Basic combinatorial topology

CHAPTER 2

Basic combinatorial topology In this chapter we present some elementary combinatorial results and apply these t o get nontrivial information about the topology of the Euclidean spaces Rn, n E N. The main result is the Brouwer Fixed-Point Theorem.

2.1. Affine notions In this section we will introduce simplicial complexes and present some basic results on them. So-called subdivisions of simplicial complexes are important in the proof of Brouwer’s Fixed-Point Theorem as well as in ANRtheory. For completeness sake, we begin by reviewing some elementary Linear Algebra. The proofs of these results are left to the reader. Let V denote a fixed vector space. Let F be a finite subset of V , say F = (211,. . .,w,}. An afine combination of ul,. . . ,u, is a vector u that can be written in the form Cy=ltivi with t l , . , . ,t , E R and Cr=lti = 1. Such a u is called geometrically dependent on F . A linear (.fine) subspace of V is a subset of V closed under the formation of linear (affine) combinations.

Theorem 2.1.1. Let A be a subset of V and let a E A. Then A is an affine subspace if and only if A - a is a linear subspace.

If S 5 V , then the intersection of all affine (linear) subspaces of V containing S is the smallest affine (linear) subspace of V containing S . This subset is called ufine hull (linear hull) of S and is denoted by aff(S) (lin(S)). Theorem 2.1.2. aff(S) is the set of all affine combinations of elements of S. Moreover, for every a E S the following equality holds: aff(S) - a = lin(S - a ) . We see that if S g V then aff(S) is a translated linear subspace of V , since for every a E S the equality aff (S)= a

+ lin(S - a )

111

2. BASIC COMBINATORIAL TOPOLOGY

112

holds. We say that S spans the affine subspace aff(S). Let v1, . . . ,v, E V. Then

(1) the vectors 211,. . . ,Y, are said to be geometrically independent if for all elements tl, . . .,tn E R with CZ, ti = 0 and Cy=ltivi = 0 we have tl = - .. = t, = 0, (2) a subset S C V is linearly (geometrically) independent if and only if every finite subset is.

Theorem 2.1.3. Let S g V . The following statements are equivalent: (1) S is geometrically independent, (2) no x E S is geometrically dependent on a finite subset F g S \ {x}, (3) for every s E S, {x - s : x E S, 5 # s} is linearly independent, (4) for some s E S, {x - s : x E S , x # s} is linearly independent.

Corollary 2.1.4. Let S C Rn be geometrically independent. Then S contains at most n 1 points.

+

Consider the affine subspace aff(S) and fix an arbitrary a E S. We saw that aff(S) = a + lin(S - a). So there exists a subset T C S such that lin(S - a) = lin(T - a) while moreover T - a is linearly independent. By Theorem 2.1.3 it follows that T' = { a } U T is geometrically independent. Then aff(T') = a

+ lin(T' - a) = a + lin(T - a) = a + lin(S - a )

= af€(S).

So we conclude that S and its geometrically independent subset T' span the same affine subspace. An affine subspace of a linear space is called m-dimensional if it is spanned by m 1 geometrically independent vectors.

+

Proposition 2.1.5. Let S 2 V be geometrically independent. If A , B C S then aff(A) n aff(B) = aff(A n B). The following result is a nice test for proving that subsets of V are geometrically independent.

Theorem 2.1.6. Let ao, a l , . . . ,a, be elements of V such that ai+1

for every i

< n.

6 afl({ao, > a i l ) f

* *

Then {ao, . . . ,a,} is geometrically independent.

A function between affine subspaces of linear spaces is called afine if it preserves affine combinations. Images and preimages of affine sets under affine functions are again affine.

2.1. AFFINE NOTIONS

113

Theorem 2.1.7. Let V1 and V2 be linear spaces, let A1 G V1 and A2 C V2 be affine subspaces and let f : A1 -+ A2 be a function. Then the following statements are equivalent: (1) f is affine, (2) the composition

A1 - a1 3 A1

& A2 3 A2 - a2

i.e., the function 5: A1 - a1 5(2) =

(a1

E Al,az = f(al)),

+ A2 - a2 defined by

f (2+ a1) - a2r

is linear. From Theorem 2.1.7 it follows that an affine function, the domain and range of which are both contained in a finite dimensional normed linear space, is continuous. This can be seen as follows. We first claim that for arbitrary n,m E N,linear functions

Rn 3Rm are continuous. First observe that each linear function f : R + R is a multiplication and hence is continuous. Assume that every linear function f : Ri + R is continuous for i 5 TI, and let F : Rnfl+ R be linear. If f1 = F Rn x (0) and f2 = F (0) x ( 0 ) x . . . x (0) xR. n-1

then they are continuous by assumption. But then F is continuous by linearity, being the sum of the continuous functions (21,.

and

*fib1

. .,%+1)

7 .

. .,z n ,0)

* f2(0,...,0,2,+1).

(21,...,%+1)

Consequently, each linear function f : R" + Rm is continuous since Rm is endowed with the product topology. By Exercise 1.1.18 it follows that each finite dimensional normed linear space is topologically isomorphic to some Rn . So we conclude that a linear function between finite dimensional normed linear spaces is indeed continuous. This gives us what we want since by Theorem 2.1.7 each afine function f , the domain and range of which are both contained in a finite dimensional normed linear space, is of the form

<

f =toForl,

are translations and hence homeomorphisms (Exerwhere both and cise 1.1.3),and F is linear.

114

2. BASIC COMBINATORIAL TOPOLOGY

Simplices. Let V be a linear space. An n-simplex in V is a geometrically independent subset of V having precisely n 1 points. Simplices are denoted by Greek letters. If u and r are simplices and u C r then u is called a face of r ; to indicate that u is a face of T we shall also use the notation: u r. An n-simplex in V is sometimes also called an n-dimensional simplex.

+

+

Theorem 2.1.8. Let u be a simplex in V and let A = aff(u). Then every t, v element x E A can be written uniquely as an f i n e combination CuE0 of u. In addition, the functions av: A + IR defined by a,(x) = t, are &ne.

-

Corollary 2.1.9. Let u and r be simplices in V such that r

A = aff(u),

+ u and let

B = af€(r).

If x E B then a,(x) = 0 for every v E u \ r. The real numbers a,(x) for v E u are called the afine coordinates of x with respect to u. We call the a, the coordinate functions of aff(u). This notation will remain in force throughout the book. Observe that by the above remarks and Theorem 2.1.8 it follows that the coordinate functions of aff (u)are continuous on aff (u). A geometric simplex is the convex hull of a simplex. We use IuI as an abbreviation for conv(u) and sometimes say that lul is the geometric simplex spanned by u. If x E then it can be written as a convex combination of the elements of u (Lemma 1.1.1). But a convex combination is a special case of an affine combination, and affine coordinates are unique by Theorem 2.1.8. So we conclude that the affine coordinates of x E \u\are non-negative. The elements of u are called the vertices of 101. If r is a face of u then 171 is also called a face of IuI. The union of all proper faces of IuI is called the (geometric) boundary dl01 of lul and its complement is the (geometric) interior l ~ l Oof In[. It is clear that the geometric interior of a simplex is non-empty. Just observe that a point in 1u1 all whose affine coordinates are strictly positive belongs to the geometric interior of lul. Since a geometric simplex IuI is the convex hull of the finite set u it follows that 101 is compact (Lemma 1.1.1(2)); this remark will be used without explicit reference in the remaining part of this book.

Theorem2.1.10. Ifuisasirnplexandal,az C_ u then lallnluzl = 1u1nu2.21. The diameter of a geometric simplex is attained at its vertices, as the next result shows.

Theorem 2.1.11. Let (V,11.11) be a normed linear space, and let u a simplex. Then diam(u) = diam(lu1).

C V be

115

2.1. AFFINE NOTIONS

Proof. We prove that diarn(la1) 5 diam(u). Take arbitrary p , E In[. Then a,@>= 1 we get since CUEu

from which it follows by using standard properties of the norm that IlP - 411 5

c

au(P>IIV -

Qll

UEU

= max 1 1 - 411. VEU

As a consequence we obtain for every w E a that 114 - wI(5 maxvEu1 1 ~ 1 wII. so IlP - Qll

5 zg 1 b - Qll 5 v%.u

110 -

41, 0

as required.

Figure 6. be the vector all coordinates of which For each 0 5 i 5 k let di E are 0, except the (i 1)-th coordinate which equals 1. The vectors do, . . . ,dk are clearly linearly independent (hence geometrically independent). For z in aff((d0,. . . ,d k } ) we have that the affine coordinates of 2 = (20,.. ., ~ k ) with respect to 4 , .. . ,d k are equal to ZO,. . .,Zk,respectively. Put

+

‘Tk = { d o , .

. . , dk}.

2. BASIC COMBINATORIAL TOPOLOGY

116

Observe that [ T ~ I = {x E Ik+l : C;=,xi = 1) is a closed and bounded subset of Rkfland hence is compact (alternatively, use that Irk I is a geometric simplex).

Theorem 2.1.12. Let (V,11

11)

be a normed linear space, and let

0={wo,

be a Ic-simplex. Then the function f : 101

f(.)

cv + I T ~ I defined by

..., wk}

= (Quo (a),. . . 7 Qula

(4)

is a homeomorphism.

Proof. First observe that the functions aui are affine (Theorem 2.1.8). Consequently, f is continuous by the remark following Theorem 2.1.7, and oneto-one by the definition of affine coordinates (Theorem 2.1.8). Moreover, f is clearly surjective since if y E then f(z) = y, where

IT~I

n

x = c y i . wi. 2=

By compactness of ercise A.5.9).

