NicolasDelanoue-S´ebastienLagrange Classiﬁcationofmappingsfrom R to R

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Introduction to classification Stable mappings of the plane and their singularities

Interval analysis and mappings fromR²toR². Computing the Apparent Contour Conjecture and conclusion

Classification of mappings from R ² to R ²

Nicolas Delanoue - S´ ebastien Lagrange

SWIM 2013 - Small Workshop on Intervals Methods - Brest http://www.ensta-bretagne.fr/swim13/

Nicolas Delanoue - S´ebastien Lagrange Classification of mappings fromR²toR²

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Outline

1

Introduction to classification Objects, Equivalence, Invariants Discretization - Portrait of a map

2

Stable mappings of the plane and their singularities Stable maps

Withney theorem - Normal forms

Compact simply connected with boundary

3

Interval analysis and mappings from R

²

to R

²

.

4

Computing the Apparent Contour

5

Conjecture and conclusion

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Objects, Equivalence, Invariants The one dimentional case Discretization - Portrait of a map

Objects

The set of square matrices of order n (denoted by M

_n

)

Equivalence

We can defined an equivalence relation ∼ between elements of M

_n

with

A ∼ B ⇔ ∃P ∈ GL, A = PBP

⁻¹

Invariant

The set of eigenvalues is an invariant because A ∼ B ⇒ sp(A) = sp(B )

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Objects

The set of square matrices of order n (denoted by M

_n

) Equivalence

We can defined an equivalence relation ∼ between elements of M

_n

with

A ∼ B ⇔ ∃P ∈ GL, A = PBP

⁻¹

A ∼ B ⇒ sp(A) = sp(B )

(5)

Objects

The set of square matrices of order n (denoted by M

_n

) Equivalence

We can defined an equivalence relation ∼ between elements of M

_n

with

A ∼ B ⇔ ∃P ∈ GL, A = PBP

⁻¹

Invariant

The set of eigenvalues is an invariant because A ∼ B ⇒ sp(A) = sp(B )

(6)

Invariant

The set of eigenvalues is not a strong enough invariant since

∃A, B ∈ M

_n

, A 6∼ B and sp(A) = sp(B )

A =

0 0 and B =

0 0 A really strong invariant

Let us call by J the Jordan method, we have

A ∼ B ⇔ J(A) = J(B)

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Invariant

The set of eigenvalues is not a strong enough invariant since

∃A, B ∈ M

_n

, A 6∼ B and sp(A) = sp(B ) Example

A =

0 0 0 0

and B =

0 1 0 0

A really strong invariant

Let us call by J the Jordan method, we have A ∼ B ⇔ J(A) = J(B)

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Invariant

The set of eigenvalues is not a strong enough invariant since

∃A, B ∈ M

_n

, A 6∼ B and sp(A) = sp(B ) Example

A =

0 0 0 0

and B =

0 1 0 0

A really strong invariant

Let us call by J the Jordan method, we have

A ∼ B ⇔ J(A) = J(B)

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Objects Equivalence Invariants

Square matrices A ∼ B ⇔ ∃P ∈ GL, A = PBP

⁻¹

Eigenvalues,

Jordanisation

Real bilinear A ∼ B ⇔ ∃U, V ∈ SO , A = UBV Singularvalues forms

Smooth maps f ∼ f

⁰

if there exist diffeomorphic ? changes of variables (g , h) on X and Y such that f = g ◦ f

⁰

◦ h

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Objects Equivalence Invariants

Square matrices A ∼ B ⇔ ∃P ∈ GL, A = PBP

⁻¹

Eigenvalues, Jordanisation

Smooth maps f ∼ f

⁰

if there exist diffeomorphic ?

changes of variables (g , h) on X

and Y such that f = g ◦ f

⁰

◦ h

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Objects Equivalence Invariants

Square matrices A ∼ B ⇔ ∃P ∈ GL, A = PBP

⁻¹

Eigenvalues, Jordanisation

Real bilinear A ∼ B ⇔ ∃U, V ∈ SO , A = UBV Singularvalues forms

Smooth maps f ∼ f

⁰

if there exist diffeomorphic ? changes of variables (g , h) on X and Y such that f = g ◦ f

⁰

◦ h

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Objects Equivalence Invariants

Square matrices A ∼ B ⇔ ∃P ∈ GL, A = PBP

⁻¹

Eigenvalues, Jordanisation

Real bilinear A ∼ B ⇔ ∃U, V ∈ SO , A = UBV Singularvalues forms

Smooth maps f ∼ f

⁰

if there exist diffeomorphic ?

changes of variables (g , h) on X

and Y such that f = g ◦ f

⁰

◦ h

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Global picture

One wants a global “picture” of the map which does not depend on a choice of system of coordinates neither on the configuration space nor on the working space.

