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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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[t]↦(cos⁡2πt,sin⁡2πt) 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/Z⟶C,h([t])=(cos⁡2πt,sin⁡2πt),

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

Facts & Assumptions

Given: The quotient projection p:R→R/Z and the unit circle C⊆R2.

[L1]

If q:X→Y is a quotient map and a continuous function f:X→W is constant on every fibre of q, then there is exactly one continuous function fˉ:Y→W with fˉ∘q=f (For a quotient map q:X→Y, 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 sin⁡0=0 and cos⁡0=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 m≥1, (X,d) is a metric space, A⊆X, and f:A→Rm, 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↦(cos⁡s,sin⁡s) is a bijection from [0,2π) onto C (t↦(cos⁡t,sin⁡t) 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:R→R/Z induces the quotient topology and satisfies p(x)=p(y) exactly when x−y∈Z (The circle as S1=R/Z with basepoint [0]).

[L12]

For every real x there is exactly one integer m with m≤x<m+1 (Integer part: for every real x there is exactly one integer m with m≤x<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
1.1L2L3L4L5L6L10L12L15L16L17

Define F(t)=(cos⁡2πt,sin⁡2πt). By [L2] and [L3], sine and cosine are continuous; by [L10], t↦2πt is continuous; hence their composites are continuous by [L17], and [L4] makes F:R→R2 continuous. For t=m+r with m=⌊t⌋ and 0≤r<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:R→C continuous. Finally [L5] gives F(t+n)=F(t) for every integer n.

2.1step 1.1L1L11

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/Z→C satisfying h∘p=F, namely h([t])=F(t).

3.1step 2.1L5L6L11L12L16algebra

To prove surjectivity, let z∈C. By [L6], z=(cos⁡s,sin⁡s) 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,n∈Z 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 x−y=m−n∈Z, so [L11] gives [x]=[y]. Therefore h is bijective.

4.1step 2.1step 3.1L2L7L8L9L13L14∎

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])=(cos⁡0,sin⁡0)=(1,0).

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