trees, fundamental groups and homology groups

Annals of Pure and Applied Logic 111 (2001) 185–201www.elsevier.com/locate/apal

Trees, fundamental groups and homology groupsKatsuya Eda ∗, Masasi Higasikawa

School of Science and Engineering, Waseda University, Tokyo 169-0072, Japan

Received 22 December 1999; received in revised form 25 April 2000

Communicated by T. Jech

Abstract

For a tree T of its height equal to or less than !1, we construct a space XT by attaching acircle to each node and connecting each node to its successors by intervals. H is the Hawaiianearring and HT

1 (X ) denotes a canonical factor of the 2rst integral singular homology group. Thefollowing equivalences hold for an !1-tree T : (1) �1(X!1 ) is embeddable into �1(XT ), if andonly if HT

1 (X!1 ) � ��!1Z is embeddable into HT

1 (XT ), if and only if T is not an Aronzajntree. (2) �1(XT ) is embeddable into ××!Z � �1(H) if and only if HT

1 (XT ) is embeddable intoZ! � HT

1 (H) if and only if T is a special Aronzajn tree. (3) �1(XT ) has a retract isomorphic toan uncountable free group, if and only if HT

1 (XT ) has a summand isomorphic to an uncountablefree abelian group, if and only if T has an uncountable anti-chain. c© 2001 Elsevier ScienceB.V. All rights reserved.

MSC: 03E75; 20E05; 20F34; 54F50; 55N10; 55Q20; 55Q52

Keywords: Free �-product; �-Word tree; Aronzajn tree; Fundamental group; Singular homology

1. Introduction

A tree (T;4) is a partially ordered set whose initial segments {s∈T : s≺ t} are well-ordered, which has a root element r for which r4 t holds for every t ∈T and whichalso satis2es the following property (*). The order type of {s∈T : s≺ t} is denoted byo(t). In case o(t) is a successor or a limit, we call t to be a successor element or alimit element, respectively. The property (*) is

For limit elements s and t of T , {u∈T : u≺ s}= {u∈T : u≺ t} implies s = t.

In the sequel an element r ∈T is always the root of T . A subtree of a tree T is anon-empty subset of T which is closed under taking initial segments. (Consequently

∗ Corresponding author.E-mail addresses: [email protected] (K. Eda), [email protected] (M.

Higasikawa).

0168-0072/01/$ - see front matter c© 2001 Elsevier Science B.V. All rights reserved.PII: S0168 -0072(01)00025 -2

186 K. Eda, M. Higasikawa /Annals of Pure and Applied Logic 111 (2001) 185–201

a subtree of T contains r.) The set of all successor elements in T is denoted by T s.A branching point of a tree is an element which has at least two immediate successors.The height of a tree T is sup{o(t) + 1 : t ∈T}. An !1-tree is a tree of height !1 suchthat {t ∈T : o(t)¡�} is countable for each �¡!1. A chain is a subset of a tree whichis totally ordered by ≺. An anti-chain in T is a subset of T whose distinct elements areincomparable with respect to ≺. An Aronzajn tree is an !1-tree without an uncountablechain. A special Aronzajn tree is an Aronzajn tree which is a union of countably manyanti-chains. The trees we investigate in this paper are of height at most !1. We remarkthat ordinals under the natural ordering are trees.

A tree has a natural topology, which is called the tree topology, i.e. a basic open setis of the form {r} or {s′ ∈T : s≺ s′ 4 t} for s≺ t. To relate a tree T to a fundamentalgroup and a homology group, we connect each element of T and each of its successorsby an interval and also attach a circle to each element. We naturally extend the treetopology to this extended space XT .

A subspace Y of a space X is a retract, if there exists a continuous map r : X →Y(which is called a retraction), such that the restriction of r to Y is the identity. We alsouse the same notion for groups. A subgroup H of a group G is a retract, if there existsa homomorphism h : G→H (which is also called a retraction), such that the restrictionof h to H is the identity. When a group is abelian, we say summand for retract as usual.The Hawaiian earring is the plane continuum H=

⋃∞n=1{(x; y) : (x−1=n)2 +y2 = 1=n2}:

For notational convenience, we regard H as {o}∪⋃∞n=1 In for copies In of the unit

open interval (0; 1) and a point o corresponding to the origin.Following are the main results of this paper. For homology groups, we recall a

canonical factor of a singular homology group HT1 (X ) from [8]. As far as HT

1 (XT )is concerned, we may take the following as the de2nition of HT

1 (XT ): an element xof HT

1 (XT ) is a function from T to Z corresponding to a loop such that x(t) is thewinding number of the loop around the circle attached to t ∈T . Unde2ned notions willbe explained at the end of this section.

Theorem 1.1. Let T be a tree of height equal to or less than !1. Then the followinghold:

(1) �1(X!1 ) is embeddable into �1(XT ); if and only if HT1 (X!1 )��

!1Z is embeddable

into HT1 (XT ); if and only if T has an uncountable chain. When the cardinality of

a tree T is less than the least measurable cardinal; the negation of any of thesethree equivalent statements is also equivalent to the Z-re7exivity of HT

1 (XT ).(2) �1(XT ) is embeddable into ××n¡!Fn; where each Fn is a free group; if and only

if HT1 (XT ) is embeddable into �n¡!An; where each An is a free abelian group;

if and only if T is a union of countably many anti-chains.(3) Let � be an uncountable cardinal. Then there exists a retract of �1(XT ) which

is isomorphic to the free group of the cardinality �; if and only if there existsa summand of HT

1 (XT ) which is isomorphic to the free abelian group of thecardinality �; if and only if T has an anti-chain of the cardinality �.

K. Eda, M. Higasikawa /Annals of Pure and Applied Logic 111 (2001) 185–201 187

Corollary 1.2. Let T be an !1-tree. Then the following hold:(1) �1(X!1 ) is embeddable into �1(XT ); if and only if HT

1 (X!1 )��!1Z is embeddable

into H1(XT ); if and only if HT1 (XT ) is not Z-re7exive; if and only if T is not an

Aronzajn tree.(2) �1(XT ) is embeddable into ××!Z � �1(H); if and only if HT

1 (XT ) is embeddableinto Z! �HT

1 (H); if and only if T is a special Aronzajn tree.(3) �1(XT ) has a retract isomorphic to an uncountable free group; if and only if

HT1 (XT ) has a summand isomorphic to an uncountable free abelian group; if and

only if T has an uncountable anti-chain.

Theorem 1.3. Let S and T be trees of height equal to or less than !1. If �1(XS) isisomorphic to a retract of �1(XT ); then the subspace S ′ of S consisting of all limitpoints is homeomorphic to a closed subspace of T; which consists of limit points.Moreover if �1(XS) is isomorphic to �1(XT ); then S ′ is homeomorphic to the closedsubspace T ′ of T consisting of all limit points.

