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On Brouwer’s proof of invariance ofdimension

Bachelor Thesis

Giuliano Basso

Monday 7th October, 2013

Advisor: Prof. Dr. Michael Eichmair

Department of Mathematics, ETH Zurich

Abstract

In this thesis we give an elementary proof of the Invariance of Dimen-sion Theorem, i.e. the statement that for distinct positive integers nand m no non-empty open subset of Rn is homeomorphic to an opensubset of Rm. We closely follow Brouwer’s paper [2, p.161-165] andwork out his geometric lines of thoughts. Throughout the thesis we tryto stay as self-contained as possible, building up the prerequisite mate-rial from convex analysis and introduce some useful notions in orderto reformalize Brouwer’s arguments.

i

Acknowledgement

I would like to thank my advisor Michael Eichmair for suggesting thetopic of this thesis and for guiding me through it. Especially, I amdeeply grateful for our fruitful discussions, which I truly enjoyed; forhis willingness to share his experience; and last but not least for hisconstructive criticism, which was of great benefit to me.

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Contents

Contents iii

1 Triangulations 11.1 Basic properties of simplices . . . . . . . . . . . . . . . . . . . 11.2 Standard triangulation of the n-dimensional unit cube . . . . 7

2 Invariance of dimension 132.1 Fundamental Lemma . . . . . . . . . . . . . . . . . . . . . . . . 14

2.1.1 Linear regularization . . . . . . . . . . . . . . . . . . . . 142.1.2 Non collapsed sets and points . . . . . . . . . . . . . . 162.1.3 Degree modulo 2 . . . . . . . . . . . . . . . . . . . . . . 192.1.4 Proof of the Fundamental Lemma . . . . . . . . . . . . 25

2.2 Proof of the Invariance of Dimension Theorem . . . . . . . . . 26

A Auxiliary results 29A.1 Various facts from topology and analysis . . . . . . . . . . . . 29A.2 Technical Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Bibliography 38

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Chapter 1

Triangulations

The goal of this chapter is to describe a canonical way to divide the n-dimensional unit cube into nice subsets that do only intersect, if at all, alongtheir boundaries. We introduce the notion of a simplex and explain some ofits basic properties. Simplices turn out to be easy to handle; therefore, weuse them as building blocks for our subdivision of the n-dimensional unitcube. Such a subdivision into simplices is called a triangulation.

1.1 Basic properties of simplices

This is an introductory section setting up the basic definitions. We need todefine the notions of a convex hull, simplex, vertex order and the sign of asimplex. All of these notions are essential. Without them we are not able toprove Brouwer’s Invariance of Dimension Theorem later on.

Definition 1.1 (Convex hull) Let A = {x1, . . . , xk} be a finite set of points inRn. The set

conv(A) :=

{k∑i=1

λixi : λi > 0,k∑i=1

λi = 1

}(1.1)

is called the convex hull of A.

Recall that a set C ⊂ Rn is said to be convex if for all points x, y ∈ C andλ ∈ [0, 1] we have that λx + (1 − λ)y is contained in the set C. To begin, weshow that the convex hull of a finite set deserves its name and is actuallyconvex.

Lemma 1.2 The convex hull of a finite set A = {x1, . . . , xk} of points in Rn isconvex and compact.

Proof Let A = {x1, . . . , xk} denote an arbitrary finite subset of Rn. At first,we deal with the compactness of the set conv(A). We want to find a contin-uous map f and a compact set B such that conv(A) is the image of B under

1

1.1. Basic properties of simplices

f. The existence of such a map f implies immediately the compactness ofconv(A). This is well-known from analysis. We define the set

∆k :=

{(λ0, . . . , λk) ∈ Rk+1 :

k∑i=0

λi = 1, ∀i ∈ {0, 1, . . . , k} : λi > 0

}. (1.2)

In order to use ∆k as our set B we need to show that ∆k is compact. Let usdefine the map

g : Rk+1 → R(λ0, . . . , λk) 7→ λ0 + · · ·+ λk.

The continuity of g implies the closedness of the set g−1(1). Given thatthe half spaces H+

i = {(λ0, . . . , λk) ∈ Rk+1 : λi > 0} are closed for everyi ∈ {0, . . . , k} we obtain that

∆k = g−1(1) ∩ (

k⋂i=0

H+i ) (1.3)

as an intersection of closed sets is closed as well. For ∆k is bounded we mayconclude that it is compact. We define the map

f : ∆k → conv(A)

(λ0, . . . , λk) 7→ λ0x0 + · · ·+ λkxk.(1.4)

The continuity of f is immediate. The equality f(∆k) = conv(A) follows fromthe definition of conv(A). Thus, conv(A) is compact.

Now we deal with the convexity of conv(A). Note that the set ∆k as anintersection of convex sets, see Equation (1.3), is convex as well. The map fis the restriction of the affine linear map

f : Rk+1 → Rk

(λ0, . . . , λk) 7→ λ0x0 + · · ·+ λkxk(1.5)

to ∆k. We may invoke Lemma A.1 to conclude that conv(A) as the image ofa convex set under an affine linear map is convex, as desired. �

Sometimes it is useful to define the convex hull of a finite subset A of Rn as⋂B⊂Rn convex, closed

A⊂B

B. (1.6)

Using that the set conv(A) is convex and compact, see Lemma 1.2, we seethat (1.6) is a subset of conv(A). On the other hand, the Definition 1.1 of

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conv(A) and the fact that the set (1.6) is convex and contains A imply theopposite inclusion, i.e. conv(A) ⊂ (1.6). Hence the two definitions (1.1) and(1.6) of the convex hull of a finite set in Rn are equivalent.

Now we built up all the prerequisite material we need to define the k-dimensional simplex. We follow loosely [4, p. 1] and establish the followingdefinition of a simplex.

Definition 1.3 (k-dimensional simplex) The convex hull of the points x0, . . . , xkcontained in Rn is said to be a k-dimensional simplex, if x1 − x0, . . . , xk − x0 arelinearly independent. We refer to x0 as the base point. The set {x0, . . . , xk} is calledvertex set.

Clearly k cannot be strictly greater than n. If the dimension of a k-dimensionalsimplex S with vertices in Rn does not matter, we simply call S a simplex.

Lemma 1.4 (Independence of base point) Let S denote a k-dimensional sim-plex with vertex set {x0, . . . , xk} consisting of points of Rn. Then for every i in{0, . . . , k} the vectors x0 − xi, . . . , xi − xi, . . . , xk − xi are linearly independent.

Proof If k is equal to zero, there is nothing to prove. We may choose x1 asbase point. Assume there exist α0, α2, α3, . . . , αk ∈ R such that

α0(x0 − x1) + α2(x2 − x1) + · · ·+ αk(xk − x1) = 0. (1.7)

By adding zero we may convert Equation (1.7) into

α2(x2 − x0) + · · ·+ αk(xk − x0) + (−α0 − α2 − · · ·− αk)(x1 − x0) = 0. (1.8)

As x0 is the base point of the k-dimensional simplex S the coefficients in thesum (1.8) all vanish. So for every i in {2, . . . , k} the coefficient αi of xi − x0in Equation (1.8) is equal to zero. We conclude the proof by noticing that α0vanishes, for it is the only non zero constituent of the coefficient of x1 − x0in Equation (1.8). �

To state this in a different manner, Lemma 1.4 implies that the definitionof S is independent of the base point. Using the independence of the basepoint, i.e. Lemma 1.4, it is immediate that every subset with cardinality1 6 l 6 k+ 1 of the vertex set of a k-dimensional simplex is again the vertexset of an (l − 1)-dimensional simplex. This (l − 1)-dimensional simplexis called an (l − 1)-face of the k-dimensional simplex. We have alreadyseen in the proof of Lemma 1.2 that there is a natural relationship betweenpoints in the set ∆k, see Equation (1.11), and the points in the convex hullof a finite set consisting of k + 1 points. In the next lemma we prove thatthis correspondence is unique provided that the finite set in question is ak-dimensional simplex.

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Lemma 1.5 (Barycentric coordinates) Let S = conv({x0, x1, . . . , xk}) be a k-dimensional simplex with vertices in Rn. Then for every point x ∈ S there existunique λi > 0, where i ∈ {0, 1, . . . , k}, such that

x =

k∑i=0

λixi andk∑i=0

λi = 1. (1.9)

Proof Due to the fact that x is contained in S, existence is guaranteed. Nowwe deal with uniqueness. Assume we have two representations (1.9) for x.Then,

x =

k∑i=0

λixi =

k∑i=0

λ ′ixi ⇐⇒k∑i=1

(λi − λ′i)(xi − x0) = 0. (1.10)

In Equation (1.10) we used the fact that the coefficients λi and λ ′i sum uprespectively to one. Seeing that the points x1 − x0, . . . , xk − x0 are linearlyindependent, it follows that the coefficients λi and λ ′i have to coincide forevery i ∈ {1, . . . , k}. Furthermore, again on account of the fact that thecoefficients λi and λ ′i sum up respectively to one, the points λ ′0 and λ0 haveto coincide as well, thereby the lemma follows. �

To put it differently, Equation (1.9) implies that for every k-dimensional sim-plex S in Rn there exists an injective continuous map β, which we callbarycentric map, from the set ∆k to S. The assignment of the barycentricmap β is given by

(λ0, . . . , λk) 7→ λ1(x1 − x0) + · · ·+ λk(xk − x0) + x0. (1.11)

Note that β is affine linear. Since ∆k is compact and Rn is a Hausdorff space,the map β is even a homeomorphism between ∆k and S. For every point xin the simplex S we call the point β−1(x) ∈ ∆k the barycentric coordinates ofx.

The fact that the image of a convex set under an affine linear map is againconvex played a crucial role in the proof of Lemma 1.2. It is natural to askgiven an affine linear map whether the convex hull of the images of thevertices of a simplex is again a simplex of the same dimension.

Lemma 1.6 Suppose S is a k-dimensional simplex with vertices {x0, . . . , xk} in Rnand f : S→ Rn is the restriction of an injective affine linear map. Then the convexhull of the points f(x0), . . . , f(xk) is again a k-dimensional simplex.

Proof All we have to check is the linear independence of the points f(x1) −f(x0), . . . , f(xk) − f(x0). Let A ∈ Rn×n and b ∈ Rn such that f(x) = Ax + b.The injectivity of f implies that A is injective as well. Let α1, . . . , αk ∈ R begiven such that

α1(f(x1) − f(x0)) + · · ·+ αk(f(xk) − f(x0)) = 0. (1.12)

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The cancellation of the b’s in all of the terms f(xi)−f(x0) = Axi+b−(Ax0+b) = A(xi − x0) and the linearity of A make it possible to rewrite Equation(1.12) as follows

A(α1(x1 − x0) + · · ·+ αk(xk − x0)) = 0.

Since A is injective, we deduce that

α1(x1 − x0) + · · ·+ αk(xk − x0) = 0.

Now we can conclude the proof by noticing that α1 = . . . = αk = 0, for thepoints x1 − x0, . . . , xk − x0 are linearly independent. �

Since every k-dimensional simplex is homeomorphic to the set ∆k it de-serves a special name.

Definition 1.7 (Standard simplex) For every k > 0 we call the set ∆k, see Equa-tion (1.11), the k-dimensional standard simplex.

It would be nice if the k-dimensional standard simplex is actually a k-simplex.Indeed, this follows immediately from the previous Lemma 1.6, given thatthe barycentric map β is the restriction of an injective affine linear map. Thefollowing theorem is a standard result from convex analysis.

