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Each continuous mapping between compact oriented manifolds of the same dimension has a degree, which is an integer number.

Let $f$ be a continuous mapping from the $n$-dimensional ball $\mathbb{B}^n$ to itself. Suppose $f$ maps the boundary of $\mathbb{B}^n$ to itself. So we can define a mapping $g$, which is the restriction of $f$ to the $(n-1)$-dimensional boundary of $\mathbb{B}^n$.

Is there any relation between the degree of $f$ (as an $n$-dimensional mapping) and the degree of $g$ (as an $(n-1)$-dimensional mapping)?

In particular, is it true that $\deg(f)=\deg(g)$?

Erel Segal-Halevi
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  • How do you define the degree of a manifold with boundary? – Thomas Rot Dec 02 '16 at 18:47
  • @ThomasRot Good question. Here is an attempt. Let $\mathbb{B}'^n$ be a mirror-image of $\mathbb{B}^n$, and let $f': \mathbb{B}'^n\to \mathbb{B}'^n$ be the corresponding mirror-image of $f:\mathbb{B}^n\to \mathbb{B}^n$. If we glue $\mathbb{B}^n$ and $\mathbb{B}'^n$ at their boundary, we get the sphere $\mathbb{S}^n$. Since $f$ maps the boundary to the boundary, we now have a continuous mapping of $\mathbb{S}^n$ to $\mathbb{S}^n$. So let's define $\deg(f)$ as the degree of that mapping. – Erel Segal-Halevi Dec 03 '16 at 18:40
  • So in the one-dimensional case, $f$ is a function from the interval $[-1,1]$ to itself. There are two cases: if $f(-1)=f(1)$ then the degree of the glued function on $S^1$ is 0, otherwise its degree is 1 or -1. – Erel Segal-Halevi Dec 03 '16 at 20:47

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