Definition - Equivalence

Let f and f

⁰

be two smooth maps. Then f ∼ f

⁰

if there exists diffeomorphisms g : X → X

⁰

and h : Y

⁰

→ Y such that the diagram

X −−−−→

^f

Y



 y

^g

x



^h

X

⁰ ^f

0

−−−−→ Y

⁰

commutes.

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Global picture

One wants a global “picture” of the map which does not depend on a choice of system of coordinates neither on the configuration space nor on the working space.

Definition - Equivalence

Let f and f

⁰

be two smooth maps. Then f ∼ f

⁰

if there exists diffeomorphisms g : X → X

⁰

and h : Y

⁰

→ Y such that the diagram

X −−−−→

^f

Y



 y

^g

x



^h

X

⁰ ^f

0

−−−−→ Y

⁰

commutes.

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Example

R −−−−→

^2x+6

R



 y

^g

x



^h

R −−−−→

^x+1

R

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Example

R −−−−→

^2x+6

R



 y

^x+2

x



^2y

R −−−−→

^x+1

R

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Examples

1

f

1

(x) = x

²

, f

2

(x) = ax

²

+ bx + c, a 6= 0 f

1

∼ f

2

f

₁

(x) = x

²

+ 1, f

₂

(x) = x + 1, f

₁

6∼ f

₂

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Examples

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Examples

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Proposition

Suppose that f ∼ f

⁰

with

x

₁

−−−−→

^f

y

₁



 y

^g

x



^h

x

₂

−−−−→

^f⁰

y

₂

then f

⁻¹

({y

₁

}) is homeomorphic to f

⁰⁻¹

({y

₂

}).

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Proposition

Suppose that f ∼ f

⁰

with

x

₁

−−−−→

^f

y

₁



 y

^g

x



^h

x

₂

−−−−→

^f⁰

y

₂

then rank df

_x₁

= rank df

_x⁰₂

. Proof

Chain rule, df = dh · df

⁰

· dg

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Definition

Let us defined by S

f

the set of critical points of f : S

_f

= {x ∈ X | df (x) is singular }.

Corollary

f ∼ f

⁰

⇒ S

f

' S

f⁰

where ' means homeomorphic.

i.e. the topology of the critical points set is an invariant.

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This is not a strong enough invariant,

there exists smooth maps f , f

⁰

: [0, 1] → [0, 1] such that S

_f

' S

_f⁰

and f 6∼ f

⁰

.

Figure : Singularity theory.

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This is not a strong enough invariant,

there exists smooth maps f , f

⁰

: [0, 1] → [0, 1] such that S

_f

' S

_f⁰

and f 6∼ f

⁰

.

Figure : Topology of X .

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Definition - Abstract simplicial complex

Let N be a finite set of symbols {(a

⁰

), (a

¹

), . . . , (a

ⁿ

)}

An abstract simplicial complex K is a subset of the powerset of N satisfying : σ ∈ K ⇒ ∀σ

₀

⊂ σ, σ

₀

∈ K

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K =

(a

⁰

), (a

¹

), (a

²

), (a

³

), (a

⁴

), (a

⁰

, a

¹

), (a

¹

, a

²

), (a

⁰

, a

²

), (a

³

, a

⁴

),

(a

⁰

, a

¹

, a

²

)

This will be denoted by a

⁰

a

¹

a

²

+ a

³

a

⁴

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Definition

Given abstract simplicial complexes K and L, a simplicial map F : K

⁰

→ L

⁰

is a map with the following property :

(a

⁰

, a

¹

, . . . , a

ⁿ

) ∈ K ⇒ (F (a

⁰

), F (a

¹

), . . . , F (a

ⁿ

)) ∈ L.