Theorem 1.8 (Caratheodory’s Theorem) Let A = {x0, . . . , xk} be a finite set ofpoints in Rn. Suppose that the convex hull of A is not a k-dimensional simplex.Then every point in conv(A) lies in an r-simplex with vertices in A, where 0 6 r <k.

Proof This proof closely follows the proof of Caratheodory’s Theorem asgiven in [9, p. 155].

To begin, we observe that k must be greater than or equal to 1, since onepoint sets are always 0-dimensional simplices. Let x ∈ conv(A) be arbitrary.Then, x can be written as:

x = λ0x0 + · · ·+ λkxk, wherek∑i=0

λk = 1, ∀i : λi > 0. (1.13)

Recall that we say that x can be expressed as a convex combination of k + 1points in A if Equation (1.13) holds. The points x1 − x0, . . . , xk − x0 arelinearly dependent, since conv(A) is not a k-dimensional simplex. Hencethere exist ξ1, . . . , ξk ∈ R, not all equal to zero, such that

ξ1(x1 − x0) + · · ·+ ξk(x1 − x0) = 0. (1.14)

We can multiply Equation (1.14) by ξ ∈ R and add it to Equation (1.13) toobtain:

x = (λ0−ξ · (ξ1+ · · ·+ξk))x0+(λ1+ξ ·ξ1)x1+ · · ·+(λk+ξ ·ξk)xk. (1.15)

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Define ξ0 = −ξ1 − · · ·− ξk. Since ξ was arbitrary we can choose

− ξ = min{λiξi

: i ∈ {0, . . . , k}, ξi 6= 0}. (1.16)

In fact, since the ξi’s are not all equal to zero, the minimum in Equation(1.16) is obtained. Hence there exists an i ∈ {0 . . . , k} such that we can write−ξ = λi

ξi. Observe that for all j ∈ {0, . . . , k} the number λj + ξ · ξj is non

negative andk∑j=0

λj + ξ · ξj = 1. (1.17)

In particular, λi + ξ · ξi = 0. We deduce that x can be expressed as a convexcombination of k points in A. If the convex hull of these points is a (k − 1)-dimensional simplex, we are done. Otherwise we repeat the constructionuntil we reach the point that the convex hull of the remaining points isa simplex. At worst, we are left with two points. If they are equal, weconclude that x is contained in a 0-dimensional simplex with vertices in A.If the two points are distinct, we see that x is contained in an 1-dimensionalsimplex with vertices in A. Thus, the lemma follows. �

Caratheodory’s Theorem plays a crucial role in many proofs of the next chap-ter and gives furthermore a connection between non collapsed points, seeDefinition 2.10, and non collapsed sets, see Definition 2.7. This connectionis stated in Lemma 2.12.

It turns out to be useful later on to be able to speak about the orientationof a simplex. The first step towards a proper setting for the orientation of asimplex is the following definition.

Definition 1.9 (Vertex order) Let A = {x0, . . . , xn} be a finite set in Rn. Every(n + 1)-tuple in the set {(xσ(0), . . . , xσ(n)) : σ ∈ Sn+1} is called vertex order ofthe finite set A.

The vertex order of a finite set A = {x0, . . . , xn} is uniquely determined byσ ∈ Sn+1. Brouwer uses the name indikatrix for something almost similarto what we call vertex order. See [1, p. 100] for the proper definition ofBrouwer’s notion of an indikatrix.

Definition 1.10 (Sign of a finite set) Suppose A = {x0, . . . , xn} is a finite setin Rn with vertex order (xσ(0), . . . , xσ(n)). The signum of the determinant ofthe matrix that has the vectors xσ(1) − xσ(0), . . . , xσ(n) − xσ(0) (in this order) ascolumns is called the sign of A. We denote this number symbolically by sgn(A).

The sign of A depends on the vertex order. Suppose we are given a setB = conv({x0, . . . , xn}) and a vertex order for the finite set A = {x0, . . . , xn},

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1.2. Standard triangulation of the n-dimensional unit cube

then we breach notation and call the number sgn(A) the sign of B. As-sume A happens to be an n-dimensional simplex. As the definition of Ais independent of the base point, see Lemma 1.4, we know that for everyσ ∈ Sn+1 the points xσ(1)−xσ(0), . . . , xσ(n)−xσ(0) are linearly independentand therefore their determinant is never equal to zero. Hence the sign ofevery n-dimensional simplex is always either −1 or 1.

1.2 Standard triangulation of the n-dimensional unit cube

In this subsection we construct a standard method to triangulate the n-dimensional unit cube, see Definition 1.15, into simplices. Furthermore, weeven canonically assign to each simplex of this triangulation a vertex orderand therefore also a sign. We denote by (Sn, ◦) the group of bijections of theset {1, . . . , n}. Recall that Sn acts linearly on Rn by the law

σ.(x1, . . . , xn) = (xσ(1), . . . , xσ(n)).

Symbolically we write Sn y Rn. This action serves as a tool to addressthe different simplices of our triangulation. In fact, every element of Sncorresponds to a simplex of the triangulation. We begin with the set thatcorresponds to the neutral element of the group (Sn, ◦).

Lemma 1.11 The set Sid = {(x1, . . . , xn) ∈ Rn : 0 6 x1 6 . . . 6 xn 6 1} is ann-dimensional simplex with vertex set Vid = {(0, . . . , 0), (0, . . . , 0, 1), . . . , (1, . . . , 1)}.

Proof To begin, we introduce a handy notation. We denote (0, . . . , 0) in Vidby v0 and for each i in the set {1, . . . , n} the point (0, . . . , 0, 1, . . . , 1) ∈ Vid,where the first i−1 coordinates are zero and all other coordinates are equal toone, by vi. The points v0, . . . , vn are clearly contained in Sid. Furthermore,

det

1 0 . . . 0...

. . . . . ....

.... . . 0

1 · · · · · · 1

= 1. (1.18)

Hence the points v1 − v0, . . . , vn − v0 are linearly independent. Now we areleft to show that convex hull of the points v0, . . . , vn is equal to Sid. Letx = (x1, . . . , xn) ∈ Sid be arbitrary. Define λn := x1 > 0 and for everyi ∈ {1, . . . , n − 1} define λi := xn−i+1 − xn−i. Note that λi > 0. Defineλ0 := 1 − λn − · · · − λ1 = 1 − xn. Note that λ0 > 0 and 1 = λ0 + · · · + λn. Itfollows that

x =

n∑i=0

λivi.

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Thus, we have shown that x ∈ conv(v0, . . . , vn). Let x, x ′ ∈ Sid be two dis-tinct points. Elementary estimates show that every point on the line segment[x, x ′] is contained in Sid. Therefore conv(v0, . . . , vn) ⊂ Sid, for Sid is convexand contains the points v0, . . . , vn. The Lemma follows. �

The fact that Sid is a simplex enables us to deduce that for every σ containedin Sn the set σ.Sid is again a simplex.

Corollary 1.12 For each σ ∈ Sn the set Sσ := {(x1, . . . , xn) ∈ Rn : 0 6xσ(1) 6 . . . 6 xσ(n) 6 1} is an n-dimensional simplex with vertex set Vid ={σ.v0, . . . , σ.vn}.

Proof As in the proof of Lemma 1.11 we have to show that the convex hullconv(σ.v0, . . . , σ.vn) is equal to Sσ. By definition Sσ = σ.Sid. Since the actionSn y Rn is linear, we may deduce, using the fact that in finite dimensionalvector spaces convexity is preserved under linear maps, that the set σ.Sid =Sσ is convex. Thus, we have that conv(σ.v0, . . . , σ.vn) ⊂ Sσ. Let x ∈ Sσ bearbitrary. The point y = σ−1.x has a representation as

y =

n∑i=0

λivi

where λi > 0,n∑i=1

λi = 1. Hence we can conclude the proof by observing

that

x =

n∑i=0

λiσ.vi �

Note that every vi ∈ Vid has a distinct number of ”1 “’s and that this numberis preserved under the action of every σ ∈ Sn. Hence we obtain for everytwo distinct σ, τ ∈ Sn that

σ.vi = τ.vj ⇒ i = j. (1.19)

Equation (1.19) is helpful in the proof of Lemma 1.19.

Lemma 1.13 The sign of the simplex Sσ with vertex type (σ.v0, . . . , σ.vn) is equalto sgn(σ).

Proof Let e1, . . . , en be the canonical basis of Rn. Define the n × n matrixPσ := [δσ(i)j]. It follows immediately that σ.x = Pσx, for every x ∈ Rn.Thus,

det(σ.v1| · · · |σ.vn) = detPσ · 1 = sgn(σ),

as the determinant is a multiplicative map and Equation (1.18) holds. OurLemma follows. �

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Note that Lemma 1.13 implies that the linear map σ.(·) preserves volumes.For later purposes we introduce the following convention.

Convention 1.14 Let σ be contained in the set Sn. By convention, we use thevertex order (σ.v0, . . . , σ.vn) for the simplex Sσ. Hence we canonically assumethat our simplices Sσ have signs equal to sgn(σ).

Sometimes it is handy to have a short term expression for the n-dimensionalunit cube.

Definition 1.15 (n-dimensional unit cube) The set In = [0, 1]n ⊂ Rn is calledthe n-dimensional unit cube.

Note that In = conv({0, 1}n). Let O(n) denote the orthogonal group, i.e. thegroup of distance-preserving endomorphisms of Rn. Recall that for a fixedbasis of Rn there is a bijective correspondence between the set O(n) and theset of orthogonal matrices with entries over R; therefore, we may work fromnow on with the set of orthogonal matrices, which we call O(n) again forconvenience. The group O(n) acts linearly on Rn by matrix multiplication,i.e. by the law

A.x = Ax,

where x is contained in Rn. The action O(n)yRn gives rise to a homomor-phism

ϕ : O(n)→ Aut(Rn)A 7→ (A. : x 7→ A.x).

(1.20)

Here Aut(Rn) denotes the group of automorphisms of Rn. Define the affineorthogonal group AO(n) as the semi direct product of Rn and O(n) withrespect to ϕ, which is given by Equation (1.20). In symbols: AO(n) :=RnoϕO(n). Every element contained in AO(n) is of the form (b,A). Weinterpret b in Rn as a translation and A in O(n) as a rotation. Note that thegroup AO(n) acts linearly on Rn by the law

(b,A).x = b+Ax (x ∈ Rn) (1.21)

We can use the action AO(n) y Rn, given by (1.21), to define the generalnotion of a cube.

Definition 1.16 (n-dimensional cube) Let (b,A) be contained inAO(n) = RnoϕO(n), and let r > 0 be a positive number. The set (b,A).(r · In) is called n-dimensional cube with edge lenght r, center b, and orientation A.

The next definition formalizes the idea of a triangulation from the beginningof this chapter. Definition 1.17 is merely a reformulation of Definition 2.2.1in [4, p. 54].

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Definition 1.17 (Triangulation) A triangulation of a finite set A ⊂ Rn is a setS consisting of n-dimensional simplices S with vertices in A such that:

• the union of these simplices is equal to conv(A)

• the intersection of each distinct pair of these simplices is either empty or acommon l-face, where l ∈ {0, . . . , n− 1}

Since In = conv({0, 1}n) we misuse Definition 1.17 and call a triangulation of{0, 1}n a triangulation of the n-dimensional unit cube In. In the followingLemma 1.19 we want to prove that our sets Sσ are in fact a triangulation ofIn. In order to prove this we need the notion of an extremal point, see [9, p.162].