When trees are ordinals, we can decide more precisely.

Corollary 1.4. Let � and � be countable ordinals. Then; �1(X�) is isomorphic to aretract of �1(X�) if and only if � is homeomorphic to a closed subspace of �.

In Section 5 we shall prove this corollary and determine which ordinals can be closedsubspaces of other ordinals. We state a precise de2nition of XT in the following. Insteadof a circle we use two edges by identifying end points.

For t ∈T , let At; Bt ; Ct be copies of the open interval (0; 1) such that all At; Bu; Cv arepairwise disjoint. (We do not use Ct for a limit t ∈T .) For 0¡ ¡1, we let At(0; ) andAt(1− ; 1) be copies of (0; ) and (1− ; 1) in At , respectively, and also for Bt and Ct

similarly. Let the underlying set XT be T ∪⋃{At ∪Bt ∪{te} : t ∈T}∪⋃ {Ct : t ∈T s}.We call the set T ∪⋃{Ct : t ∈T s} a stem and de2ne [s; t] to be the subset {u∈T : s4 u4 t}∪⋃ {Cu : s≺ u4 t; u∈T s} of a stem for s≺ t. For each point x∈At ∪Bt ∪Ct , basicneighborhoods are given from (0; 1). Each point te has the topology as the end point ofboth At and Bt , that is, {te}∪At(1− ; 1)∪Bt(1− ; 1) are basic open neighborhoods for0¡ ¡1. We de2ne a neighborhood base for t as follows. A basic open neighborhoodof the root r is

{r} ∪ Ar(0; )∪Br(0; )

∪⋃

{Cu(1 − ; 1) : u is an immediate successor of r}

for ¿0. If t is a successor, a basic open neighborhood of t is

{t} ∪ At(0; ) ∪ Bt(0; ) ∪ Ct(0; )

∪⋃

{Cu(1 − ; 1) : u is an immediate successor of t};


for ¿0. If t is a limit, a basic open neighborhood of t is

[s; t] ∪ At(0; ) ∪ Bt(0; )

∪⋃

{{ue} ∪ Au ∪ Bu : s 4 u ≺ t}∪Cs(0; ) ∪

⋃{Cv(1 − ; 1) : s 4 U 4 t; v is an immediate successor of u}

for ¿0 and s≺ t with s∈T s.We investigate the fundamental group of the space XT for a tree T . Since !1 itself

is a tree, X!1 is a space consisting of the long ray with !1-many circles. The spaceX!+1 is homotopy equivalent to the Hawaiian earring H, whose fundamental group isisomorphic to ××!Z: (See Remark 1.5.) As another canonical tree, let B(�) be a treeconsisting of the root and its �-many immediate successors. Then the space XB(�) ishomotopy equivalent to the bouquet with � + 1-many circles and so the fundamentalgroup is isomorphic to the free group on � + 1-many generators.

Since the fundamental groups of spaces satisfying the properties of Proposition 2.1are subgroups of a free �-product of copies of Z, our investigation can be seen asthat of subgroups of a free �-product of copies of Z, ××�

I Z, which is the funda-mental group of the Hawaiian earring of I -many circles HI [4, Appendix]. On theother hand, HT

1 (HI )��I Z holds by [8, Theorem 4:6], where ��

I Z= {x∈�IZ : supp(x)is countable}.

For basic terminology and unde2ned notions, we refer the reader to [14] for algebraictopology, [12] for group theory and [11] for set theory.

Remark 1.5. (1) The space X!+1 homeomorphic to a space consisting of Y in theproof of Lemma 4.5 and a circle. Since a space consisting of the Hawaiian earring anda circle is again the Hawaiian earring, X!+1 is homotopy equivalent to the Hawaiianearring H. See also [5, Remark 3:12].

(2) We remark that XT and XS may not be homeomorphic for homeomorphic trees Sand T . For instance, Let S be a tree with two copies of !+1 as successors of the root.Then S and ! + ! + 1 are homeomorphic, but XS and X!+!+1 are not homeomorphic.

2. Edge-path spaces and Proof of Theorem 1.3

A loop is a path f : [a; b]→X such that f(a) = f(b). If we do not specify anything,the domains of paths and loops are the unit interval [0; 1].

A space X is semi-locally simply connected at x, if there exists a neighborhood Uof x such that any loop in U with base point x is null-homotopic relative to {0; 1} inX . For a path f : [0; 1]→X , we de2ne a path f− by: f−(t) = f(1 − t). For paths fand g with f(1) = g(0), let f g : [0; 1]→X be a path de2ned by

fg(t) =

{f(2t) for 0 6 t 6 1=2;

g(2t − 1) for 1=2 6 t 6 1:


For a loop f with base point x∈X , [f]∈ �1(X; x) denotes the homotopy class relativeto {0; 1} containing f, which is an element of �1(X; x).

We recall a theorem from [7].

Proposition 2.1 (Cannon and Conner [7; Theorem 1:1]). Let X be a space satisfyingthe following properties: There exist arcs Pi (i∈ I) with Pi = {ui; vi} and a closed setD such that(1) X = D∪⋃

i∈I Pi and D∩Pi = Pi for each i;(2) Pi\Pi is open and ui �= vi for each i;(3) D contains no arc.Then; the fundamental group of X is isomorphic to a subgroup of ××�

i∈IZi :

We call an edge-path space for any space satisfying the properties of Proposition 2.1.Since XT is an edge-path space with D = {t; te : t ∈T}, the fundamental group of XT isisomorphic to a subgroup of ××�

i∈IZi : Since we need this embedding for further uses,we review the embedding precisely.

A path f : [a; b]→X with f(a); f(b)∈D is proper, if the following hold: Forany a¡c¡b with f(c)∈Pi\Pi = int(Pi); there exist a6u¡v6b such that u¡c¡v,f(u); f(v)∈ Pi, f(u) �= f(v) and Im(f � (u; v))⊂int(Pi):

Lemma 2.2 (Cannon and Conner [7; Lemma 2:1]). Let X be a space satisfying theproperties of Proposition 2.1. Then; for any path f : [0; 1]→X with f(0); f(1)∈D;there exists a proper path g which is homotopic to f relative to {0; 1}.

For a proper path f : [s; t]→X with f(s); f(t)∈D; we de2ne a word Wf as fol-lows: Let O =

⋃i∈I int(Pi) and f−1(O) =

⋃m¡/(am; bm); where (am; bm)∩ (an; bn) = ∅

for distinct m; n and /6!: Let MWf = / and m≺ n if am¡an for m; n∈ /. Let

Wf(m) =

{i if Imf � [am; bm] = Pi and f(am) = ui;

−i if Imf � [am; bm] = Pi and f(am) = vi;

where i is the generator of Zi : For x∈D, {Wf : f is a loop with base point x} is asubgroup of ××�

i∈IZi, which is isomorphic to �1(X; x). (When the index set I is !, weuse a symbol 0n for a generator of Zn to avoid a confusion.)