101 it

1

therefore follows that f is a homeomorphism (Ex-

Triangulation. A simplicial compZex is a countable collection S of nonempty, finite sets such that: (SC)abstact

if

0

E

S and

The elements of the set

0# T

then

T

E S.

s=us

are called the vertices of the simplicial complex S. Although S need not be a subset of a normed linear space, without loss of generality we may assume that this is the case. In fact, we can think of S as being a linearly independent subset of some normed linear space. This can be seen as follows. For every n E N let zn E l2 be the vector all whose coordinates are 0 except for the n-th coordinate which equals 1. Observe that the set D = {x, : n E N} is linearly independent. Let f : S -+ D be an injection, and identify every simplex E S with f[0] 12. Observe that S, now regarded to be a subset of 12, is linearly independent and hence geometrically independent. By Theorem 2.1.10 it therefore follows that for every 0 1 , 0 2 E S we have 1011 n b221 = 101 n 021. (SC) In the sequel, when dealing with a simplicial complex S, we shall always require that S = U S is contained in a normed linear space L , and that (SC) holds. The set IS1 = U{ 101 : 0 E S} is called the geometric realization of S

2.1. AFFINE NOTIONS

117

and IS1 is said to be triangulated by S (or S is a triangulation of ISl). Observe that the collection S I I= {((TI : (T E 5) consists of geometric simplices in L and has the following properties: (SC)geometric

IT[

E 811, (1) if 101 E S I Iand T is a face of (T then n 1021 is a (2) for all lm1,Iml E SIIwith lm1n la1# 0, face of In1 I as well as 102 1.

Let [(TI be a geometric simplex in L. The collection of all faces of lgl is clearly a simplicial complex which triangulates 101. It is called the standard triangulation of 101 and is denoted by 3 ( ( ~ ) . Topologizing a simplicial complex. From this moment on, we will no longer distinguish between an abstract simplicial complex and its geometric realization. So a simplicial complex in a normed linear space L is a collection S of geometric simplices (which by abuse of terminology we will call simplices for short in the sequel) in L such that for all ( T I , ( TE~ S we have

(1) if 0 E S and T is a face of (T then T E S, (2) (TI n C T ~# 0 + 01 n 02 is a face of 01 as well as ( ~ 2 .

A subcollection 7 of S which is also a simplicial complex is called a subcomplex of S. Let 7 be a simplicial complex in L . For each m 2 0 define 7(m) = {s E 7 : (T is at most m-dimensional}.

So 7(')is the set of vertices of all simplices in 7, 7(l)is the collection of all at most one-dimensional faces of all simplices in 7 ,etc. The elements of 7 ( O ) are called the vertices of 7 . It is a triviality that 7J(m) is a subcomplex of 7 ; we call it the m-skeleton of 7 . Let L be a normed linear space and let 7 be a simplicial complex in L. We are interested in useful topologies on the set X = u7. There are several natural candidates for such a topology. First, X is a subset of L and therefore carries the subspace topology inherited from L. It turns out that this is in general an interesting but not a useful topology. We shall now describe a better topology. To this end, let us first agree that there cannot be ambiguity about the topology that each simplex of 7 should carry. By Theorem 2.1.12, each k-dimensional simplex in L with its subspace topology is naturally homeomorphic to a standard k-dimensional simplex in Rk+l; each simplex in L from this moment on is endowed with its subspace topology. It would be very unnatural if the useful topology we want to define on X should induce a different topology on one of the simplices of 7 . So there is a natural candidate for a topology on X , namely, the largest topology that induces the 'right' topology on every simplex of 7 ,i.e.,

U C X is open iff for every simplex

T

E 7 ,U n T is open in T.

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2. BASIC COMBINATORIAL TOPOLOGY

This is called the Whitehead topologg on X and X with this topology is usually denoted by 171. Lemma 2.1.13. Let 7 be a simplicial complex in L and let X = U 7. Then

(1) The Whitehead topology on X is a topology and is finer than the topology that X inherits from L , i.e., the inclusion

171 L)

xEL

is continuous, (2) On every simplex of 7 the Whitehead topology and the topology on L induce the same subspace topology.

Proof. The proof of (1) is a triviality the verification of which we leave to the reader. For (2), let T be a simplex of 7. It suffices to prove that for every T there exists a Whitehead-open V 171 such relatively open subset U that V n T = U . We claim that V = U U (171\ T) is as required. It is clear that V n T = U , and so it remains to prove that V is Whitehead-open. To this end, let o E 7 and put V' = V n (T U o). Then V' is open in T U o since its complement is equal to T \ U which is compact and hence closed in T U o. 0 Consequently, V' n CT = V n o is open in o,as required. Observe that a subset A of 171 is closed if and only if A n T is closed for every T E 7. This immediately implies that 171 is a Tl-space, i.e., a space in which every singleton is closed (this follows also from Lemma 2.1.13(1)). Due to our self-chosen (sometimes unpleasant) restriction to deal with separable metrizable spaces exclusively for the time being, there is a small problem with the Whitehead topology since it need not be metrizable. How71 is ever, there is a simple combinatorial condition on 7 that ensures that 1 ' locally finite if every separable and metrizable; call a simplicial complex T vertex of 7 is contained in at most finitely many simplices of 7. In turns out that for our considerations it always suffices to consider locally finite simplicia1 complexes and these are precisely the simplicial complexes that are separable metrizable when given the Whitehead topology, see Proposition 2.1.21 71 but below. For the moment, we shall not worry about the metrizability of 1 we shall prove a few easy but important lemmas from which metrizability in the locally finite case will follow rather easily later. Lemma 2.1.14. Let 'T be a simplicial complex with subcollection S. Then

(1) US is a closed subspace of 171, (2) if S is a subcomplex then the topology that US inherits from 1 71 coincides with the Whitehead topology on U S.

2.1. AFFINE NOTIONS

119

Proof. For (l),for convenience, put Y = US. Since each simplex of 7 has finitely many faces only, and 7 is a simplicial complex, it is clear that for every T E 7, Y n T is closed in T . By definition of the Whitehead topology, we therefore conclude that Y is closed in 171.

c

For (2) observe that since S T, the topology that U S inherits from 171 is coarser than its Whitehead topology. Now let A IS1 be closed. We shall prove that A is a closed subset of 171. To this end, let T be an arbitrary simplex in T. Since 7- has finitely many faces only, there is a finite subcollection 3 of S such that

c

U31-7 = USn T

7.

Consequently, A n T = ( An US) n T = ( An U 3)n T . Now since A & IS1 is closed, for every (T E F, A n (T is closed in (T,and consequently, A n (T n T is closed in T . Since 3 is finite, it therefore follows that ( An U 3)n T is closed 0 in T . We conclude that A n T is closed in T . From the above lemma it follows that if 7 is a simplicial complex and if S is a subcomplex of T then we can identify IS1 and the subspace US of 171. It will be convenient to do that.

Corollary 2.1.15. Let 7 be a simplicial complex. Then lT(o)lcarries the discrete topology. Proof. Since T(O) is a subcollection of T, by Lemma 2.1.14(1) each subset of 7 ( O ) is closed in 171. So the subspace topology of UTo is the discrete 0 topology. An application of Lemma 2.1.14(2) therefore does the job. This corollary easily shows that the Whithehead topology on a simplicial complex is usually different from the topology it inherits from its ambient linear space.

Example 2.1.16. The collection 7 = { {0}, {l},{ . . . } consists of { 0-dimensional simplices of 1w and hence is a simplicial complex. The Whitehead topology on U 7 is the discrete topology by Corollary 2.1.15. But the subspace topology on U 7 is not discrete. For each x E 171 define the star, St x, and the carrier, carx, of x by St x = 171 \

U { T

E 7:x

$z T } ,

and

carx = n { T E 7 :2 E T I , respectively. We shall prove that carx E 7 and it is therefore the smallest simplex of T that contains x.

Corollary 2.1.17. Let 7 be a simplicial complex. Then for every x E 171, (1) St x is open in IT[,

120

2. BASIC COMBINATORIAL TOPOLOGY

(2) carx belongs to 7 . In addition, the family {St x : x E 7 ( O ) } covers 19’). Proof. For (l),let S = { T E 7 : x T}. From Lemma 2.1.14 we obtain hence St x is open. that (JS is closed in IT], For (2), first take T E 7 such that x E T. Since for every CT E 7 such that x E (T we have CT n T 3 T ,and T has only finitely many faces, car x E 7 . Take an arbitrary y E (71.If carp is a zero-dimensional simplex then y is a vertex of 7 and hence belongs to U{St x : x E ’T(O)}. Assume that cary is not zero-dimensional and let x be a vertex of cary. Observe that cary is the smallest simplex containing y, hence y does not belong to the boundary of cary. We claim that y E Stx. If not, then there is a simplex -r E 7 containing y but not containing x. Then T n car y is a proper face of car y containing y, contradiction.

Lemma 2.1.18. Let 7 be a simplicial complex and Let K & 171 be compact. Then K is metrizable and is contained in the union of finitely many simplices of 7 . Proof. That K is metrizable follows immediately from Lemma 2.1.13(1) and Exercise A.5.9. Suppose that K is not contained in the union of finitely many simplices of 7. Then we can find an infinite collection S = { T ~: n E N} of simplices in 7 and for every n E N a point (*>

2 ,

E

(Kn T

\

n

U

~ + ~ ) T ~ .

i= 1

Since K is compact, and metrizable, we may by Theorem A.5.1 assume without loss of generality that x = limn-,ooxn exists (and belongs to K ) . Now let T E be an arbitrary simplex. Since rnn T is either empty or a face of T, there exists m E N such that

6

Trim=

n=l

u m

Tnn-r,

n=l

from which it follows that

{ x , : n ~ N } n ~(21, & ...,xm}. By the definition of the Whitehead topology, this implies that {x, : n E N} is closed in 171 (recall that 191 ’ is TI). Since the sequence (xn)n converges, it therefore has to be eventually constant, which contradicts (*). 0 One of the reasons that the Whitehead topology on a simplicial complex is relatively simple to deal with is that it is easy to check that certain functions are continuous.

2.1. AFFINE NOTIONS

121

Lemma 2.1.19. Let 7 be a simplicial complex and let X be a space. A function f : (713 X is continuous if and only if the restriction off to every simplex r E T is continuous on r.

