Multivariable calculus/analysis problem

[tim_lou]

Hi guys... Haven't been in the forum for a couple years now.
I have an old analysis problem that I never manage to solve. Would be nice if someone can shed some light on this.let f be a C^1 function from \mathbb{R}^n \rightarrow \mathbb{R}^n, n>1. df is invertible except at isolated points (WLOG assume only at 0), prove that f is locally injective, i.e. there is a neighborhood around each point in the domain such that f is injective.

thoughts about this problem: contraction principle doesn't work at all, since df gets really small around 0. The theorem is false when n=1 (like y=x^2), this makes me think the problem should involve some (if not mostly) topology of R^n

I asked my topology professor but he said he can't think of a solution right away, he told me to consider the eigenvectors of the derivative... would be nice if anyone can put this problem to "sleep" once and for all.

 

[quasar987]

Well, the inverse function theorem implies that f is locally bijective everywhere except perhaps at 0. You must prove that f is injective at 0 also?

 

[tim_lou]

yes, precisely what the problem is about... i think it may be best to look at special cases like \mathbb{R}^2 first.

 

[quasar987]

Is it true for f(x,y)=x²+y² (the 2-dimensinal version of y=x²)??

 

[tim_lou]

i think you forgot that f has to be a function from \mathbb{R}^n to \mathbb{R}^n

 

[tim_lou]

In case anyone is still interested in this problem...in the 2D case, i can show (i think) that there exists a neighborhood U of 0, such that f is injective on U-{0} using Jordon curve theorem. However, I cannot proceed any further.

I don't know if my reasoning is 100% correct. Generally, I first pick a U around 0, small enough such that |df| is bounded by some upper bound. f(U) is bounded, by uniform continuity. I suppose x,y\neq 0 and f(x)=f(y), take a path from x to y, call it g:I\rightarrow \mathbb{R}^2, with non-vanishing derivative. it's image under f will be a loop, call it h. if the loop is self-intersecting, we can cut out the self-intersecting part (the pre-image will become line segments, but that is fine). This cutting process must end because we can cover the image of g by finitely many balls under which f is a homeomorphism by inverse function theorem and compactness, so that the length of each sub-loop (the self-intersecting part) must have a non-zero lower bound. the resulting loop is piecewise C^1 and has non-vanishing derivative (when it exists) since dh=df\cdot dg. Then f(U-\{\textrm{finitely many line segments}\})=f(U)-\{\textrm{the modified loop}\} has at least two components by Jordan curve theorem, yet U-\{\textrm{finitely many line segments}\} is path connected, a contradiction.