http://www.elsevier.com/locate/aim

Advances in Mathematics 189 (2004) 268–300

A discrete homotopy theory for binaryreflexive structures

Benoit Larosea,b,1 and Claude Tardifc,�,2

aDepartment of Mathematics and Statistics, Concordia University, 1455 de Maisonneuve West,

Montreal, Que., Canada H3G 1M8bDepartment of Mathematics, Champlain Regional College, 900 Riverside Drive, St-Lambert, Que.,

Canada J4P 3P2cDepartment of Mathematics and Computer Science, Royal Military College of Canada, P.O. Box 17000,

Station ‘‘Forces’’, Kingston, Ont., Canada K7K 7B4

Received 19 June 2003; accepted 26 November 2003

Communicated by Laszlo Lovasz

Dedicated to Andre Joyal on the occasion of his 60th birthday

Abstract

We present a simple combinatorial construction of a sequence of functors sk from the

category of pointed binary reflexive structures to the category of groups. We prove that if the

relational structure is a poset P then the groups are (naturally) isomorphic to the homotopy

groups of P when viewed as a topological space with the topology of ideals, or equivalently, to

the homotopy groups of the simplicial complex associated to P: We deduce that the group

skðX ; x0Þ of the pointed structure ðX ; x0Þ is (naturally) isomorphic to the kth homotopy group

of the simplicial complex of simplices of X ; i.e. those subsets of X which are the homomorphic

image of a finite totally ordered set.

r 2003 Elsevier Inc. All rights reserved.

MSC: primary 06B30; secondary 05C20; 55Q99

Keywords: Binary structures; Reflexive digraphs; Homotopy; Posets

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�Corresponding author.

E-mail addresses: larose@mathstat.concordia.ca (B. Larose), claude.tardif@rmc.ca (C. Tardif).

URLs: http://www.cicma.mathstat.concordia.ca/faculty/larose/, http://www.rmc.ca/academic/math cs/

tardif/.1Research is supported by a grant from NSERC.2Supported by grants from NSERC and ARP.

0001-8708/$ - see front matter r 2003 Elsevier Inc. All rights reserved.

doi:10.1016/j.aim.2003.11.011

1. Introduction

Let G be a finite set of finitary relations on a nonempty finite set A: The (restricted)constraint satisfaction problem (CSP for short) CSPðGÞ is the following decisionproblem: given a pair /V ;CS where V is a set of variables and C is a finite set ofconstraints, i.e. pairs ðs; yÞ where s is a tuple of variables from V and yAG; determineif there exists a function f : V-A such that, for each constraint ðs; yÞ we have thatf ðsÞAy: These include the well-known problems of graph q-colouring, 3-satisfiability,graph unreachability and so on. This class of decision problems has received a greatdeal of attention lately, especially in the direction of a dichotomy result: it is expectedthat CSPs fall into two distinct families, those that have polynomial-time complexityand those that are NP-complete (see [4,3]). Bulatov et al. [3] have shown that toeach such CSP problem one may associate a finite relational structure S with thefollowing properties:

(1) CSPðGÞ is polynomial-time equivalent to the homomorphism problem for S; i.e.given a structure R of the same type as S; deciding whether there exists ahomomorphism from R to S;

(2) every operation on S is idempotent, i.e. every homomorphism f : Sn-Ssatisfies f ðx;y; xÞ ¼ x for all xAS:

The same authors have conjectured that the complexity of the homomorphismproblem for S is determined by the presence or absence of a certain type of well-behaved operation on S; namely Taylor operations. More precisely, it is known thatif there is no Taylor operation onS then the homomorphism problem forS isNP-complete ([11], see also [3]). The conjecture states that if a Taylor operation is presentthen the homomorphism problem is in P (consult [1,2] for recent results in thisdirection).

Taylor operations first appeared in [19] in connection with topological algebras;they have since then come to play an important role in the classification of varietiesof universal algebras (see [8]). Inspired by Taylor’s results, Larose and Zadori provedthe following: if a finite connected poset admits a Taylor operation, then all its

homotopy groups are trivial [10]. Here the poset is viewed both as a relationalstructure and as a finite topological space (see Section 3). It appeared that this resultmight serve as a sieve to identify NP-complete CSPs; indeed, it is known that allCSPs are polynomial-time equivalent to a poset retraction problem [4]. However, itseems desirable to have at hand a more flexible class of finite structures than partiallyordered sets. Unlike the case of posets, there is no topology on general reflexivestructures that makes homomorphisms coincide with continuous maps, and so analternative notion of homotopy had to be developed. We now briefly discuss thisapproach and return to its applications in a moment.

In the present paper, we define a series of functors sk from the category of pointedbinary reflexive structures to the category of groups. Roughly speaking, if ðX ; yÞ is abinary reflexive relational structure, the group s1ðX ; x0Þ is a ‘simpleminded’ analogof the classical fundamental group of topological spaces, if one thinks of a loopbased at x0 in the structure X as a sequence of elements x0; x1;y; xn; x0 such that

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300 269

x0yx1; x2yx1; x2yx3 and so on. The notion of homotopy is replaced by connectivityin the ‘loop space’ FðX ; x0Þ: to accomodate loops of different lengths, one simplydefines a loop based at x0 as a homomorphism from the one-way infinite fence F

(which plays the role of the unit interval) which has value x0 at one end and is equalto x0 almost everywhere. It is a simple affair to define higher ‘homotopy’ groups in

this way via the spaces F kðX ; x0Þ (see Section 2). This allowed the first author toprove a generalisation of the result for posets in this setting: if a finite, connected,

pointed binary reflexive relational structure ðX ; x0Þ admits a Taylor operation, then all

the groups skðX ; x0Þ are trivial [9].If one is to use the groups sk to detect the absence of a Taylor operation one

requires methods to compute them. In the present paper we show that for everypointed reflexive binary structure ðX ; x0Þ; there is a pointed topological space ðC; c0Þsuch that skðX ; x0Þ is isomorphic to the kth homotopy group of ðC; c0Þ: Moreprecisely, let us call a subset of the structure X a simplex if it is the homomorphicimage of a finite totally ordered set. It is clear that the simplices of X form a

simplicial complex; then skðX ; x0Þ is isomorphic to pkð dSðXÞSðXÞ; fx0gÞ where dSðXÞSðXÞ is thegeometric realisation of this complex (Corollary 5.6); this isomorphism is actually anatural equivalence (Theorem 5.7).

We illustrate the above discussion with a few examples:

Example 1 (Digraph retraction problems). Consider the binary relation ðH; yÞdepicted in Fig. 1: an arrow from x to y indicates that xyy; and an edge indicates thatboth xyy and yyx hold (loops are omitted). Thus H is a reflexive digraph. Considerthe retraction problem RetðHÞ: given a reflexive digraph G containing a copy of H;determine whether G retracts onto H: Obviously, the simplices of H are f0; 1; 2g;f0; 1; 3g; f0; 2; 3g; f1; 2; 3g and their nonempty subsets; so the complex triangulatesthe 2-sphere. It follows from the above that s2ðH; 0ÞCZ and hence there is noTaylor operation on the digraph H: Consequently, the decision problem RetðHÞ isNP-complete (see [9]).

Example 2 (Solving systems of equations over a group). Let G be a group (or asemigroup, or in fact any universal algebra). We’ll take G ¼ S3 the symmetric groupon 3 letters. Consider the following decision problem: given a system of (finitelymany) equations of the form

a1X1a2X2?anXnanþ1 ¼ b1Y1b2Y2?bmYmbmþ1

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0

2

1

3

Fig. 1. The graph H:

B. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300270

with ai; biAG; determine whether the system admits a solution. It is not difficult tosee that this problem is equivalent to the homomorphism problem for the structureS ¼ /G; g; fggðgAGÞS where

g ¼ fðx; y; zÞ : xy ¼ zg:

The operations onS are precisely the idempotent group homomorphisms f : Gn-G:Obviously S is not a binary reflexive structure; however, every idempotent operationon S must leave invariant the subset of G

B ¼ fxAG : x2 ¼ 1g;

and hence the binary relation

y ¼ fðx; yÞ : y ¼ bx for some bABg:

Since 1AB this is a binary reflexive relation on G and every idempotent operation onS is an operation on /G; yS:3 It is easy to see that the relational structure obtainedwhen G is the group S3 is the one depicted in Fig. 2: clearly the simplicial complexassociated has dimension 1 and is not a tree, so its fundamental group is nontrivial;we conclude as before that the decision problem is NP-complete.

(This example is a bit contrived, but is small enough to illustrate a general strategywithout undue details. In fact, one may argue as follows for any non-Abelian group:if f is an idempotent operation on the structure S above, then following Taylor, G isa group in the variety generated by the algebra A ¼ /G; fS; in [19] Taylor provesthat all groups in an idempotent variety admitting a Taylor term are Abelian; hencef cannot be a Taylor operation and so solving systems of equations over G is NP-complete. See [5] for an alternative proof, and [13] for the case of monoids.)