Proof. Suppose that f : 171 + X is such that the restriction of f to every simplex r E 7 is continuous. Let U 5 X be open. For each r E 7 ,

f-*[u] n = (f 1 T)-'[u] 7

is open in r. By the definition of the Whitehead topology, it therefore follows 0 that f-l[U] is open in 171.

Corollary 2.1.20. Let 7 be a simplicial complex. Then 171 is a normal topological space, i.e., for every pair of disjoint closed subsets A and B of 171 there exists a continuous function f : 1 71 + 1such that f / A 0 and f tB E 1. Proof. Since 7 is countable, we can enumerate it as {rn : n E N}. By + 1 such that Lemma A.4.1 there exists a continuous function f1:

fi / (rln A ) 0, f (rln B ) = 1. By induction on n, we shall construct continuous functions such that (1) fn t ( A n U;=l ri) 0, (2) f n t ( B n u;=1 Ti) 1, (3) if n > 1 then fn extends

=

fn :

uyLl

ri

+1

fn-l.

Since we already defined f1, assume fi to be constructed for 1 5 i 5 n. We shall construct fn+l. Let A' = A n U?=',' ri and B' = B n Uy=',' ri, respectively. Define 9: A' U B' U U;=l ri -+ II by g(.)

=

{Y

fn(.)

.( E A'), .( E B'), .( E

u;=,

Ti).

;:u;

Then g is clearly continuous and extends fn. Since ri is compact, it is metrizable by Lemma 2.1.18. By Corollary 1.2.5 there consequently exists ~i + 1 of g. It is clear that fn+l is as a continuous extension f n + l : required. fn. Then f is a well-defined function from 1 71 into 1 Now put f = having the property that f A E 0 and f t B f 1. Moreover, f is continuous by Lemma 2.1.19. 0

UyL;

UEl

These simple results enable us to prove the following: Proposition 2.1.21. Let 7 be a locallyfinite simplicial complex in the linear space L. Then 17)is a locally compact separable metrizable space.

122

2. BASIC COMBINATORIAL TOPOLOGY

Proof. We shall first prove that each point of 171 has an open neighborhood which is separable, metrizable, and has compact closure in (71.To this end, take an arbitrary x E [TI. By Corollary 2.1.17(1), S t x is open. Since 7 is locally finite, St x is contained in the union of finitely many simplices of 7 . This can be seen as follows. First observe that St x is contained in the union of all simplices that contains x. But, as we will show, there are only finitely many of such simplices. Striving for a contradiction, assume that S is an infinite subcollection of 7 such that x E 5. Fix u E S. Then for u' E S \ {u} we have that u'nu is a face of u. So there are an infinite subcollection S' of S and a face r of u such that for every u' E 5' we have u' n u = r. But then any vertex of r is contained in infinitely many simplices of 7 which contradicts the local finiteness condition. Consequently, St z is contained in a subspace of 171 which is compact and therefore metrizable (Lemma 2.1.18). We conclude that 1 71 is locally compact. Since 7 is countable, 1 71 is the union of countably many compact subspaces. So J'TIis Lindelof and from the above it therefore follows that IT1 can be covered by a countable family 'B consisting of open, separable metrizable subspaces of 171. For each B E 'B let 3 ( B )be a countable base for B consist3(B)is countable, consists of open ing of open subsets of B . Then 3 = UBEB subsets of 171, and is easily seen to be a base for 171. So by Corollary 2.1.20 it follows that \!TI is a second countable regular 2'1-space and is therefore a separable metrizable space (see Page 465). 0

n

In Exercise 2.1.15 we shall see that if 1 71 is metrizable then 7 is locally finite. We shall now prove that the Whitehead topology on 171 is in fact 'intrinsic', i.e., that it has nothing to do with the topology on L. To begin with, let us say that a subset F of 7('),the O-skeleton of 7 ,spans a simplex in 7 if there exists a simplex r = ~ { x o ,...,xk}1 E 7 such that F = {xo,. . . ,xk}. Lemma 2.1.22. Let 7 be a simplicial complex in L and let F be a subset of 7(').Then F spans a simplex in 7 if and only if St x # 0.

nzEF

Proof. Suppose that F spans a simplex r in 7 and let b be a point in 1rIo.We St x. If not, then there exists x E F such that b # St x, claim that b E or, equivalently, there is a simplex u E 7 containing b but not containing x. Then u n r is a face of r. However, b is not contained in a proper face of r , which implies that u n r = r and hence that x E r g u, contradiction. Conversely, assume that F is such that there exists y E St x. We claim that F spans a face of cary. If for certain x E F , x # cary then by definition, St x n cary = 0, which is a contradiction. Consequently, F C cary. Since 7 is a simplicial complex, and F g 7('),each x E F is a vertex of cary. We conclude that F spans a simplex in 7 . 0

nzEF

nrEF

2.1. AFFINE NOTIONS

123

Now let L and E be normed linear spaces and let 7 and S be simplicial complexes in L and E , respectively. We say that 7 and S are combinatorially equivalent if there exists a bijection f : T(O) -+S(O) such that F 7 ( O ) spans a simplex in 7 if and only if f[F] spans a simplex in S. Let us also say 71 and IS1 are simplicially homeomorphic if there exists a homeomorthat 1 phism h: 171 + IS[ such that for all T E 7 and CJ E S, we have h [ ~E] S as well as h-l [CJ] E 7.

c

Proposition 2.1.23. Let L and E be normed linear spaces and let 7 and S be simplicial complexes in L and E , respectively. The following statements are equivalent: (1) 7 and S are combinatorially equivalent, (2) 171 and IS1 are simplicially homeomorphic.

c

Proof. For (1) + (2), let f : 7(') + S(O) be a bijection such that F 7 ( O ) spans a simplex in 7 if and only if f[F]spans a simplex in 5. So for every simplex T = I(x0,. . . , x k } l E 7, the set {f(xo),. . . ,f ( z k ) } spans a simplex CJ of S. Define a homeomorphism f T : T + CJ by k

k

i=O

i=O

here ao(x),. . . , ~ v k ( z ) denote the affine coordinates with respect to T of an arbitrarily chosen point x E T of course. It is easily seen that f7 is indeed a homeomorphism, cf. Theorem 2.1.12. Now define f:1 71 + IS1 by TE3

By unicity of affine coordinates (Theorem 2.1.8), f is well-defined. In addition, f is continuous since the restriction of f to every simplex of 7 is continuous (Lemma 2.1.19). It is easily seen that f is one-to-one and surjective and that f-' is continuous for the same reason f is. We conclude that f is a homeomorphism. Clearly both f and f-l are 'simplex preserving'. Conversely, let f: 171 + IS/ be a homeomorphism such as given by (2). The function is clearly as required.

0

From the above proposition we conclude that the topology of a simplicial complex depends only on the combinatorial properties of its vertex set. For that reason when discussing a simplicial complex, it is no longer necessary to mention the normed linear space it is a subset of. Let X be a space. We say that X is a polytope if there exists a locally finite simplicial complex 7 such that X and 171 are homeomorphic. In case 7

124

2. BASIC COMBINATORIAL TOPOLOGY

is finite we say that X is a polyhedron. Observe that each polyhedron is compact. Polytopes are very important since they are the 'bridge' between abstract topological spaces and concrete ones.

If X is a polytope, say X is homeomorphic to 171, then we will find it sometimes convenient to not distinguish between X and 171. This will never cause confusion because the triangulation under consideration will always be explicitly defined but the reader should keep in mind that there usually are many different triangulations of the same polytope. If 171 is a polytope then a subset Y g 171 is called a subpolytope if there is a subcomplex S E 7 such that Y = 1st. Similarly for subpolyhedron. Theorem 2.1.24. Each polyhedron is an ANR. Proof. Let 7 be a finite simplicial complex. By induction on the cardinality of 7 we shall prove that 171 is an ANR. If 7 consists of one simplex only then IT1 is an AR by Theorems 2.1.12 and 1.2.9. Suppose that for every simplicial complex 7 of cardinality at most n - 1 1 1, (71is an ANR, and let T be a simplicial complex of cardinality n. Since 'J is finite, there exists a simplex T E "such that for every cr E T with T =$ cr, cr = T (let T be any simplex of maximal dimension). Pu t S = 7 \ { T } . Then S is a simplicial complex and 171 = T U 181. In addition, clearly T n IS1 = I{T n cr : cr E S}l. So by our inductive assumptions and by Theorem 1.2.16(1) it follows that 171 is an ANR. 0

In Corollary 4.3.6 we shall prove that every polytope is an ANR, which generalizes Theorem 2.1.24. Exercises for $2.1. 1. Prove Theorem 2.1.1.

2. Prove Theorem 2.1.2. 3. Prove Theorem 2.1.3. 4. Prove Proposition 2.1.5.

5 . Prove Theorem 2.1.6. 6. Prove Theorem 2.1.8.

7. Prove Theorem 2.1.10. 8. Let A be an affine subspace of a finite dimensional linear space L. Prove that A is closed in L. b9. Prove that if u is an n-simplex in

10. Let D be a simplex and let p E D. if p is one of the vertices.

Rn then Int(lu1) is equal to 1 0 1 ~ . Prove that u \ { p } is convex if and only

2.2. BARYCENTERS AND SUBDIVISIONS

125

11. Let a be a simplex and let C be a convex set which is contained in da. Prove that C is contained in a proper face of u. 12. Let X be a set triangulated by the simplicial complex S. Prove that the collection of all geometric interiors of elements of IS1 is a partition of X . 13. Let

U be a countable collection of sets. Prove that the collection { 3 : 3 g U is finite and 03 # 0)

is a simplicial complex.

14. Let 7 be a simplicial complex. Prove that IT1 is paracompact (we do not assume that T is locally finite). 15. Let 'J be a simplicial complex that is not locally finite. Prove that 171 is not metrizable (this is the converse to Proposition 2.1.21). b16. Let 7 be a simplicial complex and S a subcomplex of 7. Prove that the

topological interior and topological boundary of IS( in (71are given by

{z E IS[ : z E a E T* a E S} and respectively.

