Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $D$ be a region in $R^n$. If $f:D\to R^n$ is continuous, nonzero on $\partial D$ and of Brower degree 0, does there exists a continuous function $g=f$ on $\partial D$ and $g\neq 0$ on $D$?

share|improve this question
    
The Brower degree is usually defined for maps $f:M\to N$ that send $\partial M$ into $\partial N$ and does not seem to extend to this setting. What is the degree of the map $f:[-1,1] \to \mathbb R$ which sends linearly $[-1,0]$ to $[-1,0]$ and $[0,1]$ to $[0,-1/2]$? –  Bruno Martelli Nov 24 '10 at 11:46
    
Is the region an open subset with smooth boundary? Is it open? Is it closed? Is its boundary smooth? Does it have non-empty interior? –  André Henriques Nov 24 '10 at 19:58
1  
Bruno, I think the OP means degree w.r.to the point $p=0$: the setting is the Brouwer degree attached to a triple $(D,f,p),$ where $D$ is a bounded open subset of $\mathbb{R}^n;$ $f:\bar D\to \mathbb{R}^n$ and $p\in\mathbb{R}^n\setminus f(\partial D)$. –  Pietro Majer Nov 24 '10 at 23:15

2 Answers 2

If $D$ is a smooth compact manifold with boundary, then any $f: \partial D \to \mathbb{R}^n \setminus 0$ of degree $0$ can be extended to a map $D \to \mathbb{R}^n \setminus 0$. Proof: A theorem of Hopf asserts that homotopy classes of maps $\partial D \to S^{n-1}$ are determined by their degree. Thus your map $f|_{\partial D}$ is nullhomotopic. Use the nullhomotopy to extend $f$ over a small interior collar of $\partial D$ inside $D$. On the interior boundary of the collar, it is constant and can be extended by the constant map to all of $D$.

share|improve this answer
    
Yes, thank you. It still seems not to be completely clear to me what happens if $\partial D$ is not smooth; then there might not be a neighborhood of $\partial D$ in $D$ homeomorfic to $\partial D\times [0,1]$, but this can be probably solved. –  Peter Franek Nov 24 '10 at 22:49

I assume that the setting is as described by Pietro in the comments above, so $D$ is a connected open set and $f(\partial D)$ does not contain $0$. You can perturb $f$ so that $f|_D$ is smooth and $0$ is a regular value. Then $f^{-1}(0)$ consists of finitely many points $x_1,\ldots, x_{2n}$ contained in the open domain $D$ and $f$ is a local diffeomorphism at each $x_i$. Since the degree is zero, half of these local diffeomorphisms are orientation-preserving, say on $x_1,\ldots, x_n$. The others are orientation-reversing. Choose $n$ disjoint smooth arcs in $D$ that connect $x_i$ to $x_{i+n}$. Take a small regular neighborhood of each arc: it is an open ball with smooth boundary. On each such ball $B$ the map $f|_B$ has degree zero and you can apply Ebert's argument to modify $f|_B$ such that it avoids $0$.

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.