Besides the applications to the complexity of CSP problems, it seems that thefunctors sk might have an intrinsic interest. For instance, it is not difficult to showthat the functor skþ1 is naturally equivalent to the functor which assigns to each

pointed structure ðX ; x0Þ the group s1ðF kðX ; x0Þ; x0Þ; however, FkðX ; x0Þ is infiniteeven when X is finite. Instead, one may consider for each even integer nX2; the

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(12) (13) (23)

id (123) (132)

Fig. 2. The graph associated to S3:

3More generally, the subsets of Gr that can be expressed in the form f %x : ( %yfð %x; %yÞg where f is a

conjunct of atomic formulas in the language of S are precisely those that are invariant under the

operations on S (see for example [14]).

B. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300 271

group s1ðF kn ðX ; x0Þ; x0Þ: the correspondence is itself functorial, the group

skþ1ðX ; x0Þ is the direct limit of these groups, and the structure Fkn ðX ; x0Þ is finite

if X is. In particular, each of these groups admits a finite presentation. It would alsobe interesting to see to what extent our functors are related to axiomatic homotopytheory [16]. We mention finally that our results are also closely related to recent workby Hardie et al. [6,7] on the relationship between finite topological spaces andpolyhedra.

We now outline the contents of the paper. In Section 2 we define the groupsskðP; p0Þ: Sections 3 and 4 contain the main technical result, the proof that for anypointed poset ðP; p0Þ; the group skðP; p0Þ is isomorphic to the kth homotopy grouppkðP; p0Þ of P when viewed as a finite topological space (Theorem 4.11); this is thencombined with McCord’s theorem (Theorem 3.2) which states that the grouppkðP; p0Þ is isomorphic to the kth homotopy group of the simplicial complex of P

(Theorem 4.12). In Section 5, we prove that for every reflexive structure X the groupsskðX ; x0Þ and skðSðXÞ; fx0gÞ are isomorphic, where SðXÞ is the poset of simplices ofX ordered by inclusion (Theorem 5.6).

2. Discrete homotopy

In this section we define the functors sk from the category of pointed binaryreflexive structures to the category of groups.

2.1. Basic definitions

A binary reflexive structure is a nonempty set X equipped with a reflexive binary

relation y; i.e. a subset of X 2 which contains ðx; xÞ for every xAX : In the following,we shall usually omit the symbol y and write x-y instead of ðx; yÞAy; or moresuccinctly, we will say that xy is an edge of X : We will often use structure instead ofthe more cumbersome reflexive binary structure. A pointed binary reflexive structureis a pair ðX ; x0Þ where X is a binary reflexive structure and x0AX : If X and Y arebinary structures, a map f : X-Y is a homomorphism if f ðxÞ-f ðyÞ in Y wheneverx-y in X : Let f : X-Y be a homomorphism where ðX ; x0Þ and ðY ; y0Þ are pointedstructures. We say that f is base-point preserving if f ðx0Þ ¼ y0:

Let X and Y be two binary structures. We define a binary structure HomðX ;YÞ asfollows: its base set consists of all homomorphisms from X to Y ; if f and g are twosuch homomorphisms, we define f-g if f ðxÞ-gðyÞ whenever x-y: The product oftwo structures X and Y is the structure X � Y on the (set) product of the base sets of

X and Y and we define ðx; yÞ-ðx0; y0Þ if x-x0 and y-y0: We shall use X k to denotethe kth power of a structure, i.e. the product X � X �?� X of X with itselfk times.

Definition 2.1. The one-way infinite fence F is the binary reflexive structure defined asfollows (Fig. 3): its base set is the set of nonnegative integers, and the relation is as

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300272

follows: n-m if n ¼ m or n is even and jn � mj ¼ 1: For nX1 let Fn denote thesubstructure of F with universe f0; 1;y; ng:

A (weak) path from x to y in a structure X is a sequence of elements of X

x ¼ t0; t1;y; tn ¼ y such that ti-tiþ1 or tiþ1-ti for all iX0: Notice that there existsa path from x to y if and only if there exists a homomorphism f : F-X and aninteger NX0 such that f ð0Þ ¼ x and f ðnÞ ¼ y for all nXN: Two elements of astructure are in the same connected component if there is a weak path between them.A structure is connected if all its elements are in the same connected component.

Definition 2.2. Let ðX ; x0Þ be a pointed binary reflexive structure. For all kX1; let

FkðX ; x0Þ be the set of all homomorphisms f : F k-X such that there exists NX0with f ðt1;y; tkÞ ¼ x0 if there is an i such that ti ¼ 0 or tiXN:

The set FkðX ; x0Þ is of course a pointed binary reflexive structure in the obvious

way: the relation is the one inherited from the structure HomðF k;X Þ and the basepoint is the constant map with value x0; which we will denote by x0:

For fAF kðX ; x0Þ and 1pipk let Niðf Þ denote the least nonnegative even integerN such that f ðt1;y; tkÞ ¼ x0 if tiXN and let Nðf Þ denote the least nonnegative eveninteger N such that f ðt1;y; tkÞ ¼ x0 if there is an i such that tiXN (thusNðf Þ ¼ max Niðf Þ).

Let ekn be the natural embedding of F k

n into F k: Let rn denote the retraction of F

onto Fn defined by

rnðtÞ ¼t if tpn;

n otherwise:

�

Let rkn denote the retraction of Fk onto F k

n defined by

rknðt1;y; tkÞ ¼ ðrnðt1Þ;y; rnðtkÞÞ

for all tiAF : Clearly rkn3e

kn is the identity on F k

n :We shall also need the following slight modification in a later section:

Definition 2.3. Let ðX ; x0Þ be a pointed binary reflexive structure. For all kX1; and

for all nX1; let F kn ðX ; x0Þ be the set of all homomorphisms f : Fk

n -X such that

f ðt1;y; tkÞ ¼ x0 if there is an i such that ti ¼ 0 or ti ¼ n:

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0 2 4 6

1 3 5 7

Fig. 3. The one-way infinite fence F :

B. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300 273

Finally, here is a simple result we shall require later:

Lemma 2.4. For any structure X ; the composition operation

3 : HomðX ;X Þ2-HomðX ;XÞ

which maps ðf ; gÞ to f 3g is a homomorphism.

Proof. Straightforward. &

2.2. Definition of the functors

Definition 2.5. Let skðX ; x0Þ denote the set of connected components of the structure

FkðX ; x0Þ: For fAFkðX ; x0Þ; ½f shall denote the connected component of F kðX ; x0Þcontaining f :

Definition 2.6. We define a group structure on skðX ; x0Þ as follows: let

f ; gAFkðX ; x0Þ and let NXN1ðf Þ be an even integer. Define ðf ; gÞNAFkðX ; x0Þ by

ðf ; gÞNðt1; t2;y; tkÞ ¼f ðt1; t2;y; tkÞ if t1pN;

gðt1 � N; t2;y; tkÞ otherwise:

�

Then let

½f � ½g ¼ ½ðf ; gÞN1ðf Þ :

Before we prove that this is indeed a group operation we introduce some technicaldefinitions that will be needed throughout this paper.

Definition 2.7. Let nX0 and define a map mn : F-F as follows: for all tAF let

mnðtÞ ¼t if tpn;

n if t ¼ n þ 1;

t � 2 otherwise:

8><>:

It is easy to see that the maps mn are homomorphisms, and that mn-mnþ1 for n

even and mn’mnþ1 for n odd.

Lemma 2.8. Let fAFkðX ; x0Þ; nX0 and 1pipk: Consider the map gn :ðt1;y; tkÞ/f ðt1;y; ti�1; mnðtiÞ; tiþ1;y; tkÞ:

(1) The map gn is in FkðX ; x0Þ:(2) If nXNðf Þ then f ¼ gn:(3) The maps gn and f are in the same connected component of FkðX ; x0Þ:

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300274

Proof. The proofs of (1) and (2) are simple verifications. Now let m be the maximumof Nðf Þ and n: By the remarks above, the maps mj ðj ¼ n;y;mÞ form a path joining

the maps mn and mm: Thus by (1) the maps gi ði ¼ n;y;mÞ form a path from gn to gm:By (2) this last map is f so we are done. &

Lemma 2.9. The product ½f � ½g ¼ ½ðf ; gÞN1ðf Þ is well defined.

