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✓ 29 results · all verified · 11 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 18 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Rank Theorems and Embedded Submanifolds

1 · Prerequisites

2 · Summary

This page transports the Euclidean inverse-function and constant-rank normal forms into manifold charts, then uses those local models to build the embedded and immersed submanifold theory needed later. It keeps the manifold-level normal forms, embedding criteria, level-set and preimage theorems, and the diagonal/graph constructions on one proof spine.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-31Open item page →

The rank of a smooth map at a point

Definition

Let F:M→N be a smooth map and let p∈M. The rank of F at p is

rank⁡pF:=rank⁡(dFp),

the rank of the linear map dFp:TpM→TF(p)N (The differential of a smooth map).

This is chart-independent because the differential changes under coordinates by composition with linear isomorphisms, so its rank is unchanged.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Immersions, submersions, and constant-rank maps

Definition

Let F:Mm→Nn be a smooth map.

  • F is an immersion at p∈M when dFp is injective.
  • F is a submersion at p∈M when dFp is surjective.
  • F has constant rank r on A⊆M when rank⁡pF=r for every p∈A (The rank of a smooth map at a point).

The map is an immersion or submersion without qualification when the corresponding pointwise condition holds at every point of M.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Regular and critical points and values

Definition

Let F:M→N be a smooth map.

  • A point p∈M is a regular point of F when F is a submersion at p, and a critical point otherwise (Immersions, submersions, and constant-rank maps).
  • A point q∈N is a regular value of F when every point of F−1(q) is a regular point. This includes the empty-fibre case.
  • A point q∈N that is not a regular value is a critical value.
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

The immersion and submersion loci are open

Statement

Let F:Mm→Nn be a smooth map. The set of points where F is an immersion is open in M, and the set of points where F is a submersion is open in M.

Facts & Assumptions

Given: A smooth map F:Mm→Nn.

[F1]

F is an immersion at p exactly when rank⁡pF=m, and it is a submersion at p exactly when rank⁡pF=n (Immersions, submersions, and constant-rank maps).

[L1]

For a smooth Euclidean map, the locus where the differential has rank at least a fixed integer is open (Differential rank is lower semicontinuous).

[L2]

Every manifold chart is a diffeomorphism onto an open Euclidean set (Chart maps are diffeomorphisms onto Euclidean open sets).

Proof

technique · direct
1.1F1L1L2given

Fix an immersion point p. Choose charts around p and F(p) as in [L2], and let f be the coordinate representative of F. Then rank⁡Df(φ(p))=m by [F1]. Since the differential of a map Rm→Rn cannot have rank above m, [L1] gives a Euclidean neighbourhood on which the rank stays at least m, hence exactly m.

1.2F1L1L2given

The same argument with n in place of m shows that near any submersion point the rank stays equal to n, so the submersion locus is open. Here the upper bound rank⁡Df≤n is the relevant maximal-rank bound.

2.1step 1.1L2

Translating step 1.1 back through the source chart, every point of a smaller neighbourhood of p is again an immersion point. Hence the immersion locus is open.

3.1step 2.1step 1.2∎

Steps 2.1 and 1.2 prove the two openness claims.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

The smooth inverse function theorem on manifolds

Statement

Let F:M→N be a smooth map and let p∈M. If dFp:TpM→TF(p)N is an isomorphism, then there are open neighbourhoods U of p and V of F(p) such that F∣U:U→V is a diffeomorphism.

Facts & Assumptions

Given: A smooth map F:M→N and a point p∈M with dFp an isomorphism.

[F1]

The differential of a smooth map is the induced linear map on tangent spaces (The differential of a smooth map).

[L1]

Differentials satisfy the chain rule (The chain rule for differentials of smooth maps).

[L2]

A Euclidean C1 map with invertible derivative at a point has a local C1 inverse (The Euclidean inverse function theorem).

[L3]

If the original Euclidean map is smooth, then its local inverse is smooth (A local inverse of a Ck regular map is Ck).

[L4]

Chart maps and their inverses are smooth diffeomorphisms onto open Euclidean sets (Chart maps are diffeomorphisms onto Euclidean open sets).

Proof

technique · direct
1.1F1L1L4given

Choose smooth charts (U0,φ) at p and (V0,ψ) at F(p) as in [L4], and let f:=ψ∘F∘φ−1. By [L4], the chart maps φ−1 and ψ are diffeomorphisms, so their differentials are linear isomorphisms. Applying [L1] gives Df(φ(p))=dψF(p)∘dFp∘d(φ−1)φ(p). Because dFp is an isomorphism by [F1], the Euclidean differential Df(φ(p)) is invertible.

2.1step 1.1L2L3

By [L2], after shrinking to open neighbourhoods U~⊆φ(U0) and V~⊆ψ(V0) of φ(p) and ψ(F(p)), the restriction f∣U~:U~→V~ is a bijection with C1 inverse. Since f is smooth, [L3] upgrades that inverse to a smooth one.

3.1step 2.1L4construct

Put U:=φ−1(U~) and V:=ψ−1(V~). Then F∣U=ψ−1∘(f∣U~)∘φ, so F∣U:U→V is bijective. Its inverse is φ−1∘(f∣U~)−1∘ψ, which is smooth by [L4] and step 2.1.

4.1step 3.1L4∎

Thus F∣U is a bijective smooth map with smooth inverse, hence a diffeomorphism of neighbourhoods.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

The constant-rank theorem for manifolds

Statement

Let F:Mm→Nn be smooth, let p∈M, and suppose F has constant rank r on a neighbourhood of p. Then there are smooth charts φ:U→Rm at p and ψ:V→Rn at F(p) such that

ψ∘F∘φ−1(u,v)=(u,0)

for (u,v)∈Rr×Rm−r near φ(p), with the zero in Rn−r.

