COINCIDENCE AND THE COLOURING OF MAPS

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1 COINCIDENCE AND THE COLOURING OF MAPS JAN M. AARTS AND ROBBERT J. FOKKINK ABSTRACT In [8, 6] it was shown that for each k and n such that 2k n, there exists a contractible k-dimensional complex Y and a continuous map φ: n Y without the antipodal coincidence property, that is, φ(x) φ( x) for all x n. In this paper it is shown that for each k and n such that 2k n, and for each fixed-point free homeomorphism f of an n-dimensional paracompact Hausdorff space X onto itself, there is a contractible k-dimensional complex Y and a continuous map φ: X Y such that φ(x) φ( f (x)) for all x X. Various results along these lines are obtained. 1. Introduction The classical antipodal coincidence theorem of Borsuk and Ulam states that for every map φ: n k with k n, there is an antipodal coincidence point, that is, a point x such that φ(x) φ( x). By Conner and Floyd [3], it was shown that k can be replaced by a k-dimensional manifold if k n. S c epin [8], Izydorek and Jaworowski [6] showed that for each k and n with 2k n, there exists a contractible k-dimensional complex Y and a map φ: n Y with no antipodal coincidence point, φ(x) φ( x) for all x n. Furthermore, it was shown in [7] that the bound 2k n is sharp. These results shed some light on the role of the codomain k in the Borsuk Ulam coincidence theorem. In this paper we investigate the role of the domain in the results of S c epin, Izydorek and Jaworowski. Here is our main result. THEOREM 1. Let X be a paracompact Hausdorff space such that the co ering dimension dim X n. Suppose that f is a fixed-point free homeomorphism of X onto itself. Then for each k with 2k n, there exists a contractible k-dimensional complex Y and a continuous mapping φ: X Y such that φ(x) φ( f (x)) for all x X. Briefly, the conclusion of Theorem 1 is the existence of a k-dimensional contractible complex Y and a mapping φ: X Y such that φ has no f-coincidence point. We introduce some terminology. Let f: X X be a fixed-point free map. We say that f can be coloured with k colours if there is a closed cover C,,C k such that no C i contains a pair x, f(x) or, equivalently, C i f(c i ) for each i 1,, k. The elements of are called colours, and we shall say that is a colouring of f. It is easily seen that a fixed-point free continuous map of a compact Hausdorff space has a colouring. Furthermore, a fixed-point free closed continuous map f of a normal Hausdorff space to itself has a colouring if and only if its C ech Stone extension βf is fixed-point free (compare [4]). In [1, Theorem 2] it was shown that under the hypotheses of Theorem 1, the homeomorphism f can be coloured with n 3 colours. Here we use the techniques of map colouring from [1] to prove the following result. Received 21 November Mathematics Subject Classification 55M10, 54C05. Bull. London Math. Soc. 30 (1998) 73 79

2 74 JAN. M. AARTS AND ROBBERT J. FOKKINK THEOREM 2. Let X be a normal Hausdorff space such that the co ering dimension dim X n. Suppose that f: X X is a closed continuous map which has a colouring. Then for each k with 2k n, there exists a contractible k-dimensional complex Y and a continuous mapping φ: X Y such that φ(x) φ( f(x)) for all x X. As any homeomorphism of a paracompact Hausdorff space onto itself has a colouring, by [4, Theorem 1.1], Theorem 1 is a special case of Theorem 2. The proofs of Theorems 1 and 2 have two parts. First, in Section 2, we prove the theorems under the extra hypothesis that the space X is metrizable and compact metrizable, respectively. In Section 3 we explain how the extra hypothesis can be removed. Recall the following facts [2]. If A i i I and B i i I are covers of X, then is called a shrinking of and is called a swelling of if B i A i, for every i I. Every finite closed cover F i i I of a normal space has an open swelling U i i I which is combinatorially equivalent, that is, F i F i F im if and only if U i U i U im for every finite set i,, i m of indices of I. Every finite open cover of a normal space has a closed shrinking. We point out that in view of these facts, the colouring of a map could be defined via open covers as well. The closure of a set V is denoted by V, and its boundary by V. If V i i I is an indexed collection, then ord p ( ) is the number of indices i in I such that p V i and ord V sup p ord p. 2. The metrizable case In this section we assume that all spaces are metrizable. The following lemma was proved in [1, Lemma 11]. Note that in this lemma the homeomorphism may have fixed points. LEMMA 3. Let X be a metrizable space such that dim X n, and let f be a homeomorphism of X onto itself. Suppose that U i 1 i m is an open swelling of the closed co er K i 1 i m of X. Then there is a closed co er L i 1 i m such that K i L i U i for each i, and for each function δ: 1,, m 0, 1, we ha e f δ (i ) ( L i ) f δ (i n+ )( L in+ ) (1) whene er 1 i i n+ m. Here we use the con entions f id and f f. In the following proposition we isolate the key to the proof of the metrizable version of Theorem 1. PROPOSITION 4. Let X be a metrizable space such that dim X n. Suppose that f is a fixed-point free homeomorphism of X onto itself. Then for each k with 2k n, there is a colouring C,,C l of f such that (1) ord k 1, (2) ord ( f( )) n 2. Proof. By [4, Theorem 1.1], f has a colouring, say K i 1 i m.ina standard fashion, we obtain an open swelling U i 1 i m of such that