UP\ S) n PI,

2.2. Barycenters and subdivisions

Througout this section, L is a fixed normed linear space. All simplices under consideration are subsets of L. If cr is a simplex in L then u denotes the set of its vertices.

Barycenters. Let o be a simplex. The burycenter b, of is the point in cr whose affine coordinates (with respect to 6 ) are all equal: if

u = {vo, . . . , v,} then

b, =

1 Cn n+l . vi. i=O

Observe that b, belongs to the geometric interior cro of cr.

Lemma 2.2.1. Let S be a simplicial complex. Then i f cr, cr' E S and b, E o' then (T is a face of o'. Proof. By (SC), cr n 0'is a face of o. This face contains b, so it must be cr itself. 0

Lemma 2.2.2. Let 71 c 72 c . + .c rk: be a strictly increasing collection of faces of a simplex cr. Then the barycenters b,, , b,, , . . . ,b,, axe geometrically independent.

2. BASIC COMBINATORIAL TOPOLOGY

126

Proof. By Theorem 2.1.6 it suffices to show that for i bTi+l

@

aff({bTI

>

< k,

bT2, * . ' 7 b T i } ) '

To this end, take an arbitrary vertex w E

\

ii+l i i .

Then

Since {bT1 7 bT2> * * ' 7 h i }

E

we get

G aff(+i). aff({bT1,b~~,...,bTi}) This and Corollary 2.1.9 imply that for every x E aff ( { b T l , bT2, . . . ,b T i } ) we have a,(x)= 0. This is clearly as required. 0 Let u be a simplex and let X = { T I , T ~ ,. . . ,~ h be } a chain of faces of u (by a chain we mean of course a chain with respect to inclusion) with corresponding chain of vertices k = {+I, i 2 , . . . ,i h } . If X is a maximal chain of faces then clearly u E X and consequently Uk = ir. This rather trivial remark will be used without explicit reference a few times in the remaining part of this section.

For every w E U k define its height ht(w) to be the first i 5 k with w E ti. Define a quasi-order '=$'on U k by putting: v =$ w e ht(w) 5 ht(w). Lemma 2.2.3. Let u be an n-simplex and let X be a maximal chain of faces of u with corresponding quasi-order =$.Then

+

(1) X has cardinality n 1 and the corresponding quasi-order =$ is a linear order on ir = Uk. (2) If /3 is the set of all barycenters of elements of X then it is geometrically independent and 1/31 = {x E

0

: (VW,W

E 6)(w 3 w

* CY,(x) 2 a , ( 2 ) } .

(3) If w 3 w and for certain x E l,Bl we have a,(x) = a,(x) then for every T E X with w E i and w # i we have that the affine coordinate of I in 1/31 with respect to bT is equal to 0. Proof. We claim that if ct is the direct successor of /3 in X then d! singleton: if it contains two distinct points w and w then

. . .,P, IB u {W}I,Q,.

\

is a

..

would be a chain extending X;likewise one sees that the first element of X is a singleton. Thus X = {TO,7 1 , . . . ,T ~ where } i i has i 1 points. Now write ir = {wo, w l , . . .,vn} such that for every 0 5 i 5 n, i i = { t ~ o , v l , .. . ,wi}. Then wi 3 wj if and only if i 5 j. This proves (1).

+

2.2. BARYCENTERS AND

127

/3 is geometrically independent by Lemma 2.2.2. 1/31 can be written as

For ( 2 ) , observe that An arbitrary element x E n

x=

SUBDIVISIONS

n

t i . bTi

where ti

2 0 for every i and xti = 1 i=O

i=O

(Lemma 1.1.1). Observe that

For every 0 5 i 5 n define

and observe that n

c s i i=O

=c[ti.C-] =Ctz n

i

i=O

j=O

1 2+1

n

i=O

= 1.

From this we conclude that the si are the affine coordinates of x (Theorem 2.1.8) and satisfy SO 2 s1 2 . . - 2 5,. Conversely, let p be a point in c whose affine coordinates the property that so 2 s1 2 . . . 2 sn. Then the numbers

ti = (i + l ) ( ~-i ~ i + l ) tn = (n 1)sn

+

(0 5 i

SO,. . .

,sn have

< n),

are non-negative and have the property that n

i=O

c n

=

i=O =p.

si. ui

In addition, CF0ti = Cy=osi = 1. The ti are consequently the affine coordinates of p in 1/31 (Theorem 2.1.8). This proves ( 2 ) . For (3), simply observe that if for p above we have that if si = s j for 0 certain i < j 5 n then all t k for i 5 k < j are equal to 0.

2. BASIC COMBINATORIAL TOPOLOGY

128

Lemma 2.2.4. Let (T be a simplex and let X1 and X 2 be chains of faces of 8 . Let 81 and P 2 denote the sets of barycenters of elements of X1 and X 2 , respectively Then

lPll n lP2l = IP1 r l P 2 1 . n lP2l is convex and contains P1 n P 2 , it is clear that

Proof. Since

nP2) c lPll n IP2I. For the reverse inclusion, take an arbitrary 3 E 1/31 I n IPzl. Since

IP1 np21=

(X1 \ X 2 >n ( X 2 \ Xl) = 0, without loss of generality we may assume that (T # XI \ X 2 . We assume without loss of generality that X1 \ X 2 # 0 for otherwise we get what we want from Lemma 2.2.2. It is clear that there exists a linear order $1 on the set u such that for every element of T E XI we have that i is an initial $1segment. This order corresponds to a maximal chain of faces of (T. Pick an arbitrary element T E XI \ X Z ; observe that T # (T.The collection X 2 can easily be extended to a maximal chain C of simplices such that T $ C (Exercise 2.2.1). Consequently, i is not an initial segment with respect to the corresponding linear order $2 (Lemma 2.2.3(1)), i.e., there exist distinct elements v, w E c i such that v E i,w # i and w $2 v; consequently, a w b )

(Lemma 2.2.3(2)). Observe that v

$1

2 adz) w from which it follows similarly that

2 aw(2). By Lemma 2.2.3(3) we therefore conclude that the affine coordinate of the simplex with respect to b, is equal to 0.

2

in 0

Subdivisions. Let X be a subset of L which is triangulated by S. A triangulation 7 of X is called a subdivision of S if for every simplex (T E S we have that the collection T(f7)= { T E

:T

c

(7)

is finite and is a triangulation of u.

Lemma 2.2.5. Let S and 7 be triangulations of X . If 7 is a subdivision of S then for every T E 7 and u E S such that b, E u we have T C_ (T. Proof. Take an element T’ E 7(a)such that b, E 7’. Then by Lemma 2.2.1 it follows that T T’ (T,as required. 0

c

Theorem 2.2.6. Let S, 7 and X be simplicial complexes. IfS is a subdivision of 7 and 7 is a subdivision of X then S is a subdivision of 3.

2.2. BARYCENTERS AND SUBDIVISIONS

129

Proof. We have to prove that S ( p ) is a finite triangulation of p E 32. That the collection S ( p ) is a simplicial complex is clear since condition (SC) holds for all simplices in S. Let (T E S ( p ) . We claim that there exists an element T, E T ( p ) such that u g T,. This is easy. Indeed, since U T ( p ) = p there exists T, E T ( p ) such that b, E 7,. So by Lemma 2.2.5 it follows that T, is as required. Since T ( p ) is finite and for every T E ' T ( p ) , S ( T ) is finite, we obtain that the collection S(p) is finite. To conclude the proof, observe that U S ( p ) = p is trivial since U T ( p ) = p and U S ( T ) = T for every T E T ( p ) . 0 Let cr be a geometric simplex in L. We shall define a special triangulation of (T,the so-called burycentric triangulation. Let B ( a ) denote the set of all barycenters of faces of (T.We shall define a simplicial complex B(u) consisting of subsets of B ( a ) as follows:

A nonempty subset /? of B ( a ) belongs to B(a) if and only if the faces of (T the barycenters of which belong to ,B form a chain.

c

It is clear that if P E 'B((T)and y /3 then y E B((T).So B((T)is an abstract simplicial complex. By Lemma 2.2.4 it follows that the collection I'B(a)I= {IS1 : P E 'B(a)) satisfies condition (SC) (see Page 116). So the geometric realization of B((T) is IB(cr)I and, in particular, is contained in (T.In order to prove that (B(o)I is a subdivision of (T,it therefore remains to prove that Let

2

E

Without loss of generality we have

(T.

ow, ).(

c i = (210,.

. . ,w,}

and

5 . . . i Qlwo (.I.

5 QltJ,-l(.)

This ordering corresponds to a (maximal) chain of faces in (T,which in turn corresponds t o a (maximal) simplex /3 E B(u). By Lemma 2.2.3(2), x E 1/31. Let X be a set in L which is triangulated by the simplicial complex S. As above, for every (T E S let B(o) be the barycentric triangulation of (T.The collection B(S) = 1. ( E I'B(ff)I: B E S} is called the burycentric subdivision of S . That B(S) is a collection geometric simplices with union X is clear. We claim that it is a simplicial complex and is a subdivision of S. For that we only need to verify condition (SC) (see Page 116) because for every (T E S,

1.1(

E lB(S)I : 171

and IB(a)I is a finite triangulation of

c 0) = I'B(g)I

(T.

2. BASIC COMBINATORIAL TOPOLOGY

130

Let Pi E 'B(ai)for ai E S, i = 1,2. If a1 n u2 = 0 then there is nothing ~. to prove. If (TI n a2 # 0 then D = a1 n 02 is a face of 01 as well as C T Now Put yi = pi n D (i = 1,2). Then y1 and

72

belong to 'B(a) and we claim that

1~~y= i J pi\n 0

(i = 172).

fl D is clear. Take an arbitrary z E That I7il element /3 E B(a) such that z E [PI. Observe that

lpil f l D . There exists an

P E 'B(Q) n B(a2). Consequently, zE

IPil n IPI = IPi n PI c IPi n Dl = IYiI.