Definition 1.18 (Extremal point) Suppose C is a convex set in Rn. A pointx ∈ C is called extremal if there exist no distinct points y, z ∈ C and 0 < t < 1such that

x = ty+ (1− t)z.

The fact that the set S consisting of all simplices Sσ, see Corollary 1.12 forthe definition, is a triangulation of In is the matter of the following Lemma1.19.

Lemma 1.19 For every τ, σ ∈ Sn the intersection Sτ ∩ Sσ is either empty or equalto a common l-face, where l ∈ {0, . . . , n}. Furthermore the union of all Sσ, whereσ ∈ Sn, is equal to In. In other words, the set S of simplices given by {Sσ : σ ∈ Sn}is a triangulation of the n-dimensional unit cube In.

Proof At first, we will show that every point x = (x1, . . . , xn) ∈ In is con-tained in some simplex Sτ. Let σ ∈ Sn be the permutation that arrangesx1, . . . xn in an increasing order, i.e

xσ−1(1) 6 . . . 6 xσ−1(n).

So, x is contained in Sσ−1 . Let τ, σ ∈ Sn be arbitrary. The set Sτ ∩ Sσ iscompact and convex. It is immediate that if x ∈ Sτ ∩ Sσ is extremal thenσ−1.x is also an extremal point of Sid. Since no other points of Sid thanv0, . . . , vn are extremal, it follows that the only extremal points of x ∈ Sτ∩Sσlie in the set {σ.vi, τ.vi}

ni=0. Therefore, using Equation (1.19), we conclude

that for every extremal point p of Sτ ∩ Sσ there exists an i is contained in{0, . . . , n} such that p = σ.vi = τ.vi. Define

I := {i ∈ {0, . . . , n} : σ.vi is an extremal point of Sσ ∩ Sτ}

Since σ.vi = τ.vi for every i ∈ I, we deduce that conv({σ.vi}i∈I) is equalto conv({τ.vi}i∈I). Thus, the set conv({σ.vi}i∈I) is a common (|I| − 1)-face ofthe simplices Sτ and Sσ. Given that every convex compact set is equal tothe convex hull of its extremal points, see Theorem (A.2), we obtain thatSτ ∩ Sσ = conv({σi}i∈I). We can conclude the result. �

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We call this triangulation S of In, where we use Convention 1.14 for thevertex orders, the standard triangulation with grid length one.

For every m > 1 we can write In as the union of mn smaller squares that doonly intersect, if at all, on their boundaries. Namely,

In =⋃

j∈{0,...,m−1}n

[j1m,j1 + 1

m

]× · · · ×

[jn

m,jn + 1

m

].

Define Inj := [ j1m ,j1+1m ] × · · · × [ jnm ,

jn+1m ]. The image of every simplex Sσ of

the standard triangulation under the injective affine linear map

ψj : In → Inj

x 7→ 1

mx+ j,

is again a simplex with verticesψj(σ.v0), . . . , ψj(σ.vn), see Lemma 1.6. There-fore the simplices {ψj(Sσ)}σ∈Sn triangulate the cube Inj . It is natural to use(ψj(σ.v0), . . . , ψ(σ.vn)) as the canonical vertex order of S, if S is the imageof the simplex Sσ. Thus, we have a canonical way to triangulate Inj . The setof simplices ⋃

j∈{0,...,m−1}n

{ψj(Sσ)

}σ∈Sn (1.22)

is a triangulation of the unit cube In. We denote the set (1.22) with ξ(m; In)in order to make its dependence on m visible.

Definition 1.20 For all m > 1 the set (1.22) denoted by ξ(m; In) is called stan-dard triangulation with grid length m. For every j in {0, . . . ,m−1}n we canonicallyuse the vertex order (ψj(σ.v0), . . . , ψj(σ.vn)) for ψj(Sσ) contained in ξ(m; In).

The (n − 1)-faces of the simplices of every standard triangulation ξ(m; In)are called the main faces of ξ(m; In). If a main face happens to lie entirely onthe topological boundary of In, we call it boundary face of ξ(m; In). For each0 6 l < n − 1 every l-face of every simplex of the standard triangulation iscalled main edge of ξ(m; In). We denote the set of main edges of ξ(m; In)by Fe. For the sake of clarity we omit to indicate the underlying standardtrianglation. The set of zero dimensional faces is called the set of vertices ofξ(m; In).

We denote the complement in In of the union of the main edges of thestandard triangulation ξ(m; In) by Θnm. Let m = 1. Even if we remove themain faces from Θn1 and denote this set by Θn1 , we see that

Θn1 =⊔σ∈Sn

σ.S◦id, (1.23)

11


since S = {Sσ : σ ∈ Sn} is a triangulation of In. Hence S◦id is a fundamentaldomain of the action Sn y Θn1 . The identity σ.S◦id = S◦σ even implies thateach S◦σ is a fundamental domain.

12

Chapter 2

Invariance of dimension

In the year 1911 L. E. J. Brouwer published in his paper ”Beweis der In-varianz der Dimensionenzahl” [2, p. 161-165] a strongly geometrically ori-entated proof of the fact that for every point in the image of an injectivecontinuous map, going from an non-empty open subset of Rn to Rn; andfor every open neighborhood in Rn of this point, there exists a non-emptyopen set that is contained completely in the intersection of the image andthe neighborhood. See Theorem 2.18 for the exact statement. Brouwer usesthis statement to provide an elementary proof of the sophisticated Theoremof Invariance of Dimension, which says that for distinct positive integers nand m no non-empty open subset of Rn is homeomorphic to an open subsetof Rm, see Theorem 2.19. Brouwer’s paper [2, p. 161-165] contains essen-tially the techniques of the mapping degree and simplicial approximation.At the time the paper was considered as clever but inaccessible; cf. [10, p.265]. Consult [6, p. 954 ff.] for a detailed historical overview.

The goal of this chapter is to work out carefully all steps of Brouwer’s proofin [2, p.161-165]. Basically, Brouwer approximates the injective continuousmap in question with maps that are defined with respect to the standardtriangulations we have constructed in the first chapter. Moreover, looselyspeaking Brouwer establishes a Lemma, which we call Fundamental Lemma2.1, which guarantees that a map ϕ : In → In having only small displace-ments cannot have an image that does not contain a n-dimensional cube,see Definition 1.16. In fact, setting up the prerequisite material we need inorder to prove the Fundamental Lemma 2.1 turns out to be the major partof this chapter. Having the Fundamental Lemma 2.1 on hand, the proof ofthe Invarianve of Dimension Theorem is found rather fast.

13

2.1. Fundamental Lemma

2.1 Fundamental Lemma

The goal of this section is to develop an adequate language which enables usto give a proof from first principles of the subsequent Fundamental Lemma2.1.

Lemma 2.1 (Fundamental Lemma) Suppose ϕ : In → In is continuous and thenumber

d := maxx∈In‖ϕ(x) − x‖Rn (2.1)

is strictly smaller than 12 . Then every cube Ke concentric to In with edge length

2e < 2(12 − d) is entirely contained in ϕ(In).

By construction, all cubes are homeomorphic to each other. Therefore, if wesubstitute the unit cube In in the Fundamental Lemma 2.1 with an arbitrarycube with edge length 2s, and the number 12 in the Fundamental Lemma 2.1with the number s, we can deduce that this adjusted Fundamental Lemma2.1 remains valid. Loosely speaking, the Fundamental Lemma is valid foran arbitrary cube.

Before diving into the proof of the Fundamental Lemma 2.1 we need a bunchof new concepts and notations. We develop them in the following subsec-tions and eventually come back to the Fundamental Lemma 2.1 and providea proof in the subsection 2.1.4. As everything we develop serves the ultimategoal to allow a proof of the Fundamental Lemma 2.1, it seems appropriateto hold some notions fixed. For the whole section we fix a cube Ke concen-tric to In with edge length 2e and a map ϕ : In → In; both with propertiesas stated in the Fundamental Lemma 2.1. This means in particular that ϕsuffices Equation (2.1) and 2e < 2(12 − d). Furthermore, we fix a standardtriangulation ξ(m; In) and an enumeration x1, . . . , xN of its vertices. NoteN = (m + 1)n. Recall that for every simplex S in the standard triangulationξ(m; In) we canonically have a (n + 1)-tupel k = (k0, . . . , kn) contained inthe set {1, . . . ,N}n+1 such that the simplex S has vertex order (xk0 , . . . , xkn).

2.1.1 Linear regularization

In this subsection we define the notion of a linear regularization. Roughlyspeaking, a linear regularization is a map from In to Rn that maps anysimplex of ξ(m; In) into the convex hull of points in Rn that have beenassociated to the vertices of the simplex.

Recall that we already fixed a standard triangulation ξ(m; In) and an enu-meration x1, . . . , xN of its vertices. For each point y = (y1, . . . , yN) in Rn·Nwe want to define a map σy : In → Rn. We define the map σy on each sim-plex of the fixed standard triangulation ξ(m; In) separately. The assignment

14


of the map σy restricted to the simplex S in ξ(m; In) is given by

x =

n∑i=0

λi · xki 7→n∑i=0

λi · yki , (2.2)

where x is a point of S and (k0, . . . , kn) ∈ {1, . . . ,N}n+1 is the (n + 1)-tupelsuch that the simplex S has vertex order (xk0 , . . . , xkn). Since ξ(m; In) is atriangulation of the unit cube In the map σy is well defined.

Definition 2.2 We call the map σy : In → Rn, given by Equation (2.2) on eachsimplex S, the linear regularization associated to the point y in Rn·N.

Note that the map

‖ · ‖? : Rn·N → R(y1, . . . , yN) 7→ max

i∈{1,...,N}‖yi‖Rn

(2.3)

is a norm on Rn·N. We use this norm to make the notation less unwieldy.

Lemma 2.3 The map σy is continuous. Furthermore, there exists for every simplexS contained in ξ(m; In) a constant CS > 0, such that for all u, v ∈ S:

‖σy(u) − σy(v)‖Rn 6 CS‖y‖|?‖u− v‖Rn . (2.4)

Proof Let the simplex S in ξ(m; In) be arbitrary. Let (xk0 , . . . , xkn) be thevertex order of the simplex S. Denote with β the barycentric coordinates ofS. The map

p : ∆n → Rn

(λ0, . . . , λn) 7→n∑j=0

λjykj

is a polynomial in each component; thus, the map p is in particular con-tinuous. Hence on the simplex S the map σy = p ◦ β−1 is continuous, asa composition of continuous maps. Since all simplices S contained in thestandard triangulation ξ(m; In) are closed and their union is equal to In, wecan apply the Pasting Lemma, see A.11, and deduce that σy is continuous.