Facts & Assumptions

Given: A smooth map F:Mm→Nn, a point p∈M, and constant rank r near p.

[F1]

The rank of F at a point is the rank of its differential, and constant rank means that same rank at every point of the chosen set (The rank of a smooth map at a point, Immersions, submersions, and constant-rank maps).

[L1]

Differentials satisfy the chain rule (The chain rule for differentials of smooth maps).

[L2]

A nonzero rank minor supplies an explicit local source-coordinate map Φ that is a Ck diffeomorphism; the identity handles rank zero (A nonzero rank minor supplies the source coordinates for the constant-rank theorem).

[L3]

Chart maps are smooth diffeomorphisms onto open Euclidean sets (Chart maps are diffeomorphisms onto Euclidean open sets).

[L4]

A local inverse of a smooth Euclidean map with invertible differential is smooth (A local inverse of a Ck regular map is Ck).

[L5]

In the source rank coordinates y=Φ(x), after shrinking to a product neighbourhood the map has the form g(u,v)=(u,h(u)) (In source rank coordinates, the remaining components depend only on the rank coordinates).

Proof

technique · direct
1.1F1L1L3given

Choose charts (U0,α) at p and (V0,β) at F(p) as in [L3], and let f:=β∘F∘α−1. By [L3], the maps α−1 and β are diffeomorphisms, so their differentials are linear isomorphisms. Therefore [L1] gives Df(α(x))=dβF(x)∘dFx∘d(α−1)α(x) for every x near p, and the rank of Df(α(x)) equals the rank of dFx. Using [F1], after shrinking U0 the map f has constant rank r on an open Euclidean neighbourhood of α(p).

2.1step 1.1L2L4L5construct

After permuting source and target coordinates, use [L2] to form the explicit source map Φ from the first r components of the smooth map f and the remaining source coordinates. Thus Φ is smooth. Its local inverse is unique and is Ck for every finite k by [L4], hence is smooth. Put g:=f∘Φ−1. By [L5], after shrinking to a product neighbourhood, g(u,v)=(u,h(u)); here h is smooth because g is smooth. The target shear β~(z,w):=(z,w−h(z)) and its explicit inverse (z,w)↦(z,w+h(z)) are smooth. Therefore β~∘f∘Φ−1(u,v)=(u,0) near the distinguished point.

3.1step 2.1L3construct

Compose the original source chart with Φ and the target chart with the shear: φ:=Φ∘α and ψ:=β~∘β. By [L3] these are smooth charts, and their coordinate representative for F is the normal form from step 2.1.

4.1step 3.1∎

Therefore F has the claimed local slice form around p. When r=0 or r=min⁡{m,n}, the empty blocks in the displayed formula are interpreted in the usual way.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Local normal form for immersions

Statement

Let F:Mm→Nn be a smooth immersion at p∈M. Then there are charts near p and F(p) in which the coordinate representative of F is

u⟼(u,0)

near the distinguished point.

Facts & Assumptions

Given: A smooth map F:Mm→Nn that is an immersion at p.

[L1]

The immersion locus is open (The immersion and submersion loci are open).

[L2]

A constant-rank-m map has local normal form (u,v)↦(u,0) (The constant-rank theorem for manifolds).

Proof

technique · direct
1.1L1given

Because F is an immersion at p, the linear map dFp:TpM→TF(p)N is injective. Its domain has dimension m, so its rank is m. Therefore [L1] supplies a neighbourhood of p on which F has constant rank m.

2.1step 1.1L2

Apply [L2] with r=m. The source has no normal coordinates left, so the local model is exactly u↦(u,0).

3.1step 2.1∎

This is the asserted immersion normal form.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Local normal form for submersions

Statement

Let F:Mm→Nn be a smooth submersion at p∈M. Then there are charts near p and F(p) in which the coordinate representative of F is

(u,v)⟼u

near the distinguished point.

Facts & Assumptions

Given: A smooth map F:Mm→Nn that is a submersion at p.

[L1]

The submersion locus is open (The immersion and submersion loci are open).

[L2]

A constant-rank-n map has local normal form (u,v)↦(u,0) in adapted coordinates (The constant-rank theorem for manifolds).

Proof

technique · direct
1.1L1given

Since F is a submersion at p, the linear map dFp:TpM→TF(p)N is surjective. Its target has dimension n, so its rank is n. Thus [L1] gives a neighbourhood on which F has constant rank n.

2.1step 1.1L2

Apply [L2] with r=n. The target normal factor has dimension 0, so the normal form reads (u,v)↦u.

3.1step 2.1∎

This is the claimed submersion normal form.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Every immersion is locally an embedding

Statement

Let F:Mm→Nn be a smooth immersion and let p∈M. Then some neighbourhood U of p is sent homeomorphically onto an embedded m-dimensional submanifold of N, and F∣U is still an immersion.

Facts & Assumptions

Given: A smooth immersion F:Mm→Nn and a point p∈M.

[L1]

Near p, suitable coordinates identify F with the coordinate inclusion u↦(u,0) (Local normal form for immersions).

Proof

technique · direct
1.1L1given

By [L1], after shrinking about p and F(p) there are charts in which F becomes u↦(u,0).

2.1step 1.1

The coordinate inclusion is injective, is a homeomorphism onto the slice Rm×{0} with the subspace topology, and that slice is an embedded submanifold of the ambient Euclidean space. Pulling this description back through the charts gives the stated neighbourhood U.

