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The trigonometric loops give $π_{1} ({(x, y) : x^{2} + y^{2} = 1}, (1, 0)) ≅ Z$

Statement

Let $C = {(x, y) \in R^{2} : x^{2} + y^{2} = 1}$ with basepoint $(1, 0)$ . Then

$π_{1} (C, (1, 0)) ≅ (Z, +) .$

Under this isomorphism, the loop

$t ⟼ (\cos 2 π n t, \sin 2 π n t)$

corresponds to $n$ for every $n \in Z$ .

Facts & Assumptions

Given: The quotient-circle homeomorphism $h ([t]) = (\cos 2 π t, \sin 2 π t)$ and its inverse.

[L1]

$[t] \mapsto (\cos 2 π t, \sin 2 π t)$ is a homeomorphism from $R / Z$ to the unit circle and sends $[0]$ to $(1, 0)$ ( $[t] \mapsto (\cos 2 π t, \sin 2 π t)$ is a homeomorphism from $R / Z$ to the unit circle).

[L2]

$Deg : π_{1} (R / Z, [0]) \to (Z, +)$ is an isomorphism ( $Deg : π_{1} (R / Z, [0]) \to (Z, +)$ is an isomorphism).

[L3]

Every pointed continuous map $f$ induces a well-defined group homomorphism $f_{*}$ ; moreover, for pointed continuous maps, ${id}_{*} = id$ and $(g \circ f)_{*} = g_{*} \circ f_{*}$ (Induced fundamental-group maps are well defined, functorial and invariant under based homotopy).

[L4]

$\deg (ω_{n}) = n$ for every integer $n$ ( $\deg (ω_{n}) = n$ for every integer $n$ ).

[L5]

An isomorphism is a bijective group homomorphism (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

Proof

technique · direct

1.1L1L3L5

The based homeomorphism $h$ and its inverse induce homomorphisms $h_{*}$ and $(h^{- 1})_{*}$ . By [L3], their composites are the induced maps of the two identity maps, so they are mutually inverse. Hence $h_{*}$ is a group isomorphism in the sense of [L5].

2.1step 1.1L2algebra

Compose $(h^{- 1})_{*} : π_{1} (C, (1, 0)) \to π_{1} (R / Z, [0])$ from step 1.1 with the degree isomorphism [L2]. The composite is an isomorphism from $π_{1} (C, (1, 0))$ to $(Z, +)$ .

3.1step 2.1L1L4∎

The homeomorphism sends $ω_{n} (t) = [n t]$ to $h ([n t]) = (\cos 2 π n t, \sin 2 π n t)$ by [L1]. Under the isomorphism of step 2.1 this geometric loop is sent back to $[ω_{n}]$ and then to $n$ by [L4]. This includes $n = 0$ and negative integers.

The trigonometric loops give $π_{1} ({(x, y) : x^{2} + y^{2} = 1}, (1, 0)) ≅ Z$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The trigonometric loops give π1({(x,y):x2+y2=1},(1,0))≅Zπ1​({(x,y):x2+y2=1},(1,0))≅Z

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The trigonometric loops give $π_{1} ({(x, y) : x^{2} + y^{2} = 1}, (1, 0)) ≅ Z$