Proof. Let N ¼ N1ðf Þ: First we show that if MXN then ½ðf ; gÞM ¼ ½ðf ; gÞN : It issufficient to show this for M ¼ N þ 2: It is easy to see that ðf ; gÞMðt1;y; tkÞ ¼ðf ; gÞNðmNðt1Þ; t2;y; tkÞ for all t1;y; tk: Thus by Lemma 2.8(3) we are done. Next,

suppose ½f ¼ ½f 0 and ½g ¼ ½g0 : Hence there are homomorphisms g; d : F-F kðX ; x0Þand an integer l such that gð0Þ ¼ f ; dð0Þ ¼ g; gðtÞ ¼ f 0 and dðtÞ ¼ g0 for all tXl: LetW be an even integer such that Fðt1;y; tkÞ ¼ x0 whenever tiXW for some 1pipk;and this for all FAfgðtÞ; dðtÞ : tX0g: It is easy to see that the map from F to

FkðX ; x0Þ defined by

t/ðgðtÞ; dðtÞÞW

is a homomorphism and hence there is a path in F kðX ; x0Þ from ðf ; gÞW to

ðf 0; g0ÞW : &

Theorem 2.10. The set skðX ;x0Þ together with the operation � is a group with

unit ½x0 :

Proof. We first show the product is associative. Let f1; f2; f3AFkðX ; x0Þ and fori ¼ 1; 2; 3 let Ni be an even integer such that fiðt1;y; tkÞ ¼ x0 whenever t1XNi:Clearly ½f1 � ð½f2 � ½f3 Þ ¼ ½ðf1; ðf2; f3ÞN2

ÞN1 : Next notice that ðf1; f2ÞN1

ðt1;y; tkÞ ¼ x0

whenever t1XN1 þ N2: Thus ð½f1 � ½f2 Þ � ½f3 Þ ¼ ½ððf1; f2ÞN1; f3ÞN1þN2

: It is easy to see

that ðf1; ðf2; f3ÞN2ÞN1

¼ ððf1; f2ÞN1; f3ÞN1þN2

:

It is obvious that ½x0 is a unit for this product. Now let fAF kðX ; x0Þ and letN ¼ N1ðf Þ: Define

f 0ðt1; t2;y; tkÞ ¼f ðN � t1; t2;y; tkÞ if t1pN;

x0 otherwise:

�

We claim that ½f � ½f 0 ¼ ½x0 : Indeed, ½f � ½f 0 ¼ ½ðf ; f 0ÞN where

ðf ; f 0ÞNðt1;y; tkÞ ¼f ðt1;y; tkÞ if t1pN;

f ð2N � t1;y; tkÞ if Npt1p2N;

x0 otherwise:

8><>:

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For 0pipN � 1 define

giðt1;y; tkÞ ¼

f ðt1; t2;y; tkÞ if 0pt1pN � i � 1;

f ðN � i; t2;y; tkÞ if N � ipt1pN þ i;

f ð2N � t1; t2;y; tkÞ if N þ i þ 1pt1p2N;

x0 if t1X2N

8>>><>>>:

and let gN ¼ x0: It is routine to verify that the gi form a path in F kðX ; x0Þ and thatg0 ¼ ðf ; f 0ÞN : Thus ½f � ½f 0 ¼ ½ðf ; f 0ÞN ¼ ½x0 : &

Definition. Let ðX ; x0Þ and ðY ; y0Þ be two pointed structures and let F : X-Y

be a homomorphism such that Fðx0Þ ¼ y0: Define a correspondence F# :

skðX ; x0Þ-skðY ; y0Þ by F#ð½f Þ ¼ ½F3f for all fAF kðX ; x0Þ:

Theorem 2.11. The correspondence which assigns to a pointed structure ðX ; x0Þ the

group skðX ; x0Þ and to a homomorphism F the map F# defines a covariant functor

from the category of pointed reflexive binary structures to the category of groups with

homomorphisms.

Proof. The fact that the correspondence is functorial is straightforward. We showthat F# is well defined and a group homomorphism, where F : X-Y : Indeed, it is

clear that F3fAFkðY ; y0Þ if fAF kðX ; x0Þ: If ½f ¼ ½g where f ; gAFkðX ; x0Þ; let f ¼f0;y; fl ¼ g be a path in FkðX ; x0Þ: Then clearly F3f0;y;F3fl is a path in FkðY ; y0Þfrom F3f to F3g: Thus F is well defined. Certainly F#ð½x0 Þ ¼ ½y0 : Let

f ; gAFkðX ; x0Þ and let N ¼ N1ðf Þ: Then certainly ðF3f Þðt1;y; tkÞ ¼ y0 if t1XN:Thus ½F3f � ½F3g ¼ ½ðF3f ;F3gÞN : It is easy to see that ðF3f ;F3gÞN ¼ F3ðf ; gÞN so

that F#ð½f � ½g Þ ¼ F#ð½f Þ � F#ð½g Þ: &

3. Preliminaries on posets and topology

The reader may consult [10,18] for basic results and terminology concerning thetopological structures associated to posets, and [17] for standard results in algebraictopology. We now review the definitions and concepts we shall need later. In whatfollows, I ¼ ½0; 1 denotes the unit interval.

A poset (partially ordered set) is a binary reflexive structure P ¼ ðX ; yÞ where y isantisymmetric and transitive. In a poset P we usually write xpy instead of x-y: It iseasy to verify that if P and Q are posets then so is HomðP;QÞ and that the product ofposets is also a poset. In particular, since the one-way infinite fence F is a poset, so

are Fk and F kn for all n; k; furthermore, F kðP; p0Þ is a poset for any pointed poset

ðP; p0Þ: A map f : P-Q is order-preserving if f ðxÞpf ðyÞ whenever xpy; i.e. if f is ahomomorphism of binary structures.

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Let P be a poset. A subset JDP is an ideal (down-set, order ideal) of P if xAJ

whenever xpy where yAJ: We shall use the notation kp ¼ fxAP : xppg to denotethe (principal) ideal generated by pAP: It is easy to see that we may view a poset as atopological space whose open sets are precisely the order ideals. These topologicalspaces, the so-called T0 A-spaces, are characterized by the fact that they satisfy theT0 separation axiom and that an arbitrary intersection of open sets is open (see [12]).The poset P is connected as a relational structure if and only if it is connected (path-connected) as a topological space. It is easy to see that the continuous maps betweenposets coincide with the order-preserving ones. It is not difficult to verify thefollowing: if X is an arbitrary topological space, a map f : X-P is continuous if and

only if for each xAX ; there exists an open set U containing x such that fðyÞpfðxÞ for

all yAU :Let P be a poset. A chain (totally ordered set) in P is a subset CDP such that xpy

or ypx for all x; yAC: Any poset P defines a simplicial complex whose simplices arethe finite chains of P:

Definition. The geometric realization of the simplicial complex of a poset P is the

space PD½0; 1 P defined as follows:

P ¼ fh : P-½0; 1 :XpAP

hðpÞ ¼ 1; suppðhÞ is a finite chain of Pg;

where the support of h is the set suppðhÞ ¼ fpAP : hðpÞa0g: The set is equipped withthe coherent topology (see [17, p. 111]).

Definition. Let ðX ; x0Þ and ðZ; z0Þ be pointed topological spaces and let Y be asubspace of X : We say that two continuous maps f ; g : X-Z that coincide on Y arehomotopic relative to Y and write fBg (rel Y ) if there exists a continuous mapF : I � X-Z such that

(i) Fðt;x0Þ ¼ z0 for all tAI ;(ii) Fð0; xÞ ¼ f ðxÞ for all xAX ;(iii) Fð1; xÞ ¼ gðxÞ for all xAX ;(iv) Fðt; yÞ ¼ f ðyÞ for all yAY :

Let @Ik denote the set of all k-tuples with entries in I with at least one coordinateequal to 0 or 1.

Definition. Let P be a poset and p0AP: Then CkðP; p0Þ shall denote the set of all

continuous maps f : Ik-P such that f ð@IkÞDfp0g:

Definition. As usual pkðP; p0Þ will denote the set of equivalence classes of CkðP; p0Þunder the relation fBg (rel @Ik). If fACkðP; p0Þ let ½f denote the equivalence classof f under this relation.

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Definition. Let f ; gACkðP; p0Þ: Let /f ; gS denote the map

/f ; gSðx1;y; xkÞ ¼f ð2x1; x2;y; xkÞ if 0px1p1=2;

gð2x1 � 1; x2;y;xkÞ if 1=2px1p1:

�

Then of course /f ; gSACkðP; p0Þ and the product in the group pkðP; p0Þ is given by

½f %½g ¼ ½/f ; gS :

The correspondence which assigns to every pointed space ðX ; x0Þ the group pkðX ; x0Þis the usual kth homotopy group functor; it assigns to a continuous map f : X-Y

the group homomorphism ½f /½f3f : If f induces a bijection from the set ofconnected components of X onto those of Y and for all xAX the homomorphisminduced from pkðX ; xÞ to pkðY ;fðxÞÞ is an isomorphism for all kX1 we say that f isa weak homotopy equivalence.

Definition 3.1. Let P be a poset. The map aP : P-P is defined by putting

aPðhÞ ¼ min suppðhÞ:

Theorem 3.2 (McCord [12]). Let P be a poset. Then the map aP is a weak homotopy

equivalence.

Definition. Let f : P-Q be an order-preserving map between posets. The simplicial

lifting of f is the map #f : P-Q; where

#fðhÞðqÞ ¼XpAP

fðpÞ¼q

hðpÞ:

We also define for every pAP an element pAP defined by

pðxÞ ¼1 if x ¼ p;

0 otherwise:

�

The simplicial lifting of an order-preserving map is well defined and continuous.Furthermore, the following holds:

Lemma 3.3. Let f : P-Q be an order-preserving map between posets. Then

aQ3#f ¼ f3aP:

Proof. Let hAP: Then suppð #fðhÞÞ ¼ fðsuppðhÞÞ; and since f is order-preserving,

fðmin suppðhÞÞ ¼ min fðsuppðhÞÞ i.e., aQ3#f ¼ f3aP: &

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300278

Lemma 3.4. Let f : P-Q be an order-preserving map between posets. Then for every

hAP we have

fðhÞ ¼XpAP

hðpÞ dfðpÞfðpÞ

(where functions are added pointwise).