3 COINCIDENCE AND THE COLOURING OF MAPS 75 f(u i ) U i for each 1 i m (thus is an open colouring of f ). By the previous lemma there is a colouring L i 1 i m such that K i L i U i for each i, and equation (1) of Lemma 3 holds. Now we form a covering G i 1 i m as follows: (1) G L, (2) G i L i (L L i ) for 2 i m. By [1, Lemma 13], is a colouring of f such that (3) is a shrinking of, (4) G i G j G i G j for 1 i j m, (5) if G i G ij, then L i L ij, whenever j 2 and 1 i i j m. We shall show that ord ( f( )) n 2. Suppose that x belongs to s members of f( ). Suppose that x G j for p indices j, and x f(g j ) for q indices j. Note that p q s, because is a colouring of f. By (4) and (5) we conclude that x L j for at least p 1 indices j, and x f( L j ) for at least q 1 indices j. By equation (1) of Lemma 3, we obtain p 1 q 1 n. Thus s n 2. The result is that the condition (2) of the proposition is satisfied by the G i instead of the C i. The cover f( ) has an open swelling which is combinatorially equivalent. Then there is an open cover such that (6) G i V i V i U for 1 i m, i (7) f(v i ) V, i (8) ord ( f( )) n 2, where V i 1 i m. Now let K x ord x k 1. Note that K is closed. By contradiction, we shall show that K f(k). Assume that there is an x in K f(k). Then ord x ( ) k 1 and ord f(x) ( ) k 1. It follows that ord f(x) ( f( ) ) 2k 2 n 2, contradicting (8). There is an open set V such that K V and f(v) V. Let V i K 1 i m V. Note that ord k 1, because ord V i K 1 i m k. Let be a closed shrinking of. Note that is a colouring of f and ord k 1. Now we can repeat the first part of the proof with and replaced by and, respectively. The resulting colouring satisfies the conditions of the proposition. There is a straightforward relation between the colouring of a map f and maps into complexes with no f-coincidence. Let C i 1 i l be a colouring of the map f of the space X. Let U i 1 i l be an open swelling of such that and are combinatorially equivalent and f(u i ) U i for all i. For each i we define a Urysohn map ψ i : X [0, 1] such that ψ i (C i ) 1 and ψ i (X U i ) 0. Let φ i (x) ψ i (x) l j= ψ j (x). We shall call such a family φ i 1 i l a partition of unity related to the colouring C i 1 i l. The simplex (x,, x n : n i= x i 1, x i 0 in n is denoted by (n) or. Its k- skeleton is denoted by S k ( (n)) or S k ( ). LEMMA 5. Let f: X X be a map which has a colouring C i :1 i l of order k 1. Let φ i :1 i l be a partition of unity related to this colouring. The product map Φ: X l defined by x (φ (x),, φ l (x)) sends X into the k-skeleton S k ( (l)) and has no f-coincidence: Φ(x) Φ( f(x)) for all x X.