Now we prove the estimate (2.4). As S is a simplex, we know that the pointsxk1 − xk0 . . . , xkn − xk0 are linearly independent and form a basis B of Rn.Let ‖ · ‖1 be the 1-norm with respect to the basis B. Recall that on the spaceRn all norms are equivalent, for this reason there exists a constant CS2 > 0

such that ‖ · ‖1 6 CS2 ‖ · ‖Rn . Now let the points u, v in the simplex S be

15


arbitrary. Denote with λ0, . . . , λn the barycentric coordinates of u and withλ′0, . . . , λ

′n the barycentric coordinates of v. We can conclude

‖σy(u) − σy(v)‖Rn = ‖n∑j=1

(λj − λ′j)(ykj − yk0)‖Rn

6 2 · ‖y‖?n∑j=1

|λj − λ′j| 6 ‖y‖?CS‖u− v‖Rn ,

(2.5)

since the coordinates of u− v with respect to B are (λ1 − λ′1, . . . , λn − λ′n).�

2.1.2 Non collapsed sets and points

From now on we fix an enumeration S1, . . . , Sn!·mn of the simplices of thestandard triangulation ξ(m; In). Recall that we have already fixed an enu-meration x1, . . . , xN of the set of vertices of ξ(m; In). Let ki = (ki0, . . . , k

in)

in {1, . . . ,N}n+1 be the (n+1)-tupel such that the ith simplex of the enumer-ation has the vertex order (xki0 , . . . , xkin).

For each i contained in the set {1, . . . , n! · mn} let dki : Rn·N → R be themap that assigns to each point y = (y1, . . . , yN) in the set Rn·N the deter-minant of the matrix that has the points yki1 − yki0 , . . . , ykin − yki0

(in thisorder) as columns. Note that the real number dki (y1, . . . , yN) is exactlyequal to zero if the points yki1 − yki0 , . . . , ykin − yki0

are linearly dependent.So, if dki (y1, . . . , yN) is equal to zero, then the convex hull of the pointsyki0

, . . . , ykin is not an n-dimensional simplex.

Definition 2.4 We call

Mξ = Rn·N \

n!·mn⋃i=1

d−1ki

(0)

the set of non collapsed maps associated to ξ(m; In).

Lemma 2.5 The set Mξ of non collapsed maps associated to ξ(m; In) is open anddense in Rn·N.

Proof Let the simplex S in ξ(m; In) and the point y contained in Rn·N be ar-bitrary. Furthermore, let k = (k0, . . . , kn) contained in the set {1, . . . ,N}n+1

be the (n+1)-tupel such that the simplex S has the vertex order (xk0 , . . . , xkn).Clearly, the set d−1k ((0,∞)) is non-empty. We take a point z in the non-emptyset d−1k ((0,∞)). Consider the map

γ : [0, 1]→ Rt 7→ dk(ty+ (1− t)z).

16


The map γ is a polynomial and on account of γ(0) 6= 0 we may deduce thatγ is not the zero polynomial. But, a non zero polynomial has only finitelymany zeros, hence for each open neighborhood Uy of y there must existpoints that are contained in the intersection of the set Rn·N \ d−1k (0) andthe set Uy. This is immediate, since the intersection W of the line segment[z, y] and the open neighborhood Uy of the point y is an uncountable set;therefore, not all the images of the points contained in the intersection Wunder the map dk can be equal to zero, for the map γ has only finitely manyzeros. We have shown that the open set Rn·N \ d−1k (0) is dense in Rn·N.Consequently, using Lemma A.3, we can conclude the proof by noticingthat the set Mξ must be dense and open, as it is the finite intersection ofdense and open sets. �

Definition 2.6 Let S be a simplex in ξ(m; In) and y be a point in Rn·N, whereN is as usual the number of vertices of the standard triangulation ξ(m; In). Weuse the notation Sy for the set σy(S). We call Sy the image of S under the linearregularization σy.

By construction, the image of S under the linear regularization σy, i.e. theset Sy, is equal to the convex hull of the points yk0 , . . . , ykn . Recall that(xk0 , . . . , xkn) is the vertex order of the simplex S ∈ ξ(m; In). If the point yhappens to lie in the set Mξ of non collapsed maps associated to ξ(m; In),the sets Sy, where the simplex S is contained in standard triangulationξ(m; In), are n-dimensional simplices with vertex sets VSy = {yk0 , . . . , ykn}.The fact that for every S in ξ(m; In) the image Sy is an n-dimensional sim-plex if y is contained inMξ, is why we callMξ the set of non collapsed mapsassociated to ξ(m; In). Consequently, if the set Sy is not an n-dimensionalsimplex, we call it collapsed. Note that this can only happen if y /∈Mξ.

Definition 2.7 (Non collapsed set) Let y be a point in Rn·N and S be a simplexof the standard triangulation ξ(m; In). The set Sy is called non collapsed if it is ann-dimensional simplex.

It is natural that if the simplex S in ξ(m; In) has vertex order (xk0 , . . . , xkn),to calculate the sign of Sy with respect to the vertex order (yk0 , . . . , ykn).Recall that the sign of Sy is equal to the signum of the determinant of thematrix that has the points yk1−yk0 , . . . , ykn−yk0 (in this order) as columns,see Definition 1.10. In the proof of Lemma 2.14 we need the notion of asso-ciated sets.

Definition 2.8 (Associated sets) Let y, z be points in Rn·N and Si, Sj be sim-plices in ξ(m; In). We call the sets Syi and Szj associated if i is equal to j.

As usual let Fe denote the set of main edges of the standard triangulation

17


ξ(m; In). Let y be a point in Rn·N. We define the set

µy =

(Rn \

⋃F∈Fe

σy(F)

)⋂K◦e. (2.6)

The set µy consists of those points in K◦e that are not contained in the imageof a main edge.

Lemma 2.9 The set µy is open and connected in the subspace topology of K◦e.

Proof Every main edge is a closed subset of a plane of codimension strictlygreater than one. The notion of a plane is defined in Definition A.4. Theimage of this plane under the map σy is again a plane of codimension strictlygreater than one. Let F denote the complement of the union of the imagesof theses planes under the map σy. The set K◦e is convex and not a measurezero set. By considering the first part of the proof of Lemma A.6 closely wesee that all we need to make the proof work is that Rn is convex and not ameasure zero set; as a result, Lemma A.6 can be adapted to conclude thatK◦e intersected with F is path connected. Furthermore, we see that the setK◦e ∩ F is open and dense in the subspace topology of K◦e. Recall that theset µy is the complement in K◦e of the image of the union of the main edgesunder the map σy. We obtain the set µy by adding open sets to the set F∩K◦e.Hence µy is open and since F ∩ K◦e ⊂ µy ⊂ Ke = F ∩ K◦e, we deduce that µyis connected, as it is contained in closure of a connected set. �

Definition 2.10 (Non collapsed point) Let y be a point in Rn·N and the pointz be contained in µy. The point z is called non collapsed if the set σ−1y (z) containsno points that lie on a main face of the standard triangulation ξ(m; In).

Conversely, if a point z in µy happens to lie in the image of a main faceunder the map σy, the point z is said to be collapsed. Observe that everymain face of ξ(m; In) is a set of measure zero in In, as it is contained in an(n− 1)-dimensional hyperplane and all planes of codimension greater thanor equal to one have measure zero, see Lemma A.5. The map σy is Lipschitzcontinuous on each simplex of ξ(m; In), see Lemma 2.3, therefore the imageof every main face is also a set of measure zero. Hence the union of theseimages is also of measure zero. Due to the fact that K◦e has a non zeromeasure, there must exist points in K◦e that are not contained in the unionof the images of the main faces. To put this differently, the subset of µy thatconsists of the non collapsed points is non-empty. For later references westate this as a lemma.

Lemma 2.11 For all y ∈ Rn·N the subset of non collapsed points in µy is non-empty.

18


In the next lemma we state the crucial observation that every non collapsedpoint in µy cannot be contained in a collapsed image set of a simplex undera linear regularization.

Lemma 2.12 Suppose y ∈ Rn·N and let z ∈ µy be a non collapsed point. Further-more, let S ∈ ξ(m, In) be arbitrary. If z ∈ Sy, then the set Sy is non collapsed.

Proof Assume, for the sake of a contradiction, that the point z in µy iscontained in a collapsed image simplex Sy with vertex order (yk0 , . . . , ykn).On one hand, since z is non collapsed, we know that in every expression

z =

n∑i=0

λiyki , wheren∑i=0

λi = 1, (2.7)

all the λi’s must be positive. Otherwise z would be a collapsed point. Onthe other hand, on account of Theorem 1.8, we can express z as

z =

n∑i=0

λ′iyki , (2.8)

where the λ′i’s sum up to one, as desired, but at least one of them is equal tozero. This is possible due to the fact that Sy is collapsed and therefore notan n-dimensional simplex. This contradiction concludes the proof. �

2.1.3 Degree modulo 2

We have seen in Lemma 2.11 that for each y in Rn·N the set µy containsnon collapsed points. Furthermore, every non collapsed point in µy is onlycontained, if at all, in those images Sy that happen to be n−dimensionalsimplices. Recall that we canonically assign to each Sy a sign, induced bythe sign of the simplex S. Non collapsed points are therefore somewhatspecial for they always exist and are only covered, if at all, by those sets ofthe Sy’s that are simplices. These observations serve as motivation for thenext definition.

Definition 2.13 We define for every point y in Rn·N and every non collapsed pointz contained in µy the non negative integers p+(z), p−(z). Denote by p+(z) > 0the number of those sets of Sy1 , . . . , S

yn!·mn that are positively signed simplices and

contain z. Analogously, let p−(z) > 0 denote the number of these sets that containz and are negatively signed.

Note that the non negative integers p+(z), p−(z) depend a priori on thenon collapsed z in µy and on y in Rn·N as well. In the progress of thissubsection we see that if y ∈ Rn·N lies in a certain subset of Rn·N and z is anon collapsed point in µy, the number

p+(z) + p−(z) mod(2) (2.9)

19


does not depend on the choice of z. Hence for every y contained in a certainsubset of Rn·N Equation (2.9) produces a unique number, which we call n(y).Taking this for granted, observe that we constructed a map

n: dom(n)→ {0, 1}

y 7→ n(y).(2.10)

We call the number n(y) degree modulo 2 of the point y in dom(n) ⊂ Rn·N.

In the remaining part of this subsection we deal with the question whetherthe degree modulo 2 is well defined and define the set dom(n) properly. First,we start with the construction of dom(n). Recall that the cube Ke has edgelength 2e strictly smaller than 1. Moreover, we have fixed a triangulationξ(m; In) and an enumeration x1, . . . , xN of the elements of the set of vertices.We need that the number n(y) is independent of the non collapsed point inµy, we use for its computation. It turns out to be useful to restrict ourselvesto points y that are near to the vertices of ξ(m; In). Let y be a point inRn·N such that each component of y has distance smaller than 1

2 − e fromthe corresponding component of x = (x1, . . . , xN). Such a point y is nearto the point x with respect to the norm ‖ · ‖?. Furthermore, note that forsuch a point y the images of the boundary faces, i.e. main faces lying onthe topological boundary of In, under the map σy do not intersect with thecube Ke. Indeed this property suffices to characterize dom(n). We define

dom(n) :=

{y = (y1, . . . , yN) ∈ Rn·N : ∀i ∈ {1, . . . ,N} : ‖yi − xi‖ <

1

2− e

}.

The set dom(n) is equal to the ‖ · ‖?-ball around x with radius 12 − e. Notethat dom(n) depends on the set of vertices (and its enumeration) of ξ(m; In).Nevertheless, we omit this dependence in the notation for the sake of clarity.

Next, we deal with the well-definedness of the degree modulo 2. We dothis in two independent steps. At first, we fix a non collapsed point z in µyund observe that for arbitrary close neighborhoods of y we find a point y′

contained in it such that the point z remains non collapsed with respect toy′ and even the numbers (2.11) of y and y′, both calculated with respect toz, coincide.

