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.