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Construct a surjective map $S^n \rightarrow S^n$ of degree zero for ah $n\ge 1$.

I’ve been struggling with this exercise from hatcher. I know that if the map is not surjective then the degree is zero, but I have no idea how to approach this one.

topology master
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1 Answers1

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Here is an explicit construction that uses (extremely minimal) knowledge of higher homotopy groups (Hatcher also proves it purely axiomatically for homology): The map $1 + -1: S^n \rightarrow S^n$ is surjective and nullhomotopic.

Here is a non explicit construction using point set topology:

The Hahn–Mazurkiewicz theorem tells us that every sphere is the image of the unit interval. Let p denote the projection of $S^n$ onto its first coordinate. Then we can compose p with a surjective map guaranteed by the Hahn-Mazurkiewicz theorem to get a surjective map $S^n \rightarrow S^n$ that factors through a nullhomotopic map. This means it is nullhomotopic.

Here is a hint at an explicit construction not using complicated point set topology or anything about adding maps of spheres:

Project $S^n$ onto $D^n$ then just manipulate the stereographic projection to get a surjective map. It should just be defined piecewise in two parts.

Connor Malin
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  • No, I am unfortuntely just getting started on homotopy groups. Anyway the problem is from the homology chapter – topology master May 17 '19 at 22:11
  • I added other ways. – Connor Malin May 17 '19 at 22:23
  • Take $n = 1$. Then $-1$ is represented by $r(z) = \overline{z}$. But then $1 + (-1)$ does not map to $S^1$. – Paul Frost May 17 '19 at 22:25
  • @PaulFrost I meant the inverse in the homotopy group. How is it usually called? (I understand that -1 usually means antipodal map). – Connor Malin May 17 '19 at 22:32
  • For $n=1$ the coordinate flip is $r(z) = \overline{z}$. This means $(1 + (-1))z) = z + \overline{z} = 2\text{Re}(z)$. Anyway, the essence of you answer is that there is a surjection $D^n \to S^n$ which is true. – Paul Frost May 17 '19 at 22:37
  • Yes, I understand for odd dimensions antipodal map has degree 1 (thank you for pointing it out my answer was misleading), but how should you refer to the inverse of the identity? – Connor Malin May 17 '19 at 22:38
  • The antipodal map has degree $(-1)^{n+1}$, so $-1$ ifff $n$ is even. But then $(1 + (-1)(z) = 0 \notin S^n$. – Paul Frost May 17 '19 at 22:41
  • In fact , $-1$ is given by a coordinate flip. But your answer reduces to the last sentence. – Paul Frost May 17 '19 at 22:43
  • I think there should be a proof using suspension since the suspension of a circle is a sphere and then maybe one can induct – topology master May 18 '19 at 16:36