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?
Best Answer
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.