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When I was studying analysis some time ago, the next problem arose:

Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be a differentiable function, not necessarily $C^1$, such that $\det(f'(a))=0$, for all $a$ in $\mathbb{R}^n$, where $f'(a)$ stands for the Jacobian matrix at $a$. Is $f$ not injective?

I do not have a proof, but I do have the conviction that this must be true. If $f$ is $C^1$, we can find, for each point, a direction in wich the directional derivative is 0, and the directions change smoothly. Then we could find an integral curve where all the points have the same image. I don't have a formal proof of this idea either.

I think there must be a simpler argument that solves the general case. We do not need a whole curve of points with the same image, just two of them, so maybe the argument can be simplified. Any sugestiond through a solution or even a complete proof would be highly apreciated.

Andrés Ibáñez Núñez
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    Thank you, I made some changes! – Andrés Ibáñez Núñez Apr 21 '17 at 13:38
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    Andrés, don't be discouraged about the downvotes. Some members are fairly anal about "not showing your work" in questions. It gets better. – Martin Argerami Apr 21 '17 at 14:33
  • @AndrésIbáñezNúñez Much better! You will get good results I think. – rschwieb Apr 21 '17 at 14:49
  • @MartinArgerami I think the tone of your earlier comment rises to the level of an inappropriate overreaction to healthy use of downvotes. At any rate, the purpose of my earlier (now self-deleted) suggestions and earlier downvote (now reversed) has been fulfilled as the post is now in good condition. That said, the same comment would be appropriate on a question where there are lots of downvotes, or expressed frustration by the user. (Neither seems to appear here: It is not even clear there were ever a plural number of downvotes.) – rschwieb Apr 21 '17 at 14:57

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The answer by PhoemueX at A continuously differentiable function with vanishing determinant is non-injective? doesn't actually need continuity of the derivative. If $f$ were injective, it would be an open map by invariant of domain. So its image would have positive measure. Sard's theorem holds for differentiable functions $f:\mathbb R^n\to\mathbb R^n,$ that is, $\det(f'(a))=0$ for all $a$ implies that the image of $f$ has zero measure. See Is the image of a null set under a differentiable map always null?

Dap
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