# surjective map $S^n \rightarrow S^n$ of degree zero

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.

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.

• No, I am unfortuntely just getting started on homotopy groups. Anyway the problem is from the homology chapter – topology master May 17 at 22:11
• I added other ways. – Connor Malin May 17 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 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 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 at 22:37