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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 8 results · all verified · 4 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 4 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Smooth Manifolds and Smooth Maps: Examples and Counterexamples

1 · Prerequisites

2 · Summary

These examples exhibit the standard atlas constructions that drive the page: Euclidean spaces, spheres, projective space, products, and countable disjoint unions. The counterexamples isolate exactly where second countability, two-sided compatibility, and smooth inverses are needed.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-29Open item page →

The circle from two stereographic charts

Example

Let

S1:={(x,y)∈R2:x2+y2=1}.

Removing the north pole N=(0,1) and south pole S=(0,−1), define stereographic charts

σN(x,y)=x1−y,σS(x,y)=x1+y.

Their inverses are

σN−1(u)=(2u1+u2,u2−11+u2),σS−1(u)=(2u1+u2,1−u21+u2).

These two charts form a smooth atlas on S1, so they exhibit the circle as a smooth 1-manifold.

Facts & Assumptions

Given: The circle S1, the poles N,S, and the two stereographic maps σN,σS.

[F1]

A smooth atlas is a covering family of pairwise smoothly compatible charts (Smooth atlases).

[F2]

Smooth compatibility requires both transition maps on the overlap to be smooth (Smoothly compatible charts and the smoothness of Euclidean transition maps).

[F3]

A smooth manifold is a topological manifold equipped with a smooth structure (Smooth manifolds and their smooth charts).

Verification

technique · direct
1.1given

The domains S1∖{N} and S1∖{S} are open in S1 [given] and cover it. The displayed inverse formulas show that each σN and σS is a homeomorphism onto R: composing the inverse with the chart returns the original point, and composing the chart with the inverse gives u. So these are genuine charts.

2.1F1F2step 1.1

On the overlap S1∖{N,S} the transition maps are

σS∘σN−1(u)=1u,σN∘σS−1(u)=1u,

defined on R∖{0}, hence smooth. Therefore the two charts are smoothly compatible by [F2], and [F1] makes them a smooth atlas. [F1, F2, step 1.1]

3.1F3step 2.1∎

This smooth atlas equips S1 with a smooth structure, so [F3] makes the [F3, step 2.1] circle a smooth 1-manifold.

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-29Open item page →

The n-sphere with its standard smooth atlas

Example

For n≥1, let

Sn:={(x1,…,xn+1)∈Rn+1:x12+⋯+xn+12=1}.

With north and south poles N=(0,…,0,1) and S=(0,…,0,−1), the stereographic charts

σN(x)=(x1,…,xn)1−xn+1,σS(x)=(x1,…,xn)1+xn+1

define a smooth atlas on Sn. Their overlap transition is u↦u/∥u∥2 on Rn∖{0}.

Facts & Assumptions

Given: The sphere Sn and the two stereographic maps σN,σS.

[F1]

A smooth atlas is a covering family of pairwise smoothly compatible charts (Smooth atlases).

[F2]

Smooth compatibility means that both transition maps are smooth on the overlap (Smoothly compatible charts and the smoothness of Euclidean transition maps).

[F3]

A smooth manifold is a topological manifold equipped with a smooth structure (Smooth manifolds and their smooth charts).

Verification

technique · direct
1.1given

The domains Sn∖{N} and Sn∖{S} are open and cover [given] Sn. The inverse formulas σN−1(u)=(2u1+∥u∥2,∥u∥2−11+∥u∥2),σS−1(u)=(2u1+∥u∥2,1−∥u∥21+∥u∥2) show that both maps are homeomorphisms onto Rn.

2.1F1F2step 1.1

On the overlap the transition maps are σS∘σN−1(u)=u∥u∥2,σN∘σS−1(u)=u∥u∥2, defined on Rn∖{0}, so they are smooth rational maps there. Hence the two stereographic charts are smoothly compatible by [F2], and [F1] makes them a smooth atlas on Sn.

3.1F3step 2.1∎

Therefore Sn is a smooth manifold by [F3].

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The torus as a product smooth manifold

Example

The torus

T2:=S1×S1

is a smooth 2-manifold with the product smooth structure.

Facts & Assumptions

Given: The circle S1 with its two-chart smooth atlas and the product T2=S1×S1.

[F1]

The circle is a smooth 1-manifold (The circle from two stereographic charts).

[F2]

Products of smooth manifolds carry canonical product smooth structures (Products of smooth manifolds have a canonical product smooth structure).

Verification

technique · direct
1.1F1

By [F1], each factor S1 is a smooth 1-manifold.