Proof. Simple verification. &

The following lemma is due to Quillen [15].

Lemma 3.5 (Quillen [15]). Let X be a topological space and P be a poset. Let f ; g :X-P be continuous maps such that fpg and f ; g coincide on some subset Y of X :Then fBg (rel Y ).

Proof. Define a map f : ½0; 1 � X-P by putting

fðt; xÞ ¼f ðxÞ if to1;

gðxÞ if t ¼ 1:

�

For an ideal J of P; we have f�1ðJÞ ¼ ½0; 1½�f �1ðJÞ,½0; 1 � g�1ðJÞ which is open,so f is continuous and defines a homotopy between f and g: &

We now exhibit explicit homeomorphisms between powers of the unit interval andthe geometric realization of products of fences.

Let n; k be positive integers, n even. Let F denote the map

F : cFknFkn - ðcFnFnÞk;

h / ðh1;y; hkÞ;

where

hiðpÞ ¼X%tAFk

nti¼p

hð%tÞ

for all pAI :The map F is a homeomorphism (see for example [20, p. 184]).

Definition 3.6. Let nX2 be an even integer and kX1: We define a map Fkn : cFk

nFkn -Ik

by

FknðhÞ ¼

X%pAFk

n

hð %pÞp1

n;y;

pk

n

for all hAcFknFkn :

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300 279

Lemma 3.7. Let nX2 be an even integer and kX1: Then the map Fkn is a

homeomorphism.

Proof. It is easy to see that F1n is a homeomorphism. Now let F be the

homeomorphism F : cFknFkn -ðcFnFnÞk defined earlier. It is easy to see that

Fkn ¼ ðF1

nÞk3F;

where ðF1nÞ

kðh1;y; hkÞ ¼ ðF1nðh1Þ;y;F1

nðhkÞÞ for all hiAcFnFn: Since this map is also a

homeomorphism we are done. &

Definition 3.8. Let nX2 be an even integer and kX1: Let Ckn : ½0; 1 k-cF k

nF kn denote the

inverse of the map Fkn defined above, and let ak

n : cFknFkn -Fk

n denote the map aF kn:

Definition 3.9. Let nX2 be an even integer and kX1: A subset B of ½0; 1 k is an n-box

if B ¼Qk

i¼1 Bi where each Bi is of the form ai=n; bi=n½ or ½0; bi=n½ or ai=n; 1 whereaiobi are odd integers in f0; 1;y; ng:

Definition 3.10. Let P be a poset. A continuous map f : ½0; 1 k-P is n-simple if

f �1ðkpÞ is a union of n-boxes for all pAP:

Lemma 3.11. Let %q ¼ ðq1;y; qnÞAF kn : Then

ðaknC

knÞ

�1ð %qÞ ¼Yk

i¼1

Bi;

where

Bi ¼

qi

n

n oif qi is odd;

½0; 1n½ if qi ¼ 0;

n � 1

n; 1 if qi ¼ n;

qi � 1

n;qi þ 1

n½ otherwise:

8>>>>>>>>><>>>>>>>>>:

Proof. Let hAcFknFkn and put ak

nðhÞ ¼ %q: Then

FknðhÞ ¼

X%pAFk

n

hð %pÞp1

n;y;

pk

n

:

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300280

If qi is odd, then it is maximal in Fn so for all %pAsuppðhÞ we have pi ¼ qi and thus the

ith coordinate of FknðhÞ is qi=n: If qi is even, then let us consider the case where

0oqion (the remaining cases are similar). Now for all %pAsuppðhÞ we have piAfqi � 1;

qi; qi þ 1g; and since hð %qÞ40 the ith coordinate of FknðhÞ will be in qi�1

n; qiþ1

n½: Thus we

have proved that

ðaknC

knÞ

�1ð %qÞDYk

i¼1

Bi:

Now to get equality use the fact that akn is onto and that for distinct values of qi the

corresponding sets Bi must be disjoint. &

Lemma 3.12. Let nX4: Let %q ¼ ðq1;y; qnÞAFkn : Then

ðaknC

knÞ

�1ðk %qÞ ¼Yk

i¼1

Ci;

where

Ci ¼

qi � 2

n;qi þ 2

n½ if qi is odd and distinct from 1 and n � 1;

½0; 3n½ if qi ¼ 1;

n � 3

n; 1 if qi ¼ n � 1;

qi � 1

n;qi þ 1

n½ if qi is even and distinct from 0 and n;

½0; 1n½ if qi ¼ 0;

n � 1

n; 1 if qi ¼ n:

8>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>:

Proof. By Lemma 3.11 we have that

ðaknC

knÞ

�1ðk %qÞ ¼[%pp %q

Yk

i¼1

Bi; %p;

where the Bi; %p are of the specified form. Suppose that qi is even. Then qi ¼ pi for all

%pp %q so that Bi; %p ¼ Bi; %q for all %pp %q: If qi is odd, then for all %pp %q pi is one of

fqi � 1; qi; qi þ 1g: Thus

[%pp %q

Yk

i¼1

Bi; %pDYk

i¼1

Ci;

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300 281

where the Ci are as described in the statement of the lemma. Furthermore, if

%xAQ

Ci; it is clear that one may choose %pp %q in such a way that %xAQ

Bi; %p: &

Lemma 3.13. Let nX4: Let %qAF kn and put

X ¼ ðaknC

knÞ

�1ð %qÞ and X 0 ¼ ðaknC

knÞ

�1ðk %qÞ:

If Y is a union of n-boxes such that XDY then X 0DY :

Proof. Immediate by the two preceding lemmas and the definition of n-box. &

Lemma 3.14. (1) For nX4 the maps aknC

kn are n-simple. (2) An n-simple map is

sn-simple for every odd integer sX1:

Proof. (1) Immediate by Lemma 3.12. (2) It clearly suffices to prove that if s is oddthen an n-box is an sn-box and this is obvious. &

Lemma 3.15. Let ðP; p0Þ be a pointed poset and let nX4: A map f : Ik-P is n-simple

if and only if there exists an isotone map g : F kn -P such that g3ak

nCkn ¼ f ; i.e. the

following diagram commutes:

Ik !f P

Cknk mgcFknFkn ak

n

! Fkn

If this map exists, it is unique. Furthermore, if f ð%tÞ ¼ p0 for all tuples %t whose ith

coordinate is 0 or 1, then gð%tÞ ¼ p0 for all tuples %t whose ith coordinate is 0 or n; in

particular if fACkðP; p0Þ then gAFkn ðP; p0Þ:

Proof. If g exists then it is unique: indeed, we have that f 3Fkn ¼ g3ak

n and akn is onto.

Now we prove the last statement. Let %tAF kn ; clearly ak

nCkn

%tn

� �¼ %t so that

gð%tÞ ¼ f %tn

� �: If ti ¼ 0 then gð%tÞ ¼ p0 and if ti ¼ n then ti

n¼ 1 so gð%tÞ ¼ p0:

ð)Þ Let f be n-simple, fix %pAFkn : By Lemma 3.11 we have that ðak

nCknÞ

�1ð %pÞ ¼QXi where

Xi ¼

pi

n

n oif pi is odd;

½0; 1n½ if pi ¼ 0;

n � 1

n; 1 if pi ¼ n;

pi � 1

n;pi þ 1

n½ otherwise:

8>>>>>>>>><>>>>>>>>>:

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300282

Now pick %xAQ

Xi; let b ¼ f ð %xÞ and since f is n-simple, f �1ðkbÞ is a union of

n-boxes. Thus there exists an n-box B ¼Q

Bi such that %xABDf �1ðkbÞ:Fix i such that pi is even, and suppose that pi is different from 0 and n: Then

xiAXi ¼ pi�1n; piþ1

n½: Since B is an n-box and xiABi we have that XiDBi: Similar

considerations for the remaining cases show that XiDBi for all i:

ThusQ

XiDB i.e. ðaknC

knÞ

�1ð %pÞDf �1ðkbÞ: This means that for every

%yAðaknC

knÞ

�1ð %pÞ we have that f ð %yÞpf ð %xÞ: Since %x was arbitrarily chosen we conclude

that f is constant on ðaknC

knÞ

�1ð %pÞ:Thus the map g : Fk

n -P that sends p to f ð %xÞ where %xAðaknC

knÞ

�1ð %pÞ is well defined.Now we show that this map is isotone. Let %pp %q in Fk

n ; then gð %pÞ ¼ f ð %xÞ and

gð %qÞ ¼ f ð %yÞ for some %xAðaknC

knÞ

�1ð %pÞ and %yAðaknC

knÞ

�1ð %qÞ: But %pAk %q implies that

%xAðaknC

knÞ

�1ðk %qÞ: Since f is n-simple, f �1ðkf ð %yÞÞ is a union of n-boxes that contains

ðaknC

knÞ

�1ð %qÞ: By Lemma 3.13 we conclude that f �1ðkf ð %yÞÞ contains ðaknC

knÞ

�1ðk %qÞ so%xAf �1ðkf ð %yÞÞ and thus gð %xÞ ¼ f ð %xÞpf ð %yÞ ¼ gð %yÞ:ð(Þ Let g be isotone, let f satisfy f ¼ g3ak

nCkn : Then f is obviously continuous and

by Lemma 3.14 aknC

kn is n-simple. Let pAP: Then

f �1ðkpÞ ¼ ðaknC

knÞ

�1ðg�1ðkpÞÞ:

But g is isotone so g�1ðkpÞ is an ideal in Fkn i.e.

g�1ðkpÞ ¼[t

j¼1

kpj;

where the pjAF kn : Thus

f �1ðkpÞ ¼[t

j¼1

ðaknC

knÞ

�1ðkpjÞ

which is a union of n-boxes. &

Lemma 3.16. Let P be a poset, let X be a topological space and let fO1;y;Omg be an

open cover of X : Let

f : f1; 2;y;mg-P

be a function such that, if Oi-Oja| then f ðiÞ and f ðjÞ are comparable. Then the map

g : X-P defined by gðxÞ ¼ minff ðiÞ : xAOig is continuous.