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Example - Simplicial map

K = a

₀

a

₁

+ a

₁

a

₂

+ a

₂

a

₃

, L = b

₀

b

₁

+ b

₁

b

₂

b

b b

F : a

⁰

7→ b

⁰

a

¹

7→ b

¹

a

²

7→ b

²

a

³

7→ b

¹

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Example - NOT a Simplicial map

K = a

₀

a

₁

+ a

₁

a

₂

+ a

₂

a

₃

, L = b

₀

b

₁

+ b

₁

b

₂

b

b b

F : a

⁰

7→ b

⁰

a

¹

7→ b

¹

a

²

7→ b

²

a

³

7→ b

⁰

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Example - Simplicial map

K = a

₀

a

₁

a

₂

+ a

₁

a

₂

a

₃

, L = b

₀

b

₁

b

₂

b

F : a

⁰

7→ b

⁰

a

¹

7→ b

¹

a

²

7→ b

²

a

³

7→ b

⁰

(39)

Definition

Let f and f

⁰

be continous maps. Then f and f

⁰

are topologically conjugate if there exists homeomorphism g : X → X

⁰

and h : Y → Y

⁰

such that the diagram

X −−−−→

^f

Y



 y

^g

x



^h

X

⁰ ^f

0

−−−−→ Y

⁰

commutes.

Proposition

f ∼ f

⁰

⇒ f ∼

₀

f

⁰

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Definition

Let f be a smooth map and F a simplicial map, F is a portrait of f if

f ∼

₀

F

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Example - Simplicial map The simplicial map

b

b b

is a portrait of [−4, 3] 3 x 7→ x

²

− 1 ∈ R

-4 -3 -2 -1 1 2 3

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Introduction Stable maps

Withney theorem - Normal forms Compact simply connected with boundary

Proposition

For every closed subset A of R

ⁿ

, there exists a smooth real valued function f such that

A = f

⁻¹

({0})

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Proposition

For every closed subset A of R

ⁿ

, there exists a smooth real valued function f such that

A = f

⁻¹

({0}) We are not going to consider all cases . . .

(44)

Definition

Let f be a smooth map, f is stable if their exists a nbrd N

_f

such that

∀f

⁰

∈ N

_f

, f

⁰

∼ f

Examples

1

g : x 7→ x

²

is stable,

2

f

0

: x 7→ x

³

is not stable, since with f : x 7→ x(x

²

− ),

6= 0 ⇒ f 6∼ f

0

.

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Withney theorem

Let X and Y be 2-dimentional manifolds and f be generic. The critical point set S

_f

is a regular curve. With p ∈ S

_f

, one has

T

_p

S

_f

⊕ ker df

_p

= T

_p

X or T

_p

S

_f

= ker df

_p

Geometric representation

1

if T

p

S

_f

⊕ ker df

p

= T

p

X ,

2

if T

_p

S

_f

= ker df

_p

,

(46)

Geometric representation

1

T

p

S

_f

⊕ ker df

p

= T

p

X : fold point

2

T

_p

S

_p

= ker df

_p

: cusp point

(47)

Geometric representation

1

T

p

S

_f

⊕ ker df

p

= T

p

X : fold point

2

T

_p

S

_p

= ker df

_p

: cusp point

(48)

Normal forms

1

If T

p

S

f

⊕ ker df

p

= T

p

X , then there exists a nbrd N

p

such that

f |N

_p

∼ (x, y ) 7→ (x, y

²

).

2

If T

_p

S

_f

= ker df

_p

, then there exists a nbrd N

_p

such that

f |N

_p

∼ (x, y) 7→ (x , xy + y

³

).

(49)

Definition

Let f a smooth map from X → R

²

with X a simply connected compact subset of R

²

with smooth boundary ∂X . The apparent contour of f is

f (S

_f

∪ ∂X )

The topology of the Apparent contour is an invariant.

(50)

Definition

Let f a smooth map from X → R

²

with X a simply connected compact subset of R

²

with smooth boundary ∂X . The apparent contour of f is

f (S

_f

∪ ∂X )

The topology of the Apparent contour is an invariant.