By Lemma 2.2 we may deal only with proper paths for edge-path spaces. A pathf : [a; b]→X is reduced, if any null-homotopic loop f � [u; v] is constant for u6v, i.e.f � [u; v] is constant in case f(u) = f(v) and f � [u; v] is null-homotopic relative to{u; v}.

By [7, Lemmas 2:2 and 2:4] we have

Lemma 2.3. Let f; g : [0; 1]→X be proper paths in an edge-path space X with endpoints in D.• f is a reduced path if and only if Wf is a reduced word;• f and g are homotopic relative to {0; 1}; if and only if Wf = Wg.


A loop f with base point x0 is essentially in U , if there exist a loop g with basepoint x and a path p from x to x0 such that Im(g)⊂U and f = p−gp. For an elementa of �1(X; x0), we say that a is essentially in U if there exists a loop f with basepoint x0 such that [f] = a and f is essentially in U .

Lemma 2.4. Let X be an edge-path space. For a∈ �1(X; x0); there exists a reducedloop f with base point x satisfying the following: If b is a conjugate to a; then thereexists a path p from x to x0 such that b = Wp−fp and Wp−fp is reduced.Consequently; if [g] is conjugate to a and g is essentially in U; then Im(f)⊂U .

Proof. There exist a reduced loop f and a reduced path q such that Wq−WfWq isreduced, a = [q−fq] and WfWf is reduced [5, Lemma 4:4]. Let x be a base pointof f. If b is a conjugate to a, then there exists a loop h with base point x0 suchthat b = [h−]a[h]. Let p be a reduced path homotopic to qh relative to {0; 1}. Thena = Wp−fp holds. The reduced-ness of Wp−fp follows from the fact that WfWf isreduced.

Let [g] be conjugate to a, g = p−g′p and Im(g′)⊂U . We may assume that p is aproper path. Then, there exists a loop h with base point x0 such that a = [h−][g][h]. LetV be a reduced word of WpWh, then Wf = V−1Wg′V holds. Since WfWf is reduced,each letter in Wf appears in Wg′ [5, Lemma 4:4] and hence Im(f)⊂U .

For the next lemma, we refer the reader to [7].

Lemma 2.5. Let Y be an edge-path space with y∈D and h : �1(H; o)→ �1(Y; y) be ahomomorphism. Then; there exists a continuous map f :H→Y and u∈ �1(Y; y) suchthat h(a) = u−1f∗(a)u for a∈ �1(H; o). In addition if h(0n) �= e for in;nitely many n;the point f(o) is uniquely determined.

Proof. Since the 2rst half is a combination of Theorems 4:2 and 4:4 in [7], we provethe uniqueness of f(o). Let g :H→Y be a continuous map and v∈ �1(Y; y) suchthat h(a) = v−1g∗(a)v for a∈ �1(H; o). Suppose that f(o) �= g(o). Then uv−1 �= e holds,since u and v are determined by paths connecting y with f(o) and g(o); respectively,cf. [7, Lemma 2:2]. (We remark that f(o) and g(o) belong to D, since Im(f∗) andIm(g∗) are in2nitely generated.) Since f∗(0n) = uv−1g∗(0n)vu−1 �= e for in2nitely manyn’s, at least one of f∗ and g∗ is not a standard homomorphism, which contradicts[7, Theorem 4:2].

For a homomorphism h : �1(X; x0)→G, a point x∈X is wild with respect to h, iffor any neighborhood U of x there exists a loop f such that h([f]) �= e and f isessentially in U . The set of all points wild respect to h is denoted by Xwh. In case his the identity, we denote Xwh by Xw. Then, for a path-connected space X , Xw is theset of all x∈X such that X is not semi-locally simply connected at x.

Let X be a path-connected and 2rst countable space and Y be an edge-path space.For x∈Xwh, take a path p from x to x0 and loops fn with base point x such that


[fn]∈ �1(X; x) is non-trivial for each n and the images of fn’s converge to x. Let ’p

be the canonical isomorphism from �1(X; x)→ �1(X; x0); i.e. ’p([f]) = [p−fp], where[p−fp] = [(p−f)p] = [p−(fp)]. There exists a continuous map g :H→X such thatg(o) = x and the restriction of g to In is a copy of the restriction of fn to (0; 1), sincethe images of fn’s converge to x.

Now apply Lemma 2.5 to h ◦’p ◦ g∗, then we have a continuous map f :H→Y .We let F(x) = f(o). Next we show that F is continuous on Xwh, when a space X islocally path-connected.

Lemma 2.6. Let X be a path-connected; locally path-connected and ;rst countablespace and Y be an edge-path space. Let h : �1(X; x0)→ �1(Y; y0) be a homomorphism.Then; F : Xwh →Y is continuous and Im(F)⊂Yw.

Proof. Suppose that F is not continuous on Xwh. Then, there exist xn; x∞ ∈Xwh andan open neighborhood U of F(x∞) such that limn→∞ xn = x∞, but F(xn) =∈ MU .

By the local path-connectivity, we can choose paths pn from xn to x∞ so thatIm(pn)’s converge to x∞ and also choose a path p from x∞ to x. Since xn ∈Xwh, wechoose gn :H→X satisfying:1. gn(o) = xn;2. the restriction of gn to Im corresponds to a loop l for each n such that h ◦’pnp ◦

gn∗([l]) �= e;3. for h ◦’pnp ◦ gn∗ (=h ◦’p ◦’pn ◦ gn∗) there is a continuous map fn :H→Y such

that fn∗ satis2es the required properties in Lemma 2.5.Without any loss of generality, we may assume that Im(gn) converge to x∞. Sincefn(o) = F(xn) =∈ MU , there exists a sequence (mn : n¡!) such that fn(Ik)∩ MU = ∅ fork¿mn. De2ne g∞ :H→X so that g∞(o) = x∞ and g∞∗(0n) = ’pn ◦ gn∗([0mn ]). ApplyLemma 2.5 to h ◦’p ◦ g∞∗. Then we have a continuous map f∞ :H→Y such that(f∞)∗ is conjugate to h ◦’p ◦ g∞∗.

Then Im(h ◦’p ◦ g∞∗) is not 2nitely generated and hence for a suOciently large nthe image of the essential part of h ◦’p ◦ g∞∗(0n) is contained in U , which contradictsthat the image of the essential part of h ◦’p ◦ g∞∗(0n) = h ◦’p ◦’pn ◦ gn∗(0mn) does notintersect with MU . Hence F is continuous. Since F(x) is a limit of F(xn)’s, F(x)∈Yw

and the second proposition holds.

Let X be a path-connected, locally path-connected and 2rst countable space andY be an edge-path space. For a homomorphism h : �1(X; x0)→ �1(Y; y0), we de2ne acontinuous map h : Xwh →Yw according to Lemma 2.4. The following theorem is avariant of [3, Theorem 1:3].