Proof. Let xAX : By the comments at the beginning of this section it suffices to findan open set U containing x such that gðUÞDkgðxÞ: By definition of g there is aniAf1; 2;y;mg with xAOi and f ðiÞ ¼ gðxÞ: Let yAOi: Then gðyÞpf ðiÞ ¼ gðxÞ: &

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300 283

Theorem 3.17. Let ðP; p0Þ be a pointed poset. Let f : Ikþ1-P be a continuous map

and let nX2 be an even integer. Then there exist an odd integer s and an sn-simple

function g : Ikþ1-P such that gpf : Furthermore, if f ðl; xÞ ¼ p0 for all lAI

and all xA@Ik; or if fACkþ1ðP; p0Þ; then g can be chosen with the same

property.

Proof. For every xAIkþ1 there exists rx40 such that, if jjy � xjjorx then f ðyÞpf ðxÞby continuity of f : Hence for every xAIkþ1 there exists an open set Bx containedin the open ball of radius rx centered at x such that Bx is a product of subintervalsof I ; and such that f ðyÞpf ðxÞ for all yABx: In fact, for each fixed x; we maychoose some odd integer sx large enough so that Bx is a product of intervals of the

form ½0; 1sxn½ or sxn�1

sxn; 1 or a

sxn; b

sxn½ where a; b are odd integers. Hence for every

xAIkþ1; Bx is an sxn-box. Notice that we may also assume that if the ith coordinateof x is neither 0 nor 1 then the ith factor of Bx is an interval of the third type;in particular, if the ith coordinate of x is 0 or 1 and xABy then the ith coordinate

of y is 0 or 1.

Clearly if Bx-Bya| then

jjx � yjjorx þ ry

2pmaxðrx; ryÞ

and thus f ðxÞ and f ðyÞ are comparable.

By compactness of Ikþ1 there exists a finite subcover fBx1;y;Bxl

g of fBx :

xAIkþ1g which covers Ikþ1 and satisfies the conditions of Lemma 3.16. Let g be themap thus obtained, gðxÞ ¼ minff ðxiÞ : xABxi

g; obviously gpf : Let pAP: It is easyto see that

g�1ðkpÞ ¼[

fBxi: xiAf �1ðkpÞg:

Let s denote the least common multiple of sx1;y; sxl

: Then s is odd, and the Bxiare

sn-boxes so g is sn-simple. Finally, if xA@Ik and ðl; xÞABxithen xi ¼ ðl; yÞ where

yA@Ik; hence gðl; xÞ ¼ p0 for all lAI and all xA@Ik: A similar argument shows that

if fACkþ1ðP; p0Þ then gACkþ1ðP; p0Þ: &

4. Main theorem for posets

We now describe a correspondence between skðP; p0Þ and pkðP; p0Þ whichwe will ultimately prove is a group isomorphism. In fact, the correspondence thatassigns to a pointed poset ðP; p0Þ the isomorphism Dk turns out to be a natural

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300284

equivalence.

Fix a pointed poset ðP; p0Þ: Let

Dk : skðP; p0Þ-pkðP; p0Þ

be defined as follows : if fAFkðP; p0Þ then

Dkð½f Þ ¼ ½f 3ekn3a

knC

kn ;

where n is a positive even integer such that nXNðf Þ:

In the following we shall drop the index k and write simply D instead of Dk:

4.1. D is well defined

Lemma 4.1. The correspondence D is well defined.

Proof. Fix fAF kðP; p0Þ: Let nXNðf Þ be an even integer.

(a) We prove first that f 3ekn3a

knC

knACkðP; p0Þ: Obviously the map is continuous, so

it suffices to prove that its value on %t is equal to p0 if some ti is equal to 0 or 1. Put

fn ¼ f 3ekn : Let %tAIk such that ti ¼ 0: Let h ¼ Ck

nð%tÞ; i.e.

%t ¼X%yAFk

n

hð %yÞy1

n;y;

yk

n

so that

0 ¼ ti ¼X

%yAsuppðhÞhð %yÞ

yi

n;

thus yi ¼ 0 for all %yAsuppðhÞ and so the ith coordinate of aknðhÞ is also 0. Thus

fn3aknðhÞ ¼ p0:

Now suppose that ti ¼ 1: Then

1 ¼ ti ¼X

%yAsuppðhÞhð %yÞ

yi

n;

if there exists %z with zi

no1 then

Phð %yÞyi

noP

hð %yÞ ¼ 1; a contradiction. Thus yi ¼ n

for all %yAsuppðhÞ and so the ith coordinate of aknðhÞ is n and we’re done.

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300 285

(b) We show that Dð½f Þ does not depend on the value of nXNðf Þ: It suffices toshow that, for all mXnXNðf Þ;

bfnfn3CknB bfmfm3Ck

mðrel @IkÞ;

where fn ¼ f ekn and fm ¼ f ek

m; indeed, if this is the case then we have by Lemma 3.3

fn3aknC

kn ¼ aP3

bfnfn3CknBðrel @IkÞaP3

bfmfm3Ckm ¼ fm3ak

mCkm:

Put s ¼ m=n: Consider the continuous map F : Ikþ1-Ik defined by

Fðl; %xÞ ¼1

s� 1

� �lþ 1

� �%x

for all lAI and %xAIk: Let

C ¼ bfmfm3Ckm3F : Ikþ1-P:

It satisfies

Cð0; %xÞ ¼ bfmfm3Ckmð %xÞ and Cð1; %xÞ ¼ bfmfm3Ck

m

%x

s

� �:

To finish the proof of (b) it suffices to prove that (i) bfmfm3Ckmð %xsÞ ¼ bfnfn3Ck

nð %xÞ and (ii)

Cðl; %xÞ ¼ bp0p0 for all %xA@Ik:

Proof of (i): let %xAIk and let h ¼ Cknð %xÞ i.e. %x ¼

P%yAFk

nhð %yÞðy1

n;y; yk

nÞ: Define

h0 : F km-I as follows:

h0ðt1;y; tkÞ ¼hðt1;y; tkÞ if tipn for all i;

0 otherwise:

�

Obviously suppðh0Þ is a chain in Fkm so h0AcF k

mF km: We claim that Ck

mð %xsÞ ¼ h0: Indeed, by

definition of h0 we have that

Fkmðh0Þ ¼

X%yAFk

m

h0ð %yÞy1

m;y;

yk

m

¼X

%yAsuppðhÞhð %yÞ

y1

m;y;

yk

m

¼ 1

s

X%yAFk

n

hð %yÞy1

n;y;

yk

n

¼ %x

s:

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300286

It follows that bfmfm3Ckmð %xsÞ ¼ bfmfmðh0Þ: Fix pAP: Then

bfmfmðh0ÞðpÞ ¼X

fh0ðyÞ : fmðyÞ ¼ pg

¼X

fh0ðyÞ : f ðyÞ ¼ p; yipm for all i ¼ 1;y; kg

¼X

fhðyÞ : f ðyÞ ¼ p; yipn for all i ¼ 1;y; kg

¼X

fhðyÞ : fnðyÞ ¼ pg

¼ bfnfnðhÞðpÞ:

Proof of (ii): Let %xA@Ik; suppose that xi ¼ 0: Then

Cðl; %xÞ ¼ bfmfm3Ckm

1

s� 1

� �lþ 1

� �%x

�¼ bfmfm3Ck

mð %yÞ;

where yi ¼ 0: Let Ckmð %yÞ ¼ h; i.e. %y ¼

P%zAFk

mhð%zÞðz1

m;y; zk

mÞ: Then as in (a) above, we

find that zi ¼ 0 for all zAsuppðhÞ: By Lemma 3.4

bfmfmðhÞ ¼X

%zAsuppðhÞhð%zÞ dfmð%zÞfmð%zÞ ¼

X%zAsuppðhÞ

hð%zÞ bp0p0 ¼ bp0p0:

