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[t](cos2πt,sin2πt) is a homeomorphism from R/Z to the unit circle

Statement

Let

C={(x,y)R2:x2+y2=1}

with the Euclidean subspace topology. The function

h:R/ZC,h([t])=(cos2πt,sin2πt),

is a homeomorphism. Thus [t](cos2πt,sin2πt) is a homeomorphism from R/Z to the unit circle and sends [0] to (1,0).

Facts & Assumptions

Given: The quotient projection p:RR/Z and the unit circle CR2.

[L1]

If q:XY is a quotient map and a continuous function f:XW is constant on every fibre of q, then there is exactly one continuous function fˉ:YW with fˉq=f (For a quotient map q:XY, a map out of Y is continuous iff its composite with q is; a continuous map on X constant on the fibres of q factors uniquely through q; and a composite of quotient maps is a quotient map).

[L2]

The functions sin and cos are differentiable on R, with sin0=0 and cos0=1 (The derivatives of sine and cosine are cosine and minus sine).

[L3]

A real function differentiable on a set is continuous at every point of that set (A function differentiable at c is continuous at c).

[L4]

If m1, (X,d) is a metric space, AX, and f:ARm, then f is continuous exactly when all its coordinate functions are continuous (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions).

[L5]

Both sine and cosine have period 2π, and no smaller positive number is a common period (The zero sets of sine and cosine and the least positive common period 2 pi).

[L6]

The map s(coss,sins) is a bijection from [0,2π) onto C (t(cost,sint) is a bijection from [0,2π) onto the real unit circle).

[L7]

R/Z is compact and path-connected (R/Z is compact and path-connected).

[L10]

Every constant real-valued function and the identity are continuous, and finite sums, products, and scalar multiples of continuous functions are continuous (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function).

[L11]

The quotient projection p:RR/Z induces the quotient topology and satisfies p(x)=p(y) exactly when xyZ (The circle as S1=R/Z with basepoint [0]).

[L12]

For every real x there is exactly one integer m with mx<m+1 (Integer part: for every real x there is exactly one integer m with mx<m+1).

[L15]

A map into a subspace is continuous if and only if its composite with the inclusion into the ambient space is continuous (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[L16]

π>0 and π/2 is the smallest positive zero of cosine (Pi as twice the smallest positive zero of cosine).

Proof

technique · direct
1.1

Define F(t)=(cos2πt,sin2πt). By [L2] and [L3], sine and cosine are continuous; by [L10], t2πt is continuous; hence their composites are continuous by [L17], and [L4] makes F:RR2 continuous. For t=m+r with m=t and 0r<1 from [L12], periodicity [L5] gives F(t)=F(r), while [L6] applied to 2πr[0,2π), using [L16], shows F(r)C; [L15] therefore makes F:RC continuous. Finally [L5] gives F(t+n)=F(t) for every integer n.

L2L3L4L5L6L10L12L15L16L17
2.1

By [L11], the fibres of p are precisely the integer-translation classes, so step 1.1 says that F is constant on every fibre. The quotient universal property [L1] gives a unique continuous h:R/ZC satisfying hp=F, namely h([t])=F(t).

step 1.1L1L11
3.1

To prove surjectivity, let zC. By [L6], z=(coss,sins) for a unique s[0,2π); since [L16] gives 2π>0, the real r=s/(2π) lies in [0,1) and h([r])=z. For injectivity, suppose h([x])=h([y]). Write x=m+r and y=n+q with m,nZ and r,q[0,1) using [L12]. Periodicity [L5] gives F(r)=F(q), and the injectivity in [L6] on [0,2π) gives 2πr=2πq, hence r=q. Thus xy=mnZ, so [L11] gives [x]=[y]. Therefore h is bijective.

step 2.1L5L6L11L12L16algebra
4.1

The source is compact by [L7]. By [L13], R2 is metrizable; by [L14], its subspace C is metrizable, and [L9] makes C Hausdorff. Thus the continuous bijection from steps 2.1 and 3.1 is a homeomorphism by [L8]. Finally [L2] gives h([0])=(cos0,sin0)=(1,0).

step 2.1step 3.1L2L7L8L9L13L14

Depends on

Used by

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