By Lemma 2.2.4 we therefore conclude that

lPll n lPzl = lPll n I P S 1 n o = 1,711 n 1721 = I71 n Y 2 1 = IP1 nP217 which is as required. Sometimes it will be convenient to denote S by sd(O)Sand B(S) by sd(')S. We define the second barycentric subdivision sd(2)Sof S by 'B('B(S)). Similarly, one defines sd(")S, the n-th barycentric subdivision of S. Notice that by Theorem 2.2.6, sd(")S is a subdivision of S for all n. Let S be a triangulation of a subset of L. The mesh of S, denoted by mesh(S), is defined to be the number sup diam(a ) . UES

We allow mesh(S) to be equal to

CQ.

Theorem 2.2.7. Let D be an n-simplex in the normed linear space L and let %(a) be the barycentric triangulation of CT.Then

Proof. First observe that for every p E 'B(D) we have diam(/3) = diam(lp1) (Theorem 2.1.11). Fix ,Ll E ' B ( D ) and let d , c E P. We will estimate the distance between d and c. Choose an enumeration {wo,w1,.

. .,%I

of Lr such that c is the barycenter of (210, w1,.. . ,wk} and d is the barycenter of {wo,wl,. . . ,urn}, respectively. We may assume without loss of generality

2.2. BARYCENTERS AND SUBDIVISIONS

that k

131

5 m. Then

Moreover,

m <-' -m+l

diam(u).

In the last inequality we used that one of the terms (if i = j ) is equal to 0. We conclude that m diam(@) 5 -. diam(u). m+l Finally observe that m

5 n.

0

Corollary 2.2.8. Let S be a triangulation of a set X in a normed linear ' is the barycentric space L such that every u E S is at most n-dimensional. If B subdivision of S then n mesh(%) 5 -. mesh(S). n+l Corollary 2.2.9. Let 171 be a geometric simplex in a normed linear space L. Then for every E > 0 there is an m E W such that mesh ( ~ d ( ~ ) 3 ( < r )E).

132

2. BASIC COMBINATORIAL TOPOLOGY

Proof. Assume that T is an n-simplex. By Corollary 2.2.8 it follows that for every m E N we have

. diam(.r). mesh ( ~ d ( ~ ) 3 5 ( ~(A)"' )) nS1 We conclude that for a sufficiently large m, rnesh(sd(")3(~))< E .

0

We shall now formulate a very important property of polyhedra.

Theorem 2.2.10. Let 7 be a finite simplicia1complex. Then for every open cover U of 171 there exists m E N such that sd(rn)7refines U,i.e., for every simplex u E sd(")7 there exists an element U E U with c E U .

Proof. Fix T E 7 for a moment. By Corollary 2.2.9 and Lemma A.5.3 there exists n ( ~E) N such that every u E sd("('))9(7) is contained in an element of U. Let m = max{n(r) : T E 7). Now take an arbitrary n E sd(")7. By the definition of sd(rn)7it is clear that there exists T E 7 with n E ~ d ( ~ ) 3 ( 7 ) . Since every element of ~ d ( ~ ) 3is (contained ~ ) in an element of ~ d ( ~ ( ' ) ) 3 ( ~ ) , 0 the simplex K is contained in an element of U. The above theorem can be generalized as follows: for every polytope 171 and for every open cover U of 171 there exists a subdivision S of 7 such that each simplex u E S is contained in an element of U. However, for noncompact polytopes, S generally cannot be chosen to be an iterated barycentric [408]. subdivision. For details, see WHITEHEAD Exercise for 52.2. b1. Let u be a simplex and let X be a chain of faces of u. Prove that if 7 is a proper face of o such that 7 $ X then there is a maximal chain of faces

L such that 3c

L but r

$!

L.

2.3. The nerve of an open covering

Let X be a space and let U be a countable open cover of X. We say that a simplicial complex 7 is a nerve of U if 7(') can be indexed as { z ( U ) : U E U} such that for every n 2 0,

(*I

z(u0), . . . ,z(un)spans a simplex in 7 iff

n

(l # 0. ~i

i=O

If 7 is a nerve of U then it is convenient to adopt the notation { z ( U ): U E U } for T(O),where it is implicitly assumed that the 'indexing' is such that (*) holds.

2.3. T H E NERVE OF AN O P EN COVERING

133

Proposition 2.3.1. Let X be a space. Each countable open cover of X has a nerve.

W

cover

nerve Figure 7.

Proof. Let U = {U, : n E N} be a countable open cover of X. For each n let z, E l2 be the vector all coordinates of which are 0, except for the n-th is linearly independent, hence coordinate which equals 1. The sequence (z,), geometrically independent. Define

3 = { F g N : F is finite and

n

Un #

0}.

nEF

For F E 3 let T ( F )be the simplex in l2spanned by { x n : n E F } . P u t 'T= { T ( F ) : F E F}.

It is easy to verify that 7 is a simplicial complex and that it moreover is a nerve for U. 0 Clearly, any two nerves of the same open cover are combinatorially equivalent (hence 'isomorphic'), so by Proposition 2.1.23 we can now speak of the nerve N ( U ) of the cover U. If L is a normed linear space containing a simplicial complex 'T which is combinatorially equivalent to the nerve N ( U ) then we say that N ( U ) can be realized in L. Let N( U ) be the nerve of the locally finite open cover U (recall that U is countable, see Page 486). We shall use the /+functions with respect to U to

2. BASIC COMBINATORIAL TOPOLOGY

134

define a canonical continuous function 6 :X + IN(U)I. This function is the ‘bridge’ between the ‘abstract’ space X and the ‘concrete’ space IN(U)I and is called the 6-function of the cover U. Indeed, define n: X

+ IN(U)I by 6(2)=

c

6U(Z)

*.(U).

UEU

We claim that

K

is well-defined. To check this, for x E X put

3(2)= {U E u : 2 E U } .

Observe that by the local finiteness condition, 3(z) is finite, that nu(z) = 0 if U # 3 ( s ) ,and that CUEF.(x) ~ ( x=)1. This implies that

.(x) =

c

ICu(2)- x ( U ) .

UEF(2)

Hence the infinite sum in the definition of 6 reduces to a finite sum. In addition, the { w ( x ) : u E 3k)) are the affine coordinates of ~ ( xin) the simplex T ( Z ) spanned by the vertices

{ Z ( U ) E U : u E qx)} of N ( U ) . This shows that ~ ( zis) well-defined, and also that ~ ( xE)~ ( 2 ) . Lemma 2.3.2. Let X be a space and let U be a locally finite open cover ) N(U). of X . Then for each x E X, T ( X ) is the carrier of ~ ( z in

Proof. As remarked above, ~ ( zis) a simplex of N ( U ) that contains ~ ( x ) As . remarked above, the ~ ~ ( 2 )are ’ sthe affine coordinates of ~ ( z with ) respect to the simplex in N ( U ) spanned by the vertices { x ( U ) : U E 3(2)}. Now ) 0, T(X) is the smallest simplex since for every U E S ( x )we have that n ~ ( z # 0 containing ~ ( z )i.e., , ~ ( z=) car ~ ( x ) . Theorem 2.3.3. Let X be a space and Jet U be a locally finite open cover of X . Then the 6-function 6 :X + IN(U)I has the followingproperties:

(1) K. is continuous; (2) for every U E U:6-l [St x ( U ) ]= U . Proof. We shall first prove (2). Observe that for every x E X and U E U,

# F(x) e x(U)# T ( X ) = carK(x) (Lemma 2.3.2). Now if z ( U ) # T ( X ) then since ~ ( xE) T ( x ) ,~ ( x$!) St x ( U ) . x#U

U

This proves that &-‘[St x ( U ) ]g U . Conversely, if n(x) $! St x ( U ) then there ) not s ( U ) . Since carrc(s) g T , is a simplex T E N ( U ) which contains ~ ( xbut this gives us that x ( U ) $2 carr;(s) = T ( Z ) and hence that z $2 U . This proves that 471 g Stz(U).

2.3. THE NERVE OF AN OPEN COVERING

135

For (l),let L be a normed linear space which realizes N ( U ) (e.g., 12, see the proof of Proposition 2.3.1). Take an arbitrary x E X . There is a neighborhood W of x meeting only finitely many elements of U. Since the infinite sum in the definition of n for points of W clearly reduces to a finite sum, and the functions KU are all continuous, by the continuity of the algebraic operations on L it follows that K regarded as a mapping from X into L is continuous. Now let W = {U E U : U n W # 0) and let Y be the union of all simplices of N ( U ) the vertices of which correspond to elements of the collection W . Observe that Y is a finite union of elements of N ( U ) and hence is compact. Since 2 E W it follows that 3(2)C W and so K(X) E T(X)

g Y.

Since Y is compact, by Lemma 2.1.13(1) it follows that the topology that Y inherits from L is the same as the topology that Y inherits from IN(U)I. We conclude that n: X -+ IN(U))is continuous at all points of W , and hence at the point x. 0 Let X and Y be spaces and let U be an open cover of X . A continuous function f : X + Y is called a U-mapping if there is an open cover V of Y such that f-'[V] < U.This concept is related to the concept of a &-mapping on Page 34. For let f : X -+ Y be an &-map, where X is compact, and let U be the open cover of X consisting of all open sets of diameter less than E . We claim that f is a U-mapping. To this end, first observe that if

Y E v =Y\f[XI

then V is an open neighborhood of y such that f-l[V] = 0 is contained in every U E U.Next, if y E f [ X ]then let U be an open neighborhood of f-'(y) such that diam(U) < E . Since by compactness of X the function f is closed (Exercise A.5.5), there is a neighborhood V of y such that f-'[V] g U (Exercise A.1.15). So we conclude that f is a U-mapping. We see that the concept of an &-map is indeed the 'compact' version of the concept of a U-map.

Corollary 2.3.4. Let X be a space and let U be a locally finite open cover of X . Then K : X -+ IN(U)I is a U-mapping. Proof. This follows from Theorem 2.3.3(2) and Corollary 2.1.17.