Lemma 2.14 Suppose the point y is contained in dom(n) and let z in µy be noncollapsed. Then there exists an ε > 0 such that the set B‖·‖?ε (y) ∩ dom(n) ∩Mξ isnon-empty and if we calculate the number

p+(z) + p−(z) mod(2), (2.11)

once with respect to y and then with respect to some point y′ contained in the setB‖·‖?ε (y) ∩ dom(n) ∩Mξ, then these two numbers are well defined and coincide.

20


Proof This proof is merely an application of the Technical Lemma, seeLemma A.16. Let y in dom(n) and the non collapsed z in µy be arbitrary.Let Fy denote the image of an arbitrary main face of ξ(m; In) under thelinear regularization σy. Let F be the set consisting of all Fy. Due to thefact that all elements of F are compact and the fact that the point z is noncollapsed we may deduce that the number

ε := minFy∈F

dist(z, Fy) (2.12)

is strictly greater than zero, since the non collapsed z cannot lie in a col-lapsed set Sy, see Lemma 2.12, and by the compactness of all the elementsof the set F this is the only possibility for the distances in Equation (2.12) toget zero. Here we have used Theorem 1.8, which implies that a collapsedsimplex is equal to one of its main faces. Using Lemma A.10 we can deducethat the set Mξ ∩ dom(n) is dense in dom(n) with respect to the subspacetopology induced by the standard topology. Since on Rn·N all norms areequivalent, there exists at least one y′ contained in Mξ ∩ dom(n) such that‖y − y′‖? < ε. Therefore we may choose a y′ in B‖·‖?ε (y) ∩ dom(n) ∩Mξ

arbitrarily.

Now we show that z is contained in µy′ and furthermore that z is non col-lapsed. Thus, the number (2.11) calculated with respect to y′ is well defined.Choose S in ξ(m; In) arbitrarily. Let Sy

′be the associated set to Sy, see Def-

inition 2.8. If we can show that z cannot be contained on the boundary ofSy′, we have shown that z ∈ µy′ and that it is non collapsed. This is the case

since if z is collapsed, it is contained in the image of a main face of ξ(m; In)under σy′ , and therefore also on the boundary of some Sy

′. We make a dis-

tinction of cases on the sign of the set Sy, which can either be zero or nonzero.

At first, we assume that the sign of Sy is equal to zero. Note that the setsA := VSy and A′ := VSy′ satisfy the hypotheses of the Lemma A.17 forε. Additionaly, we know by the construction of ε that the point z has adistance greater than or equal to ε from every main face of Sy. Hence wecan apply Lemma A.17 and deduce that z is neither contained in Sy, whichwas evident before, and more importantly that the point z is not containedin Sy

′. Hence in particular the point z is not contained in the image of a

main face of S under σy′ .

Next, let the sign of Sy be non zero. Note that the sets Sy and Sy′

satisfy thehypotheses of the Technical Lemma A.16 for ε. As before, we know fromthe construction of ε that the point z has a distance greater than or equalto ε from every main face of Sy. Using both conclusions of the TechnicalLemma we see that the point z is, if at all, only contained in the interiorof the simplex Sy

′and not on its boundary. Hence z is non collapsed with

respect to σy′ .

21


Now we are left to show that if we calculate the number (2.11) with regardto y and y′ respectively the result remains the same. If Sy is collapsed, weknow already that z can neither be contained in Sy nor in its associated setSy′. Thus, we only have to consider the pairs (Sy, Sy

′), where Sy is non

collapsed. Let v be an arbitrary vertex of a non collapsed set Sy, so thereexists exactly one main face Fv of Sy that does not contain v. Note that Fv iscontained in a (n − 1)-dimensional plane of Rn. See Definition A.4 for thenotion of a plane. Let the point v′ contained in the vertex set of Sy

′denote

the associated vertex to v. By construction of ε the points v and v′ lie onthe same side of Fv. So since v was arbitrary, we deduce that if Sy is noncollapsed, then Sy and its associated set Sy

′have the same sign. Now we

may conclude the result, for the Technical Lemma guarantees us that thepoint z is either contained in both of the sets Sy, Sy

′or in none of them. �

Lemma 2.15 Let y be a point in dom(n) ∩Mξ. Then for all z ∈ µy, which arenon collapsed, the number given by

p+(z) + p−(z) mod(2) (2.13)

is constant.

Proof Let z1, z2 in µy be two non collapsed points. Lemma A.8 and Lemma2.9 imply that z1, z2 can be connected by a piecewise linear continuous mapγ : [0, 1] → µy, having only finitely many break points. By the use of molli-fiers we can ”smooth out” the break points of γ; therefore, we may assumethat γ is smooth. Furthermore, we are able to redefine γ : [0, 1] → µy, ifnecessary, in a way such that the endpoints stay unchanged and the traceof γ : [0, 1] → µy does only contain finitely many points that lie on imagesof main faces. The existence of a redefined map γ′ : [0, 1] → µy with theseproperties is an immediate consequence of the Transversality Theorem, seeTheorem A.15. Choose 3ε > 0 to be strictly smaller than the distance be-tween the trace of γ and the union of the images of the main edges underσy; and the distance of the trace of γ and the boundary of the cube Ke. Weset (t, s) ∈ [0, 1]×Bε(0) 7→ F(t, s) = γ(t)+s. Clearly: F(t, s) ∈ µy for all (t, s).Let Z′ denote the image of an arbitrary main face F under the map σy. Theset Z′ minus the closure of the ε neighborhood of the union of the images ofthe main edges that lie on the main face F is a boundaryless submanifold ofRn of dimension (n − 1). Let us denote this new set by Z. Now, F and therestriction of F to the boundary of [0, 1]×Bε(0) are smooth and transversal toZ. See Definition A.14 for the notion of transversality. Hence we can applythe Transversality Theorem A.15 to deduce that for almost every s in Bε(0)the map t 7→ γs(t) = γ(t) + s is transversal to Z. We see by construction ofγ and as a result of the Transversality Theorem that the trace of the map γsdoes only contain finitely many images of the main face F under the mapσy. Since this holds for almost every s in Bε(0) and on account of the fact

22


that we have chosen the main face F arbitrarily, we may choose an s in Bε(0)such that the trace of map t 7→ γs(t) = γ(t) + s does only contain finitelymany points that lie on images of main faces. Let γ′ : [0, 1] → µy be thecontinuous map whose trace is equal to the union of the trace of γs andthe line segment [z1, γs(0)] and the line segment [γs(1), z2]. We see that themap γ′ is the desired redefinition of the map γ. Hence, we may assumefrom now on that the trace of the map γ does only contain finitely manypoints that lie on images of main faces. The linear regularization σy doesnot map the boundary faces of the standard triangulation ξ(m; In) into thecube Ke. Hence, the trace of γ does not contain points that lie on imagesof boundary faces and by definition of the set µy neither points that lie onimages of main edges.

Let t1 < · · · < tm in [0, 1] be the preimages of those points in the traceof γ that lie on images of main faces. Note that the number (2.13) is notdefined for the points γ(t1), . . . , γ(tm). Define t0 := 0 and tm+1 := 1.Clearly t1 6= t0, tm 6= tm+1 and the number (2.13) is constant on the setsγ([t0, t1)), γ((t1, t2)), . . . , γ((tm−1, tm)), γ((tm, tm+1]), for (2.13) can onlychange along γ if the image of a main face is passed.

Since the point y is contained in the setMξ of non collapsed maps associatedto the standard triangulation ξ(m; In), the image of every main face, that isnot a boundary face, is the common face of two distinct simplices Syi , S

yj

and lies in a unique (n − 1)-dimensional hyperplane of Rn. The notion ofa (n − 1)-dimensional hyperplane is defined in Definition A.4. Let i be anarbitrary element of the set {1, . . . ,m}. The point γ(ti) is contained in k > 1many images F1, . . . , Fk of main faces. Denote with Hl the unique (n − 1)-dimensional hyperplane that contains Fl. For each l in the set {1, . . . , k} wehave that the set Rn\Hl has two open connected components H+

l , H−l . Thus,

the trace γ((ti−1, ti)) is either completely contained in H+l or H−

l , for everyl ∈ {1, . . . , k}. The same applies for the trace of γ((ti, ti+1). Let Sl1, S

l2 ∈

{Sy1 , . . . , S

yn!·mn} be the unique simplices that have the face Fl in common.

Note that the elements of C := {Sl1, Sl2}l∈{1,...,k} are pairwise distinct.

Let l ∈ {1, . . . , k} be arbitrary. We have the following two possibilities:

1. the interior of the simplices Sl1, Sl2 lie on the same side of Hl or

2. the interior of the simplices Sl1, Sl2 lie on different sides of Hl.

To conclude the result we are obligated to conduct an utterly boring distinc-tion of cases. Before we begin with this distinction of cases, note that foreach set Sy not contained in C = {Sl1, S

l2}l∈{1,...,k} we have

γ((ti−1, ti)) ∈ Sy ⇐⇒ γ((ti, ti+1)) ∈ Sy, (2.14)

since by definition of C no face of Sy can be touched or crossed on the linesegment [ti−1, ti+1].

23


Assume now that the first case is valid and that the traces γ((ti−1, ti)) andγ((ti, ti+1)) are both contained in the same connected component of Rn\Hl.The fact that Hl is a (n− 1)-dimensional hyperplane implies

γ((ti−1, ti)) ∈ (Sl1 ∩ Sl2)⇐⇒ γ((ti, ti+1)) ∈ (Sl1 ∩ Sl2). (2.15)

If the traces γ((ti−1, ti)) and γ((ti, ti+1)) are contained in different con-nected components of Rn \Hl, we have the equivalence

γ((ti−1, ti)) /∈ (Sl1 ∩ Sl2)⇐⇒ γ((ti, ti+1)) ∈ (Sl1 ∩ Sl2). (2.16)

Next, suppose we are in the second case and assume that the traces γ((ti−1, ti))and γ((ti, ti+1)) are both contained in the same connected component ofRn \Hl. We have for every m ∈ {1, 2}

γ((ti−1, ti)) ∈ Slm ⇐⇒ γ((ti, ti+1)) ∈ Slm. (2.17)

And finally, if γ((ti−1, ti)) and γ((ti, ti+1)) are contained in different con-nected components of Rn \Hl, we have for every m ∈ {1, 2} the equivalence

γ((ti−1, ti)) ∈ Slm ⇐⇒ γ((ti, ti+1)) /∈ Slm. (2.18)

Since l contained in {1, . . . , k} was arbitrary, the Equations (2.14), (2.15),(2.16), (2.17), (2.18) imply that the numbers p+(u)+p−(v) and p+(u)+p−(u)for the points u in γ((ti−1, ti)) and v in γ((ti, ti+1)) can only differ by a mul-tiple of 2. Hence we obtain that (2.13) is constant on γ((ti−1, ti+1)) \ γ(ti).We have chosen i contained in {0, . . . ,m} arbitrarily; thus, we can concludethat (2.13) is constant on γ((t0, tm+1)). Since the number (2.13) is constanton γ([t0, t1)) and γ((tm, tm+1]), we may obtain that the numbers (2.13) ofthe two points z1, z2 are equal, as desired. �

As a result of Lemma 2.15 the degree modulo 2 of all points contained inthe intersection of Mξ and dom(n) is well defined. By consulting Lemma2.14 we know that every point y of dom(n) has a ‖ · ‖?-neighborhood onwhich the numbers (2.11) of all points that are contained additionally in theset Mξ coincide with the number (2.11) of the point y, if they are calculatedboth with respect to the same non collapsed point. To put this differently,for each non collapsed z in µy the number (2.11) of y is equal to the degreemod 2 of the points contained in the intersection of the ‖ · ‖?-neighborhoodof y and the set Mξ. Hence two non collapsed points z1 and z2 in µythat lead to different numbers (2.11) cannot exist, for the density of Mξ

in dom(n), consider Lemma 2.5 and Lemma A.10; the well definedness ofthe degree modulo 2 of all points contained in the intersection of Mξ anddom(n), see Lemma 2.15; and the consequences of Lemma 2.14, as mentionedin the previous two sentences; imply that there exists a point in Mξ∩dom(n)having degree modulo two equal to both the numbers (2.11) of z1 and z2.Hence we have proven the subsequent Lemma.