3.1step 1.1step 2.1∎

The restricted map remains an immersion because it is locally identified with the coordinate inclusion.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Every submersion is an open map

Statement

Every smooth submersion F:M→N is an open map.

Facts & Assumptions

Given: A smooth submersion F:M→N.

[L1]

Around each point of M, a submersion is locally the projection (u,v)↦u (Local normal form for submersions).

[L2]

Products carry the usual product topology, so if (u0,v0) lies in an open set of a product, some product neighbourhood of (u0,v0) lies inside that open set (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice).

Proof

technique · direct
1.1L1given

Let W⊆M be open and let y∈F(W). Choose x∈W with F(x)=y. By [L1], after shrinking around x and y, the map is identified with a coordinate projection (u,v)↦u.

2.1step 1.1L2

Since W is open and contains x, [L2] gives a product neighbourhood U1×U2⊆W in those coordinates. The projection sends this product neighbourhood onto the open set U1. Therefore y has an open neighbourhood contained in F(W).

3.1F1step 2.1∎

Because every point of F(W) is interior, F(W) is open. Thus F is an open map in the sense of [F1].

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

A smooth map of locally maximal rank has locally constant rank

Statement

Let F:M→N be smooth and let p∈M. Suppose there is a neighbourhood U of p such that rank⁡xF≤rank⁡pF for every x∈U. Then F has constant rank on some neighbourhood of p.

Facts & Assumptions

Given: A smooth map F:M→N, a point p∈M, and a neighbourhood U on which the rank never exceeds rank⁡pF.

[F1]

rank⁡xF is the rank of the differential at x (The rank of a smooth map at a point).

[L1]

For Euclidean smooth maps, the set where the differential has rank at least a fixed value is open (Differential rank is lower semicontinuous).

[L2]

Charts identify neighbourhoods in manifolds with open Euclidean sets (Chart maps are diffeomorphisms onto Euclidean open sets).

Proof

technique · direct
1.1F1L2given

Choose charts around p and F(p) as in [L2], and let f be the coordinate representative of F. By [F1], Df(φ(p)) has rank r:=rank⁡pF, and the hypothesis says nearby ranks are at most r.

2.1step 1.1L1

By [L1], the locus where Df has rank at least r is open. Since φ(p) lies in that locus and nearby ranks are never above r, there is a smaller Euclidean neighbourhood on which the rank is both at least r and at most r, hence exactly r.

3.1step 2.1L2∎

Pulling that neighbourhood back through the source chart, F has constant rank r on a neighbourhood of p.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Embedded submanifolds and slice charts

Definition

Let Mm be a smooth manifold, let k∈N satisfy 0≤k≤m, and let S⊆M. One says that S is an embedded k-dimensional submanifold of M when for every p∈S there is a smooth chart φ:U→φ(U)⊆Rm with p∈U such that

φ(S∩U)=φ(U)∩(Rk×{0}).

Such a chart is a slice chart for S at p. The topology on S is the subspace topology inherited from M (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).

The next lemma proves that the restricted slice charts define a smooth k-manifold structure on S.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

Slice-chart restrictions form a smooth atlas

Statement

Let S⊆Mm be an embedded k-submanifold. For each slice chart φ:U→φ(U), restrict φ to U∩S and identify φ(U)∩(Rk×{0}) with an open subset of Rk by projection onto the first k coordinates. These restricted charts are smoothly compatible and generate exactly the subspace topology on S.

Facts & Assumptions

Given: An embedded k-submanifold S⊆Mm.

[F1]

In a slice chart, S is cut out by the coordinate slice Rk×{0} (Embedded submanifolds and slice charts).

[L1]

Chart maps are homeomorphisms onto open Euclidean sets (Chart maps are diffeomorphisms onto Euclidean open sets).

Proof

technique · direct
1.1F1L1given

Let φ:U→φ(U) and ψ:V→ψ(V) be slice charts. By [F1], after projecting away the zero normal coordinates, the overlap transition on S is x↦πk ⁣(ψ∘φ−1(x,0)), where πk denotes projection onto the first k coordinates. Because ψ∘φ−1 is a smooth map between open Euclidean sets and restriction to (x,0) plus projection are smooth, the restricted transition maps are smooth.

2.1F1step 1.1

The restricted charts cover S because the slice charts do. Their images are open in Rk: indeed φ(U)∩(Rk×{0}) equals (πk(φ(U)∩(Rk×{0})),0), and the slice condition in [F1] says that every point of this set has an ambient product neighbourhood whose first-factor projection stays inside the image.

3.1F2L1step 2.1

By [L1], each ambient chart map φ is a homeomorphism, so its restriction identifies U∩S with φ(U)∩(Rk×{0}). Therefore the restricted charts make U∩S open exactly when it is open in the subspace topology from [F2]. The atlas therefore generates precisely the subspace topology on S.

4.1step 1.1step 2.1step 3.1∎

Steps 1.1-3.1 prove smooth compatibility and the topology claim.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Smooth embeddings

Definition

Let F:M→N be a smooth map. Then F is a smooth embedding when

  1. F is injective,
  2. F is an immersion (Immersions, submersions, and constant-rank maps), and
  3. F:M→F(M) is a homeomorphism, where F(M) carries the subspace topology from N (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological, 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).
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

The smooth structure of an embedded submanifold is unique

Statement

Let S⊆M be an embedded k-submanifold. The smooth structure obtained by restricting slice charts is the unique smooth k-manifold structure on the underlying set S for which the inclusion i:S↪M is a smooth embedding.

Facts & Assumptions

Given: An embedded submanifold S⊆M.

[L1]

The restricted slice charts define a smooth atlas on S with the subspace topology (Slice-chart restrictions form a smooth atlas).