(51)

Definition

Let f a smooth map from X → R

²

with X a simply connected compact subset of R

²

with smooth boundary ∂X . The apparent contour of f is

f (S

_f

∪ ∂X )

The topology of the Apparent contour is an invariant.

(52)

Definition

Let f a smooth map from X → R

²

with X a simply connected compact subset of R

²

with smooth boundary ∂X . The apparent contour of f is

f (S

_f

∪ ∂X )

The topology of the Apparent contour is an invariant.

(53)

Definition

Let f a smooth map from X → R

²

with X a simply connected compact subset of R

²

with smooth boundary ∂X . The apparent contour of f is

f (S

_f

∪ ∂X )

The topology of the Apparent contour is an invariant.

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Theorem (Global properties of generic maps)

Let X be a compact simply connected domain of R

²

with

∂X = Γ

⁻¹

({0}). A generic smooth map f from X to R

²

has the following properties :

1

S

_f

is regular curve. Moreover, elements of S are folds and

cusp. The set of cusp is discrete.

(55)

Theorem

3

3 singular points do not have the same image,

4

2 singular points having the same image are folds points and they have normal crossing.

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Theorem

5

3 boundary points do not have the same image,

6

2 boundary points having the same image cross normally.

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Theorem

7

3 different points belonging to S

_f

∪ ∂X do not have the same image,

8

If a point on the singularity curve and a boundary have the same image, the singular point is a fold and they have normal crossing.

(58)

Theorem

9

if the singularity curve intersects the boundary, then this point is a fold,

10

moreover tangents to the singularity curve and boundary

curve are different.

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Cusp Fold - Fold Boundary - Boundary Boundary - Fold

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Proposition

Let f be a smooth generic map from X to R

²

, let us denote by c the map defined by :

c : X → R

²

p 7→ df

p

ξ

p

(1)

where ξ is the vector field defined by ξ

p

=

∂

2

det df

p

−∂

₁

det df

_p

.

If c(p) = 0 and dc

p

is invertible then p is a simple cusp. This

sufficient condition is locally necessary.

(61)

Interval Newton method

c : X → R

²

p 7→ df

p

ξ

p

(2)

(62)

Interval Newton method

c : X → R

²

p 7→ df

p

ξ

p

(3)

(63)

2 different folds

S

^∆2

= {(x

₁

, x

2

) ∈ S × S − ∆(S) | f (x

1

) = f (x

2

)}/ ' where ' is the relation defined by

(x

1

, x

2

) ' (x

₁⁰

, x

₂⁰

) ⇔ (x

1

, x

2

) = (x

₂⁰

, x

₁⁰

).

Method

Bisection scheme on X × X .

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Method

Bisection scheme on X × X .

(65)

Method

Bisection scheme on X × X .

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Method

Bisection scheme on X × X .

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Method

Bisection scheme on X × X .

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Method

Bisection scheme on X × X .

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Method

Bisection scheme on X × X .

(70)

Method

Bisection scheme on X × X .

(71)

Method

Bisection scheme on X × X .

(72)

Method

Bisection scheme on X × X .

(73)

Method

Bisection scheme on X × X .

(74)

Method

Bisection scheme on X × X .

(75)

[x

₁

] 6= [x

₂

]

[x

1

] = [x

2

]

(76)

Let us define the map folds by

folds : X × X → R

⁴

x

₁

y

1

,

x

₂

y

2

7→







det df (x

₁

, y

₁

) det df (x

₂

, y

₂

) f

1

(x

1

, y

1

) − f

1

(x

2

, y

2

) f

₂

(x

₁

, y

₁

) − f

₂

(x

₂

, y

₂

)







One has

S

^∆2

= folds

⁻¹

({0}) − ∆S / ' .

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For any (α, α) in ∆S , the d folds is conjugate to







a b 0 0

0 0 a b

a

11

a

12

a

11

a

12

a

21

a

22

a

21

a

22







which is not invertible since det

a

₁₁

a

₁₂

a

₂₁

a

₂₂

= det df (α) = 0. In other words, as any box of the form [x

₁

] × [x

₁

] contains ∆S , the interval Newton method will fail.

One needs a method to prove that f |S ∩ [x

₁

] is an embedding.

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One needs a method to prove that f |S ∩ [x

1

] is an embedding.

