Degree of a circle endomorphism

Let $𝑓 : 𝕊 1 \to 𝕊 1$ be continuous. Then there is a unique continuous $𝜑 : [0, 1] \to ℝ$ with the property $𝜑 (0) = 0$ and $𝑓 e x = 𝑓 (1) \cdot e x 𝜑$ , homotopy so the following diagram commutes¹

$\begin{tikzcd} {[0,1]} && {\mathbb R} \\ \\ {\mathbb S^1} && {\mathbb S^1} \arrow["{\mathrm{ex}}"', from=1-1, to=3-1] \arrow["{f(1)\mathrm{ex}}", from=1-3, to=3-3] \arrow["f"{description}, from=3-1, to=3-3] \arrow["{f\mathrm{ex}}"{description}, from=1-1, to=3-3] \arrow["\varphi", dashed, from=1-1, to=1-3] \end{tikzcd}$

Then the degree $d e g 𝑓$ of $𝑓$ is given by

d e g 𝑓 = 𝜑 (1)

which is always a whole number.

Proof

Without loss of generality we may assume $𝑓 (1) = 1$ , since otherwise we may use $𝑔 (𝑥) = 𝑓 (1) - 1 𝑓 (𝑥)$ First we will show that if such a $𝜑$ exists it is necessarily unique. Let $𝜑, 𝜓 : [0, 1] \to ℝ$ with $𝜑 (0) = 𝜓 (0) = 0$ and $e x 𝜑 = e x 𝜓 = 𝑓 e x$ . Then
$e x 𝜑 e x 𝜓 (𝑡) = e x (𝜑 (𝑡) - 𝜓 (𝑡)) = 1$
for all $𝑡 \in [0, 1]$ , which may be the case iff $𝜑 (𝑡) - 𝜓 (𝑡) \in ℤ$ for all $𝑡 \in [0, 1]$ . Since $𝜑$ and $𝜓$ are continuous so is $𝜑 - 𝜓$ , and thus $𝜑 - 𝜓$ is a constant map. Thus $(𝜑 - 𝜓) (𝑡) = (𝜑 - 𝜓) (0) = 0$ for all $𝑡 \in [0, 1]$ , i.e. $𝜑 = 𝜓$ .

Since $𝑓 e x : [0, 1] \to 𝕊 1$ is continuous it is uniformly continuous by the Heine-Cantor theorem, we can divide $[0, 1]$ by $0 = 𝑡 0 < 𝑡 1 < \dots < 𝑡 𝑘 = 1$ with finite $𝑘$ so that
$𝑡 \in [𝑡 𝑗, 𝑡 𝑗 + 1] ⟹ | 𝑓 e x (𝑡) - 𝑓 e x (𝑡 𝑗) | < 2$
for all integers $0 \leq 𝑗 \leq 𝑘 - 1$ . We write $𝑗 (𝑡)$ to denote the value of $𝑗$ corresponding to some $𝑡$ . whence it follows that $𝑓 e x (𝑡)$ and $𝑓 e x (𝑡 𝑗 (𝑡))$ are not antipodes, namely
$𝑓 e x (𝑡) 𝑓 e x (𝑡 𝑗 (𝑡)) \neq - 1$
and therefore the Main branch of the complex logarithm $L n (𝑓 e x (𝑡) / 𝑓 e x (𝑡 𝑗 (𝑡)))$ is well-defined. We define $𝜑$ as follows
$𝜑 (𝑡) = 1 2 𝜋 𝑖 𝑗 (𝑡) \sum 𝑛 = 1 L n 𝑓 e x (𝑡 𝑛) 𝑓 e x (𝑡 𝑛 - 1) + L n 𝑓 e x (𝑡) 𝑓 e x (𝑡 𝑗 (𝑡))$
which is continuous by properties of the Main branch of the complex logarithm. Additionally, $𝜑 (0) = 0$ and clearly $e x 𝜑 = 𝑓 e x$ .

All that’s left to show is that $𝜑 (1) \in ℤ$ . This is true since by definition $𝑓 (1) e x 𝜑 (1) = 𝑓 e x (1) = 𝑓 (1)$ and hence $e x 𝜑 (1) = 1 ⟹ 𝜑 (1) \in ℤ$ .

Generalisation to closed path

If $𝛼 : 𝕊 1 \to ℂ$ is a closed continuous path, then we may define the winding number of $𝛼$ around $𝑧$ as

𝑛 (𝛼; 𝑧) = d e g 𝑓 𝛼, 𝑧

where $𝑓 𝛼, 𝑧 : 𝕊 1 \to 𝕊 1$

𝑓 𝛼, 𝑧 (𝜁) = 𝛼 (𝜁) - 𝑧 | 𝛼 (𝜁) - 𝑧 |

Ring isomorphism

$𝗁 𝖳 𝗈 𝗉 (𝕊 1, 𝕊 1)$ is a ring with function multiplication as addition and composition as multiplication. Then $d e g : (𝗁 𝖳 𝗈 𝗉 (𝕊 1, 𝕊 1), \cdot, \circ) \to (ℤ, +, \cdot)$ is a ring isomorphism, since Circle endomorphisms are homotopic iff they are of equal degree and $d e g 𝑧 𝑛 = 𝑛$ for all $𝑛 \in ℤ$ ,

Examples

$d e g 𝑐 𝑇$ for any Constant map
$d e g i d = 1$
$d e g 𝑧 𝑛 = 𝑛$

Properties

Circle endomorphisms are homotopic iff they are of equal degree

tidy | en | SemBr

2010, Algebraische Topologie, pp. 37–41 ↩

Table of Contents

Degree of a circle endomorphism

Generalisation to closed path

Ring isomorphism

Examples

Properties

Graph View

Backlinks

Table of Contents

Degree of a circle endomorphism

Generalisation to closed path

Ring isomorphism

Examples

Properties

Footnotes

Graph View

Backlinks