Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $U$ be an open set in $\mathbb{R}^2$ and $F:U\to \mathbb{R}^3$ a one-to-one differentiable function such that its inverse from $F(U)$ to $\mathbb{R}^2$ is also continuous. Is it possible that $dF$, the derivative of $F$ as a matrix, is not of rank 2 at some point of $U$?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Yes. Consider the function $F:\mathbb{R}^2\to\mathbb{R}^3$ defined by $F(x,y) = (x^3,y^3,0)$. The inverse of the function is $(a,b,0)\mapsto (a^{1/3},b^{1/3})$ which is continuous (but not differentiable at 0) and the rank of $dF$ at the origin is zero.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.