Theorem 2.7. Let X and Y be locally path-connected; path-connected; ;rst countableedge-path spaces. If the fundamental group of X is isomorphic to a retract of thefundamental group of Y; that is; i : �1(X; x0)→ �1(Y; y0) is injective and r : �1(Y; y0)→�1(X; x0) and r ◦ i is the identity; then r ◦ i is the identity on a space Xw and hence


Xw is homeomorphic to a retract i(Xw) of Yw. In particular if the fundamental groupof X and Y are isomorphic; then Xw and Yw are homeomorphic.

Proof. To show r ◦ i(x) = x for x∈Xw by contradiction, suppose the negation. Thereexists an open neighborhood U of r ◦ i(x) such that x =∈ MU . Then, there exists an openneighborhood V of x such that r ◦ i(V )⊂U and V ∩U = ∅. Since x∈Xw, there existsa reduced loop f with base point x0 such that f is essential in V and satis2es theproperty in Lemma 2.4. Then [r ◦ i ◦f] is conjugate to [f] and essential in U , whichcontradicts Lemma 2.4.

Proof of Theorem 1.3. Since the spaces XS and XT are locally path-connected, path-connected, 2rst countable edge-path spaces, we apply Theorem 2.7 to these spaces.Then Xw

S and XwT are S ′ and T ′ respectively. Since a retract of a space is closed and

T ′ is closed in T , Theorem 1.3 is now clear.

3. Presentation of �1(XT ) and HT1 (XT ) and preliminary facts

According to the presentation of �1(XT ) in Section 2, we use letters for all At; Bt ;Ct’s for t ∈T s, but as in the case of edge-path groups for graphs [14, p. 141] wehave a more tight presentation, which we adopt here. Since any continuous image ofthe unit interval in T ∪⋃{{te}∪Bt : t ∈T}∪⋃{Ct : t ∈T s} is contractible, by a similarreasoning to [7, Lemma 2:6], a homomorphism obtained by deleting letters for Bt’s andCt’s is injective on {Wf : f is a loop with base point x}. Therefore, we regard Wf is aword for ××�

t∈TZt , for which t corresponds to a path from t to te through At . Here, weexplain what element of �1(XT ; r) corresponds to t explicitly. Let f be a loop consistingof paths from r to t through [r; t], from t to te through At , from te to t through Bt , andfrom t to r through [r; t]. Now, the element [f]∈ �1(XT ; r) corresponds to t under ourpresentation, i.e. the word Wf is t. In the sequel, we identify �1(XT ; r) with a subgroupof ××�

t∈TZt under this correspondence. Similarly we identify �1(H; o) with ××n¡!Zn.Let q : XT →HT be a map which maps T ∪⋃{{te}∪Bt : t ∈T}∪⋃{Ct : t ∈T s} as onepoint o. Then this presentation Wf can be seen as a presentation of q ◦f.

Since we only use letters for generators and their inverses for in2nite cyclic groups,we adopt the notion “g-reduced words” (abbreviation of “generator-wise reduced”)from [5, p. 300], which is an in2nitary version of reduced words for free groups.The diSerence between the reduced-ness and the g-reduced-ness is small and confusionbetween them is not serious in the present paper, but to sustain exactness we use thisterminology.

Next we state a presentation of HT1 (XT ). To see that qT

∗ : HT1 (XT )→HT

1 (HT ) isinjective, let f∈C(91; XT )∩Z1(XT ) with qT

∗ ([f]T ) = 0. To show that [f]T = 0, weneed a notion. For L⊂T , let YL = XT\

⋃{At ∪{te}∪Bt : t =∈L} and rL : XT →YL be aretraction such that rL(At ∪{te}∪Bt) = {t} for t =∈L.

Since the topology of Z1(XT ) is induced from the compact open topology on C(91;XT ), we take compact Ki⊂91 and open Ui⊂XT such that f(Ki)⊂Ui for 16i6m.


Since f(Ki) is compact, there exist at most 2nitely many t ∈T such that t =∈Ui andte ∈f(Ki). Then, L =

⋃mi=1{t ∈T : t =∈Ui & te ∈f(Ki)} is 2nite and rL ◦f(Ki)⊂Ui for

every 16i6m, that is, rL ◦f belongs to the basic neighborhood determined by Ki’sand Ui’s. Since qT

∗ ([f]T ) = 0 and L is 2nite, rL ◦f∈B1(XT ) and we have shownf∈B1(XT ), i.e. [f]T = 0. Since HT

1 (HI ) � ��I Z by [8, Theorem 4:6], HT

1 (XT ) isnaturally a subgroup of ��

t∈TZt . Since any member of ��¡!1

Z� is realized by a loopin X!1 , we obtain HT

1 (X!1 ) = ��¡!1

Z�.Since Im(f) is compact and connected for a path f, Im(f)⊂T is some particular

subset. Actually, we get

Lemma 3.1. Let f be a loop in XT with base point r. Then Im(f)∩T is a countablesubtree of T which has only ;nitely many branching points. In particular A∩ Im(f)is ;nite for an anti-chain A of T .Conversely; for any countable subtree S of T which has only ;nitely many branch-

ing points; there exists a loop in XT with base point r such that S = Im(f)∩T .

Proof. Suppose that there exist in2nitely many branching points in Im(f), tn (n¡!).Since there exists a subsequence of (tn : n¡!) which converges to a point, we mayassume tn ≺ tn+1. Then, there exists sn ∈ Im(f) such that tn ≺ sn but tn+1 and sn arenot comparable. Then {sn : n¡!} becomes an in2nite anti-chain. Since an anti-chainhas no accumulation point, this is a contradiction. The converse is easy to see.

Lemma 3.2. Let S and T be trees of height equal to or less than !1. If a tree S ishomeomorphic to a subspace of a tree T; then �1(XS) is isomorphic to a subgroup of�1(XT ). If a tree S is homeomorphic to a closed subspace of a tree T; then �1(XS)is isomorphic to a retract of �1(XT ). Similar statements also hold for HT

1 (XS) andHT

1 (XT ).