0

It is unfortunately not the case that a cover U of a space X is locally finite if and only if the simplicial complex N ( U ) is locally finite. Since, as we pointed out above, our main interest is in locally finite simplicial complexes, it is natural t o wonder when N(U) is locally finite. First a definition. An open cover U of X is said to be star-finite if for every U E U the set {V E U : V n U # 0)

2. BASIC COMBINATORIAL TOPOLOGY

136

is finite. Observe that a star-finite open cover is locally finite. Consequently, every star-finite cover is countable. This definition leads us to the following

Theorem 2.3.5. Let X be a space and let U be an open cover of X . Then (1) there exists an open refinement V of U such that V is star-finite, (2) N ( U ) is locally finite if and only if U is star-finite.

Proof. For (l),we may assume without loss of generality that X is a subspace of the Hilbert cube Q (Corollary A.4.4). For every U € U let U C Q beopensuchthatUnX=Uandputc={U:UEU}. ThenV=Ucis an open subset of Q and hence is locally compact and a-compact. There are compacta F, C V for n E N such that

F1 ~ I n t F z E F 2 C _ I n t F 3 C . . . C F n - 1c I n t F , C F , C . . . while moreover V = U =,: F,. Put Fo = 0. For every n let &, be a finite subcollection of with Fn C U &a. NOWput

c

W = { U n (Q \ Fn-1) n Int Fn+l : U E &,,n E N}. It is clear that W covers V, is star-finite and that W the collection V = W 1X is as required.

<

c. As a consequence,

For (2), first assume that N ( U ) is locally finite. If there exists U E U such that the set {v E U : v n U # 0) is infinite then the vertex x(U)of N(U) is contained in infinitely many onedimensional simplices of N ( U ) , which contradicts the local finiteness condition on N ( U ) . The verification of the reverse implication is a triviality which 0 we leave to the reader. We shall now prove that every space can be 'approximated arbitrarily closely' by a polytope. Corollary 2.3.6. Let X be a space. For every open cover U of X there exists a polytope P and a U-mapping f : X + P. If X is compact then P can be chosen to be a polyhedron. Proof. Let U be an open cover of X . By Theorem 2.3.5(1), there exists a star-finite open cover V of X which refines U. Let IC:

x + lN(V)(

be the /+mapping of V. Since V is star-finite, IN(%')( is locally finite by Theorem 2.3.5(2). By Corollary 2.3.4, IC is a V-mapping. Since V < U, we are done. If X is compact then V is a finite subcover of U and proceed as above. 0

2.3. THE NERVE OF AN OPEN COVERING

137

Freudent hal's Approximation Theorem. We now come to an interesting 'Approximation Theorem'. See 51.10 for results that are in the same spirit. Freudenthal's Approximation Theorem 2.3.7. Every compact space is homeomorphic to the inverse limit of an inverse sequence consisting of polyhedra. Proof. We start as in Example 1.10.5. By Corollary A.4.4 we may restrict ourselves to a compact subspace X of Q. For every n let where 7,: Q defined by

+ 9"

x, = n,[XI, is the projection. In addition, let fn(x1r...,x,+1)

f,:

Xn+l

+ X,

be

= (x1,...rGJ

Then l p ( X , , f,), = X (Example 1.10.5). Unfortunately, the sets X , need not be polyhedra. It will require a little extra work to take care of this problem.

Claim 1. Let X be a compact subspace of J" for some n. Then for every neighborhood U of X there exists a polyhedron P such that X C P U .

c

Proof. By Corollary A.5.4 there exists 6 > 0 such that if A C: Jn,diamA < 6 and AnX # 0 then A E U . Since J" is a polyhedron, it has by Theorem 2.2.10 a triangulation S such that for every a E S we have diama < 6. Let P be the union of all simplices in S that intersect X . An easy check show that P is as required. 0 Now for every n E N we shall define a polyhedron P, in 9" with X , 2 P,, as follows. Let PI = J. By Claim 1 there exists a polyhedron P2 in J 2 such that X2 C P2 B ( X 2 , By the Dugundji Theorem 1.2.2, we can extend the function f 1 : X2 + X1 C_ PI to a continuous function [ I : P2 + P I . For P3 we have to do a little extra work. Since P2 is an ANR (Theorem 2.1.24), there is a neighborhood U of X 3 such that the function f 2 : X 3 + X Z + P2 can be extended to a continuous function q : U + P2. Pick a neighborhood V C U of X 3 such that

c

c

c

V[Vl B ( X 2 , 1 / 3 ) , G O dV1 B P I , 1/31. By Claim 1, we may pick a polyhedron P3 such that X3 be the function 1 P3.

P3

E V . Let &

Continuing in this way inductively, yields an inverse sequence of polyhedra (P,,&) such that for every n, X , P, 2 J", [ Xn+l = f, and

(*I

n t, 03

m=n

c

.. .

tm[P,+~I = x,.

138

2. BASIC COMBINATORIAL TOPOLOGY

f

We think of @(Pn, En) and 'm(X,, fn) = X as subspaces of claim that @(P,, En) = @( , fn) = X . It is clear that @ ( X n , fn)

n:=,

9". We

2 @(Pn, En).

Now take an arbitrary x E @(Pn,En) and fix N E N. Then,

n 03

XN

E

EN

0

... o E

~ [ P ~ + ~ I

m=N

and hence X N E X N by (*). Since implies that x E l&(X,, fn).

Cn

extends

fn

for every n, this easily 0

Exercises for $2.3. 1. Let X be a space and let U and V be open covers of X such that V < U. U ( V ) . This defines a For every V E V pick U ( V ) E U such that V function fo : N(V)(O)+ N(U)(O).

Extend this function 'linearly' over each simplex of N ( V ) and prove that the resulting function from IN(V)I + IN(U)I is continuous.

2. Prove that the following statements for a space X are equivalent: (1) X is compact. (2) Every locally finite open cover of X is star-finite.

2.4. Simplices in Rn In this section we shall formulate and prove some fundamental properties of simplices in Rn. Among other things, we shall present a proof of the Brouwer Fixed-Point Theorem and some of its applications.

Lemma 2.4.1. Let u1 and 6 2 be two (geometric) n-simplices in Rn which intersect in a common ( n - 1)-face r. Then u1 U 02 is a neighborhood of the baxycenter of r. Proof. Observe that the lemma is trivial if n = 1; we may therefore assume that n > 1. Let 61 = {ao, al, . . . ,an-l, af} and b2 = {ao, a l , . . . ,an-l, a;}, where T = u1 n C T ~= I{ao,al,.. . The points {ao, a1 , . . . ,a,-l, ax} are geometrically independent, from which it follows by Theorem 2.1.3 that the points a1

- ao,a2 - ao, . . . ,an-1

1

- ao,a, - a0

2.4. SIMPLICES IN

W"

139

are linearly independent. Now consider the standard basis el, e2,. . . ,e, in R" and a linear isomorphism g1 : Rn + Rn with the properties:

{

gl(ei) = ai - a0 gl(en) = at - ao.

(i < n),

This isomorphism followed by the translation x I+ x + a0 is an affine isomorphism (Theorem 2.1.7) and hence a homeomorphism (see Page 113) which we denote by f l . Observe that f has the following properties:

{

f"Q) = ao, fl(ei) = ai fl(e,) = ah.

(1 5 i

5 n - I),

In the standard basis we now change e, into -en. By a similar argumentation as above we obtain an affine isomorphism f : Rn + Rn such that

{

f2(Q) = ao, f2(ei) = ai f2(-en) = a:.

(1 5 i 5 n - I),

Observe that f ' and f 2 agree on the plane P = {x E Rn : xn = 0). Also observe that f ' 1 P = f 2 1 P is an afine isomorphism between P and the hyperplane spanned by the elements of i = {ao,a l , . . . ,un-l}. Now define a homeomorphism f : Rn + Rn as follows

Let b be the barycenter of and the set

T

and let c =

n n2 is a neighborhood of c of which of which it is geometrically obvious that it is mapped by f into t71 U t72 (see Exercise 2.4.1). We conclude that 01 U t72 is a neighborhood of b. 0 The next result is the key in the proof of the Brouwer Fixed-Point Theorem.

Theorem 2.4.2. Let T be an n-simplex in Rn, and let S be a finite triangulation of T which subdivides its standard triangulation. Finally, let p E S be an (n - 1)-simplex and let b be the barycenter of p. (1) If b E dT then p C_ dr and p is a face of precisely one n-simplex in S. (2) If b # 67 then p is a face of precisely two n-simplices in S.

Proof. Assume that b E 67. We will show that p 67. Since S subdivides the standard triangulation of T there is a simplex t7 E S with b E t7 such

140

2. BASIC COMBINATORIAL TOPOLOGY

that D is contained in a proper face of T , i.e., D g &. By Lemma 2.2.1 we obtain that p is a face of D so that p g 67. Now we shall prove that depending on whether b E drr or not, p is a face of at most one or at most two n-simplices in 5.

To this end, first assume that b E 87. If there exist two n-simplices in S having p as a common face then the union of these two simplices is a neighborhood of b (Lemma 2.1.5). But this contradicts the fact that b E 67 which is equal to the topological boundary of T (Exercise 2.1.9). If b # d~ and there exist three distinct n-simplices, say 61, ~2 and ( having p as a common face then since S is a simplicia1 complex,

~ E 3

S,

(al u g2)n (c2u c3)n ( g 3 u ol) = (gl n g 2 ) u ( g 2 n 0 3 )u ( g 3 n gl) = p.

For the last equality observe e.g., that p is an ( n - 1)-simplex contained in the simplex 01 n 02. If (TI n ( ~ 2is an n-simplex then ~1 = C J ~which is a contradiction. So 01 n 6 2 is an (n - 1)-simplex since it contains p as a face, hence it must be equal to p. By Lemma 2.4.1 the above intersection is also a neighborhood of b in R" . Consequently, p is a neighborhood of b in Rn which is impossible because p is contained in an ( n - 1)-dimensional hyperplane.

For the remaining part of the proof, put

u = T\

U{DE s : b #

0).

