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Complexes of Curves and Injective Homomorphisms
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6/01 cJClRC/DateDuepes-p. 15

SUPERINJECTIVE SIMPLICIAL MAPS OF
COMPLEXES OF CURVES AND INJECTIVE
HOMOMORPHISMS OF MAPPING CLASS

GROUPS
By

Elmas Irmak

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

2002

ABSTRACT

SUPERINJECTIVE SIMPLICIAL MAPS OF
COMPLEXES OF CURVES AND INJECTIVE
HOMOMORPHISMS OF MAPPING CLASS GROUPS

By

Elmas Irmak

In this thesis, we prove two main results. First, we study the complex of curves,
C (S), on a closed, connected, orientable surface, S , of genus at least 3. We prove
that a simplicial map, A : C (S ) —+ C (S), preserves nondisjointness (i.e. if a and B
are two vertices in C(S) and z'(a,fl) gé 0, then i(A(a),/\(B)) 74 0) if and only if it
is induced by a homeomorphism of S. We call these simplicial maps superinjectz’ve

simplicial maps of the complex of curves of S.

Second, we study injective homomorphisms of finite index subgroups of the ex-
tended mapping class group, M ad}, for a closed, connected, orientable surface of
genus at least 3. Using our first result, we prove that if K is a finite index subgroup
of M 0d; and f : K —> M 0d§ is an injective homomorphism, then f is induced by
a homeomorphism of S and f has a unique extension to an automorphism of the

extended mapping class group of S .

To my family

iii

ACKNOWLEDGMENTS

I would like to thank Dr. John D. McCarthy, my advisor, for his guidance, help

and support throughout this work and reviewing the details of my thesis.

I am grateful to Dr. Nikolai Ivanov for his support and his suggestions about the

work in this thesis.

I would also like to thank the members of my advisory committee Dr. Ronald

Fintushel, Dr. John Hall and Dr. Efstratia Kalfagianni for reviewing my results.

Finally, I thank Bulent Buyukbozkirli for having very helpful discussions with me.

iv

TABLE OF CONTENTS

LIST OF FIGURES vii

0 Introduction 1

1 Complexes of Curves and Extended Mapping Class Groups 4
1.1 Introduction .................................. 4
1.2 The Complex C (R) .............................. 4
1.3 Extended Mapping Class Groups, M 0d}: .................. 6
1.4 Superinjective Simplicial Maps of Complexes of Curves .......... 8
2 Induced Map On Complex Of Arcs 22
2.1 Introduction .................................. 22
2.2 The Complex B(R) .............................. 23
2.3 Pairs of Disjoint Arcs and Neighborhoods ................. 24
2.4 Properties of Superinjective Simplicial Maps ................ 29
2.5 Arcs and Encoding Simplices ........................ 35
2.6 Induced Map ................................. 41

3 Injective Homomorphisms of Subgroups of Mapping Class Groups 50
3.1 Introduction .................................. 50

3.2 Injective Homomorphisms and Centralizers ................. 51

3.3 Injective Homomorphisms of Finite Index Subgroups ........... 54

3.4 Induced Map On Complex Of Curves .................... 55

BIBLIOGRAPHY 58

vi

LIST OF FIGURES

1.1 Circles, a, b, c, d, and the corresponding subcomplex ............ 5
1.2 Four possible cases for P0 .......................... 11
1.3 Pants decompositions with separating circles ................ 12
1.4 Adjacent circles ................................ 13
1.5 Nonseparating circles, x, y, z, bounding a pair of pants .......... 19
1.6 Circles intersecting once ........................... 21
2.1 Unlinked disjoint arcs of type 2 ....................... 25
2.2 The graph of a, b and the boundary components .............. 25
2.3 Disjoint arcs on pairs of pants ........................ 27
2.4 Disjoint arcs and neighborhoods ....................... 28
2.5 “Horizontal” and “vertical” circles ..................... 30
2.6 Sphere with five holes ............................ 31
2.7 Circles on torus with two boundary components .............. 33
2.8 Splitting the arc j along the end of z' .................... 38
2.9 Sphere with four holes . ............................ 42
2.10 Circles associated to type 2 arcs ....................... 43
2.11 Arcs and their encoding circles, I ...................... 46
2.12 Arcs and their encoding circles, II ...................... 47

vii

CHAPTER 0

Introduction

In this thesis, our starting point was to study the injective homomorphisms of finite
index subgroups of the extended mapping class group, M odfg, of a closed, connected,
orientable surface S. Our main results are motivated by the following theorems of

Ivanov and the theorem of Ivanov and McCarthy.

N.V. Ivanov, [Ivl], proved that if the genus of S is at least 2 and S is not a
closed surface of genus 2 and I‘l,I‘2 are finite index subgroups of M 0d}, then, all

isomorphisms I‘1 —> F2 have the form :r —> mtg-1,9 E M odfg.

Ivanov got this result as a corollary of another result in which he proved that
if the genus of S is at least 2, then, all automorphisms of the complex of curves,
C (S), of S are given by elements of M odg. More precisely, if S is not a closed
surface of genus 2, then Aut(C (S )) = M odg. If S is a closed surface of genus 2, then
Aut(C(S)) = Mod§/Center(Mod§).