Let %xA@Ik with xi ¼ 1; then

Cðl; %xÞ ¼ bfmfm3Ckm

1

s� 1

� �lþ 1

� �%x

�¼ bfmfm3Ck

mð %yÞ;

where yi ¼ ð1s� 1Þlþ 1X1

s: Let Ck

mð %yÞ ¼ h; i.e. yi ¼P

%zAsuppðhÞ hð%zÞzi

m: We show that

ziXn for all %zAsuppðhÞ: Indeed, since suppðhÞ is a chain in Fkm; then fzi : %zAsuppðhÞg

is a chain in Fm; it follows that if one of the zi is strictly less than n then all of themmust be at most n: Thus

n

m¼ 1

spyi ¼

X%zAsuppðhÞ

hð%zÞzi

mpt

n � 1

mþ ð1� tÞn

m¼ n � t

m

for some 0otp1; a contradiction. Thus

bfmfmðhÞ ¼X

%zAsuppðhÞhð%zÞ dfmð%zÞfmð%zÞ ¼

X%zAsuppðhÞ

hðzÞ bp0p0 ¼ bp0p0:

(c) Finally, we must show that if ½f ¼ ½g then ½fn3aknC

kn ¼ ½gm3ak

mCkm for all

nXNðf Þ and all mXNðgÞ: By (b) we may assume that n ¼ mXNðf Þ; NðgÞ: We may

further assume that fpg: But then fnpgn and so fn3aknC

knpgn3ak

nCkn : By Lemma 4.2

(with Y ¼ @Ik) we are done. &

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300 287

4.2. D is onto

Lemma 4.2. D is onto.

Proof. Let ½g ApkðP; p0Þ where gACkðP; p0Þ: Then by Theorem 3.17 and Lemma 3.5

there exist nX1 and an n-simple function g0ACkðP; p0Þ such that g0Bg ðrel @Ik). By

Lemma 3.15 there exists fAFkn ðP; p0Þ such that f 3ak

nCkn ¼ g0: Hence we have that

f rknAFkðP; p0Þ and of course Nðf rk

nÞpn: Then

Dð½f rkn Þ ¼ ½f rk

n3ekn3a

knC

kn

¼ ½f 3aknC

kn

¼ ½g0

¼ ½g : &

4.3. D is one-to-one

To prove that D is one-to-one, we shall require some further lemmas.Let sX1 be an odd integer and let nX2 be an even integer. For every integer tX0

there exists a unique integer a such that either (i) t ¼ as or (ii) t ¼ as þ b where a iseven and �s þ 1pbps � 1:

Definition 4.3. Let sX1 be an odd integer and let nX2 be an even integer. Thefunction ds;n : F-F is defined by

ds;nðtÞ ¼a if tpsn;

t � nðs � 1Þ otherwise;

�

where a is defined as above.

Lemma 4.4. Let sX1 be an odd integer and let nX2 be an even integer. Then

ds;n ¼ ðmnÞs�12 3ðmn�2Þs�1

3?3ðm2Þs�13ðm0Þ

s�12 :

Proof. We claim that for every integers i; qX0 we have that

ðmiÞqðtÞ ¼

t if 0ptpi � 1;

i if iptpi þ 2q;

t � 2q if i þ 2qot:

8><>:

The claim is easily verified by induction on q: The lemma follows by induction on n

(tedious, but straightforward). &

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300288

Definition. Let sX1 be an odd integer and let nX2 be an even integer. Let dks;n denote

the map

dks;nðt1;y; tkÞ ¼ ðds;nðt1Þ;y; ds;nðtkÞÞ:

Lemma 4.5. Let sX3 be an odd integer and let nX2 be an even integer. We have that

ðrkn3d

ks;n3e

ksnÞ3ak

snCksn ¼ ak

nCkn ;

i.e. the following diagram commutes:

Ik ��!aksnC

ksn

Fksn

aknC

knk kdk

s;neksn

F kn ’

rkn

F k

Proof. Let %t ¼ ðt1;y; tkÞAF ksn: Then

Fksnðb%t%tÞ ¼ t1

sn;y;

tk

sn

implies that ak

snCksnðt1

sn;y; tk

snÞ ¼ %t: By Lemma 3.14(1) and (2), the map ak

nCkn is

sn-simple; thus by Lemma 3.15 there exists a map g : F ksn-F k

n such that g3aksnC

ksn ¼

aknC

kn : Hence gð%tÞ ¼ ak

nCknðt1

sn;y; tk

snÞ: For each i write ti ¼ ais þ bi where ai is even

when bia0: If ai is odd then bi ¼ 0 so ti

snAfai

ng: If ai is even then one verifies easily that

ti

snA ai�1

n; aiþ1

n½: We conclude by Lemma 3.11 that

t1

sn;y;

tk

sn

Aðak

n3CknÞ

�1ða1;y; akÞ

i.e.

akn3C

kn

t1

sn;y;

tk

sn

¼ ða1;y; akÞ;

which is by definition equal to rknd

ks;ne

ksnðt1;y; tkÞ: &

Lemma 4.6. D is one-to-one.

Proof. Let f ; gAFkðP; p0Þ such that Dð½f Þ ¼ Dð½g Þ:We must show that there exists a

path in F kðP; p0Þ between f and g: By hypothesis, there exists a continuous map

G : Ikþ1-P such that

(i) Gð0; %xÞ ¼ Fð %xÞ for all %xAIk;

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300 289

(ii) Gð1; %xÞ ¼ Gð %xÞ for all %xAIk;(iii) Gðl; %xÞ ¼ p0 for all %xA@Ik and all lAI ;

where F ¼ f ekna

knC

kn and G ¼ gek

naknC

kn for some even nXmaxðNðf Þ;NðgÞÞ:

By Theorem 3.17 there exists an odd sX1 and an sn-simple map bpF with

property (iii). By Lemma 3.15 there exists a map h : Fkþ1sn -P which by (iii) satisfies

hð%tÞ ¼ p0 whenever tiAf0; sng for some 2pipk þ 1; and such that h3akþ1sn Ckþ1

sn ¼ b;i.e. the following diagram commutes:

Ikþ1 !b P

Ckþ1sn k mhdFkþ1

snFkþ1sn akþ1

sn

�! Fkþ1sn

Claim.

hð0; %tÞpf dks;ne

ksnð%tÞ

and

hðsn; %tÞpgdks;ne

ksnð%tÞ

for all %tAF ksn:

Proof. We prove the first inequality, the second is identical. By Lemma 4.5 we havethat

F ¼ f 3ekn3a

knC

kn ¼ f 3ek

n3ðrkn3d

ks;n3e

ksnÞ3ak

snCksn:

However since nXNðf Þ it is clear that f eknr

kn ¼ f ; so we get that

F ¼ f 3dks;n3e

ksn3a

ksnC

ksn:

Let %tAFksn: Then clearly

akþ1sn Ckþ1

sn 0;%t

sn

� �¼ ð0; %tÞAFkþ1

sn

and

akþ1sn Ckþ1

sn 1;%t

sn

� �¼ ðsn; %tÞAFkþ1

sn

and also

aksnC

ksn

%t

sn

� �¼ %tAFk

sn:

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300290

Hence

hð0; %tÞ ¼ b 0;%t

sn

� �pG 0;

%t

sn

� �¼ F

%t

sn

� �¼ f dk

s;neksna

ksnC

ksn

%t

sn

� �¼ f dk

s;neksnð%tÞ:

For every iAFsn let hi : F ksn-P be defined by hið%tÞ ¼ hði; %tÞ for all %tAFk

sn: We have

that each hi is in FksnðP; p0Þ; furthermore, if ipj then hiphj so we have a path

h0;y; hsn in F ksnðP; p0Þ such that h0Xf dk

s;neksn and hsnXgdk

s;neksn by the last claim.