24


Lemma 2.16 The degree modulo 2 given by Equation (2.10) is a well defined map.

Now we are able to show that the map n is indeed constant.

Lemma 2.17 The map n: dom(n)→ {0, 1} is a constant function with value 1.

Proof The point x = (x1, . . . , xn) is trivially contained in dom(n). To begin,we show that n(x) is equal to 1. Due to the fact that ξ(m; In) is a triangu-lation of the unit cube and since the linear regularization σx is the identitymap from In to In we can deduce that a non collapsed point of µx is con-tained in exactly one of the simplices Sx1, . . . , S

xn!·mn . Thus, n(x) = 1.

Let x′ be an arbitrary point in dom(n). The set dom(n), being a ‖ · ‖?-ballwith a certain radius centered at the point x, is convex. So the line segment[x, x′] is entirely contained in the set dom(n). For every point v on the linesegment [x, x′] there exists an εv > 0, see Lemma 2.14, such that for all w inB‖·‖?ε (v)∩dom(n)∩Mξ the well defined numbers, see Lemma 2.16, n(v) and

n(w) coincide. By shrinking εv, if necessary, we may assume that the ‖ · ‖?-ball centered at v with radius εv is entirely contained in dom(n). Now wecover the line segment [x, x′] with these ‖ · ‖?-balls. The Lebesgue NumberLemma guarantees (cf. Lemma 27.5 in [8, p. 175-176]) the existence of aδ > 0 such that every subset of the compact normed space [x, x′] (endowedwith the norm ‖ · ‖?) with diameter less than δ is contained in some memberof the cover. Hence, using the fact that on finite dimensional vectorspacesnorms are equivalent, there exists a k > 1 such that for every j in {1, . . . , k}

the function n is constant on the set [x + (x − x′) j−1k , x′ jk ]. Since for every j

contained in {1, . . . , k−1} the sets [x+(x−x′) j−1k , x′ jk ] and [x+(x−x′) jk , x

′ j+1k ]

intersect non trivially, we may conclude n(x) = n(x′). Hence n is a constantfunction with value 1, as desired. �

Lemma 2.17 is a really strong result, as it implies that the set Ke is containedin every image σy(In), where y in dom(n). This can be seen as follows. Thereare two types of points in the set µy, namely: collapsed points and noncollapsed points. By definition, collapsed points are contained in the imageof a main face under the map σy, therefore all collapsed points are containedin the set σy(In). On the other hand, since for all non collapsed points z inµy the number p+(z)+p−(z) is not equal to zero, the point z is contained ina simplex Sy and therefore in particular in the image set σy(In). We obtainthat µy ⊂ σy(In). Since µy = Ke, this implies immediately Ke ⊂ σy(In), forthe set σy(In) is closed.

2.1.4 Proof of the Fundamental Lemma

In this subsection we provide a proof of the Fundamental Lemma, stated inLemma 2.1.

25

2.2. Proof of the Invariance of Dimension Theorem

Proof Let Nm = (m + 1)n denote the number of vertices of the standardtriangulation ξ(m; In). For eachm > 1 fix an enumeration x1, . . . , xNm of theset of vertices of the standard triangulation ξ(m; In). For every m > 1 definethe point ym contained in the set Rn·Nm as ym = (ϕ(x1), . . . , ϕ(xNm)). Itfollows from the definition of ϕ, see Equation (2.1), that ym ∈ dom(n) forall m > 1. Note that ϕ is uniformly continuous, for the space In is compact.Hence we see that

σym ⇒ ϕ. (2.19)

The double arrow indicates that the convergence (2.19) is uniform. We knowthat the cube Ke is contained in every σym(I

n), since ym ∈ dom(n) for everym > 1 and the map n: dom(n) → {0, 1} is constantly equal to one. Assume,for the sake of a contradiction, that there exists a point v ∈ Ke that is notcontained in ϕ(In). Since ϕ(In) is compact and v /∈ ϕ(In), we deduce that

infu∈ϕ(In)

‖u− v‖Rn > 0. (2.20)

On account of the fact that Ke is contained in every set σym(In) and because

of the uniform convergence (2.19), we conclude that for m large enough,there must exist a point w ∈ In, such that the distance between the pointsv and ϕ(w) is strictly smaller than (2.20). This is a contradiction. Thus, allpoints in Ke are contained in ϕ(In), as desired. �

2.2 Proof of the Invariance of Dimension Theorem

Now we have everything on hand to prove Theorem 2.18 that allows us todeduce the Invariance of Dimension Theorem, see Theorem 2.19, as an imme-diate consequence. Let B denote a subset of Rn. Recall that a set A ⊂ B issaid to be nowhere dense in B, if for all non-empty open sets U contained inthe subspace topology of B the set U \ (A ∩ B) is non-empty. Equivalently, ifwe endow B with the subspace topology the set A ⊂ B is said to be nowheredense in B if the interior of the closure of the set A is empty.

Theorem 2.18 [2, p. 165] Let D ⊂ Rn be a non-empty open subset and let themap ϕ : D → Rn be injective and continuous. Then for every point in the setϕ(D) ⊂ Rn and every open set V that contains this point there exists a non-emptyopen set that is entirely contained in the set ϕ(D) ∩ V .

Note that Theorem 2.18 implies in particular that the image of an injectivecontinuous map ϕ which goes from an open subset of Rn to Rn is notnowhere dense in Rn. Moreover, note that Theorem 2.18 does not tell uswhether the set ϕ(D) is open or not. We only have for every point p in theset ϕ(D) and every open neighborhood V of this point the weaker statementthat the intersection of V and ϕ(D) does contain a non-empty open set B.But, there is no need for the set B to contain the point p. Nevertheless, the

26


stronger statement which claims that the set ϕ(D) is indeed open turns outto be true. This classical theorem, known in the literature as the Invarianceof Domain Theorem, is due to L. E. J. Brouwer; cf. [3, p. 305-313]. As we dontuse the full power of the Invariance of Domain Theorem we leave it withthis remark and omit to provide a proof here. Now we establish a proof ofthe weaker Theorem 2.18.

Proof This proof closely follows Brouwer’s proof of Theorem 1 in his paper[2, p. 165].

Let v0 be a point in the set ϕ(D) and let V be any open neighborhood ofthe point v0. Let u0 be the unique point in the set D such that ϕ(u0) = v0.Assume, for the sake of a contradiction, that for all non-empty open sets Bin Rn the set B \ V ∩ϕ(D) is non-empty. Denote with k ⊂ V a cube thatcontains v0. Let 2s denote the edge length of the cube k. Due to the factthat the map ϕ is continuous, we are able to find a cube K ⊂ D such thatthe point u0 is in K and the image of K under ϕ is entirely contained inthe interior of the cube k. Let 2e denote the edge length of the cube K. Byconstruction, the set ϕ(K) is nowhere dense in V . Since K is compact, Rn is aHausdorff space, and ϕ is an injective and continuous map, we may deducethat ϕ|K is a homeomorphism with its image ϕ(K). The compactness of theset ϕ(K) implies that for all δ > 0 there exists an ε > 0 such that for allϕ(u), ϕ(u′) ∈ ϕ(K)

‖ϕ(u) −ϕ(u′)‖Rn < ε⇒ ‖u− u′‖Rn < δ. (2.21)

Recall that e is half the edge length of the cube K. Choose 0 < δe < e

and εe > 0 accordingly such that Equation (2.21) holds. Fix m > 1 suchthat 2 · 2s·

√n

m < εe. We triangulate the cube k with the standard triangula-tion ξ(m;k). Let C be the set of simplices of ξ(m;k) that contain at least oneelement of ϕ(K). Clearly, C is non-empty. Let {x1, . . . , xM} be the set that con-tains all vertices of the simplices in C. Denote with Cxi , for all i ∈ {1, . . . ,M},the set that consists of all simplices in C that have xi as a vertex. Associateto x = (x1, . . . , xM) the point y = (y1, . . . , yM) ∈ KM, where for each i

contained in the set {1, . . . ,M} we have

ϕ(yi) ∈⋃S∈Cxi

S. (2.22)

As usual, let σy : ∪C∈CC→ Rn be the linear regularization associated to thepoint y. Note that Lemma 2.3 also holds for σy. Now we define the map

ψ : K→ K

x 7→ σy(ϕ(x)).(2.23)

Note that since for each i in {1, . . . ,M} the point yi is contained in K, themap ψ is well defined. As a composition of continuous maps the map ψ is

27


continuous. We have chosen m large enough such that:

∀v, v′ ∈⋃S∈Cxi

S : ‖v− v′‖ < εe. (2.24)

Therefore, using Equation (2.21) - (2.24), we obtain

‖x−ψ(x)‖Rn < e (∀x ∈ K). (2.25)

Since the Fundamental Lemma 2.1 holds for arbitrary n-dimensional cubes,there exists a cube Ke concentric to K that is contained in the image of ψ(K).But ϕ(K) is compact and nowhere dense in k and therefore the image ofϕ(K) under the continuous map σy is nowhere dense in K, see Lemma A.13.Hence we have obtained a contradiction. �

Finally, we state and prove the Invariance of Dimension Theorem.

Theorem 2.19 (Invariance of Dimension) Let n andm be two distinct positiveintegers and let U be a non-empty open subset of Rn and V be an open subset ofRm. Then there cannot exist a homeomorphism between U and V .

Proof This proof is basically a reformulation of the proof of Theorem 2 inBrouwer’s paper [2, p. 165]. We may assume n > m. Let U and V be asstated in the Theorem. Since U is open, we may choose a set C containedin U homeomorphic to a m-dimensional cube K. Let ψ : K → C denote thehomeomorphism between K and C. By construction, the set C is containedin a plane of codimension greater than or equal to one. Corollary A.7 impliesthat the set C is nowhere dense in U, as it is contained in a m-dimensionalplane. Assume that we are given a homeomorphism ϕ : U → V . The imageunder the map ϕ of the nowhere dense subset ψ(K◦) of U is nowhere densein V , this follows from the fact that the map ϕ is a homeomorphism, seeLemma A.13 for a complete proof. This is a contradiction to Theorem 2.18.Thus, as desired, there cannot exist such a homeomorphism ϕ. �

A few final remarks are in order. The proof of Theorem 2.19 is actuallya proof of the fact that for any two distinct positive integers n > m, anynon-empty open subset U of Rn and any open subset V of Rm there cannotexist a injective continuous map between U and V . The converse, i.e. theimpossibility of the existence of an injective continuous map between U andV for m > n, is not true. An easy counterexample is the inclusion mapx ∈ R → (x, 0) ∈ R2. But it can be shown that the image of an injectivecontinuous map going from an open subset U of Rn to an open subset V ofRm is nowhere dense in V , if n is strictly smaller than m; cf. Theorem 3 in[2, p. 165].