Observe that U is a nonempty relatively open subset of

T.

We shall now prove that p is a face of at least one n-simplices of S. Striving for a contradiction, assume that there does not exist an n-simplex in S having p as a face. According to Lemma 2.2.1 this implies that p is the only simplex in S containing b which implies that U p. Now if b E 8.r then we arrive at the desired contradiction because then U g p C_ 67 and 67 has empty interior in T . If b # 67 then U n Int(.r) is a neighborhood of b E Rn which is contained in p. But this contradicts the fact that 1.1 is contained in an ( n - 1)-dimensionalhyperplane of R" . All there remains to prove is that if b # dT then there are at least two n-simplices having p as a face. Striving again for a contradiction, assume that this is not the case. We already know from the above that there is an n-simplex (T E S having p as a face. Pick an arbitrary p E U and let y E S be such that p E y. Then b E y by the definition of U and so by the same argumentation as above, p is a face of y. But then either y = p or y = D . In either case, p E D . We conclude that U g D . So Int(7) n U is a neighborhood of b E Rn which is

c

2.4. SIMPLICES IN W n

141

contained in o. Since p is a proper face of o we also have b E do. This is a contradiction. 0 Let

T

=

. . , z k } l be an arbitrary k-simplex in R" . An rn-Sperner

I(z0,.

map for T is a function from the vertices of ~ d ( ~ ) 3 (to7 (0,. ) . . , k} such that

if v is such a vertex and

y

E

I{%, . . . , Z i e ) l

then

h ( v ) E {io, . . . ,ie}. We call a k-dimensional simplex o in ~ d ( ~ ) 3 (full 7 ) if h[b]= (0,. . . , k}.

There are always full simplices as the following result shows. As to be expected, its proof is of combinatorial nature. Sperner's Lemma 2.4.3. Let T = ~ { x o ,. ., z k } l be a k-simplex in Rk and let h be an rn-Sperner map for T. The number of full simplices in ~ d ( ~ ) 3 ( ~ ) is odd and hence non-zero.

Proof. We shall prove the lemma by induction on k. If k = 0 then T = 1{zo}l, and there is one full simplex. Assume that the theorem is true for all (k- 1)simplices, k - 1 2 0. We shall prove the theorem for T. Put

P = (20,.. . ,Zk-1), ?I = {o E sd(m)F(T): o is (k - 1)-dimensional, h[b]= (0,. . . ,k - l}}, 3'2

= {o E ~ d ( ~ ) 3: o( is ~ k-dimensional, ) (0,. . . , k - 1) C h[b]},

c

?1(0) = {aE ?l :7.r IPI}, %(I)= ? 1 \ ?l(O), ?2(0) = {(T E ?2 : h[&]= (0,. . . ,k}}, ? 2 ( 1)

= R?\ ?2(0),

respectively. Observe that ~ d ( ~ ) 3 ( Psd ) (")F( ~ ) . By the definition of an rn-Sperner map it therefore follows that the restriction of h t o the vertices of ~ d ( ~ ) 3 (isp an ) rn-Sperner map for P. As a consequence, the collection 9'1 (0) is the set of full simplices in the rn-th barycentric subdivision of IPI, so IrP,(O)l is odd by our inductive assumptions. Also, ?z(O) is the set of full simplices in the rn-th barycentric subdivision of T ;so we have to prove that I?2(0)I is an odd number. Put 3= {(IE,~E ) 1'9 x "2 : IE is a face of p } .

2. BASIC COMBINATORIAL TOPOLOGY

142

We compute the cardinality of R twice. For each n E '91 put

R[n] = { p E Ip2 : (n,p) E a}.

Claim 1.

1x1= lIp1(0)( + 2 . IIp1(1)1.

Proof. Clearly,

1x1 =

c la[4

KE91

We claim that if n E Ipl(0) then l%[n]I = 1. Simply observe that such n is a face of precisely one k-dimensional simplex in ~ d ( ~ ) ! F (by . r ) Theorem 2.4.2, and this simplex clearly belongs to 3'2. We next claim that if n E Ipl(1) then lR[n]1 = 2. For take an arbitrary IE E Ipl(1) and assume first that it is . . ,xikpl}1.Then contained in a (k - 1)-dimensionalface of T , say n C 1(q0,.

h[k] g {io, . . . ,ik-1)

which implies that {io,. . . , i h - l } = (0,. . . ,k - 1) and so K 1/31, contradiction. So such a n is not contained in a (k - 1)-dimensional face of T , and therefore, again by Theorem 2.4.2, is a (k - 1)-dimensional face of precisely 0 two k-simplices in ~ d ( ~ ) 3 ' ( 7and ) , these simplices clearly belong to "2.

For each p E

"2

put

a-"p] = {n E PI : (n,p) E a}.

+

Claim 2. 1R1 = I!P2(0)I 2 . IIp2(l)l. Proof. Clearly, 131=

c

la-l[Pll.

PE92

Take p E " ~ ( 0 ) .We claim that lX-'[p]I = 1. Indeed, h[b] = (0,. . . ,k), i.e., the function h is one-to-one on b. So p has exactly one (k - 1)-dimensional face n such that (0,. . . ,k - 1) = h[k]and this face is clearly the only element then h[b] = (0,. . . ,k - 1). Since has size k 1, of lX-'[p]I. If p E fP~(1) there exist precisely two subsets E and F of b of cardinality k with the property that h[E] = (0,. . .,k - 1) = h[F]. Consequently, there exist precisely two (k - 1)-dimensional faces n1 and n2 of p with h[k1] = (0,. . .,k - 1) = h[k2]. These simplices are obviously the only elements of [ 3-l [p]I. 0

+

We find that

IP2(0)l - I%(O)l = 2 ( I ~ 2 ( l > l -I%(1)1) is even. As was remarked at the beginning of the proof, [Ip1(0)1 is odd, hence so is ltp2(0)1. 0

2.4. SIMPLICES IN

W"

143

This result has interesting consequences.

Lemma 2.4.4. Let T be a k-simplex in Rk, say T = ~ { x o.,. . ,Xk}l. In addition, let {Fi : 0 5 i 5 k} be a collection of closed subsets of R" such that for every {io, . . . ,ie} (0,. . . ,k} we have Then

n:=oFi # 0.

I {xio . ,xic}I C Fi0 u . . . u Fie. 9

Proof. If not, then T = (T \ Fo) U . . . u (T \ Fk). Let E > 0 be a Lebesgue number for this covering (use that T is compact and apply Lemma A.5.3). By Corollary 2.2.9 we can choose rn so large that m e ~ h ( s d ( ~ ) 3 < ( ~E). )Let V be the set of vertices of elements of ~ d ( ~ ) 3 ( 7Let ) . w E V and put

I,, = (0 5 i 5 k : ai(.) > 0) (here ao(w), . . . , ak(w) are the affine coordinates of w with respect to 20,.. .,Xk of course).

Figure 8. Let I,, = {io,. . . ,it}. Then E I { ~ i ~ , . . . , ~ i ~ } lE

Fi0 U...UFi,

So for w E V we can pick h(w) E (0,. . . ,k} such that

.

> 0,

E G(7J). Then h is an rn-Sperner map for T . By Theorem 2.4.3 there exists a full 7 )h, say ir = ( 2 1 0 , . . . ,wk} with h(vi) = i. Then simplex (T E ~ d ( ~ ) 3 (for clearly wi E (T n Fi for every 0 5 i 5 k. But diam(a) < E so there exists i 0 with (T T \ Fi, i.e., (T n Fi = 8. This a contradiction. ah(,,) (0)

We now come to the main result of this section.

The Brouwer Fixed-Point Theorem 2.4.5. Let f : 1" ous. Then f has a fixed-point.

+ I["

be continu-

144

2. BASIC COMBINATORIAL TOPOLOGY

Proof. Let T = [{xo,. . . ,x,}[ be an n-simplex in Iwn. Then T is homeomorphic to In (Exercise 1.1.24) and so it suffices to prove the Theorem for T . To this end, let f : T + T be continuous. For i = 0,1,. . . ,n let Fi = {X E T : ~ i ( f ( ~ ) ) 5 Q ~ ( z ) } (here ao(x),. . . ,a,($) are the affine coordinates of x with respect to ZO, of course). Let [{xi,,, . . . ,xie}[be a face of element i 6 {io, . . . ,it},

T

. . . ,xn

and x E [{xi,, . . . ,x i c } ] . Then for an

ai(x)= 0

so that

aio(x)

+ + aip(2)= 1 >_ a i o ( f ( x ) )+ ... + ail (f(z)) *

*

a

5 C with

and hence there must be 0 5 j Qij

).(

>_ ai,( f ( 4 )*

Consequently, x E Fij. We conclude that

({xio,.. . , Z i t } (

c_ Fi, u .. . u Fie.

Each Fi is closed by continuity off and the functions ai (Theorem 2.1.8). So by Lemma 2.4.4 there exists x E Fi. Then for every 0 5 i 5 n,

ny=o

and

n

n

c a i ( x )= 1 = i=O

Cai(f(2))' i=O

This implies ai(x) = ai(f(z))for every 0 5 i equal t o f(x).

5 n and consequently, x is

0

Corollary 2.4.6. The Hilbert cube Q has the fixed-point property

Proof. Let f : Q + Q be continuous. For every n E N define

K n = { z € Q :(21,..., x,)=(f(x)l,...,f(x),)}. It is clear that for every n the set K , is closed in Q and that K,+l K,. Fix n E N, let p , : Q + 9" denote the projection and define a continuous function f,: J" + J" by

. .,%, o,o,. . .). By Theorem 2.4.5, f, has a fixed-point, say (31,. . . ,x,), from which it follows fn(z1,.

. .,2,)

that

(21,.

= (P, 0 f)(.l,.

. . ,x,, o,o,. . .)

E

K,.

2.4. SIMPLICES IN

W”

145

We conclude that the Kn’s form a decreasing collection of nonempty closed subsets of Q so that by compactness of Q ,

n 00

K=

K,

#0.

n=l

It is clear that every point in K is a fixed-point of f.

