Existence of sphere maps of arbitrary degree

Question

By definition, the degree of a map $f:S^n\rightarrow S^n$ where $n>0$ is defined to be $m$ in the equality $f(\omega)=m.\omega$, where $\omega$ is a generator of $H_n(S^n)$.

I want to show the existence of sphere maps of any arbitrary degree. What is an idea to start from? I know that the degree of $h:S^n\rightarrow S^n$ given by $h(x_1,x_2,...,x_{n+1})=(-x_1,x_2,...,x_{n+1})$ is $-1$. Can we use this fact for my question?

Answer 1

For $S^1$, we have that $z \mapsto z^n$ gives a map of degree $n$ for all $n \in \mathbb Z$.

From this, consider $\Sigma f:S^2 \to S^2$, where $\Sigma f$ is an application of the suspension functor. Can you prove that $\mathrm{deg}( \Sigma f)$ agrees with the degree of $f$?

From this, one iterates the use of the suspension functor to get it for arbitrary spheres.

See proposition 2.33 in Hatcher.

Answer 2

A different example from the one given by Andres Mejia.

Background. The wedge sum is the coproduct in the homotopy category of pointed topological spaces $\mathsf{hTop}_\ast$. This category has as objects pointed topological spaces and as morphisms pointed homotopy classes of continuous maps. Now, consider $\mathsf{hTop}_\ast$ as a cocartesian monoidal category. Then $S^n$ becomes both a comonoid and a monoid object in $\mathsf{hTop}_\ast$ in the following way. The comultiplication $\Delta\colon S^n \rightarrow S^n \vee S^n$ is given by the (pointed) homotopy class of the "pinch map" which contracts the equator of $S^n$ to a point. The multiplication $\nabla \colon S^n \vee S^n \rightarrow S^n$ is given by the (pointed) homotopy class of the "fold map" induced by the identity on both summands. Since the one-point space is the zero object in $\mathsf{hTop}_\ast$, the counit and the unit are uniquely defined. I have not checked this carefully, but this should endow $S^n$ with the structure of a bimonoid.

Construction. Given two continuous based maps $f,g\colon (S^n,x)\rightarrow (S^n,y)$, we can now consider the convolution product in said bimonoid. Namely, we can define the following operation on the set of endomorphism of $S^n$ in $\mathsf{hTop}_\ast$: $$f\star g:=\nabla \circ (f\vee g) \circ \Delta.$$ This convolution product gives an associative, unital monoid structure on the set of endomorphism of $S^n$ in $\mathsf{hTop}_\ast$. It behaves well with respect to the mapping degree (which is well-defined on homotopy classes). Namely, the mapping degree $$\operatorname{deg}\colon \big(\operatorname{End}_{\mathsf{hTop}_\ast}(S^n),\star\big)\rightarrow \big(\mathbb{Z},+\big)$$ is a monoid homomorphism. This follows from the fact that, given continuous $f\colon S^n\rightarrow S^n$ and a point $y\in S^n$ with finite preimage under $f$, the degree of $f$ is given by the sum of local degrees over the points in the fiber of $y$ under $f$. This is proved in Proposition 2.30. in Hatcher.

Let $k\in \mathbb{Z}$. Firstly, assume that $k=0$. Then any constant map $S^n\rightarrow S^n$ has degree $k=0$. Secondly, assume that $k>0$. Since the identity on $S^n$ has degree one, the convolution product of $k$-many identity maps then gives a map $S^n\rightarrow S^n$ of degree $k$. Lastly, assume that $k<0$. As you note, any reflection $r$ on a coordinate plane has degree $-1$. Thus, the convolution product of $\vert k \vert$-many reflections has degree $k$.