2.1F2step 1.1∎

Applying [F2] to the two circle factors gives a canonical product smooth [F2, step 1.1] structure on S1×S1, making it a smooth (1+1)-manifold. This is the torus T2.

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

A countable disjoint union of lines is a smooth manifold

Example

The countable disjoint union

X:=⨆m∈NRm

of countably many copies of the real line is a smooth 1-manifold.

Facts & Assumptions

Given: The countable family (Rm)m∈N of copies of the real line.

[F1]

Each copy of R is a smooth 1-manifold (Euclidean spaces and Euclidean open subsets as smooth manifolds).

[F2]

A countable disjoint union of fixed-dimensional smooth manifolds is a smooth manifold (Countable disjoint unions of fixed-dimensional smooth manifolds are smooth manifolds).

Verification

technique · direct
1.1F1

By [F1], every summand Rm is a smooth 1-manifold.

2.1F2step 1.1∎

The index set N is countable, so [F2] applies to the family [F2, step 1.1] (Rm)m∈N and yields a smooth 1-manifold structure on X.

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The long line is locally Euclidean and Hausdorff but not a manifold under the library convention

Statement refuted

Every Hausdorff locally Euclidean space is a manifold.

Facts & Assumptions

Given: The long line L and the Axiom of Countable Choice ACω.

[L1]

The A-page refutation already proves that L is Hausdorff and locally Euclidean but not second countable, hence not a manifold under the library convention (Hausdorff and locally Euclidean do not by themselves make a manifold).

Counterexample

technique · direct
1.1L1

By [L1], the long line L is Hausdorff and locally Euclidean.

1.2F1L1

The same cited refutation shows that L fails second countability, so it is not a manifold under the library convention. The structural details of [F1] identify the witness but do not change that conclusion.

2.1step 1.1step 1.2∎

Thus L is the required counterexample.

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-29Open item page →

Two noncompatible atlases on the real line

Statement refuted

Any two atlases on the same topological manifold are smoothly compatible.

Facts & Assumptions

Given: The real line with the two singleton atlases A={(R,id⁡)} and B={(R,ψ)}, where ψ(x)=x3.

[F1]

The real line is a smooth manifold, so the two displayed charts are charts on one and the same manifold (Euclidean spaces and Euclidean open subsets as smooth manifolds).

[F2]

A smooth atlas is a covering family of pairwise smoothly compatible charts (Smooth atlases).

[F3]

Smooth compatibility requires both transition directions to be smooth (Smoothly compatible charts and the smoothness of Euclidean transition maps).

Counterexample

technique · direct
1.1F1F2

Each singleton family A and B covers R, so [F1, F2] by [F2] each is an atlas provided its single chart is legitimate, and [F1] supplies that legitimacy.

2.1F3step 1.1

The transition ψ∘id⁡−1(x)=x3 is smooth, but the [F3, step 1.1] reverse transition id⁡∘ψ−1(x)=x1/3 is not differentiable at 0. Hence [F3] says the two charts are not compatible.

3.1step 2.1∎

Therefore A and B are atlases on the same manifold [step 2.1] that are not compatible, which refutes the statement.

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

A bijective smooth map with nonsmooth inverse

Statement refuted

Every bijective smooth map is a diffeomorphism.

Facts & Assumptions

Given: The bijection F:R→R, F(x)=x3.

[L1]

The A-page false statement already proves that F is smooth and bijective but that F−1(y)=y1/3 is not smooth (A bijective smooth map need not be a diffeomorphism).

[F1]

A diffeomorphism is a bijective smooth map with smooth inverse (Diffeomorphisms and local diffeomorphisms of manifolds).

Counterexample

technique · direct
1.1L1

By [L1], the map F is smooth and bijective.

1.2F1L1

The same cited refutation shows that F−1 is not smooth, so [F1] rules [F1, L1] out F being a diffeomorphism.

2.1step 1.1step 1.2∎

Hence F(x)=x3 is the desired counterexample.

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-29Open item page →

An uncountable disjoint union of points is not second-countable

Statement refuted

An arbitrary disjoint union of second-countable manifolds is second-countable.

Facts & Assumptions

Given: An uncountable disjoint union X=⨆i∈I{∗i} of one-point spaces.

[L1]

The A-page false statement already proves that such a space is discrete and admits no countable basis (An arbitrary disjoint union of second-countable manifolds need not be second-countable).

Counterexample

technique · direct
1.1L1

By [L1], every singleton of X is open and any basis of X must contain [L1] uncountably many distinct singleton sets.

2.1F1step 1.1

Therefore X is not second countable in the sense of [F1].

3.1step 2.1∎

So X is the desired counterexample.

Sources