0

Corollary 2.4.7. Let X be a compact ANR. Iff : X + X is nullhomotopic then f has a fixed-point. In particular, every compact AR has the fixed-point property.

Remark 2.4.8. In view of this result, the following question naturally arises. Let X be a compact ANR and let f : X + X be continuous. Assume that f is homotopic to a continuous function g : X + X such that g has a fixed-point. Does f have a fixed-point? The answer to this question is in the negative, see Exercise 2.4.7. So in some sense, Corollary 2.4.7 is ‘best possible’. Proof. We may assume that X is a subspace of Q (Theorem A.4.4). Since every constant function X + X is extendable to a constant function Q + X , by Corollary 1.4.3 it follows that f can be extended to a continuous function f:Q+X-Q. Since by Corollary 2.4.6, Q has the fixed-point property, f has a fixed-point, say x. Since f [ Q ] X, x belongs to X , and since f extends f, we conclude that x is a fixed-point of f . Since by Theorem A.12.4 every compact AR is contractible, the second part of the Corollary immediately follows from the first part and Proposition A.12.2.

s

Remark 2.4.9. In 53.14 we will show by elementary but remote methods that Brouwer’s Fixed-Point Theorem for 13,implies the Brouwer Fixed-Point Theorem for all dimensions. This seems to show that the ‘real’ power of the Brouwer Fixed-Point Theorem is already attained at dimension three. This is a rather curious phenomenon and it is not clear whether there is a direct route from Brouwer in I3 to Brouwer in all dimensions. This is of course only of interest for ‘philosophical’ reasons. The proof of Sperner’s Lemma 2.4.3 is by induction, so it is equally complicated in all dimensions (except for the dimensions 1 and 2). Similar remarks can be made for the various other proofs that exist of Brouwer’s Theorem in the literature. This seems to indicate that even if one finds a more direct route from Brouwer in I3 to Brouwer in all dimensions, this will not result in a simpler proof of Brouwer’s Theorem. But this does not mean that clarifying the rather curious phenomenon that we will encounter in 53.14 would not be interesting. We finish this section by presenting three applications of the techniques developed in this section.

2. BASIC COMBINATORIAL TOPOLOGY

146

Application 1: The No-Retraction Theorem. It is clear that the function r : 9" + B" defined by

is a retraction. Consequently, B" has the fixed-point property by Theorem 2.4.5 and Exercise A.12.10(2) (observe that in fact 1" and Bn are homeomorphic, cf. Exercise 1.1.24).

Theorem 2.4.10 (No-Retraction Theorem). For every n E N,SnF1is not a retract of Bn. Proof. To the contrary, suppose that SnP1is a retract of B". As was remarked above, Bn has the fixed-point property. Consequently, S n - l has the fixed-point property by Exercise A.12.10(2). However, the antipodal mapping on s"-' clearly demonstrates that s"-' does not have the fixed-point property. Contradiction. 0 Corollary 2.4.11. No 9" is contractible. Proof. Suppose that 8" is contractible. Then it is an AR by Corollaries 1.2.13 and 1.4.5. Consequently, there exists a retraction r : B" + S n - l , which contradicts Theorem 2.4.10. 0 Application 2: The Theorem on Partitions. The following result is the basis for dimension theory (see Chapter 3). The Theorem on Partitions 2.4.12. Consider 9" and for i site faces

5 n its oppo-

Ai = { 2 E J" : 2i = -l}, Bi = {2 E J" : zi = 1). If Ci is a partition between Ai and Bi for i 5 n then Ci #

n:=,

8.

Proof. To the contrary, assume that for every i 5 n, Ci is a partition beCi = 0. There are closed subsets Ei and Fi tween Ai and Bi such that in 9" for i 5 n such that

ny=l

c

c

Ai Ei, Bi Fi, Ei U Fi J", Ei n Fi = Ci. By Exercise A.4.9 there exist continuous functions & : 9" + J such that 1

t i [ ~= i ](11, t i [ ~ i=] {-I}> ti'(0) = ct. Define f : J" + 9" by f(z)= (&(z), . . . , ("(2)). Then f is continuous and the point 0 does not belong to its range. For every x E 9"\ (0) the ray from 0 through x intersects the 'boundary' B = Ai U Uyrl Bi of J" in precisely one point, say r ( z ) . The function r : 9"\ (0) + B is easily seen to be continuous. The function g = r o f : 9"+ B has the following properties:

u:=,

g [ ( - i , i)"]n ( - 1 , i ) " = 0,

2.4. SIMPLICES IN W n

and for every i

5 n,

147

c

c

Bi, g[&I Ai. Therefore, g has no fixed-point, which contradicts Theorem 2.4.5. Corollary 2.4.13. Consider the Hilbert cube Q and its opposite faces

Wc' = {Z E Q : xi = -l}, If Ci is a partition between W';

W t = {X E Q :

= 1).

Ci # 0.

and W; for every i then

Proof. For every m, define f m :9" + Q by

,

fm(Z1,... Z m )

= ($1,. . . ,Z m , O,O,.

- .).

Then fm is clearly an imbedding. It is easily seen that f;l[Ci] is a partition between Ai and Bi for every i 5 m. Consequently, by Theorem 2.4.12,

n~ m

i=l

3 z 0. ~

~

1

By compactness of Q we therefore obtain that 00

nci#B,

i=l

0

which is as desired.

Application 3: The Non-Homogeneity Theorem.

N. Then (1) if A g S"-' then Bn \ A is contractible. (2) if A g B" \ Sn-' is nonempty then Bn \ A is not contractible.

Theorem 2.4.14. Let n E

Proof. (1) Define H : (B" \ A ) x I[ + Bn \ A by H ( s ,t ) = (1 - t ) z . It is easily seen that H is a contraction. (Alternatively, apply Exercise 1.2.5 and Theorem 1.4.4.)

(2) Without loss of generality, 0 E A. Assume to the contrary that

H : (B n \ A ) x I [ + B n \ A is a contraction. Define F : Sn-' x 1 + Sn-l by

Then F contracts Sn-l to a point which contradicts Corollary 2.4.11. Since J" and B" are homeomorphic, cf. Exercise 1.1.24, this yields:

Corollary 2.4.15. Let n E N. Then I[" is not homogeneous.

0

2. BASIC COMBINATORIAL TOPOLOGY

148

Exercises for $2.4. Let X be a space. We say that a closed set A in X separates X if X \ A is not connected. ~ 1 Make . the 'it is geometrically obvious' part in the proof of Lemma 2.4.1 precise.

2. Prove that the 'No-Retraction Theorem,, the 'Brouwer Fixed-Point Theorem' and the fact that no 9" is contractible are equivalent statements, in the sense that they are easily deduced from each other. D3. Let C be a compact convex subset of a normed linear space L. Prove

that for every E > 0 there exists a map f E : C -+ C such that (1) b(fe11) < El ( 2 ) fE[C] is contained in a finite dimensional linear space subspace of L.

+ X be a map with the following properties: (1) the closure of f [ X ]is compact, ( 2 ) for each E > 0 there exists E E X with e(z, f ( 2 ) )< E . Prove that f has a fixed-point.

4. Let X be a space and let f : X

5 . Let C be a compact convex subset of a normed linear space L. Prove that C has the fixed-point property. (This is called the Schauder FixedPoint Theorem.) 6 . Prove that the antipodal map 9"

+ 9" is not nullhomotopic.

7. Give an example of a compact ANR X having a homotopy H : X x 1-+ X such that HOis the identity and HIis fixed-point free.

c

R" be compact, where n > 1. Prove that if A separates R" then there is a continuous function f : A + Sn-l which is not nullhomotopic.

D8. Let A

c

9. Let A 9" be compact, where n 2 1. Prove that if A separates 9" then S n P 1which cannot be extended there is a continuous function f : A -i over 9".

2.5. The Lusternik-Schnirelman-Borsuk theorem

The Brouwer Fixed-Point Theorem which was proved in the previous section, is one of the most central results in topology. We presented a detailed proof of it for two reasons. First of all, its proof uses subdivisions of triangulations of simplices. Getting used to this is important because that will be used extensively in ANR-theory later. Secondly, it is the basis for dimension theory. A more powerful result is the so-called Lusternik-Schnirelman-Borsuk Theorem which can be proved along the same lines using a somewhat more complicated combinatorial lemma than the one of Sperner (Lemma 2.4.3). We will not present a proof of the Lusternik-Schnireiman-Borsuk Theorem here since it is used only twice in this book and for our purposes its proof does not give more insight than the proof of Sperner's Lemma.

2.5. T H E LUSTERNIK-SCHNIRELMAN-BORSUK THEOREM

149

The Lusternik-Schnirelman-BorsukTheorem 2.5.1. If M is a n arbitrary closed covering of Bn such that IMl = n 1 then at least one element M E M contains a pair of antipodal points.

+

For a nice combinatorial proof, see DUGUNDJI and GRANAS[142, Theo-

rem 4.41.

T h e Borsuk-Lusternik-Schnireiman Theorem implies many deep theorems about the topology of the Euclidean spaces Rn. Exercises for 52.5. A map f : 9" + 9" is called antipodal-preserving provided that f (-%) = -f (x)for every x E 8". 1. Prove that the Lusternik-Schnirelman-Borsuk Theorem is sharp by showing that there exists a closed covering M of 9" such that IMI = n 2 while no M E M contains a pair of antipodal points.

+

b 2 . Prove that the following statements are equivalent in the sense that they

are easily deduced from each other: (1) The Lusternik-Schnirelman-Borsuk Theorem for 9". (2) There is no antipodal-preserving map f : 8" + r - l . (3) (Borsuk Antipodal Theorem) An antipodal-preserving map f : 9-1 + gn-1

is not nullhomotopic. (4) (Borsuk-Ulam Theorem) Every continuous

f:S"+W" sends at least one pair of antipodal points to the same point. 3. Prove that the Lusternik-Schnirelman-Borsuk Theorem implies the Brouwer Fixed-Point Theorem.