Compose all the hi with rksn to obtain a path in FkðP; p0Þ from f dk

s;neksnr

ksn to

gdks;ne

ksnr

ksn: It is easy to see that these maps are respectively equal to f dk

s;n and gdks;n: By

Lemmas 4.4 and 2.4 and the remarks that follow Definition 2.7 it is easy to see that

there is a path in FkðP; p0Þ from dks;n to ðmk

nÞw for some wX0; hence we conclude using

repeated applications of Lemma 2.8 that

½f ¼ ½f 3ðmknÞ

w ¼ ½f dks;n ¼ ½gdk

s;n ¼ ½g3ðmknÞ

w ¼ ½g : &

4.4. D is a group homomorphism

Recall from Definition 2.6 that the product in the group skðP; p0Þ is given by½f � ½g ¼ ½ðf ; gÞn for any even nXN1ðf Þ:

Lemma 4.7. Let f ; gAFkðP; p0Þ: Let n be an even integer such that nX

maxðNðf Þ;NðgÞÞ: Define a function by

ðf ; gÞ0nðt1;y; tkÞ ¼f ðt1;y; tkÞ if 0ptipn for all 1pipk;

gðt1 � n;y; tk � nÞ if npti for all 1pipk;

p0 otherwise:

8><>:

Then ðf ; gÞ0nAF kðP; p0Þ and ½ðf ; gÞ0n ¼ ½ðf ; gÞn :

Proof. The proof of the first statement is straightforward. For any hAFkðP; p0Þ suchthat nXNðhÞ define a map by

h0ðt1;y; tkÞ ¼hðt1; t2 � n;y; tk � nÞ if tiXn for all 2pipk;

p0 otherwise:

�

Using the claim in the proof of Lemma 4.4 it is simple to verify that

h0ðt1;y; tkÞ ¼ hðt1; ðm0Þn2ðt2Þ;y; ðm0Þ

n2ðtkÞÞ

for all %tAF k: By repeated applications of Lemma 2.8 we have that ½h ¼ ½h0 for all h:

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300 291

It is easy to see that

ðf ; gÞ0n ¼ ðf ; g0Þn:

Hence

½ðf ; gÞ0n ¼ ½ðf ; g0Þn ¼ ½f � ½g0 ¼ ½f � ½g ¼ ½ðf ; gÞn : &

Lemma 4.8. Let f ; gACkðP; p0Þ: Define a map by

/f ; gS0ðt1;y; tkÞ ¼f ð2t1;y; 2tkÞ if 0ptip1=2; for all 1pipk;

gð2t1 � 1;y; 2tk � 1Þ if 1=2ptip1; for all 1pipk;

p0 otherwise:

8><>:

Then /f ; gS0ACkðP; p0Þ and ½/f ; gS0 ¼ ½/f ; gS :

Proof. For any hACkðP; p0Þ we have that

½h ¼ ½h � ½p0 ¼ ½/h; p0S ;

where

/h; p0Sðt1;y; tkÞ ¼hð2t1; t2;y; tkÞ if 0pt1p1=2;

p0 otherwise:

�

Hence hB/h; p0S ðrel @IkÞ: Obviously this holds if we choose any other coordinate,i.e. if we define

hiðt1;y; tkÞ ¼hðt1;y; ti�1; 2ti; tiþ1;y; tkÞ if 0ptip1=2;

p0 otherwise;

�

then we have that hBhi ðrel @IkÞ for all 1pipk: By applying this construction

repeatedly to the coordinates i ¼ 2;y; k we conclude that hBh0 ðrel @IkÞ where

h0ðt1;y; tkÞ ¼hðt1; 2t2;y; 2tkÞ if 0ptip1=2 for all 2pipk;

p0 otherwise:

�

A similar argument using the fact that ½h ¼ ½p0 � ½h shows that hBh00 ðrel @IkÞwhere

h00ðt1;y; tkÞ ¼hðt1; 2t2 � 1;y; 2tk � 1Þ if 1=2ptip1 for all 2pipk;

p0 otherwise:

�

Then for any f ; gACkðP; p0Þ we have that ½f ¼ ½f 0 and ½g ¼ ½g00 (as defined above)

so ½/f ; gS ¼ ½/f 0; g00S : One verifies easily that /f 0; g00S ¼ /f ; gS0: &

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300292

Lemma 4.9. Let nX2 be an even integer. Then

(1) for all 0ptip1=2;

ekn3a

knC

knð2t1;y; 2tkÞ ¼ ek

2n3ak2nC

k2nðt1;y; tkÞ;

(2) for all 1=2ptip1;

ekn3a

knC

knð2t1 � 1;y; 2tk � 1Þ ¼ ðek

2n3ak2nC

k2nðt1;y; tkÞÞ � ðn;y; nÞ;

Let ðs1;y; skÞ ¼ ek2n3a

k2nC

k2nðt1;y; tkÞ: Then

(3) si ¼ n if and only if 1=2ptionþ12n;

(4) sipn if and only if tionþ12n:

Proof. We prove the first statement, the other proofs are quite similar. Let

Cknð2t1;y; 2tkÞ ¼ hAcPk

nPkn so that

2ti ¼ FknðhÞ ¼

X%xAsuppðhÞ

hð %xÞxi

n

which rewrites as

ti ¼X

%xAsuppðhÞhð %xÞ

xi

2n;

since FnCF2n the support of h is also a chain in Fk2n so that in fact we may consider

that hAcFk2nFk2n: It follows from the above that

ekn3a

knC

knð2t1;y; 2tkÞ ¼ ek

n3aknðhÞ ¼ ek

2n3ak2nðhÞ ¼ ek

2n3ak2nC

k2nðt1;y; tkÞ: &

Lemma 4.10. D is a group homomorphism.

Proof. Obviously D maps ½p0 to ½p0 : Let f ; gAFkðP; p0Þ: We must prove that

Dð½f � ½g Þ ¼ Dð½f Þ � Dð½g Þ:

Choose an even integer nXmaxðNðf Þ;NðgÞÞ: Then by Lemma 4.7 the lefthand side is

Dð½f � ½g Þ ¼Dð½ðf ; gÞnÞ

¼Dð½ðf ; gÞ0n Þ

¼ ½ðf ; gÞ0n3ek2n3a

k2nC

k2n ;

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300 293

since Nððf ; gÞ0nÞp2n: Let H denote the map ðf ; gÞ0n3ek2n3a

k2nC

k2n and let %s stand for

ek2n3a

k2nC

k2nð%tÞ; then

Hðt1;y; tkÞ ¼f ðs1;y; skÞ if 0psipn for all 1pipk;

gðs1 � n;y; sk � nÞ if npsi for all 1pipk;

p0 otherwise:

8><>:

By Lemma 4.9(4) this is the same as

Hðt1;y; tkÞ ¼f ðs1;y; skÞ if 0ptionþ1

2nfor all 1pipk;

gðs1 � n;y; sk � nÞ if nþ12npti for all 1pipk;

p0 otherwise:

8><>:

By Lemma 4.9(3), if tionþ12n

for all i but there exists some j such that tj41=2 then

sj ¼ n and Hð%tÞ ¼ p0: Similarly, if tiX1=2 for all i and there exists some j such that

tjonþ12n

then by Lemma 4.9(3) sj ¼ n so gðs1 � n;y; sk � nÞ ¼ p0 ¼ Hð%tÞ: So we may

rewrite H as

Hðt1;y; tkÞ ¼f ðs1;y; skÞ if 0ptip1

2for all 1pipk;

gðs1 � n;y; sk � nÞ if 12pti for all 1pipk;

p0 otherwise:

8><>:

By Lemma 4.9(1) and (2) this is equal to

Hðt1;y; tkÞ ¼f ðek

n3aknC

knð2t1;y; 2tkÞÞ if 0ptip1

2for all 1pipk;

gðekn3a

knC

knð2t1 � 1;y; 2tk � 1ÞÞ if 1

2pti for all 1pipk;

p0 otherwise:

8><>:

Clearly we have that

H ¼ /f 3ekn3a

knC

kn ; g3ek

n3aknC

knS

0:

By Lemma 4.8 we have that

½/f 3ekn3a

knC

kn ; g3ek

n3aknC

knS

0 ¼ ½/f 3ekn3a

knC

kn ; g3ek

n3aknC

knS

¼ ½f 3ekn3a

knC

kn %½g3ek

n3aknC

kn

¼Dð½f Þ%Dð½g Þ

and this completes the proof. &

Theorem 4.11. Let P be a poset and let p0AP: Then the correspondence Dk induces a

group isomorphism from skðP; p0Þ onto pkðP; p0Þ:

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300294

Proof. Immediate from Lemmas 4.1, 4.2, 4.6 and 4.10. &

Remark. It follows immediately from the definitions that the correspondence thatassigns to each pointed poset ðP; p0Þ the isomorphism Dk is a natural equivalencefrom the functor sk to the functor pk: Consider also the correspondence which

assigns to every pointed poset ðP; p0Þ the pointed topological space ðP; bp0p0Þ and to anisotone map f : P-Q its simplicial lifting. This is a functor [20, p. 183] which wecan compose with the kth homotopy group functor pk: It is immediate by Theorem3.2 and Lemma 3.3 that the function which assigns to each pointed poset ðP; p0Þ thehomomorphism ðaPÞ# is a natural equivalence from this composition to the functor

pk restricted to pointed posets. Hence we have the

Theorem 4.12. The functor which assigns to a pointed poset ðP; p0Þ the group skðP; p0Þis naturally equivalent to the functor which assigns to ðP; p0Þ the kth homotopy group of

the geometric realisation of the simplicial complex of P; pkðP; bp0p0Þ:

5. Bounded subsets and simplices in reflexive structures

In this section we prove that for every pointed binary reflexive structure ðX ; x0Þ;there exists a pointed poset ðP; p0Þ and a homomorphism from ðP; p0Þ to ðX ;x0Þ thatinduces isomorphisms in all the groups sk: The poset has a very simple description interms of X :

Definition. Let X be a binary reflexive structure. A nonempty subset Y of X is saidto be bounded if there exists some bAY such that y-b for all yAY : Such an elementwe will call a bound for Y : Let BðX Þ denote the poset of bounded subsets of X

ordered by inclusion.

