NicolasDelanoue-S´ebastienLagrange Classiﬁcationofmappingsfrom R to R

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Robotics - Introduction Stable mappings and their singularities Interval analysis and mappings fromR²toR². Algorithm computing an invariant Conjecture and conclusion

Classification of mappings from R

²

to R

²

Nicolas Delanoue - S´ ebastien Lagrange

IPA 2012 - Intervals Pavings and Applications - Uppsala http://www.math.uu.se/ipa2012/

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1

Robotics - Introduction Motion planning

Discretization - Portrait of a map

2

Stable mappings and their singularities Stable maps

(Genericity and Thom transversality theorem) Withney theorem

Compact simply connected with boundary

3

Interval analysis and mappings from R

²

to R

²

.

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Geometric model Motion planning Equivalence

Discretization - Portrait of a map

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Position of the end effector depends on α and β

f : X → R

²

α β

7→

2 cos(α) + cos(α + β) 2 sin(β) + sin(α + β )

2

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Given a path δ for the end-effector in the working space, find a curve γ in the configuration space such that

f ◦ γ = δ

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Given a path δ for the end-effector in the working space, find a curve γ in the configuration space such that

f ◦ γ = δ

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Given a path δ for the end-effector in the working space, find a curve γ in the configuration space such that

f ◦ γ = δ

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Given a path δ for the end-effector in the working space, find a curve γ in the configuration space such that

f ◦ γ = δ

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Given a path δ for the end-effector in the working space, find a curve γ in the configuration space such that

f ◦ γ = δ

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x y

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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.

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

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Examples

1

f

1

(x) = 2x + 1, f

2

(x) = −x + 2, f

₁

∼ f

₂

2

f

₁

(x) = x

²

, f

₂

(x) = ax

²

+ bx + c, a 6= 0 f

1

∼ f

2

3

f

1

(x) = x

²

+ 1, f

2

(x) = x + 1,

f 6∼ f

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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 - Simplicial map

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

⁰

→ L

⁰

is a map with the following property :

If (a

⁰

, a

¹

, . . . , a

ⁿ

) is an element of K then F (a

⁰

), F (a

¹

), . . . , F (a

ⁿ

)

span a simplex of 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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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

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Definition - Topologically conjugate

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

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Definition - Portrait

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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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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Introduction Stable maps

(Genericity and Thom transversality theorem) Withney theorem

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})

We are not going to consider all cases . . .

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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 . . .

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Definition - Stable mapping

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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Proposition

Suppose that f ∼ f

⁰

with

x

₁

−−−−→

^f

y

₁



 y

^g

x



^h

x

₂ ^f

0

−−−−→ y

₂

then rank df

x1

= rank df

_x⁰₂

,

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Inverse function theorem

For a differentiable map f : X → Y with dim X = dim Y = n, if the rank d f

_p

= n then there exists an open nbhd U

_p

of p such that

f |U

_p

: U

_p

→ f (U

_p

) is a diffeomorphism.

Globalisation

Does ∀p ∈ X , rank d f

_p

= n imply that f : X → Y is a

diffeomorphism ?

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Inverse function theorem

For a differentiable map f : X → Y with dim X = dim Y = n, if the rank d f

_p

= n then there exists an open nbhd U

_p

of p such that

f |U

_p

: U

_p

→ f (U

_p

) is a diffeomorphism.

Globalisation

Does ∀p ∈ X , rank d f

_p

= n imply that f : X → Y is a

diffeomorphism ?

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df (x) = 



∂

₁

f

₁

(x) . . . ∂

_n

f

₁

(x) .. . . .. .. .

∂

1

f

p

(x) . . . ∂

n

f

p

(x)





Definition - ˜ d f (X )

d f ˜ (X ) =



 

 







∂

₁

f

₁

(ξ

¹

) . . . ∂

_n

f

₁

(ξ

¹

) .. . . .. .. .

∂

1

f

p

(ξ

^p

) . . . ∂

n

f

p

(ξ

^p

)





 |ξ

¹

, . . . , ξ

^p

∈ X



 

 

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Lemma

Let X be a convex compact subset of R

ⁿ

, f : X → R

^p

a smooth mapping with n ≤ p.

If ∀J ∈ d f ˜ (X ), rank J = n then f is an embedding.