[F1]

A smooth embedding is an injective immersion and a homeomorphism onto its image with the subspace topology (Smooth embeddings).

[F2]

Embedded submanifolds are defined by slice charts (Embedded submanifolds and slice charts).

[L2]

A smooth map between manifolds of the same dimension whose differential is an isomorphism is a local diffeomorphism (The smooth inverse function theorem on manifolds).

[L3]

With the smooth structure from restricted slice charts, the inclusion i:S↪M is a smooth embedding (The inclusion of an embedded submanifold is a smooth embedding).

Proof

technique · direct
1.1L1L3F1given

By [L1], slice-chart restrictions give a smooth structure on S, and [L3] shows that its inclusion is a smooth embedding. Suppose T is another smooth k-manifold structure on the same set for which the inclusion i:(S,T)↪M is a smooth embedding. By [F1], i is a homeomorphism onto S with the same subspace topology and an immersion.

2.1F1F2L2step 1.1

Fix p∈S and a slice chart φ:U→Rm around p. Choose a chart θ:W→Rk on (S,T) at p. Since i is an immersion and its image lies in the slice Rk×{0} from [F2], the differential of πk∘φ∘i∘θ−1 is an injective endomorphism of Rk, hence an isomorphism. By [L2] this map is a local diffeomorphism. Thus the slice coordinates are smooth for T, and conversely θ is smooth for the slice-chart structure.

3.1L1step 2.1∎

Every chart of T is smoothly compatible with the restricted slice charts, so the identity map between T and the slice-chart structure is a diffeomorphism. Hence the two smooth structures coincide.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

The inclusion of an embedded submanifold is a smooth embedding

Statement

If S⊆M is an embedded submanifold equipped with the smooth structure from its restricted slice charts, then the inclusion i:S↪M is a smooth embedding.

Facts & Assumptions

Given: An embedded submanifold S⊆M with its slice-chart smooth structure.

[F1]

A smooth embedding is an injective immersion and a homeomorphism onto its image with the subspace topology (Smooth embeddings).

[L1]

The restricted slice charts form a smooth atlas on S with the subspace topology (Slice-chart restrictions form a smooth atlas).

Proof

technique · direct
1.1L1given

The inclusion i:S↪M is injective and, by [L1], a homeomorphism of S onto its image because the domain topology of S is exactly the subspace topology.

1.2L1

In a restricted slice chart on S and the ambient slice chart on M, the coordinate representative of i is u↦(u,0). Its differential is the coordinate inclusion, hence injective, so i is an immersion.

2.1F1step 1.1step 1.2∎

Steps 1.1 and 1.2 verify the clauses in [F1], so i is a smooth embedding.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

The image of a smooth embedding is an embedded submanifold

Statement

Let F:Mm→Nn be a smooth embedding. Then F(M)⊆N is an embedded m-submanifold, and the corestriction F:M→F(M) is a diffeomorphism.

Facts & Assumptions

Given: A smooth embedding F:Mm→Nn.

[F1]

A smooth embedding is an injective immersion and a homeomorphism onto its image with the subspace topology (Smooth embeddings).

[F2]

Embedded submanifolds are described by slice charts (Embedded submanifolds and slice charts).

[L1]

Near any point, an immersion has coordinates of the form u↦(u,0) (Local normal form for immersions).

Proof

technique · direct
1.1F1L1given

Fix p∈M. By [F1] the map F is an immersion, so [L1] gives charts near p and F(p) in which F becomes u↦(u,0).

2.1F1F2step 1.1

In those target coordinates, the image of a small neighbourhood of p is exactly the coordinate slice Rm×{0}. Because [F1] also says that F is a homeomorphism onto its image, we may shrink the neighbourhood in the target so that no other points of M map into that same slice patch. Hence the image is locally a slice chart in the sense of [F2].

3.1F1step 2.1∎

Since every point of F(M) has such a slice neighbourhood, F(M) is an embedded m-submanifold. The corestriction F:M→F(M) is already a homeomorphism by [F1], and in the local coordinates of step 1.1 it is the identity on the Rm factor, so it is a diffeomorphism.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

An injective immersion from a compact manifold is an embedding

Statement

Let M be compact and let N be Hausdorff. Every injective smooth immersion F:M→N is a smooth embedding.

Facts & Assumptions

Proof

technique · direct
1.1L1L4given

By [L1], F is continuous. Since it is injective, the corestriction f0:M→F(M) is a continuous bijection. By [L4], the subspace F(M)⊆N is Hausdorff.

2.1L1L2L3step 1.1

Let C⊆M be closed. Since M is compact, [L2] makes C compact. To show that f0[C] is compact in F(M), let U be an open cover of f0[C] in the subspace F(M); then {f0−1(U):U∈U} is an open cover of C, so finitely many members cover C. Applying f0 back shows that the same finite subfamily covers f0[C]. Thus f0[C] is compact, hence closed in the Hausdorff space F(M) by [L3].

3.1F1step 1.1step 2.1∎

Step 2.1 shows that f0 is a closed bijection. Therefore for every open set O⊆M, the complement M∖O is closed and f0[O]=F(M)∖f0[M∖O] is open in F(M). So f0 is an open bijection, hence a homeomorphism. Since F is an injective immersion by hypothesis, [F1] now gives that F is a smooth embedding.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Immersed submanifolds

Definition

An immersed submanifold of a smooth manifold M is a smooth manifold S together with an injective immersion i:S→M (Immersions, submersions, and constant-rank maps).

