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

Statement

Let C={(x,y)∈R2:x2+y2=1} with basepoint (1,0). Then

π1(C,(1,0))≅(Z,+).

Under this isomorphism, the loop

t⟼(cos⁡2πnt,sin⁡2πnt)

corresponds to n for every n∈Z.

Facts & Assumptions

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

[L1]

[t]↦(cos⁡2πt,sin⁡2πt) is a homeomorphism from R/Z to the unit circle and sends [0] to (1,0) ([t]↦(cos⁡2πt,sin⁡2πt) is a homeomorphism from R/Z to the unit circle).

[L2]

Deg⁡:π1(R/Z,[0])→(Z,+) is an isomorphism (Deg⁡:π1(R/Z,[0])→(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∘f)∗=g∗∘f∗ (Induced fundamental-group maps are well defined, functorial and invariant under based homotopy).

[L4]

deg⁡(ωn)=n for every integer n (deg⁡(ωn)=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)∗:π1(C,(1,0))→π1(R/Z,[0]) from step 1.1 with the degree isomorphism [L2]. The composite is an isomorphism from π1(C,(1,0)) to (Z,+).

3.1step 2.1L1L4∎

The homeomorphism sends ωn(t)=[nt] to h([nt])=(cos⁡2πnt,sin⁡2πnt) by [L1]. Under the isomorphism of step 2.1 this geometric loop is sent back to [ωn] and then to n by [L4]. This includes n=0 and negative integers.

Depends on

Used by

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Sources