Proof. Let f : S →T be a homeomorphism onto a subset of T . Without any loss ofgenerality, we may assume f(r) = r. Then f naturally induces an injective homomor-phism f∗ from ××�

s∈SZs to ××�t∈TZt , i.e. f∗(s) = f(s) for s∈ S. (We remark that S does

not generate the group ××�s∈SZs, but the restriction to S determines the homomorphism

[5, Lemma 2:5].) The injectivity of f∗ follows from that of f. So, it suOces to showthat the image of the restriction of f∗ to �1(XS) is contained in �1(XT ). Let g be aproper loop in XS with the base point r. Let

⋃{(an; bn) : n¡=}= g−1(⋃{int(As)∪{se}∪

int(Bs) : s∈ S}), where =6! and (am; bm)∩ (an; bn) = ∅ for m �= n. Then, g � [an; bn] is aloop in {s}∪As ∪{se}∪Bs. Let [c; d] be a connected component of [0; 1]\⋃{(an; bn) :n¡=}, where c = d may holds. Then, g(c) and g(d) belongs to T and g(u) is a limitpoint if u is an accumulation point of {(an; bn) : n¡=}. We de2ne a loop g′ in XT asfollows:• the restriction g′ � [an; bn] is a loop in {f(s)}∪Af(s) ∪{f(s)e}∪Bf(s) which is a

copy of a loop g � [an; bn] in {s}∪As ∪{se}∪Bs;


• the restriction g′ � [c; d] for a connected component of [0; 1]\⋃{(an; bn) : n¡=} is areduced path from f(g(c)) to f(g(d)) in the stem of XT .

The continuity of g′ follows from the use of reduced paths and the following fact:

When g(u) is a limit point of S, there exists ”¿0 such that

T ∩f(S ∩ Im(g � (u − ”; u + ”)))⊂{t ∈T : t 4f(g(u))}.

It is easily seen that f∗(Wg) = Wg′ and hence f∗(�1(XS))6�1(XT ).When the image of f is closed in T , it is necessary to show the existence of a

retraction from �1(XT ) to f∗(�1(XS)). Let h :××�t∈TZt →××�

t∈Im(f)Zt be the restriction. ItsuOces to show that h(�1(XT ))6f∗(�1(XS)). Let r : XT →XT\

⋃ {At ∪{te}∪Bt : t∈T s\Im(f)} be a retraction so that r(At ∪{te}∪Bt) = {t} for t ∈T s\Im(f). Let g : [0; 1]→XT be a proper loop with base point r. For r ◦ g, we de2ne g′ : [0; 1]→XS in a similarprocess to the above using f−1 instead of f. Let [c; d] be a connected component of[0; 1]\⋃ {(an; bn) : n¡=} in this process. Then g(c) and g(d) belong to the closure of{g(an) : n¡!}, which is contained in Im(f). We can de2ne the desired g′ and showthe continuity of g′ as before.

We omit an analogous proof for HT1 (XS) and HT

1 (XT ).

4. Proofs of Theorem 1.1 and Corollary 1.2

The 2rst lemma is easy to check and we omit the proof.

Lemma 4.1. Let T be a tree of height equal to or less than !1. A space XT is locallypath-connected; path-connected and ;rst countable.

Lemma 4.2. Let T be a tree. If there exists an uncountable subset U of T whichcontains no in;nite anti-chain; then there exists an uncountable chain in T .

Proof. Since the set {t ∈U : {s∈U : s≺ t}= ∅} is 2nite, there exists u0 ∈U such that{t ∈U : u0 ≺ t} is uncountable. We de2ne un ∈U and vn ∈U for n¿1 as follows. If{t ∈U : un−1 ≺ t} is a chain, then we obtain an uncountable chain. Otherwise, there existextensions un ∈U and vn ∈U such that un and vn are incomparable and {t ∈U : un ≺ t}is uncountable. Suppose that this procedure does not stop; then we have an in2niteanti-chain {vn : n¿1} in U , which is a contradiction.

For an element x of ××n∈!Zn, there exist g-reduced words V and W such thatW−1VW and VV are g-reduced and x = W−1VW [5, Lemma 4:4]. A letter essentiallyappears for x, if the letter appears in V . (We say t appears in a word, even when −tappears in the word.) The next lemma is contained in a lemma in [5] and we omit theproof.

Lemma 4.3 (Cannon and Conner [5; Lemma 2:9]). Let h :××n¡! Zn →××�i∈IFi be a

homomorphism; where each Fi is a free group; and Bi be the set of all n’s such


that some generator a or −a of Fi essentially appears for h(0n). Then; Bi is ;nitefor each i.

Since the next lemma is essentially contained in the proof of [5, Theorem 4:1], wejust state the outline of proof using a notion “standard homomorphism” from [5].

Lemma 4.4. Let h :××n¡! Zn →××�i∈IZi be a homomorphism such that in ∈ I essentially

appears in h(0n). Then; there exists x∈××n¡! Zn such that all in’s appear in h(x).

Proof (Outline). By [5, Theorem 2:3] there exist a standard homomorphism Mh andu∈××�

i∈IZi such that h(x) = u−1 Mh(x)u for x∈××n¡! Zn. (A standard homomorphismpreserves an in2nite operation according to a word of in2nite length as de2ned inSection 2.) Since each i∈ I appears in at most 2nitely many Mh(0n)’s, we inductivelychoose kn¡kn+1 and mn so that ikn does not appear in Mh(0j) for j¿kn+1 and mn issuOciently greater than the number of appearance of ikn in Mh(0m0

k0· · · 0mn−1

kn−1) and u. Then,

x = u−10m0k0

· · · 0mnkn

· · · u is the desired element.

Lemma 4.5. Suppose that tn ∈T (n¡!) with tn ≺ tn+1 have an upper bound in T .Then there exists an injective homomorphism ’ :××n¡!Zn → �1(XT ) such that ’(0n)= tn for each n¡!. An analogous statement presented by the same formula holdsfor HT

1 (XT ).

Proof. A subspace Y = [r; t∞]∪ ⋃ {{(tn)e}∪Atn ∪Btn : n¡!} is a retract of XT . Letf :H→Y be a continuous map such that f(o) = t∞ and the restriction of f to Inis a path consisting of paths through [tn; t∞], Atn ∪{(tn)e}∪Btn and [tn; t∞]. Thenf∗ : �1(H; o)→ �1(Y; t∞) is the required injective homomorphism under our presen-tation of H. (Actually, f gives a homotopy equivalence between H and Y .) This f∗also induces the required injective homomorphism for HT

1 (XT ).

Next we consider H1(XT ) as a subgroup of ZT . By Lemma 3.1, a∈ZT belongs toHT

1 (XT ) if and only if supp(a) is countable and any in2nite subset of supp(a) has anaccumulation point. Therefore {supp(a) : a∈HT

1 (XT )} forms an ideal on the set T .The next lemma holds for an arbitrary ideal on a set T , which we apply to a tree

T . For an ideal I on a set T , let Z〈I〉 be a group {x∈ZT : supp(x)∈ I} and let I∗ bean ideal {x⊂T : any in2nite subset of x does not belong to I}.

Lemma 4.6. Let T be a set of cardinality less than the least measurable cardinal andI be an ideal on T which contains all ;nite subsets of T . Then Hom(Z〈I〉;Z) � Z〈I∗〉

holds.