The topology of S is its own manifold topology; it need not agree with the subspace topology on the subset i(S)⊆M.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

Smoothness into an embedded submanifold is an initial property

Statement

Let S⊆M be an embedded submanifold with inclusion i:S↪M, and let G:N→S be a map from a smooth manifold N. Then G is smooth if and only if i∘G:N→M is smooth.

Facts & Assumptions

Given: An embedded submanifold S⊆M, its inclusion i, and a map G:N→S.

[F1]

Embedded submanifolds are locally cut out by slice charts (Embedded submanifolds and slice charts).

[L1]

The restricted slice charts define the smooth structure on S (Slice-chart restrictions form a smooth atlas).

[L2]

Ambient charts are diffeomorphisms onto open Euclidean sets (Chart maps are diffeomorphisms onto Euclidean open sets).

Proof

technique · direct
1.1F1L1L2given

Assume first that G is smooth. Choose a chart θ on N and a slice chart φ on M around G(x)∈S. In the corresponding restricted chart on S from [L1], the representative of G has values in Rk, and the representative of i∘G is obtained by appending m−k zero coordinates. Hence i∘G is smooth.

1.2F1L1L2given

Conversely, assume i∘G is smooth. In a slice chart on M, the image of S is Rk×{0} by [F1]. Therefore the representative of i∘G has last m−k coordinates identically zero, and its first k coordinates are exactly the representative of G in the restricted slice chart from [L1]. Those first k coordinates are smooth, so G is smooth.

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove both directions.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

Smoothness of a map on an embedded submanifold is local in the ambient space

Statement

Let S⊆M be an embedded submanifold and let f:S→N be a map to a smooth manifold N. Then f is smooth if and only if every point p∈S has an open neighbourhood U⊆M and a smooth map f~:U→N such that f~∣U∩S=f∣U∩S.

Facts & Assumptions

Given: An embedded submanifold S⊆M and a map f:S→N.

[F1]

Embedded submanifolds have slice charts (Embedded submanifolds and slice charts).

[L1]

A map into an embedded submanifold is smooth exactly when its ambient composite is smooth (Smoothness into an embedded submanifold is an initial property).

[L2]

Smooth maps are continuous (Smooth maps are continuous).

Proof

technique · direct
1.1F1L2givenconstruct

Assume f is smooth. Fix p∈S and choose a slice chart φ:U→Ω⊆Rm with φ(S∩U)=Ω∩(Rk×{0}). In these coordinates, the representative of f∣S∩U depends only on the first k variables. Extend it to all of Ω by ignoring the last m−k coordinates. Transporting this extension back gives a smooth f~:U→N with f~∣U∩S=f∣U∩S.

2.1L1step 1.1given

Conversely, suppose every point has such an ambient extension. On each U∩S, the restriction of f~ to the embedded submanifold U∩S is smooth by [L1], so f is smooth near every point of S. Smooth maps being local on the source, f is smooth on all of S.

3.1step 1.1step 2.1∎

Therefore the ambient-extension criterion is equivalent to smoothness of f.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Codimension and hypersurfaces

Definition

If S is an embedded k-submanifold of an m-manifold M, its codimension in M is m−k.

An embedded submanifold of codimension 1 is a hypersurface.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Local defining maps for embedded submanifolds

Definition

Let S⊆Mm be an embedded k-submanifold and let p∈S. A local defining map for S at p is a smooth submersion

Φ:U→Rm−k

on an open neighbourhood U of p such that

S∩U=Φ−1(0).

For a hypersurface, a local defining map is equivalently a local defining function with values in R.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Embedded submanifolds admit local defining submersions

Statement

Let S⊆Mm be an embedded k-submanifold and let p∈S. Then S admits a local defining map at p.

Facts & Assumptions

Given: An embedded k-submanifold S⊆Mm and a point p∈S.

[F1]

In a slice chart near p, S is the coordinate slice Rk×{0} (Embedded submanifolds and slice charts).

[F2]

A local defining map is a smooth submersion whose zero fibre is the local trace of the submanifold (Local defining maps for embedded submanifolds).

[L1]

Chart maps are diffeomorphisms onto open Euclidean sets (Chart maps are diffeomorphisms onto Euclidean open sets).

[L2]

Differentials satisfy the chain rule (The chain rule for differentials of smooth maps).

Proof

technique · direct
1.1F1F2givenconstruct

Choose a slice chart φ:U→Ω⊆Rk×Rm−k at p as in [F1], and let pr⁡2:Rk×Rm−k→Rm−k be the second-factor projection. Define Φ:=pr⁡2∘φ:U→Rm−k. Then S∩U=Φ−1(0) by the slice description.

2.1L1L2step 1.1

In the same coordinates, Φ=pr⁡2∘φ. By [L1], the differential dφq is an isomorphism for every q∈U. Applying [L2] gives dΦq=d(pr⁡2)φ(q)∘dφq. The Euclidean differential d(pr⁡2)φ(q) is the coordinate projection onto the normal factor, hence surjective. Therefore dΦq is surjective for every q∈U, so Φ is a submersion.

3.1F2step 1.1step 2.1∎

Therefore Φ satisfies the definition in [F2] and is a local defining map for S at p.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

A regular level set is an embedded submanifold

Statement

Let F:Mm→Nn be smooth, let q∈N be a regular value, and assume F−1(q) is nonempty. Then F−1(q) is an embedded submanifold of codimension n. Equivalently, it has dimension m−n.

Facts & Assumptions

Given: A smooth map F:Mm→Nn and a regular value q∈N with nonempty fibre.

[F1]

A regular value is one whose fibre points are all submersion points; the fibre may be empty (Regular and critical points and values).

