Suppose $S^n$ is an $n$-dimensional sphere.

Definition of the degree of a map: Let $f:S^n \to S^n$ be a continuous map. Then $f$ induces a homomorphism $f_{*}:H_n(S^n) \to H_n(S^n)$ . Considering the fact that $H_n(S^n) = \mathbb {Z}$ , we see that $f_*$ must be of the form $f_*(n)=an$ for some fixed integer $a$. This $a$ is then called the degree of $f$.

Question: For every $k \in {\mathbb Z}$ how does one construct a continuous map $f: S^n \to S^n$ with $\deg(f) = k$?


Here is another solution.

Claim 1: If $f:S^n \to S^n$ has degree $d$ then so does $\Sigma f: S^{n+1} \to S^{n+1}$

Proof: Use the Mayer-Vietrois sequence for $S^{n+1}$. Let $A$ be the complement of the North pole, and $B$ the complement of the South pole. Then $S^n \simeq A \cap B$ and the connecting map $\partial_*$ in the Mayer-Vietrois sequence is an isomorphism. We get the following commutative diagram

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\la}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} H_{n+1}(S^{n+1}) & \ra{\partial_*} & H_n\left(A \cap B\right) & \la{i_*} & H_n(S^n)\\ \da{\Sigma f_*} & & \da{} & & \da{f_*} \\ H_{n+1}(S^{n+1}) & \ra{\partial_*} & H_n\left(A \cap B\right) & \la{i_*} & H_n(S^n)\\ \end{array} $$

in which each horizontal map is an isomorphism. Thus $\Sigma f_* = \partial_*^{-1} i_* f_* i_*^{-1}\partial_*$ and applying homology shows that $\text{deg}(f) = \text{deg}(\Sigma f)$

Thus we are reduced to simply showing that there is a map $f:S^1 \to S^1$ of degree $k$. But this is just the winding number and it is (reasonably well known) that the map $z \mapsto z^k$ (where we view $S^1$ as the unit circle in $\mathbb{C}$) has degree $k$.

Finally I would direct you to have a look at Algebraic Topology by Hatcher:

  • Example 2.31 gives a direct construction of a map of arbitrary degree;
  • Example 2.32 works through the calculation of the map $f(z)=z^k$ proving it has degree $k$; and
  • Prop 2.33 gives another prove of Claim 1 above (which basically takes a different route to the same commutative diagram).
  • 2
    $\begingroup$ That is a great solution! $\endgroup$ – Asaf Karagila Feb 28 '12 at 11:06
  • $\begingroup$ @Yoav: Glad to be the one giving, rather than receiving help! $\endgroup$ – Juan S Mar 4 '12 at 22:46
  • $\begingroup$ Juan: Could you please give the definition of $\Sigma f$? $\endgroup$ – Asaf Karagila Mar 6 '12 at 2:27
  • $\begingroup$ @Asaf: This is just the suspension: en.wikipedia.org/wiki/Suspension_%28topology%29 $\endgroup$ – Juan S Mar 6 '12 at 2:44

Hint: Parameterize $S^n$ as $$g(t_1,\ldots,t_n)=\begin{pmatrix} \cos(t_1)\\ \sin(t_1)\cos(t_2)\\ \vdots\\ \sin(t_1)\cdots\sin(t_{n-1})\cos(t_n)\\ \sin(t_1)\cdots\sin(t_{n}) \end{pmatrix}$$ over $t_1,\ldots,t_{n-1}\in [0,\pi]$ and $t_n\in [0,2\pi)$, and define $f:S^n\to S^n$ by $$f(g(t_1,\ldots,t_{n-1},t_n))=g(t_1,\ldots,t_{n-1},kt_n)$$ which amounts to wrapping the sphere around itself $k$ times in the $x_nx_{n+1}$ plane. One way to show that this is the desired map is to show that it is continuous and then consider one of the simplices perpendiclar to the $x_nx_{n+1}$ plane, when you view the simplicial complex of $S^n$ as a subdivided version of $\partial I^{n+1}$ sitting in $E^{n+1}$ (for example, the top edge of a square, top face of a cube, etc).

  • $\begingroup$ I think $\cos(t)$ is suposed to be $\cos(t_1)$, right? $\endgroup$ – Integral Dec 3 '13 at 18:24
  • $\begingroup$ @Integral Yes, thanks $\endgroup$ – Alex Becker Dec 3 '13 at 21:36

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