# degree (map of spheres)

Given a non-negative integer $n$, let $S^{n}$ denote the $n$-dimensional sphere. Suppose $f\colon S^{n}\to S^{n}$ is a continuous map. Applying the $n^{th}$ reduced homology functor $\widetilde{H}_{n}(\_)$, we obtain a homomorphism $f_{*}\colon\widetilde{H}_{n}(S^{n})\to\widetilde{H}_{n}(S^{n})$. Since $\widetilde{H}_{n}(S^{n})\approx\mathbbmss{Z}$, it follows that $f_{*}$ is a homomorphism $\mathbbmss{Z}\to\mathbbmss{Z}$. Such a map must be multiplication by an integer $d$. We define the degree of the map $f$, to be this $d$.

## 0.1 Basic Properties

1. 1.

If $f,g\colon S^{n}\to S^{n}$ are continuous, then $\deg(f\circ g)=\deg(f)\cdot\deg(g)$.

2. 2.

If $f,g\colon S^{n}\to S^{n}$ are homotopic, then $\deg(f)=\deg(g)$.

3. 3.

The degree of the identity map is $+1$.

4. 4.

The degree of the constant map is $0$.

5. 5.

The degree of a reflection through an $(n+1)$-dimensional hyperplane through the origin is $-1$.

6. 6.

The antipodal map, sending $x$ to $-x$, has degree $(-1)^{n+1}$. This follows since the map $f_{i}$ sending $(x_{1},\ldots,x_{i},\ldots,x_{n+1})\mapsto(x_{1},\ldots,-x_{i},\ldots,x_{n+1})$ has degree $-1$ by (4), and the compositon $f_{1}\circ\cdots\circ f_{n+1}$ yields the antipodal map.

## 0.2 Examples

If we identify $S^{1}\subset\mathbbmss{C}$, then the map $f:S^{1}\to S^{1}$ defined by $f(z)=z^{k}$ has degree $k$. It is also possible, for any positive integer $n$, and any integer $k$, to construct a map $f\colon S^{n}\to S^{n}$ of degree $k$.

Using degree, one can prove several theorems, including the so-called ’hairy ball theorem’, which that there exists a continuous non-zero vector field on $S^{n}$ if and only if $n$ is odd.

Title degree (map of spheres) DegreemapOfSpheres 2013-03-22 13:22:12 2013-03-22 13:22:12 drini (3) drini (3) 12 drini (3) Definition msc 55M25 degree