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Let $f: U\to \mathbb{R}^n$ be smooth and proper map such that det$df_x$ does not change sign outside some compact subset of $U$. Prove that $f$ is one-to-one and $\exists$ $x \in U$ such that $f(x) = 0$.

Some help please =)

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put on hold as off-topic by Michael Albanese, Daniel Rust, Avitus, Micah, Bungo yesterday

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If you browse a bit the questions and answers on this site, you'll immediately notice that people do not write "plz" and such. This is not SMS, in fact! –  Mariano Suárez-Alvarez Apr 1 '11 at 1:39
What is $U$? Presumably it's $n$-dimensional, so that $df_x$ even might have a determinant in the first place? In any case, it doesn't seem right to me that 0 would have to be in the image of $f$, unless $f$ ends up in fact having to be surjective... –  Aaron Mazel-Gee Apr 1 '11 at 8:15
oops... you´re wright. –  Júlio César Apr 1 '11 at 16:36
This is not an answer but just a comment, I would have wrote this as a comment if I had had enough reputation. 1)You should restrict yourself to a connected and n-dimensional manifold $U$, otherwise your thesis has not sense or is obviously false. 2)For easiness start assuming that $f$ is everywhere regular. 3)Because the hypothesis are preserved adding to $f$ any constant function, your thesis really means that $f$ should be bijective. In these terms your problem sounds: for a local diffeomorphism $f:U\to\mathbb{R}^n$, where $U$ is a n-dimensional connected manifold, does properness i –  Giuseppe Tortorella Apr 1 '11 at 20:36
The comment of Giu was converted from an answer, but the end was cut off at the beginning of the word "imply". It ends with: "does properness imply bijectivity?". –  Jonas Meyer Apr 4 '11 at 18:28