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I do understand the for $f : \mathbb{S}^{1} \longmapsto \mathbb{S}^{1}$ the degree of a map (thinking $f$ as a closed curve $\gamma$ defined on $[0,1]$) can be seen as "how many times a closed curve wraps itself on $\mathbb{S}^{1}"$ in the codomain, but this seems to me a very dependant explanation due to the fact that every $n$-degree map from $\mathbb{S}^{1}$ to itself is homotopic to $z^{n}$.

Defining the degree for a $f: M \longmapsto N$ of class $ C^{\infty}$ where $M$ is compact and orientable, $N$ connected and orientable, both without boundary with dim$M$= dim$N$ as deg $f =$$\sum\limits_{x \in f^{-1}(y)}\text{sgn}df_x$, where $y \in \text{RegVal}(f)$

Are there any chance to visualize why this should at least generalize for manifolds in $\mathbb{R}^{3}$ the degree of maps from $\mathbb{S}^{1}$ to itself ? I was interested in finding a $k$-degree map from $\mathbb{S}^{n} \longmapsto \mathbb{S}^{n}$ and I found this question, which was related to this other question. The answers given by Alex Becker in the first and Jared in the second I think it would be awesome, If I just understood what they really meant.

Especially in the second one, I can't see how such map should be smooth or intuitevely understand why the degree of that map should be one with the definition given above; while in the first I don't understand how would have come to my mind to think to such a map, since I'm unable to visualize what a map of degree $k$ should look like with the definition of degree as sum.

Any help would be appreciated

jacopoburelli
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  • Here's an example to start off with. Suspend the map $z\mapsto z^k$ ($k$ an integer) on the circle to $S^2$. So you're wrapping each latitude circle around itself $k$ times and fixing the north and south poles. What's the degree? Check the computation. – Ted Shifrin Jan 13 '21 at 20:35
  • @TedShifrin I think the map from the suspension has also degree $k$. You think the better way to costruct a map of any degree is to start from $\mathbb{S}^{1}$ and iterate on the suspension ? – jacopoburelli Jan 13 '21 at 20:37
  • If so, this can be done only using with degree theory and topology? – jacopoburelli Jan 13 '21 at 20:38
  • You can certainly do that. There are lots of other things to do, too. Do something interesting on a ball so that everything outside it maps to a fixed point of $S^n$. – Ted Shifrin Jan 13 '21 at 20:39
  • I don't know what you mean by your "done only using ..." question. – Ted Shifrin Jan 13 '21 at 20:40
  • @TedShifrin I don't wether the things you suggesting me to prove requires some higher theory, which for me is algebraic topology and homology – jacopoburelli Jan 13 '21 at 20:41
  • No, no. I'm a big fan of the differential topology definition of degree. It's my preference. Do it for $S^2$ for sure, and then think inductively, suspending a map from $S^{n-1}$ of degree $k$ to one on $S^n$. – Ted Shifrin Jan 13 '21 at 20:44
  • @TedShifrin Could you please suggest me at least a reference ? Can’t find it anywhere – jacopoburelli Jan 13 '21 at 23:10
  • A reference for what? Guillemin and Pollack is the standard reference for intersection theory and degree at an introductory level. – Ted Shifrin Jan 13 '21 at 23:15

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