Definition. Let X be a reflexive structure. Let b : BðXÞ-X denote some fixedfunction that assigns to each YABðXÞ a bound bAY :

Theorem 5.1. For every pointed reflexive structure ðX ; x0Þ; the map b induces an

isomorphism of skðBðX Þ; fx0gÞ onto skðX ; x0Þ; for every kX1:

Proof. That b is a homomorphism is immediate. Consider the following construc-

tion: if fAFkðX ; x0Þ; define a map

f 0 : Fk-BðXÞ

by

f 0ðvÞ ¼ f ðkvÞ ¼ ff ðuÞ : u-vg

for all vAFk:

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300 295

Claim 0. f 0AF kðBðX Þ; fx0gÞ:

Proof. If %s-%t then f ð%sÞ-f ð%tÞ so f ð%tÞ is a bound for f 0ð%tÞ; which proves that

f 0ð%tÞABðXÞ; it is also clear that f 0ð%sÞDf 0ð%tÞ; since F k is a poset. Because 0 is minimalin F it follows that if ti ¼ 0 for some i then %s-%t implies si ¼ 0; and hence f ð%sÞ ¼ x0;consequently f 0ð%tÞ ¼ fx0g: Similarly, if nXNðf Þ is an even integer, and if tiXn forsome i; it is clear that if %s-%t then siXn and hence f 0ð%tÞ ¼ fx0g: &

Claim 1. The correspondence f/f 0 induces a group homomorphism from skðX ; x0Þ to

skðBðXÞ; fx0gÞ:

Proof. Let f-g in FkðX ; x0Þ: We claim that the map

H : %t/f 0ð%tÞ,g0ð%tÞ

is in F kðBðXÞ; fx0gÞ: Indeed, if %s-%t then f ð%sÞ-gð%tÞ and gð%sÞ-gð%tÞ so gð%tÞ is a boundfor f 0ð%tÞ,g0ð%tÞ; which proves that f 0ð%tÞ,g0ð%tÞABðXÞ: One verifies easily the rest ofthe conditions. It is clear that f 0-H and g0-H and so ½f 0 ¼ ½g0 ; we conclude thatthe map

½f /½f 0

is well defined. Now we show that this map commutes with the group operation. Let

%tAFk: First consider the case where t1pn: Then s1pn for every %s-%t so

ððf ; gÞnÞ0ð%tÞ ¼ fðf ; gÞnð%sÞ : %s-%tg ¼ ff ð%sÞ : %s-%tg ¼ f 0ð%tÞ ¼ ðf 0; g0Þnð%tÞ:

Now suppose that t14n: Then s1Xn for every %s-%t so

ððf ; gÞnÞ0ð%tÞ ¼ fðf ; gÞnð%sÞ : %s-%tg

¼fgðs1 � n; s2;y; skÞ : %s-%tg

¼fgð%zÞ : %z-ðt1 � n; t2;y; tkÞg

¼ g0ðt1 � n; t2;y; tkÞ

¼ ðf 0; g0Þnð%tÞ: &

Claim 2. (i) For every fAF kðX ; x0Þ we have that f-b3f 0; (ii) for every

gAF kðBðXÞ; fx0gÞ we have that ðb3gÞ0pg:

Proof. (i) Simply notice that if %s-%t then f 0ð%tÞ contains f ð%sÞ hence f ð%sÞ-bðf 0ð%tÞÞ:(ii) We have that ðb3gÞ0ð%tÞ ¼ fbðgð%sÞÞ : %s-%tg and if %s-%t then bðgð%sÞÞAgð%sÞDgð%tÞshows that ðb3gÞ0ð%tÞDgð%tÞ:

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300296

By Claim 2 the homomorphism ½f /½f 0 is an inverse for the homomorphisminduced by b: &

In fact, it will be sufficient to consider a much smaller poset, namely that of‘simplices’:

Definition. Let X be a reflexive structure. Let%n denote the n-element chain, i.e. the

poset on f0; 1; 2;y; n � 1g with the usual ordering. A simplex in X is a subset Y

which is the image of a homomorphism from some chain%n to X : Let SðXÞ denote the

poset of simplices of X ordered by inclusion. Clearly SðX Þ is a subposet of BðXÞ; andso we may restrict the map b to SðX Þ:

Definition. Let P be a poset. Let DP denote the poset of finite chains (totally orderedsubsets of P) ordered by inclusion. Let mP : DP-P denote the map that assigns to afinite chain C ¼ fp1pp2p?pplg its maximum element max C ¼ pl :

We shall require the following result of McCord [12]. An open cover U of atopological space is basis-like if whenever xAU-V and U ; VAU there exists aWAU such that xAWDU-V :

Theorem 5.2 (McCord [12], Theorem 6). Let X and Y be topological spaces and

let f : X-Y be a continuous map for which there exists a basis-like open cover U of

Y satisfying the following condition: for each UAU; the restriction f jf �1ðUÞ :

f �1ðUÞ-U is a weak homotopy equivalence. Then f itself is a weak homotopy

equivalence.

Lemma 5.3. Let ðP; p0Þ be a pointed poset. Then the map mP induces an isomorphism

of skðDðPÞ; fp0gÞ onto skðP; p0Þ for every kX1:

Proof. Let U be the collection of principal ideals of P; it is a basis-like open cover of

P: Let UAU; say U ¼ kp: Then obviously m�1P ðUÞ has a least element, namely fpg:

Consequently, both U and m�1P ðUÞ are contractible (see for example [18]); hence

Theorem 5.2 applies and mP is a weak homotopy equivalence. Since the functors sk

and pk are naturally equivalent we conclude that mP induces an isomorphism ofskðDðPÞ; fp0gÞ onto skðP; p0Þ for every kX1: &

Lemma 5.4. Let ðX ; x0Þ and ðY ; y0Þ be pointed structures, let f and c be base-point

preserving homomorphisms from X to Y : If f-c then f# ¼ c#:

Proof. For any hAFkðX ; x0Þ; we have by Lemma 2.4 that f3h-c3h sof#ð½h Þ ¼ ½f3h ¼ ½c3h ¼ c#ð½h Þ: &

Theorem 5.5. For every pointed reflexive structure ðX ; x0Þ; the map b : SðXÞ-X

induces an isomorphism of skðSðX Þ; fx0gÞ onto skðX ; x0Þ for every kX1:

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300 297

Proof. Consider the sequence

DSðX Þ+j

DBðX Þ !d SðX Þ+i

BðXÞ!b X ;

where j and i are the natural inclusions, and the map

d : DBðXÞ-SðXÞ

is defined by

dðCÞ ¼ fbðYÞ : YACg

for every chain C in DBðX Þ: It is easy to see that d is well defined and ahomomorphism. We shall prove that all the maps in the sequence induceisomorphisms in the sk: We start with d: it is easy to see that d3jpmSðX Þ: indeed,

if C is a chain in SðXÞ then bðYÞAYDmax C for each YAC: Hence by Lemmas 5.3and 5.4 ðd3jÞ# is an isomorphism. It follows of course that d# is onto. Similarly, we

have that i3dpmBðXÞ implies that ði3dÞ# is an isomorphism. Hence d# is one-to-one

and hence an isomorphism. This in turn shows that i# and j# are isomorphisms. Since

b : SðX Þ-X induces the homomorphism b#3i#; by Theorem 5.1 we are done. &

Combining this result with McCord’s Theorem 3.2 we obtain the following

Corollary 5.6. Let ðX ; x0Þ be a pointed reflexive structure, let P denote the poset of

its simplices, and let p0 ¼ fx0g: Then the group skðX ; x0Þ is isomorphic to pkðP; bp0p0Þ for

all kX1:

Remark. For a given structure X ; there might be more than one possibility for themap b: However, it is immediate by Lemma 5.4 that any choice will induce a single,unique group homomorphism from skðSðXÞ; fx0gÞ to skðX ; x0Þ: Furthermore, thecorrespondence which assigns to a pointed structure ðX ; x0Þ the pointed posetðSðX Þ; fx0gÞ and to a map f : X-Y the order-preserving operation Sðf Þ :SðXÞ-SðYÞ defined by Sðf ÞðCÞ ¼ ff ðxÞ : xACg is easily seen to be functorial. Wemay compose the functor sk with S; and we claim that the correspondence whichassigns to each pointed structure ðX ; x0Þ the group homomorphism b# is a natural

equivalence of this functor to the functor sk: Indeed, if f : X-Y is a

homomorphism, then we must show that for each fAF kðSðXÞ; fx0gÞ; we have that

½f3bX 3f ¼ ½bY 3SðfÞ3f ;

indeed, it is not difficult to show that

f3bX 3f-bY 3SðfÞ3f

in F kðY ; y0Þ; and the result follows from Lemma 5.4.

ARTICLE IN PRESSB. Larose, C. Tardif / Advances in Mathematics 189 (2004) 268–300298

By the last remark and Theorem 5.12 we have the following result:

Theorem 5.7. The functor sk is naturally equivalent to the functor which assigns to the

pointed structure ðX ;x0Þ the kth homotopy group of the simplicial complex of the poset

of simplices of X ; pkð dSðXÞSðXÞ; dfx0gfx0gÞ:

Acknowledgments

The authors thank Andre Joyal, Andre Lebel and Pascal Tesson for helpfuldiscussions.

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