In other words, f ∼ i where i : (x

₁

, . . . , x

_n

) 7→ (x

₁

, . . . , x

_n

, 0, . . . , 0). In other words, I is portrait of f , where I is the abstract simplicial identity map.

b

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Lemma

Let X be a convex compact subset of R

ⁿ

, f : X → R

^p

a smooth mapping with n ≤ p.

If ∀J ∈ d f ˜ (X ), rank J = n then f is an embedding.

In other words, f ∼ i where i : (x

₁

, . . . , x

_n

) 7→ (x

₁

, . . . , x

_n

, 0, . . . , 0).

b

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Lemma

Let X be a convex compact subset of R

ⁿ

, f : X → R

^p

a smooth mapping with n ≤ p.

If ∀J ∈ d f ˜ (X ), rank J = n then f is an embedding.

In other words, f ∼ i where i : (x

₁

, . . . , x

_n

) 7→ (x

₁

, . . . , x

_n

, 0, . . . , 0).

In other words, I is portrait of f , where I is the abstract simplicial identity map.

b

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Two submanifolds of M , L

1

and L

2

are said to intersect transversally if

∀p ∈ L

1

∩ L

2

, T

p

M = T

p

L

1

+ T

p

L

2

.

One denotes this by L

1

t L

2

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Transversality - Definition

Let f : X → Y be a smooth map between manifolds, and let Z be a submanifold of Y . We say that f is transversal to Z , denoted as f t Z , if

x ∈ f

⁻¹

(Z) ⇒ df

x

T

x

X + T

_f_(x)

Z = T

_f_(x)

Y

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Proposition

Let k be the codimension of Z in Y .

If f t Z , then f

⁻¹

(Z ) is a regular submanifold (possibly empty) of

X of codimension k.

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Thom transversality Theorem Let Z be submanifold of Y ,

{f ∈ C

^∞

(X , Y ) | f t Z } is residual.

In this case, one says that f is generic.

Example

Generically, for a smooth map from f : R

ⁿ

→ R

ⁿ

, one has f t {0}.

Therefore {x ∈ X | f (x) = 0} is a 0-dimensional manifold.

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Thom transversality Theorem Let Z be submanifold of Y ,

{f ∈ C

^∞

(X , Y ) | f t Z } is residual.

In this case, one says that f is generic.

Example

Generically, for a smooth map from f : R

ⁿ

→ R

ⁿ

, one has f t {0}.

Therefore {x ∈ X | f (x) = 0} is a 0-dimensional manifold.

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Thom transversality Theorem Let Z be submanifold of J

^r

(X , Y ),

{f ∈ C

^∞

(X , Y ) | j

^r

f t Z } is residual.

In this case, one says that f is generic.

Example

For a generic smooth map from f : R → R , one has j

¹

f t {y = 0, p = 0}. Therefore

{x | f (x) = 0 ∧ f

⁰

(x) = 0} = ∅

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Withney theorem

Let X and Y be 2-dimentional manifolds and f : X → Y be generic. The set S(f ) = {x ∈ X | det df

x

= 0} is a regular curve.

Let p ∈ S (f ), f (p) = q. One of the following two situations can occur :

T

p

S (f ) ⊕ ker df

p

= T

p

X or T

p

S (f ) = ker df

p

2

if T

p

S(f ) = ker df

p

, then there exists nbrds N

p

and N

q

such that

f |N

_p

∼ (x, y ) 7→ (x, xy + y

³

)

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Withney theorem

Let X and Y be 2-dimentional manifolds and f : X → Y be generic. The set S(f ) = {x ∈ X | det df

x

= 0} is a regular curve.

Let p ∈ S (f ), f (p) = q. One of the following two situations can occur :

T

p

S (f ) ⊕ ker df

p

= T

p

X or T

p

S (f ) = ker df

p

Normal forms

1

if T

p

S(f ) ⊕ ker df

p

= T

p

X , then there exits nbrds N

p

and N

q

such that

f |N

_p

∼ (x, y) 7→ (x, y

²

)

2

if T

p

S(f ) = ker df

p

, then there exists nbrds N

p

and N

q

such

that

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Geometric representation

1

if T

_p

S(f ) ⊕ ker df

_p

= T

_p

X ,

2

if T

_p

S(f ) = ker df

_p

,

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Theorem (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 = {p ∈ X | det df

_p

= 0} is regular curve. Moreover, elements

of S are folds and cusp. The set of cusp is discrete.