[F2]

Codimension means ambient dimension minus submanifold dimension, and embedded submanifolds are defined by slice charts (Codimension and hypersurfaces, Embedded submanifolds and slice charts).

[L1]

Near any submersion point, suitable coordinates put F into the form (u,v)↦u (Local normal form for submersions).

Proof

technique · direct
1.1F1given

Let p∈F−1(q) be arbitrary. By [F1], F is a submersion at p.

2.1step 1.1L1algebra

Apply [L1] at p. In suitable charts around p and q, the map becomes (u,v)↦u on Rn×Rm−n. After centering q, the fibre is {0}×Rm−n. Permuting the two source-coordinate blocks sends it to the standard slice Rm−n×{0}. Thus F−1(q) is locally an embedded submanifold.

3.1F2step 2.1∎

Since every point of the fibre has such a slice neighbourhood, F−1(q) is an embedded submanifold. Its local model has dimension m−n, so by [F2] the codimension is n.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

The tangent space of a regular level set is the kernel

Statement

Let F:M→N be smooth, let q be a regular value, and let p∈F−1(q). Then

Tp(F−1(q))=ker⁡dFp.

Facts & Assumptions

Given: A smooth map F:M→N, a regular value q, and a point p∈F−1(q).

[F1]

A regular value has only submersion points in its fibre (Regular and critical points and values).

[L1]

The fibre F−1(q) is an embedded submanifold (A regular level set is an embedded submanifold).

[L2]

Near a submersion point, suitable coordinates put F into the form (u,v)↦u (Local normal form for submersions).

[L3]

Chart maps are diffeomorphisms onto open Euclidean sets (Chart maps are diffeomorphisms onto Euclidean open sets).

[L4]

Differentials satisfy the chain rule (The chain rule for differentials of smooth maps).

Proof

technique · direct
1.1F1L1L2given

Because q is a regular value and p∈F−1(q), [F1] makes F a submersion at p. Write m:=dim⁡M, n:=dim⁡N, and ℓ:=m−n. By [L2], choose local coordinates near p and q in which the representative of F is (u,v)↦u on Rn×Rℓ, with p and q sent to the origins. Then the fibre F−1(q) is represented by the slice {0}×Rℓ. By [L1], this is the embedded-submanifold structure on the fibre near p, so its tangent vectors are exactly the vectors of the form (0,w).

2.1L3L4step 1.1

Let π(u,v)=u be the coordinate projection. Step 1.1 makes ψ∘F∘φ−1=π near the distinguished point. By [L3], the differentials dφp and dψq are isomorphisms, and [L4] gives dψq∘dFp∘d(φ−1)φ(p)=dπφ(p). Because dπφ(p) has kernel {0}×Rℓ, one gets ker⁡dFp=dφp−1({0}×Rℓ). Step 1.1 identifies the same subspace with Tp(F−1(q)), so Tp(F−1(q))=ker⁡dFp.

3.1step 1.1step 2.1∎

Therefore the intrinsic tangent space of the regular level set equals ker⁡dFp.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

The preimage theorem for submanifolds under submersions

Statement

Let F:M→N be a smooth submersion and let S⊆N be an embedded submanifold of codimension c. Then F−1(S) is an embedded submanifold of M of codimension c. For each p∈F−1(S),

Tp(F−1(S))=(dFp)−1(TF(p)S).

Facts & Assumptions

Given: A smooth submersion F:M→N and an embedded codimension-c submanifold S⊆N.

[L1]

Embedded submanifolds admit local defining submersions (Embedded submanifolds admit local defining submersions).

[L2]

A regular level set is an embedded submanifold (A regular level set is an embedded submanifold).

[L3]

The tangent space of a regular level set is the kernel of the defining differential (The tangent space of a regular level set is the kernel).

[L4]

Composites of smooth maps are smooth (Identity maps and composites of smooth maps are smooth).

[L5]

Differentials satisfy the chain rule (The chain rule for differentials of smooth maps).

Proof

technique · direct
1.1L1L4L5givenconstruct

Fix p∈F−1(S) and put y:=F(p). By [L1], there is a neighbourhood V of y and a local defining submersion Φ:V→Rc for S at y. Shrink to an open neighbourhood U of p with F(U)⊆V, and set H:=Φ∘F:U→Rc. By [L4], H is smooth, and H−1(0)=U∩F−1(S). For every x∈U, [L5] gives dHx=dΦF(x)∘dFx; both factors are surjective because F and Φ are submersions, so dHx is surjective. Thus 0 is a regular value of H.

2.1L2step 1.1

By [L2], U∩F−1(S) is an embedded codimension-c submanifold near p. Since p was arbitrary, F−1(S) is embedded of codimension c.

2.2L3L5step 1.1

Applying [L3] to the regular level set of H gives Tp(F−1(S))=ker⁡dHp. By [L5], ker⁡dHp=ker⁡(dΦy∘dFp)=(dFp)−1(ker⁡dΦy). Applying [L3] again to the regular level set Φ−1(0)=V∩S at y gives ker⁡dΦy=TyS, so Tp(F−1(S))=(dFp)−1(TyS).

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 prove the theorem.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

The diagonal is an embedded submanifold

Statement

For every smooth manifold M, the diagonal

ΔM:={(p,p):p∈M}⊆M×M

is an embedded submanifold of dimension dim⁡M.

Facts & Assumptions

Given: A smooth manifold M.

[F1]

Embedded submanifolds are characterized by slice charts (Embedded submanifolds and slice charts).

[L1]

M×M has the canonical product smooth structure (Products of smooth manifolds have a canonical product smooth structure).

[L2]

Chart maps are diffeomorphisms onto open Euclidean sets (Chart maps are diffeomorphisms onto Euclidean open sets).