28

Appendix A

Auxiliary results

A.1 Various facts from topology and analysis

Lemma A.1 Suppose f : Rn → Rm is affine linear, i.e. f is of the form f(x) =Ax+ b, where A ∈ Rm×n and b ∈ Rm. Let C be a convex set. Then the image setf(C) is convex as well.

Proof Let x ′ = Ax + b and y ′ = Ay + b be two points in f(C). Since Cis convex, the line segment [x, y] is entirely contained in C. Hence we canconclude the proof by noticing that for all t in [0, 1] the points (1−t)x ′+ty ′ =A((1− t)x+ ty) + b are contained in f(C). �

Theorem A.2 (Krein-Milman, easy version) Suppose K ⊂ Rn is convex andcompact. Let E ⊂ K be the set of its extremal points, see Definition 1.18. Assume Eis finite. Then, conv(E) = K.

Proof We use a more general version of the Krein-Milman Theorem to con-clude our easy version. Let S denote the set of extremal points of a convexcompact set T , which is contained in a vector space over the reals. S. Langshows in Theorem 1.5 in his book [7, p. 88] that T is the smallest closedconvex set containing all the elements of S.

Note that we have seen in chapter one the equivalence of the Definitions 1.1and 1.6 of the convex hull of a finite set. Since the set of extremal pointsE is assumed to be finite, this implies that the Definitions 1.1 and 1.6 ofthe convex hull of E yield the same set. Therefore, putting things together,K = conv(E), as desired. �

Lemma A.3 The finite intersection of open and dense subsets of Rn is again openand dense.

Proof Since openness is preserved under finite intersections, we are left toshow that the finite intersection of open dense subsets of Rn is dense. We

29

A.1. Various facts from topology and analysis

do this by induction. Assume {Ui}Ni=1 is a finite set consisting of open and

dense subsets of Rn. Suppose that the intersection U of those sets is openand dense. Let UN+1 be an arbitrary open and dense subset of Rn. Wehave to show that U ∩ UN+1 is dense. Let x be a point in Rn and B anopen neighborhood of x. Due to the fact that U is dense, there exists a pointx ′ contained in the intersection of U and B. Since U and B are open, theirintersection is open as well and we may use the density of the set UN+1

to obtain a point x ′′ in UN+1 contained in the intersection of U and B. Byconstruction, x ′′ is contained both in U and UN+1 and additionally in theopen neighborhood B of x. Hence U ∩UN+1 is dense, as desired. �

Definition A.4 The preimage of zero of an affine linear map f : Rn → Rm withassignment f(x) = Ax+ b, where A has rank m, is called plane of codimension m.Besides, a plane of codimension of 1 is called (n− 1)-dimensional hyperplane.

Note that the rank assumption on A implies that m 6 n.

Lemma A.5 Every plane of codimension m > 1 is a set of measure zero.

Proof This is an immediate consequence of Sard’s Theorem; nonetheless,we give a proof from first principles. We may assume m < n. Otherwise theclaim follows immediately. Let H be a plane of codimension greater than orequal to one. We may write H = p +W, where p is a point in Rn and Wis a (n −m)-dimensional subspace of Rn. This is well-known from linearalgebra. Fix a basis w1, . . . , wn−m of W. Let Φ : Rn−m → W be the uniquelinear map such that for all i in {1, . . . , n−m}: Φ(ei) = wi. Here {ei}

n−mi=1 is

the standard basis of Rn−m. Let i : Rn−m → Rn denote the inclusion mapand Tp : Rn → Rn the translation by the vector p. By construction, we haveH = Tp ◦ i ◦Φ(Rn−m). Let x, y be points in Rn−m. It follows that

‖Tp◦i◦Φ(x)−Tp◦i◦Φ(y)‖Rn = ‖Φ(x)−Φ(y)‖W 6 ‖Φ‖·‖x−y‖Rn−m . (A.1)

Here ‖ · ‖ is the induced operator norm and ‖ · ‖W denotes the restriction ofthe standard norm of Rn toW. From now on we denote the map Tp◦i◦Φ byF. Let CR denote the closed cube centered at the origin with edge length 2R,i.e. CR = [−R, R]n−m. Fix k > 1. For every x in R · 2−k · {−2k, . . . , 0, . . . , 2k −1}n−m =: Jk let cx denote the cube [x, x+ R · 2−k]n−m. Clearly

CR =⋃x∈Jk

cx (A.2)

30


by construction. Let Ln denote the n-dimensional Lebesgue measure. Now

Ln(F(CR))(A.2)6

∑x∈Jk

Ln(F(cx))(A.1)6

∑x∈Jk

√n−m

n · Ln(B1(0)) · ‖Φ‖n · Rn︸︷︷︸=:C·2−n

·(2−k)n =

2(k+1)·(n−m) · C · 2−n · 2−kn = C · 2−(k+1)m.

(A.3)

The second inequality in (A.3) follows, since the estimate (A.1) implies thatF(cx) is contained in a ball of radius

√n−m · ‖Φ‖ · R · 2−k centered at F(x).

The constant C in the estimate (A.3) is independent of k; therefore, by lettingk tend to infinity we can conclude that F(CR) is a set of measure zero. SinceF(C1) ⊂ F(C2) ⊂ · · · and

H = F(Rm) =

∞⋃l=1

F(Cl),

we can use that Ln is continuous from below and obtain

Ln(H) = liml→∞Ln(F(Cl)) = 0,

as desired. �

Lemma A.6 Let {Hk}Nk=1 be a finite set of planes in Rn of codimension strictly

greater than one. Then the set

Rn \

N⋃k=1

Hk (A.4)

is dense, open and path connected.

Proof By definition, every plane H of codimension m > 1 is equal to thepreimage of zero of an affine linear map

f : Rn → Rm

x 7→ Ax+ b,

where the m × n matrix A has rank m > 1. Let the point x be containedin the set Rn minus H. Let x ′ in H be the unique orthogonal projectionof x onto H. Let w1, . . . , wn−m be independent vectors such that the setx ′+ span(w1, . . . , wn−m) is equal to the preimage of zero of the map f. Thisis well-known from linear algebra. Denote the set span(w1, . . . , wn−m) byU. The set

Ax(H) :=⋃

t∈[−1,−∞)

x+ t(x− x ′) +U (A.5)

31


is contained in the plane x ′+span(U∪(x ′−x)). Since the plane x ′+span(U∪(x ′ − x)) has a codimension greater than or equal to 1, this implies that theset Ax(H) is a set of measure zero, see Lemma A.5.

An easy calculation shows that for all points y in the set Ax(H) the linesegment [x, y] and the plane H have a point in common. On the other hand,assume that the line segment [x, y] contains a point z of H. We can writez = x ′+ x, where the point x is contained in U. There exists a t0 ∈ [−1,−∞)such that y = x + t0(x − x

′ − x) = x + t0(x − x′) + t0x. It follows that y is

contained in Ax(H); therefore, we have shown that the line segment [x, y] isentirely contained in Rn minus H if an only if y is not contained in Ax(H).For each x contained in (A.4) let

Ax :=

N⋃i=1

Ax(Hi). (A.6)

Since a finite union of measure zero sets has measure zero, it follows thatAx is a measure zero set. Let x and y be two points contained in the set(A.4). The set Ax ∪ Ay has measure zero in Rn; as a result, there exists apoint z not contained in Ax ∪ Ay. We deduce that the line segments [x, z]and [z, y] are contained entirely in (A.4). In fact, we have shown that (A.4)is path connected.

Next, we show that Rn without a plane H of codimension m > 1 is openand dense. Note that each component function of f is a non zero polynomial.This implies that for every x contained in H there cannot exist an openneighborhood B of x that is entirely contained in H. Hence the plane His nowhere dense. Due to the fact that H is also closed, we may use theequivalence (A.7) in Corollary A.7 and deduce that Rn without H is openand dense. Lemma A.3 implies that the set (A.4) is open and dense. �

Corollary A.7 Every plane H in Rn of codimension greater than or equal to oneis closed and nowhere dense.

Proof Let U be a subset of Rn. By elementary set theory we know

U open, dense ⇐⇒ Uc closed, nowhere dense. (A.7)

Let H be a plane in Rn of codimension greater than or equal to one. In thesecond part of the proof of Lemma A.6 we have shown that H is closed andnowhere dense. In the reasoning we have not used that the codimension ofthis plane needs to be strictly greater than one, hence the same argumentsalso work for a plane of codimension one. Using (A.7) we may conclude theresult. �

Lemma A.8 If U is an open and connected subset of Rn, then every two pointsx, y ∈ U can be connected by a continuous piecewise linear map γ : [0, L] → U

with only finitely many breakpoints.

32


Proof If U is empty, there is nothing to prove. Let x be a point in U. Denoteby U1 the set consisting of those points in U that can be connected with xby a continuous piecewise linear map γ : [0, L]→ U with only finitely manybreakpoints. Let U2 denote the sets consisting of those points in U thatcannot be connected with x by a continuous piecewise linear map γ : [0, L]→U with only finitely many breakpoints. The fact that U is open and theconvexity of ‖ · ‖-balls suffice to show immediately that both U1 and U2 areopen. Since U is connected and x is contained in U1 we can conclude thatU2 is empty. Thus, the statement follows. �

Lemma A.9 The union of two connected sets A,B that have a point in common isconnected.

Proof Let C,D be two disjoint non empty open sets such that A∪B = C∪D.To put it differently, assume that A∪B is not connected. Note that neither theset A nor B can have points both in C or D, for A and B are connected. ButA and B have a point in common and therefore A and B are both containedexclusively either in C or D. This is a contradiction to the fact that both Cand D are non empty. �

Lemma A.10 [8, p. 95] Let Y be a subspace of X; let A be a subset of Y; let Adenote the closure of A in X. Then the closure of A in Y is is equal to A ∩ Y.

Proof See Theorem 17.4 in [8, p. 95 ] for a complete proof. �

Lemma A.11 (Pasting Lemma) Let {Xi}Ni=1 be a finite set of closed subsets of atopological space X, whose union is equal to X. Let Y be a topological space andsuppose that for every i in {1, . . . ,N} the restriction onto Xi of the map f : X → Y

is continuous. Then, f is continuous.

Proof This proof closely follows Munkres proof of Theorem 18.3 in his won-derful book [8, p. 108]. Let C be a closed subset of Y. Now

f−1(C) =

N⋃i=1

f−1∣∣Xi

(C)

by elementary set theory. Since for every i in {1, . . . ,N} the map f is continu-ous when restricted to Xi we may deduce that every set f−1

∣∣Xi

(C) is closed.Hence f−1(C) as a finite union of closed sets is closed, as desired. �

Definition A.12 (Nowhere dense) A subset A of a topological space B is said tobe nowhere dense in B, if the closure A of A has an empty interior.

In the special case where B is a subset of Rn endowed with the subspacetopology, we see that A ⊂ B is nowhere dense in B, if for all non-emptyopen sets U contained in the subspace topology of B the set U \ (A ∩ B) isnon-empty.

33


Lemma A.13 Let f : X → Y be a continuous map between Hausdorff spaces, andlet K be a nowhere dense subset of X.