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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 the union the singularity curve and boundary 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.

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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

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 is invertible then p is a simple cusp. This

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Interval Newton method

c : X → R

²

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Interval Newton method

c : X → R

²

p 7→ df

p

ξ

p

(3)

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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

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[x

₁

] 6= [x

₂

]

[x ] = [x ]

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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

]

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

1

] is an embedding.

[x

₁

] = [x

₂

]

Not in this case ...

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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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while P⁰6=∅do

[x₁]×[x₂]←swheres∈P⁰. P⁰←P⁰− {[x₁]×[x₂]}.

if[x1] = [x2] then

iff|S∩[x₁] is an embeddingthen

Print ([x₁]×[x₁])∩S^∆2=∅

else

Divide [x1] into [x^a₁] and [x₁^b]

P⁰←P⁰∪ {[x₁â]×[xâ₁]} ∪ {[x₁â]×[x₁^b]} ∪ {[x₁^b]×[x₁^b]};

end if else

if Interval Newton algorithm withfoldson [x₁]×[x₂] succeed then P←P∪ {[x₁]×[x₂]}

else

Divide [x₁] into [x^a₁] and [x₁^b] Divide [x₂] into [x^a₂] and [x₂^b]

P⁰←P⁰∪ {[x₁â]×[xâ₂]} ∪ {[x₁â]×[x₂^b]} ∪ {[x₁^b]×[x₂â]} ∪ {[x₁^b]×[x₂^b]};

end if

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Initialisation :P← ∅,P⁰← {X×X}, while P⁰6=∅do

[x₁]×[x₂]←swheres∈P⁰. P⁰←P⁰− {[x₁]×[x₂]}.

if[x1] = [x2] then

Print ([x₁]×[x₁])∩S^∆2=∅

else

P⁰←P⁰∪ {[x₁â]×[xâ₁]} ∪ {[x₁â]×[x₁^b]} ∪ {[x₁^b]×[x₁^b]};

end if else

else

end if end if end while

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while P⁰6=∅do

[x₁]×[x₂]←swheres∈P⁰. P⁰←P⁰− {[x₁]×[x₂]}.

if[x1] = [x2] then

Print ([x₁]×[x₁])∩S^∆2=∅

else

P⁰←P⁰∪{[x₁â]×[xâ₁]} ∪ {[x₁â]×[x₁^b]} ∪ {[x₁^b]×[x₁^b]};

end if else

else

end if

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Initialisation :P← ∅,P⁰← {X×X}, while P⁰6=∅do

[x₁]×[x₂]←swheres∈P⁰. P⁰←P⁰− {[x₁]×[x₂]}.

if[x1] = [x2] then

Print ([x₁]×[x₁])∩S^∆2=∅

else

P⁰←P⁰∪ {[x₁â]×[xâ₁]} ∪ {[x₁â]×[x₁^b]} ∪ {[x₁^b]×[x₁^b]}

end if else

else

end if end if end while

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while P⁰6=∅do

[x₁]×[x₂]←swheres∈P⁰. P⁰←P⁰− {[x₁]×[x₂]}.

if[x1] = [x2] then

Print ([x₁]×[x₁])∩S^∆2=∅

else

P⁰←P⁰∪ {[x₁â]×[xâ₁]} ∪ {[x₁â]×[x₁^b]} ∪ {[x₁^b]×[x₁^b]}

end if else

else

P⁰←P⁰∪{[x₁â]×[xâ₂]} ∪ {[x₁â]×[x₂^b]} ∪ {[x₁^b]×[x₂â]} ∪ {[x₁^b]×[x₂^b]};

end if

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∂X

^∆2

= {(x

₁

, x

2

) ∈ ∂X × ∂X − ∆(∂X ) | f (x

1

) = f (x

2

)}/ '

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[x

1

] 6= [x

₂

]

[x

₁

] = [x

₂

]

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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 / ' .

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BF = {(x

₁

, x

₂

) ∈ ∂X × S | f (x

₁

) = f (x

₂

)}

(71)

[x

1

] 6= [x

₂

]

[x

₁

] = [x

₂

]