[x

₁

] = [x

₂

]

(79)

One needs a method to prove that f |S ∩ [x

1

] is an embedding.

[x

₁

] = [x

₂

] Not in this case ...

(80)

Corollary

Let f : X → R

²

be a smooth map and X a compact subset of R

²

. Let Γ : X → R be a submersion such that the curve

S = {x ∈ X | Γ(x) = 0} is contractible. If

∀J ∈ d f ˜ (X ) ·

∂

₂

Γ(X )

−∂

₁

Γ(X )

, rank J = 1

then f |S is an embedding.

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The last condition is not satisfiable if [x

1

] contains a cusp . . . Proposition

Suppose that there exists a unique simple cusp p

₀

in the interior of X . Let α ∈ R

^2∗

, s.t. α · Im df

_p₀

= 0, and ξ a non vanashing vector field such that ∀p ∈ S , ξ

p

∈ T

p

S (S contractible).

If g = P

α

_i

ξ

³

f

_i

: X → R is a nonvanishing function then f |S is injective. This condition is locally necessary.

Here the vector field ξ is seen as the derivation of C

^∞

(X ) defined by

ξ = X

ξ

i

∂

∂x

i

.

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∂X

^∆2

= {(x

₁

, x

2

) ∈ ∂X × ∂X − ∆(∂X ) | f (x

1

) = f (x

2

)}/ '

(83)

[x

1

] 6= [x

₂

]

[x

₁

] = [x

₂

]

(84)

Let us define the map boundaries by

boundaries : X × X → R

⁴

x

₁

y

1

,

x

₂

y

2

7→







Γ(x

₁

, y

₁

) Γ(x

₂

, y

₂

) f

1

(x

1

, y

1

) − f

1

(x

2

, y

2

) f

₂

(x

₁

, y

₁

) − f

₂

(x

₂

, y

₂

)







One has

∂X

^∆2

= boundaries

⁻¹

({0}) − ∆∂X / ' .

(85)

BF = {(x

₁

, x

₂

) ∈ ∂X × S | f (x

₁

) = f (x

₂

)}

(86)

[x

1

] 6= [x

₂

]

[x

₁

] = [x

₂

]

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[x

1

] 6= [x

₂

]

X × X → R

⁴

x

1

y

1

,

x

2

y

2

7→







det df (x

₁

, y

₁

) γ(x

2

, y

2

) f

1

(x

1

, y

1

) − f

1

(x

2

, y

2

) f

₂

(x

₁

, y

₁

) − f

₂

(x

₂

, y

₂

)







[x

₁

] = [x

₂

]

X → R

²

x

₁

y

₁

7→

det df (x

₁

, y

₁

) γ(x

₁

, y

₁

)

(88)

Analysis

(89)

Synthesis

X=







e¹ e² e³ e⁴ e⁵ e⁶ e⁷ e⁸ e⁹ e¹⁰ e¹¹

a 0 0 0 0 0 0 0 0 1 1 0

b 0 0 0 0 0 0 0 1 0 0 1

c₁ 0 0 0 0 0 0 0 1 1 0 0

c₂ 0 1 1 0 0 0 0 0 0 0 0

d1 0 0 0 0 0 0 0 0 0 1 1

d₂ 0 0 1 1 0 0 0 0 0 0 0

e1 1 0 0 0 0 1 0 0 0 0 0

e₂ 0 0 0 1 1 0 0 0 0 0 0

f 1 1 0 0 0 0 1 0 0 0 0

g 0 0 0 0 1 1 1 0 0 0 0







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Synthesis

X/f =







e¹ e² e³ e⁴ e⁵ e⁶ e⁷ e⁸ e⁹ e¹⁰ e¹¹

A 0 0 0 0 0 0 0 0 1 1 0

B 0 0 0 0 0 0 0 1 0 0 1

C 0 1 1 0 0 0 0 1 1 0 0

D 0 0 1 1 0 0 0 0 0 1 1

E 1 0 0 1 1 1 0 0 0 0 0

F 1 1 0 0 0 0 1 0 0 0 0

G 0 0 0 0 1 1 1 0 0 0 0







NicolasDelanoue-S´ebastienLagrange Classiﬁcationofmappingsfrom R to R

Classification of mappings from R 2 to R 2