4 76 JAN. M. AARTS AND ROBBERT J. FOKKINK Proof. Note that the collection U i 1 i l used in the construction of a partition of unity related to C i 1 i l is an open colouring of f, and that both collections have order k 1. Thus, for all x and all i, ifx U i, then f(x) U i.it follows that if φ i (x) 0, then φ i ( f(x)) 0. Hence Φ(x) and Φ( f(x)) have disjoint carriers. It follows that the map Φ has no f-coincidence. As ord k 1, for each x X the dimension of the carrier of Φ(x) is at most k 1. It follows that X is mapped into S k ( (l)). Theorem 1 claims the existence of a map into a contractible complex. But the k-skeleton of a standard simplex is not contractible. Now we make up for that. The k-skeleton S k ( ) and the simplex have the same vertex set V. Let denote the subdivision of the simplex, obtained by adding the barycentre b to V. Similarly, let S k ( ) denote the subdivision of S k ( ) obtained by adding the set of barycentres B k of all the k-dimensional simplices to V. Define r k : V B k V b as the identity map on V and r k (b k ) b for all the barycentres in B k. The map r k induces a simplicial map ρ k : k ( ). LEMMA 6. If ρ k (x) ρ k (y) for two different elements x, y S k ( ), then both x and yha eak-dimensional carrier. Proof. Suppose that ρ k (x) ρ k (y). Since ρ k is induced by a vertex map of S k ( ), which is the identity on V, both x and y have a barycentre from B k in their carrier. It follows that the carriers of x and y are k-dimensional. PROPOSITION 7. complex. The image ρ k (S k ( ) ) of S k ( ) is a contractible k-dimensional Proof. The image ρ k (S k ( ) ) is the union of all k-dimensional simplices in of which the barycentre b is a vertex. The complex ρ k (S k ( ) ) is homeomorphic to the cone over the (k 1)-skeleton S k ( ). Hence it is a contractible complex. From Lemmas 5 and 6 and Proposition 7 we obtain a proof of Theorem 1 in the metrizable case. Proof of Theorem 1, the metrizable case. By Proposition 4, the homeomorphism f has a colouring C,,C l such that ord k 1 and ord ( f( )) n 2. In the same way as was obtained from in the proof of Proposition 4, we find an open swelling U i 1 i l such that f(u) U, ord k 1 and ord ( f( )) n 2. Let φ i 1 i l be a partition of unity related to this colouring such that φ(x) 0 if and only if x U i. By Lemma 5, the product map Φ: X l sends X into S k ( (l)) such that Φ(x) Φ( f(x)) for all x. The carriers of Φ(x) and Φ( f(x)) are disjoint. The map φ ρ k Φ maps X into the complex S k ( (l)), which is contractible by Proposition 7. By contradiction, we show that φ has no f- coincidence. Suppose that φ(x) φ( f(x)) for some x. By Lemma 6, the carriers of Φ(x) and Φ( f(x)) must be k-dimensional. It follows that ord ( f( )) 2k 2 n 2, a contradiction. We now state the metrizable version of Theorem 2.

5 COINCIDENCE AND THE COLOURING OF MAPS 77 THEOREM 8. Let X be a compact metrizable space such that dim X n. Suppose that f: X X is a fixed-point free continuous map. Then for each k with 2k n, there exist a contractible k-dimensional complex Y and a continuous mapping φ: X Y with no f-coincidence. Proof. We first consider the case that f is surjective. Let Z be the inverse limit space x i i x i f(x i+ ), i. Note that Z is a compact metrizable space such that dim Z n. The projection of Z onto the jth coordinate space is denoted by π j. Let σ: Z Z be the shift map, σ( x i i ) f(x i ) i. Note that σ is a fixed-point free homeomorphism. By Proposition 4 there is a colouring of σ such that ord k 1 and ord ( f( )) 2n 2. Let be an open swelling of f( ) which is combinatorially equivalent. Then there is an open swelling of such that is a shrinking of, f(v) V for all V, ord k 1 and ord ( f( )) n 2. For each C we denote by V(C) the element of that is related to C. Each C is the intersection of the descending sequence π (π j j (C)) j. For sufficiently large j, π (π j j (C)) V(C) for all C. Asf π j π j σ, it follows that π j ( ) isa colouring of f such that ord π j ( ) k 1 and ord (π j ( ) f(π j ( ))) n 2. Now the space Y and the mapping φ can be obtained in the same way as in the proof of Theorem 1. Now we consider the case that f is not surjective. The kernel K i f i(x) is nonempty, because X is compact. The map f: K K is surjective. As in the first part of the proof, we obtain a colouring of f such that ord k 1 and ord ( f( )) 2n 2. Then there is an open swelling of such that f(v) V for all V, ord k 1 and ord ( f( )) n 2. The set W is a neighbourhood of K. For sufficiently large k, f k(x) W. Let f k( ). Then is an open cover of X such that f(u) U for all U, ord k 1 and ord ( f( )) n 2. Let be a closed shrinking of. Then is a colouring of f such that ord k 1 and ord ( f( )) n 2. Now the space Y and the mapping φ can be obtained in the same way as in the proof of Theorem The general case In this section we show how to reduce the general case of Theorem 2 to the metrizable case of Theorem 1. The possibility of such a reduction was pointed out by K. P. Hart (see [5]). THEOREM 9. Let X be a normal space. Suppose that f is a closed continuous map of X to itself. Suppose that is a colouring of f. Then there exist a compact metrizable space Y, a continuous mapping ψ: X Y and a continuous map f of Y to itself such that the collection ψ(c) C is a colouring of f and the following diagram commutes. ψ X Y f f X Y ψ Moreo er, if dim X n, we may assume that dim Y n. If f is a homeomorphism, we may assume that f is a homeomorphism.