1. If the set K is compact, then the set f(K) is nowhere dense in Y.

2. If f is a homeomorphism, then the set f(K) is nowhere dense in Y.

Proof Throughout the proof we may assume that X and Y are non-empty.

1. We prove the contrapositive of the given statement. So, we assumethat f(K) is not nowhere dense in Y. Since K is compact, the map f iscontinuous, and Y is a Hausdorff space we obtain that f(K) is closedin Y. Since f(K) is not nowhere dense in Y there exists a non emptyopen subset U of Y such that U is completely contained in f(K) = f(K).As a result, K is not nowhere dense in X, since the non empty open setf−1(U) is completely contained in K = K.

2. If we are able to prove f(K) = f(K), we may use similar methods as inthe proof of the first part and deduce the desired result immediately.Hence we are left to show f(K) = f(K). Let g be a continuous mapbetween topological spaces. It is well-known from general topology,cf. Theorem 18.1 (2) in [8, p. 104], that for every subset A of thedomain of g, one has g(A) ⊂ g(A). Since the homeomorphism f is inparticular continuous, we obtain f(K) ⊂ f(K). On the other hand, themap f−1 : Y → X is continuous as well. Therefore, we have f−1(f(K)) ⊂f−1(f(K)) = K. By elementary set theory, we see that f(K) ⊂ f(K) andconclude the proof. �

Definition A.14 [5, p. 28] Let f : X → Y be a smooth map between smooth mani-folds, and let Z be a submanifold of Y. We say that the map f is transversal to thesubmanifold Z, we write f >∩ Z, if for all points x in X the tangent space of Y atf(x) is equal to the direct sum of the image of the tangent space of X at x under thetangent map of f and the tangent space of Z at f(x).

Theorem A.15 (Transversality Theorem) [5, p. 68] Suppose F : X× S→ Y is asmooth map of manifolds, where only X has boundary, and let Z be any boundarylesssubmanifold of Y. If both F and the restriction of F to the boundary of X × Sare transversal to Z, then for almsot every s in S, both x ∈ X 7→ F(x, s) andx ∈ ∂X 7→ F(x, s) are transversal to Z.

Proof See page 68 in [5] for a complete proof. �

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A.2. Technical Lemma

A.2 Technical Lemma

In the proofs of Lemma 2.14 and 2.15 we need to ensure that points con-tained sufficiently in the interior of a simplex also stay contained in sim-plices with vertices that have positions slightly altered from the positions ofthe original simplex. This rather technical requirement is embodied in thefollowing Technical Lemma A.16.

Lemma A.16 (Technical Lemma) Suppose that S is an n-dimensional simplexwith vertex set {x0, . . . , xn} ⊂ Rn, and assume that the set S ′ = conv({x ′0 . . . , x

′n})

is the convex hull of the finite subset {x ′0 . . . , x′n} ⊂ Rn. Fix ε > 0. We assume

‖xi − x ′i‖Rn < ε for each i in the set {0, . . . , n}. Let x be contained in Rn. Assumethat the distance between the point x and every set Ei := conv(x0, . . . , xi, . . . , xn)is at least ε, then we have the following equivalence:

x ∈ S ′ ⇐⇒ x ∈ S. (A.8)

If in addition the set S ′ happens to be a simplex and if x is contained in S, x mustlie in the interior of S ′.

Proof Unfortunately, the statement is not symmetric in S and S ′, since thepoint x only suffices conditions with respect to the boundary of S. This lackof symmetry enforces us to deal in the proof of the first statement with bothdirections separately. Observe that for the first statement the set conv(S ′) isnot required to be a simplex. We begin with the proof of the equivalence(A.8).

”=⇒“: Suppose we are given a point x ′ ∈ S ′ such that the distance betweenthe point x ′ and the set Ei is greater than or equal to ε for all i ∈ {0, . . . , n}.We need to show that x ′ ∈ S; accordingly, assume for the sake of a contra-diction that x ′ /∈ S. Consider the map

F : Rn → Rn

x0 +

n∑i=1

λi(xi − x0) 7→ x ′0 +

n∑i=1

λi(x′i − x

′0).

(A.9)

On account of the linear independence of the vectors x1 − x0, . . . , xn − x0,there exist unique numbers λ1, . . . , λn for every point x in Rn such thatx = x0 + λ1(x1 − x0) + · · · + λn(xn − x0). Thus, the map F is well defined.Observe

‖x− F(x)‖ < ε ∀x ∈ S : (A.10)

and that F is affine linear. There exists at least one x in S such that F(x) = x ′

and furthermore ‖x − x ′‖ < ε. Since x ∈ S and x ′ /∈ S, we see that thereexists a point z ∈ [x, x ′] such that z lies on the boundary of S. This isimmediate since the interior of S and the set Sc are non empty open sets.

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We deduce that d(x ′, z) < ε, since z is contained on the line segment [x, x ′]and ‖x−x ′‖ < ε. Note that with a little effort one can show that ∂S = ∪ni=0Ei.This leads to a contradiction, since it implies that z ∈ ∂S must be containedin a set Ei and therefore the distance from x ′ to Ei turns out to be strictlysmaller than ε, in contradiction to our hypotheses on x ′.

”⇐=”: Let x ∈ S be such that the distance from x to Ei is greater thanor equal to ε for all i ∈ {0, . . . , n}. Unfortunately, a reasoning as in thepreceding direction does no longer work, since our point z would lie on theboundary of S ′ and to obtain a contradiction in the same spirit as before, wewould need a point on ∂S.

Nevertheless, there is a nice geometrical idea that does work. To begin,we show with similar methods as before that Bε(x) ⊂ S. Observe that ifa point y in Bε(x) is not contained in S, there exists a point z on the linesegment [x, y] such that z is contained on the boundary of S. This leads toa contradiction, since ∂S = ∪ni=0Ei implies that z must be contained in a setEi and therefore, with the same reasoning as in the preceding direction, itfollows that the distance between the point x and the set Ei must be strictlysmaller than ε. Hence we have shown that Bε(x) ⊂ S.

Now we show that x is contained in S ′. Assume for the sake of a contradic-tion that x /∈ S ′. Using Lemma A.1 we see that F(Bε(x)) ⊂ S ′ is convex. Notethat we can apply Lemma A.1 for F is affine linear and Bε(x) is convex. Theconvexity and closedness of F(Bε(x)) imply that there exists a unique pointx ′ ∈ F(Bε(x)) such that dist(x, F(Bε(x))) = d(x, x ′). Note that x 6= x ′. Weparametrise the ray, which goes through the point x and starts at the pointx ′, as follows:

γ : R>0 → Rn

t 7→ x ′ + t(x− x ′).

An easy calculation shows that for all t ∈ R>0 the point x ′ minimizes thedistance from γ(t) to F(Bε(x)). In other words,

dist(γ(t), F(Bε(x))) = d(γ(t), x ′) ∀t ∈ R>0 : (A.11)

Since the trace of γ has points in Bε(x) and Bε(x)c, elementary geometryshows us that there exists a unique t0 > 1 such that γ(t0) ∈ ∂Bε(x). Equa-tion (A.11) implies that dist(γ(t0), F(Bε(x))) = d(γ(t0), x ′) = ε+d(x, x ′) > ε.This contradicts Equation (A.10), since F(γ(t0)) ∈ F(Bε(x)) and therefore‖γ(t0) − F(γ(t0)‖ > dist(γ(t0), F(Bε(x))) > ε. Hence x ∈ S ′, as desired. Thefirst part follows.

Now we assume that S ′ is a simplex and that x satisfies the usual hypothesesand is contained in S. We know from the first part that x must be contained

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in S ′. Since S and S ′ are simplices, the map F has an inverse, which we callF−1. It is immediate that

‖x ′ − F−1(x ′)‖ < ε ∀x ′ ∈ S ′ : (A.12)

A priori it is possible that x is contained on the boundary of S ′. If x lieson the boundary of S ′, the point F−1(x) is contained on the boundary of Sand therefore in particular contained in some of the sets Ei. Equation (A.12)implies ‖x− F−1(x)‖ < ε. This leads to a contradiction to the hypotheses onx since we have dist(x ′, Ei) < ε, where Ei is the set that contains the pointF−1(x ′). This contradiction concludes the proof. �

Lemma A.16 depends heavily on the assumption that requires conv(S) tobe an n-dimensional simplex. If we drop this assumption about conv(S) weobtain that our point, which suffices the hypotheses of the lemma, cannotbe contained in neither of the sets conv(S), conv(S ′). The exact statement isembodied in the next lemma.

Lemma A.17 Suppose that A = {x0, . . . , xn} and A ′ = {x ′0 . . . , x′n} are finite

subsets of Rn respectively, and assume that the convex hull of A is not an n-dimensional simplex. Let ε > 0, assume that for each i in the set {0, . . . , n} wehave ‖xi − x ′i‖Rn < ε. Let x be contained in Rn. Assume the the distance betweenthe point x and every set Ei := conv(x0, . . . , xi, . . . , xn) is at least ε, then x isneither contained in conv(A) nor contained in conv(A′).

Proof The proof of this statement is much easier than the one of the previ-ous Lemma A.16. Basically we only need to apply Lemma 1.8.

Let x in Rn be such that the distance from x to Ei is greater than or equalto ε for each i ∈ {0, . . . , n}. In particular, x cannot be contained in one of thesets Ei. We assume that x is an element of conv(A). Since the set conv(A) isnot an n-dimensional simplex, we can apply Lemma 1.8 and conclude that xmust be contained in at least one of the sets Ei. This leads to a contradiction.Therefore x cannot be contained in conv(A).

Now assume for the sake of a contradiction that x ′ is in conv(A ′) and x ′

suffices the hypotheses of the lemma. We define the map F as in the previousproof, see Equation (A.9). Recall that there exists at least one point x inconv(A) such that F(x) = x ′. Note that since x is contained in conv(A), thereexists a set Ei such that x is contained in it. Furthermore, we know ‖x −x ′‖Rn < ε. Hence we can deduce that in contradiction to our assumptionson x ′ the distance from x ′ to Ei must be strictly smaller than ε. Thus, wemay conclude the proof. �

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Bibliography

[1] L. E. J. Brouwer. Uber Abbildung von Mannigfaltigkeiten. Mathematis-che Annalen 71, no. 1, 1911. p. 97-115.

[2] L. E. J. Brouwer. Beweis der Invarianz der Dimensionenzahl. Mathema-tische Annalen 70, no. 2, 1911. p. 161-165.

[3] L.E.J. Brouwer. Beweis der Invarianz des n-dimensionalen Gebiets.Mathematische Annalen 71, no. 3, 1911. p. 305-313.

[4] J. A. De Loera, J. Rambau, and F. Santos. Triangulations 1st Ed. Springer,2010.

[5] V. Guillemin and A. Pollack. Differential Topology 1st Ed. Prentice Hall,1974.

[6] I. M. James. History of Topology 1st Ed. Elsevier, 1999.

[7] S. Lang. Real and Functional Analysis 3rd Ed. Springer, 1993.

[8] J. Munkres. Topology 2nd Ed. Pearson, 1999.

[9] R. T. Rockafellar. Convex Analysis 1st Ed. Princeton University Press,1970.

[10] E. Sperner. Neuer Beweis fur die Invarianz der Dimensionszahl und desGebietes. Abhandlungen aus dem Mathematischen Seminar der UniversitatHamburg, 1928. p. 265-272.

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