Proof. For a∈Z〈I∗〉 and x∈Z〈I〉, let ha(x) =∑{a(t)x(t) : t ∈T}. Since supp(a)∩

supp(x) is 2nite, ha is a homomorphism. To see Hom(Z〈I〉;Z) = {ha : a∈Z〈I∗〉}, leth :Z〈I〉 →Z) be a homomorphism. De2ne a∈ZT by: a(t) = h(et) where et(t) = 1and et(s) = 0 for s �= t. The cardinality of supp(x) is less than the least measurable


cardinal for x∈Z〈I〉. Since {y : supp(y)⊆ supp(x)}∩Z〈I〉 is isomorphic to Zsupp(x) andh(et)=ha(et) for t ∈T , h(x) − ha(x) = 0 by the LoVs theorem [9, Theorem 94:4].

Proof of Theorem 1.1(1). If T has an uncountable chain, then a maximal chain ishomeomorphic to the space !1. Hence the one direction of the proof is obvious byLemma 3.2. Since the proof for HT

1 (XT ) is easier and it gives us the line of proof,we prove it 2rst for the converse direction. Let h : HT

1 (X!1 )→HT1 (XT ) be an injective

homomorphism. Let U be the set of all elements t ∈T for which pt(h(�)) �= 0 forsome �, where pt : HT

1 (XT )→Zt is the projection. We recall the Specker theorem [5],that is, Hom(Z!;Z) is a free abelian group generated by the projections. Since anycountable subset is bounded in !1, the set {� : pt(h(�)) �= 0} is 2nite for each t ∈T byLemma 4.5. Hence, U is uncountable. Suppose that U contains an in2nite antichain L.Let rL : XT →XT\

⋃ {At ∪{te}∪Bt : t =∈L} be the retraction described in Section 3. ByLemma 3.1, rT

L∗(HT1 (XT )) is a free abelian group

⊕t∈L Zt . Again, by an extended

form of the Specker theorem and Lemma 4.5, we conclude that Im(rL∗ ◦ h) is 2nitelygenerated, which is a contradiction. Hence, U does not contain an in2nite anti-chain.By Lemma 4.2, there exists an uncountable chain in T .

The basic idea for �1(XT ) is same, but some diOculty arises from conjugacy. Leth : �1(X!1 )→ �1(XT ) be an injective homomorphism. Let U be the set of all elementst ∈T which essentially appear in some h(�) (�¡!1). Since any countable subset isbounded in !1, t ∈T essentially appears in at most 2nitely many h(�)’s by Lemmas4.3 and 4.5. Hence, U is uncountable. Suppose that U contains an in2nite antichainL. Then, by the same fact that each t ∈T essentially appears in at most 2nitely manyh(�)’s, we have an ascending sequence �n¡�n+1 such that some t ∈L essentially ap-pears in each h(�n). Again, by Lemmas 4.4 and 4.5 there exist a∈ �1(X!1 ) such thatin2nitely many letters from L appear in h(a), which contradicts Lemma 3.2.

What is remaining is the proof for the equivalence of the last statement. As weremarked, HT

1 (XT )�Z〈I〉 holds, where a∈ I if and only if a is countable and anyin2nite subset of a has an accumulation point. Now I∗ consists of all a such thatany in2nite subset of a has no accumulation point. If T has no uncountable branch,any uncountable subset of T contains an in2nite anti-chain by Lemma 4.2 and hence(I∗)∗ = I . On the other hand, if T = !1, I is the ideal of all the countable subsets of!1 and I∗ is the ideal of all the 2nite subsets of !1. Consequently (I∗)∗ is the trivialideal of all the subsets of !1 and (I∗)∗ �= I . Therefore in case T has an uncountableanti-chain, X!1 is homeomorphic to a retract of XT and hence (I∗)∗ �= I .

Lemma 4.7. Let X be a subset of a tree T which does not contain an in;nite chainwith its upper bound in X . Then; X is a countable union of anti-chains.

Proof. Let X0 = X , Yn = {t ∈Xn : {s∈Xn : s≺ t}= ∅} and Xn+1 = Xn\Yn for n¡!. Then,each Yn is an anti-chain. Suppose that there exists t ∈X which does not belong toany Yn. Then there are yn ∈Yn such that yn ≺ t for each n. Hence yn ≺yn+1 holds


and {yn : n¡!} becomes an in2nite chain with its upper bound in X , which is acontradiction.

Proof of Theorem 1.1(2). Let T be a union of pairwise disjoint anti-chains An (n¡!).Let Fn be the free group generated by An. By the continuity of f, Wf contains only2nitely many letters from each An, we can de2ne an embedding ’ : �1(XT )→××n∈!Fn

by restricting Wf to An’s. (Since ××n∈!Fn naturally becomes a subgroup of ××�t∈TZt ,

the above ’ can be regarded as the identity.)To show the converse. Let h : �1(XT )→××n¡!Fn be an embedding. For each t, choose

nt so that a generator a or −a in Fnt essentially appears for h(t). Let Xn = {t ∈T :nt =n} for n¡!. Then T =

⋃ {Xn : n¡!}. According to Lemma 4.7, it suOces toshow that each Xn does not contain an in2nite chain with an upper bound. To showthis by contradiction, suppose the negation. Then there are tm ∈Xn and t ∈Xn suchthat tm ≺ tm+1 ≺ t for m¡!. Then, by Lemma 4.5 there exists an injective homomor-phism ’ :××m¡!Zm→�1(XT ) such that ’(0m) = tm for each m¡!. Then Bn for h ◦’in Lemma 4.3 becomes in2nite, which is a contradiction.

Lemma 4.8. Let T be a tree. Then; T has no uncountable anti-chain; if and only ifevery uncountable subset of T has an accumulation point.

Proof. Since an anti-chain has no accumulation point, one direction is obvious. Let Tbe a tree without an uncountable anti-chain and X an uncountable subset of T . Weconstruct Yn’s just in the same way as in the proof of Lemma 4.7. Then, each Yn ⊂Xis countable. Take x∈X \⋃ {Yn : n¡!}, then yn ∈Yn with yn ≺ x converge to somet ∈T with t 4 x.

Lemma 4.9. Let h : �1(X�)→××�j∈J Zj be a homomorphism; where � is a countable

ordinal. If h(�) = e for all �¡�; where � is in �1(X�) under our presentation; then his trivial.

Proof. For a loop f with base point r, de2ne o(f) = sup{o(t) : t ∈T; t ∈ Im(f)}. Weprove h([f]) = e by induction on o(f). We remark o(f) = max{o(t) : t ∈T; t ∈ Im(f)}.When o(f) is 2nite, h([f]) = e clearly by the assumption. In case o(f) = � + 1, thereexist loops f1; : : : ; fn with base point r such that [f] = [f1] · · · [fn], o(f2i+1)6�, and[f2i] =±(� + 1) for 062i; 2i + 16n. Then h([fi]) = e holds by the assumption andthe induction hypothesis. The crucial case is when o(f) is limit. Then choose a co2nalascending sequence of successors �n¡�n+1 to � = o(f), where �0 = 0. Let p be a pathfrom r to � through a stem.

