct. Good job! $\endgroup$ – Sempliner Jan 21 '14 at 11:22

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