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I am looking for examples of the the Brouwer degree of a smooth map $f:M \rightarrow N$, where $M$ and $N$ are subsets of the real line $\mathbb{R}$. In particular, there is a seemingly nice example on page 244 of this pdf: https://folk.ntnu.no/gereonq/TMA4190_Lecture_Notes.pdf

The Brouwer degree, as far as I can tell, and which is consistent with this document's definition on page 239 and 242, assumes that the domain must be compact and boundryless, in other words, a "closed manifold".

One thing that troubles me with this example, and hence any example when the domain of the function is some interval $M$ of the real line, is that either the domain $M\subseteq \mathbb{R}$ is compact, in which case it has boundary points, or boundryless, in which case it will be open.

Although the example linked above seems to neverthless "work" in the sense that at any regular point $y$, the degree of $f$ is the same, I constructed an example below on which a function $f$ is defined on the compact interval $[0,1]$ but has degree of $-1 + 1=0$ at regular point $y_{1}$ but degree of $-1$ at regular point $y_{2}$.

Example

Thus, my question is two parts: 1. Can we even talk about the Brouwer degree for functions between subsets of the real line, and if so 2. Why does the one in the first example given above work but mine does not?

Thanks!

Arctic Char
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Mark
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1 Answers1

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My feeling is that, in the book, they talk about compact manifold with no boundaries (such as a circle, for instance) and the picture there is just illustrative. You are perfectly correct in your observation that the degree may depend on $y$ in your case.

However, there exists a generalization / analogy of the Brower degree for maps $f: \Omega\to R^n$ where $\Omega$ is a nice compact subset of $R^n$, such as a cube. But then the degree is defined with respect to a point $y$ in the target space (your $y_1$ or $y_2$) and it may depend on $y$ (as you see). However, it is still true that the degree, as a function of $y$, is constant in the component of $R^n\setminus f(\partial\Omega)$.

In your case, that would mean it is zero whenever $y > f(0)$ and is $1$ for $y\in [f(1), f(0)]$.

As a reference for this "less standard" degree definition, you may look to page 4 of this paper and the references there.

Peter Franek
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  • This is a fantastic answer. Thank you very much for it. To elaborate on your comment, it turns out that the result you are mentioning can be found in "Degree Theory in Analysis and Applications", Theorem 2.3, part 3. – Mark Aug 16 '20 at 16:40
  • One follow up question: If we maintain that the domain is compact with boundary, can we conclude that the degree is locally constant? – Mark Aug 16 '20 at 17:30
  • Yes, if $y$ is not in the closure of $f(\partial\Omega)$ (in such case, the degree is not well-defined), Right now I'm not sure how "wild" the set $f(\partial\Omega)$ can be, but in all reasonable cases I have ever seen, it is a nice closed set. In your case, $\partial\Omega$ is finite (two points) so here all is clear. – Peter Franek Aug 16 '20 at 18:18