Proof

technique · direct
1.1L1L2given

Fix (p,p)∈ΔM. Choose a chart φ:U→Ω⊆Rm at p. By [L1] and [L2], the product chart φ×φ:U×U→Ω×Ω is a smooth chart on M×M.

2.1F1step 1.1algebra

In coordinates (u,v)∈Ω×Ω, the diagonal becomes {(u,v):u=v}. The linear change of variables L(u,v):=(u,v−u) is a diffeomorphism from Ω×Ω onto the open set L(Ω×Ω), and L({(u,u):u∈Ω})=L(Ω×Ω)∩(Rm×{0}). Therefore ΔM is locally a coordinate slice and hence an embedded submanifold by [F1].

3.1step 2.1∎

The slice has dimension m=dim⁡M, so the diagonal has that dimension as an embedded submanifold.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

The graph of a smooth map is an embedded submanifold

Statement

Let F:M→N be smooth. Its graph

ΓF:={(p,F(p)):p∈M}⊆M×N

is an embedded submanifold of M×N of dimension dim⁡M.

Facts & Assumptions

Given: A smooth map F:M→N.

[L1]

The diagonal ΔN⊆N×N is an embedded submanifold (The diagonal is an embedded submanifold).

[L2]

A regular level set is an embedded submanifold (A regular level set is an embedded submanifold).

[L3]

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

Proof

technique · direct
1.1L1L3given

Consider the map G:M×N→N×N defined by G(p,q)=(F(p),q). In product charts on M×N and N×N, its representative has the form (u,v)↦(F^(u),v), so G is smooth. The graph satisfies ΓF=G−1(ΔN).

1.2L1L3givenconstruct

Fix (p,F(p))∈ΓF. Choose a chart ψ:V→Ω⊆Rn on N at F(p). By [L3], the product chart ψ×ψ identifies a neighbourhood of (F(p),F(p)) in N×N with Ω×Ω, and in these coordinates the diagonal from [L1] is the set {(a,b):a=b}. The difference map D:Ω×Ω→Rn, D(a,b)=b−a, is a smooth submersion with zero fibre exactly that diagonal slice.

2.1step 1.1step 1.2construct

Choose a chart φ:U→Λ on M at p with F(U)⊆V, and define H:=D∘(ψ×ψ)∘G∘(φ×ψ)−1:Λ×Ω→Rn. Then H(u,v)=v−F^(u), so the Jacobian of H has block form [−DF^(u)In] and is surjective at every point. Its zero fibre is exactly the graph in these coordinates. Thus 0 is a regular value of H.

3.1L2step 2.1∎

By [L2], the zero fibre of H is an embedded codimension-n submanifold of M×N near (p,F(p)). Since (p,F(p)) was arbitrary, ΓF is an embedded submanifold. Its ambient dimension is dim⁡M+n, so its dimension is dim⁡M.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

Transverse intersections of coordinate slices have the expected local form

Statement

Let U⊆Rm be open. Suppose S,T⊆U are given near a point a∈S∩T by independent coordinate slices: after a permutation of coordinates, there are smooth submersions Φ:U→Rc and Ψ:U→Rd with S=Φ−1(0), T=Ψ−1(0) near a, and with dΦa⊕dΨa:TaU→Rc×Rd surjective. Then S∩T is, near a, an embedded submanifold of codimension c+d; in suitable local coordinates it is a coordinate slice.

Facts & Assumptions

Given: Open U⊆Rm, submersions Φ,Ψ, and a point a∈S∩T with the stated surjectivity.

[F1]

A local defining map records a submersion whose zero fibre is the local submanifold (Local defining maps for embedded submanifolds).

[L2]

A regular level set of a smooth map is an embedded submanifold (A regular level set is an embedded submanifold).

[L3]

For a Euclidean smooth map, the locus where the differential has rank at least a fixed value is open (Differential rank is lower semicontinuous).

Proof

technique · direct
1.1F1L1given

By [L1], define H:U→Rc×Rd by H(x)=(Φ(x),Ψ(x)). Then H−1(0,0)=S∩T near a by [F1]. The differential of H at a is exactly dΦa⊕dΨa, so it is surjective by hypothesis.

2.1L3step 1.1

Since the target of H has dimension c+d, step 1.1 says rank⁡dHa=c+d. By [L3], after shrinking to an open neighbourhood W of a, every point of W has differential rank at least c+d, hence exactly c+d. Therefore every point of W∩H−1(0,0) is regular for H∣W, so (0,0) is a regular value of H∣W.

3.1L2step 2.1

By [L2], the common zero set W∩S∩T=W∩H−1(0,0) is an embedded submanifold of codimension c+d and therefore is locally a coordinate slice.

4.1step 3.1∎

This is the claimed local form of the transverse intersection.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

A discrete embedded submanifold is locally closed and countable

Statement

Let S be a discrete embedded submanifold of a smooth manifold M. Then every point of S has a neighbourhood U in M such that S∩U is closed in U. Moreover S is countable.

Proof

technique · direct
1.1F1given

Because S is discrete, its local dimension is 0. Thus by [F1], around each point p∈S there is a chart U in which S∩U corresponds to R0×{0}, that is, a single point. A singleton is closed in the chart domain, so S∩U is closed in U.

2.1L1step 1.1

The manifold M is second countable, hence so is its subspace S by [L1]. Let B be a countable basis for S. Because S is discrete, every singleton {x} is open. Applying the basis property to the open set {x} shows that some B∈B satisfies x∈B⊆{x}, hence B={x}. Thus every singleton of S is itself a member of the countable family B, so S is countable.