6 78 JAN. M. AARTS AND ROBBERT J. FOKKINK Proof. We shall employ the theory of Wallman compactifications (see [2, Chapter VI]). Let denote the family of all closed subsets of X. Inductively, we select a countable subcollection of such that the following conditions are satisfied: (1),, X ; (2) if S, T, then S T, S T ; (3) if S, T and S T, then there exists H such that H S and H T ; (4) if S, T and S T, then there exist G and H in such that G T, H S and G H X; (5) if S, then f(s) and f (S). Note that the closedness of f is needed to make (5) meaningful. Let ω and ω be the Wallman representations of and, respectively. Write Y ω. AsXis normal, ω is the C ech Stone compactification of X. We consider X as a subspace of ω. The closure in ω is denoted by an upper bar. The family G G is a base for the closed sets of ω. Note that F G if and only if F G, for all F, G. A brief description of Y ω follows. The points of ω are the maximal filters of. Let F* x ω F x. The family F* F is a base for the closed subsets of Y. We have F* G* if and only if F G, for all F, G. Now we shall define the map ψ: ω ω. This map, when restricted to X,isthe map of the theorem. A point x ω is a maximal closed filter in. The intersection x of the filter x with is a closed filter base in. It is not difficult to show that there is a unique maximal closed filter y of which contains x. Define the map ψ by y ψ(x). It is easily seen that ψ(f ) F* for each F. For a proof of the continuity of ψ it is sufficient to show that ψ (F*) is closed in ω for each F. We shall show that ω ψ (F*) is open. If x belongs to this set, then ψ(x) F*, whence F ψ(x). By [2, Lemma VI.1.6], there is a G ψ(x) such that G F. By (4) there are K, L such that K L X, K F and L G. As G* L* and ψ(x) G*, we have ψ(x) L*, whence x L. For each y L, we have y K, whence ψ(y) K*. As K* F*, it follows that K is a neighbourhood of x such that ψ(k ) F*. By the continuity of ψ, it follows that ψ(f) is dense in ψ(f ) F* for each F. Hence the closure of ψ(f) inω coincides with F*. The map f is defined in a standard fashion. Suppose x ω. The collection η H f (H) x is a filter in. There is a unique maximal filter y in which contains η. Define f (x) y. Then f is continuous, and the diagram is commutative. See [2, Theorem VI.2.5] for details. Now let be a colouring of f. We have just proved that the closure of ψ(c) in Ycoincides with C*, for each C. We show that C* C is a colouring of f. By (2), C* C covers Y. For each C, f (C*) f (ψ(c )) ψ( f(c )) ψ( f(c)) f(c)*. The result follows. It remains to prove the last statement of the theorem. The observation about the homeomorphisms is rather obvious. Control of the dimension of the space Y ω is obtained by adopting the inductive selection of. At each inductive step, finitely many sets must be selected to guarantee that in the end dim ω n. This can be done by using [2, Theorem VI.2.8].

7 COINCIDENCE AND THE COLOURING OF MAPS 79 Proof of Theorem 2. Let X and f be as mentioned in the theorem. Let be a colouring of f. Choose Y and f as claimed in the last theorem. We may assume that dim Y n and that f has a colouring. By Theorem 8 there exist a contractible k- dimensional complex Z and a continuous mapping φ: Y Z with no f -coincidence. The next formula shows that φ ψ has no f-coincidence: φ ψ( f(x)) φ( f (ψ(x))) φ ψ(x). References 1. J. M. AARTS, R. J. FOKKINK and J. VERMEER, Variations on a theorem of Lusternik and Schnirelmann, Topology 35 (1996) J. M. AARTS and T. NISHIURA, Dimension and extension (North-Holland, Amsterdam, 1992). 3. P. E. CONNER and E. E. FLOYD, Differentiable periodic maps (Springer, Berlin, 1964). 4. E. K. VAN DOUWEN, βxand fixed-point free maps, Topology Appl. 51 (1993) M. A. VAN HARTSKAMP and J. VERMEER, On colorings of maps, Topology Appl. 73 (1996) M. IZYDOREK and J. JAWOROWSKI, Antipodal coincidence for maps of spheres into complexes, Proc. Amer. Math. Soc. 123 (1995) J. JAWOROWSKI, Existence of antipodal coincidence for maps of spheres, Preprint, E. V. S C EPIN, On a problem of L. A. Tumarkin, Dokl. Akad. Nauk SSSR 217 (1974) 42 43; So iet Math. Dokl. 15 (1974) Technical University Delft Delft Hydraulics Faculty of Mathematics and Informatics Department of Estuaries and Seas P.O. Box 5031 P.O. Box GA Delft 2600 MH Delft The Netherlands The Netherlands