We work for a loop p−fp with base point �. Then we have a loop g satisfying thefollowing: There exists a pairwise disjoint family of open intervals (ai; bi) (i∈ I) suchthat• [g] = [p−fp]∈ �1(X�; �);• g(ai) = g(bi) belongs to {�}∪ {�n : n¡!};


• each g([ai; bi]) is contained in one of {�; �e}∪A� ∪B� and [�n; �n+1−1]∪⋃{{Be}∪AB ∪BB : �n6B6�n+1 − 1} for n¡!;

• {i∈ I : g([ai; bi])∩D �= ∅} is 2nite, where D is each one of A� ∪B� and⋃ {AB ∪BB :

�n6B6�n+1 − 1} for n¡!;• g([0; 1]\⋃{(ai; bi) : i∈ I}) is contained in a stem.

To see this, consider the inverse image of f of {�e}∪A� ∪B� ∪⋃{C�n+1 : 16n¡!}. It

becomes a pairwise disjoint union of open intervals. Since each of {�e}∪A� ∪B� andC�n+1 for 16n¡! has a good local property at its boundary, we can homotopicallydeform f so that the inverse image of f of each of {�e}∪A� ∪B� and C�n+1 for16n¡! becomes a union of at most 2nitely many pairwise disjoint open intervals.Then we obtain the desired g. Working in each [ai; bi] and taking other ai; bi, we mayassume g(ai) = g(bi) = � and then also g([0; 1]\⋃ {(ai; bi) : i∈ I}) = {�}. In the aboveproof we implicitly construct homotopies. The continuity of these is assured by thefact that the subsets appearing in the third and fourth items converge to �.

Without any loss of generality, we may assume that I is in2nite. Therefore we assumeI = !. We de2ne G :H→XT by: G(o) = � and the restriction of G to In is a copy ofthe restriction of g to (an; bn). Then G becomes continuous and h ◦’p− ◦G∗(0n) = eby the assumption and the induction hypothesis. Now, Lemma 2.5 implies h ◦’p− ◦G∗is trivial. Let f′ : [0; 1]→H be a loop such that f′(an + (bn − an)s) = s for s∈ In andf′([0; 1]\⋃{(an; bn) : n¡!}) = {o}. Then G∗([f′]) = [G ◦f′] = [g] holds and henceh([f]) = h ◦’p−([g]) = h ◦’p− ◦G∗([f′]) = e.

Lemma 4.10. Let h : �1(XT )→××�j∈JZj be a homomorphism. If h(t) = e for all t ∈T;

then h is trivial.

Proof. Let f be a loop with base point r. As in the proof of Lemma 4.9, we proveh([f]) = e by induction on o(f). According to Lemma 3.1, it suOces to show the casewhen o(f) is a limit �. Again by Lemma 3.1, {t ∈T : o(t) = �}∩ Im(f) is 2nite; list itas {t1; : : : ; tm}. Let Yi = [r; ti]∪

⋃{{te}∪At ∪Bt : t� ti}∪⋃{Ct : t is a immediate succe

ssor of ti; i = 1; : : : ; m} for 16i6m. Then there exist loops fj and gk for 16j6n;06k6n with base point r such that• [g0f1g1 · · ·fngn] = f;• fj is a loop in some Yi for each j;• o(gk)¡� for each k.

Since each Yi is homeomorphic to X�+1 and Yi is a retract of XT and consequently�1(Yi) is also a retract of �1(XT ), we can apply Lemma 4.9 and we obtain h([fj]) = efor 16j6n. On the other hand h([gk ]) = e by the induction hypothesis for 06k6nand we conclude h([f]) = e.

Proof of Theorem 1.1(3). Suppose that T has an anti-chain A of the cardinality �.Then, there exists an embedding from B(�) into XT whose image is {r}∪A. Since{r}∪A is closed in XT , the free group of �-many generators is isomorphic to a retractof �1(XT ) by Lemma 3.2.


We show the converse by contradiction. Suppose that h : �1(XT )→F is a retractionto a free group F of the cardinality �. First we claim that {t : h(t) �= e}= L has noaccumulation point in T . Let YL = XT\

⋃{At ∪{te}∪Bt : t =∈L} and r : XT →YL be aretraction. r : XT →YL be a retraction. Then h = h ◦ r∗ holds because of our presenta-tion of loops in XT . Suppose that there exists an accumulation point of L in T . Wechoose tn ∈L so that tn’s converge to a point in T , that is, in XT . Then as in theproof of Lemma 4.9 we have a continuous map f :H→XT so that f∗(0n) = tn forn¡!. Therefore h ◦f∗(0n)�= e for all n¡!, which contradicts the Higman theoremabout the n-slenderness of free groups [4, Corollary 3:7]. Since L has no accumulationpoint, r∗(�1(XT )) = �1(YL) is a free group. If L is of the cardinality less than �, thenr∗(�1(XT )) and also F must be of cardinality less than �. Therefore L is of cardinality�. By Lemma 4.7 L is a union of countably many anti-chains, which implies thatL contains an anti-chain of the cardinality � or � is a singular cardinal of co2nality! which is a supremum of the cardinalities of antichains. Therefore T contains ananti-chain of the cardinality � in the both cases.

Proof of Corollary 1.2. It suOces to show the following. Let T be a special Aronzajntree. Then, T is a union of pairwise disjoint anti-chains An (n¡!) where the cardinalityof each An is equal to or less than !1. Take a partition Pn (n¡!) of ! so that eachPn is in2nite. Then, choose in2nite subsets It of Pn indexed by t ∈An so that It’s forman almost disjoint family. We assume that It is ordered as a subset of !; then wecan regard It as a word for ××n∈!Zn. Let ’(t) = It for t ∈T s. By the continuity off, Wf contains only 2nitely many letters from each An, so we can de2ne ’(Wf) byconcatenating It’s. Since It’s are in2nite, for each Wf we can choose at ∈ It so thatat =∈ Is for distinct s; t which appear or whose inverses appear in Wf. Therefore, theinjectivity of ’ holds.

Remark 4.11. Lemmas 4.9 and 4.10 may look trivial at 2rst glance. We remark thatthey hold because groups of the form �1(XT ) are particular subgroups of ××�

t∈TZt . Forfree subgroups of ××�

t∈TZt which contain ∗t ∈ TZt , we cannot expect Lemmas 4.9 and4.10. For such free subgroups, we refer the reader [2, 15, 6].

5. Ordinals

To investigate �1(X�), we recall certain notions for ordinals. An ordinal � is indecom-posable, when �+B=� implies B=� for any ordinals �¡� and B. Let i(�) = max{� 6 � :� is indecomposable}, n(�) = max{n : i(�)n¡�} and t(�) = min{B : �+ B = � for some�¡�}. Then, i(�) and t(�) are indecomposable and i(�) ¿ t(�) holds. The 2rst lemmais straightforward and we omit the proof.