3.1step 1.1step 2.1∎

Therefore S is locally closed and countable.

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

An injective immersion need not be an embedding

Statement

False claim: every injective immersion is an embedding.

Facts & Assumptions

Refutation

technique · direct
1.1L1given

By [L1], M is a smooth one-manifold, and on each circle component the map F is the usual inclusion into R2, hence an immersion. The images of different components are distinct circles, so F is injective.

2.1F2step 1.1

The image F(M) is the union of the concentric circles. Every Euclidean neighbourhood of a point on the unit circle meets infinitely many outer circles, so the unit-circle component is not open in the subspace topology of F(M). But that component is open in the disjoint-union topology of M. Therefore F:M→F(M) is not a homeomorphism.

3.1F1step 1.1step 2.1∎

Step 1.1 gives an injective immersion, while step 2.1 shows that the homeomorphism clause in [F1] fails. Hence F is not an embedding, which refutes the claim.

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-31Open item page →

The image of every immersion need not be an embedded submanifold

Statement

False claim: the image of every immersion is an embedded submanifold.

Facts & Assumptions

Given: The smooth map γ:R→R2, γ(t)=(sin⁡t,sin⁡2t).

[F1]

Embedded submanifolds are locally modeled on coordinate slices, hence on a one-manifold they cannot have a self-crossing neighbourhood (Embedded submanifolds and slice charts).

[F2]

An immersed submanifold need only come from an injective immersion on its own manifold (Immersed submanifolds).

[L1]

(sin⁡t)′=cos⁡t (The derivatives of sine and cosine are cosine and minus sine), and the chain rule gives (sin⁡2t)′=2cos⁡2t (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

Refutation

technique · direct
1.1L1given

By [L1], γ′(t)=(cos⁡t,2cos⁡2t). If cos⁡t=0, then t=π/2+kπ, so cos⁡2t=−1 and the second component of γ′(t) is not zero. Thus γ′(t)≠0 for all t, and γ is an immersion.

2.1F1step 1.1

The points t=0 and t=π both map to (0,0), and the two tangent directions there are (1,2) and (−1,2). Hence the image has a transverse self-crossing at the origin. A neighbourhood of that crossing is not homeomorphic to an interval, so by [F1] the image is not an embedded one-submanifold.

3.1F2step 1.1step 2.1∎

Therefore the image of the immersion γ fails to be embedded, refuting the claim.

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-31Open item page →

A regular value need not belong to the image

Statement

False claim: every regular value of a smooth map lies in its image.

Facts & Assumptions

Given: The map f:R→R, f(x)=ex.

[F1]

A value is regular when every point of its fibre is regular, and the empty fibre is allowed (Regular and critical points and values).

[L1]

The derivative of ex is ex (The exponential function is smooth and (exp⁡)′=exp⁡), and the exponential maps R bijectively onto (0,∞) (The exponential is a continuous bijection from R onto (0,∞)).

Refutation

technique · direct
1.1given

The value 0 is not in the image of f, because ex>0 for every real x.

2.1F1L1step 1.1

The fibre f−1(0) is empty. By [F1], the regular-value condition is therefore vacuous, and [L1] is consistent with that since there are no nonregular fibre points to check.

3.1step 1.1step 2.1∎

Thus 0 is a regular value that is not attained, so the claim is false.

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-31Open item page →

Rank at one point need not determine nearby rank

Statement

False claim: if a smooth map has rank r at one point, then it has rank r on some neighbourhood of that point.

Facts & Assumptions

Refutation

technique · direct
1.1L1given

By [L1], df0=0, so the rank at 0 is 0.

2.1L1step 1.1

For every x≠0, the derivative is 2x≠0, so the rank at x is 1. Thus every neighbourhood of 0 contains points of rank 1.

3.1L2step 1.1step 2.1∎

Therefore the rank at 0 does not persist locally, which refutes the claim and shows why the extra maximal-rank hypothesis in [L2] matters.

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

An embedded submanifold need not be open in the ambient manifold

Statement

False claim: every embedded submanifold is an open subset of the ambient manifold.

Facts & Assumptions

Given: The x-axis S={(x,0):x∈R}⊆R2.

[F1]

Embedded submanifolds are locally coordinate slices (Embedded submanifolds and slice charts).

[F2]

Codimension records the dimension drop inside the ambient manifold (Codimension and hypersurfaces).

Refutation

technique · direct
1.1F1F2given

S is the global coordinate slice R×{0} in R2, so [F1] makes it an embedded submanifold. Its codimension is 1 by [F2].

2.1step 1.1

No Euclidean ball centred at a point of S lies inside S, because every such ball contains points with nonzero second coordinate. Hence S is not open in R2.

3.1step 1.1step 2.1∎

Therefore an embedded submanifold need not be open in the ambient manifold.

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-31Open item page →

The intrinsic topology of an immersed submanifold need not be the subspace topology

Statement

False claim: an immersed submanifold always carries the subspace topology of its image in the ambient manifold.

Facts & Assumptions

Refutation

technique · direct
1.1F1L1given

By [L1], the source M is a smooth one-manifold, and the map F is an injective immersion componentwise. Hence (M,F) is an immersed submanifold in the sense of [F1].

2.1F2step 1.1

In the intrinsic topology of M, each circle component is open because M is a disjoint union. In the subspace topology on F(M)⊆R2, the unit circle component is not open because every neighbourhood of one of its points meets infinitely many outer circles. Thus the two topologies differ.

3.1F1step 1.1step 2.1∎

Therefore an immersed submanifold need not carry the subspace topology of its image.

5 · Examples, counterexamples and false statements

None yet.

Sources