Lemma 5.1. Let � be an indecomposable; countable; in;nite ordinal and � be lessthan �. Then; �1(X�) � �1(X�) ∗ �1(X�+1) and �1(X�+1) � �1(X�+1) ∗ �1(X�+1).


Lemma 5.2. Let � be an ordinal. Then the minimal ordinal which is homeomorphicto � as a topological space is i(�)n(�) + t(�). Moreover �1(X�) is isomorphic to�1(Xi(�)n(�)+t(�)).

Proof. The 2rst statement proved by induction and we omit the proof. Let � = i(�)n(�)+ � + t(�)¿0. Then

�1(X�) �

�1(Xi(�)n(�)+1) ∗ �1(X�+1) ∗ �1(Xt(�)) when t(�) ¿ !;

�1(Xi(�)n(�)+1) ∗ �1(X�+1) when t(�) = 1 and � ¿ !;

�1(Xi(�)n(�)+1) ∗ �1(X�) when t(�) = 1 and �¡!:

Now, the second statement follows from Lemma 5.1.

Lemma 5.3. Let � and � be countable ordinals. Then; � is homeomorphic to a closedsubspace of � if and only if• (i(�); n(�)) 6 (i(�); n(�)) lexicographically; and• t(�) 6 t(�).

Proof. The implication from the 2rst proposition to the second is obvious fromLemma 5.2. When � is 0 or a successor, t(�) = 0 or 1 and � is a closed subspaceof any ordinal greater than �. Hence, the reverse implication holds. We may assume� = i(�)n(�) + t(�) and � = i(�)n(�) + t(�) by Lemma 5.2. When � is limit, � islimit by t(�) 6 t(�). Since i(�)n(�) + 1 is a closed subspace of i(�)n(�) + 1 unlessn(�) = 0, it suOces to show t(�) is homeomorphic to a closed subspace of t(�). Let�n (n¡!) be a strictly increasing co2nal sequence to t(�) with �0 = 0. Since t(�) isindecomposable, there exists a strictly increasing co2nal sequence �n (n¡!) to t(�)such that �n +�n+1¡�n+1. De2ne f : t(�)→ t(�) by: f(�0) = �0 and f(�n +B) = �n +Bfor �n¡�n + B6 �n+1. Then f is a closed embedding and the proof is complete.

Lemma 5.4. Let h : �1(X�+1)→ �1(X!) be a homomorphism. Then Im(h) is ;nitelygenerated.

Proof. Suppose that there exists in2nitely many � 6 � such that h(�) �= e. Then thereexists an strictly increasing sequence �n¡�n+1 whose limit is equal to or less than �.

Let h : �1(XT )→××�j∈JZj be a homomorphism. If h(t) = e for all t ∈T , then h is triv-

ial. By Lemma 4.5 there exists an injection g :××! Z→ �1(X�+1) such that g(0n) = �n.Then Im(hg) is not 2nitely generated, which contradicts the Higman theorem [10, 4].Therefore, there exist only 2nitely many �’s with h(�) �= e. Now the conclusion followsfrom Lemma 4.9.

Proof of Corollary 1.4. If � is homeomorphic to a closed subspace of �, then �1(X�)is isomorphic to a retract of �1(X�) by Lemma 3.2. Conversely, suppose that �1(X�)is isomorphic to a retract of �1(X�). By Lemma 5.2 we may assume � = i(�)n(�) +t(�)¿0. Then Xw

� is homeomorphic to a closed subspace of Xw� by Theorem 2.7.


Xw� is homeomorphic to the subspace lim(�) of � consisting of limit ordinals. We

assume �¿0.Case 1: t(�) = 1. Since lim(�) is order isomorphic to lim(i(�)+1)n(�), � = i(�)n(�)+

1 6 � and hence � is homeomorphic to a closed subspace of �.Case 2: t(�) ¿ !!. Since lim(�) is homeomorphic to �; � is homeomorphic to a

closed subspace of �.Case 3: ! 6 t(�)¡!!. Since lim(!m+1n) is homeomorphic to !mn, t(�) 6 t(�)

follows from Lemmas 5.4 and 5.3. Since lim(!m+1n+ 1) is homeomorphic to !mn+ 1for m; n ¿ 1; (i(�); n(�)) 6 (i(�); n(�)) lexicographically, which implies the conclu-sion by Lemma 5.3.

Question 5.5. We do not know whether Corollary 1.4 holds for trees or not.

Remark 5.6. Recently Cannon and Conner [1] have introduced a new notion “Big fun-damental group”, which uses arbitrary compact connected, linearly ordered sets insteadof the closed unit interval. Using this notion we can treat with trees of arbitrary heights.However, S. Shelah [13] showed that the Higman theorem does not hold for uncount-able complete free products, which is a negative answer to [4, Question 3:8]. Thereforenot all of Theorem 1.1 can be extended to the case of big fundamental groups. Butthere are many things which are still unknown.

References

[1] J.W. Cannon, G.R. Conner, The big fundamental group, big Hawaiian earrings, and the big free groups,Topology Appl. 106 (2000) 273–291.

[2] J.W. Cannon, G.R. Conner, The combinatorial structure of the Hawaiian earring group, Topology Appl.106 (2000) 225–271.

[3] K. Eda, The fundamental groups of one-dimensional spaces and spatial homomorphisms, preprint.[4] K. Eda, Free �-products and noncommutatively slender groups, J. Algebra 148 (1992) 243–263.[5] K. Eda, Free �-products and fundamental groups of subspaces of the plane, Top. Appl. 84 (1998)

283–306.[6] K. Eda, Free subgroups of the fundamental group of the Hawaiian earring, J. Algebra 219 (1999)

598–605.[7] K. Eda, The fundamental groups of certain one-dimensional spaces, Tokyo J. Math. 23 (2000) 187–202.[8] K. Eda, K. Sakai, A factor of singular homology, Tsukuba J. Math. 15 (1991) 351–387.[9] L. Fuchs, In2nite Abelian Groups, Vol. 1,2, Academic Press, New York, 1970, 1973.

[10] G. Higman, Unrestricted free products, and variety of topological groups, J. London Math. Soc. 27(1952) 73–81.

[11] K. Kunen, Set Theory, North-Holland, Amsterdam, 1980.[12] J.J. Rotman, An Introduction to the Theory of Groups, Springer, Berlin, 1994.[13] S. Shelah, L. Struengmann, The failure of the uncountable non-commutative Specker phenomenon,

preprint.[14] E.H. Spanier, Algebraic Topology, McGraw-Hill, New York, 1966.[15] A. Zastrow, The non-abelian specker-group is free, J. Algebra 229 (2000) 55–85.

trees, fundamental groups and homology groups

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