N.V. Ivanov and J .D. McCarthy proved the following result about injective homo-
morphisms of the mapping class groups (M eds ): Suppose that S and S ’ are compact,
connected, orientable surfaces, the genus of S is at least 2, S’ is not a closed sur-
face of genus 2, and the maxima of ranks of abelian subgroups of M ads and M ads:

differ by at most one. If h : M ads ——> M ads, is an injective homomorphism, then

h is induced by a homeomorphism H : S —> S’, (i.e. h([G]) = [HGH‘I], for every

orientation preserving homeo. G : S ——> S).

The two theorems of Ivanov which are given above were extended to all surfaces of
genus O and 1 with the exception of spheres with S 4 holes and tori with g 2 holes

by M.Korkmaz. These extensions were also obtained by F.Luo independently.
In this thesis, we prove the following two theorems:

Theorem A : Let S be a closed, connected, orientable surface of genus at least 3.
A simplicial map, A : C (S) —> C (S), is superinjective if and only if A is induced by a

homeomorphism of S .

Theorem B : Let S be a closed, connected, orientable surface of genus at least
3. Let K be a finite index subgroup of M odfg and f be an injective homomorphism,
f : K ——> M odfg. Then f is induced by a homeomorphism of the surface S (i.e. for
some 9 E Modfg f(k) = gkg”1 for all k E K ) and f has a unique extension to an

automorphism of M 0d}.

In chapter 1, we define the extended mapping class groups, the complexes of curves,
and give the action of the extended mapping class group on the complex of curves
for a compact, connected, orientable surface possibly with boundary. We introduce
the notion of superinjective simplicial maps of the complex of curves and prove some

properties of these maps.

In chapter 2, we give the definition of the complex of arcs of a closed, connected,
orientable surface S . We prove that a superinjective simplicial map, A, of the complex
of curves, C (S), induces an injective simplicial map of the complex of arcs, B(Sc),
where c is a nonseparating simple closed curve on S. Then we prove our first main
result: We prove that a simplicial map, A : C (S) ——> C(S), is superinjective if and only

if it is induced by a homeomorphism of S.

In chapter 3, we prove our second main result: We prove that if K is a finite index
subgroup of M odf‘g and f : K ——> M odg is an injective homomorphism, then f induces

a superinjective simplicial map of C (S) and f is induced by a homeomorphism of S.

Theroem B is deduced from Theorem A. Theorem A generalizes the closed case of
Ivanov’s first theorem which is given above for surfaces of genus at least 3. Theorem
B generalizes Ivanov’s second theorem in the case of closed surfaces and Ivanov and
McCarthy’s theorem in the case when the surfaces are the same and closed. In our
proof, some advanced homotopy results about C(S) used by Ivanov are replaced by

elementary surface topology arguments.

CHAPTER 1

Complexes of Curves and

Extended Mapping Class Groups

1 . 1 Introduction

In this chapter we give the basic definitions and the notations that we use in the
thesis. We define the complexes of curves and the extended mapping class groups for
compact, connected, orientable surfaces possibly with boundary. We give the action
of the mapping class groups on the complexes of curves. We introduce the notion of
superinjective simplicial maps of complexes of curves for closed surfaces and prove

some properties of these simplicial maps.

1.2 The Complex C (R)

A circle on a surface, R, of genus g with b boundary components is a properly
embedded image of an embedding S1 —> R. A circle on R is said to be nontrivial (or
essential) if it doesn’t bound a disk and it is not homotopic to a boundary component
of R. Let C be a collection of pairwise disjoint circles on R. The surface obtained

from R by cutting along C is denoted by RC. Two circles a and b on R are

called topologically equivalent if there exists a homeomorphism F : R —> R such that

F(a)=b.

Let A denote the set of isotopy classes of nontrivial circles on R. The geometric
intersection number i(a, 5) of (1,5 E A is the minimum number of points of a O b

where aEa and bEB.

The complex of curves, C (R), on R is an abstract simplicial complex, as given
in [Sp], with vertex set A such that a set of n vertices {01,02,...,an} forms an
n — 1 simplex if and only if a1, 0;, ..., can have pairwise disjoint representatives. This
complex was introduced by W.Harvey [H]. It is used to study the algebraic structure

of the mapping class groups.

C (R) is an infinite complex. It has a finite dimension which is equal to the maximal
number of pairwise nonisotopic, pairwise disjoint, and nontrivial circles on R minus
1. Hence, the dimension of C (R) is equal to 3g — 4 + b, except in the cases where
g=1 andb=O,andwhereg=OandOSbS3. Inthecasewheregzland
b = O, C (R) is empty. In the case where g = O and 0 S b S 3, the dimension of

C (R) is zero.

   

Figure 1.1. Circles, a, b, c, d, and the corresponding subcomplex

In Figure 1.1, we see the circles a,b,c,d on a genus three surface, S, and the
subcomplex of C (S) which corresponds to these circles. Since b, c,d are pairwise
disjoint, their isotopy classes form a 2 simplex in C (S) [a] and [d] are connected
by an edge in C (S) since a and d are disjoint. Since a intersects each of b and
c transversely once, i([a], [b]) 74 0 and i([a], [c]) # 0. Therefore, there is no edge

between [a] and [b], and [a] and [c].

1.3 Extended Mapping Class Groups, M 0d]:2

Let R be a compact, connected, orientable surface, possibly having nonempty bound-
ary. The mapping class group, M odR, of R is the group of isotopy classes of orien-
tation preserving homeomorphisms of R. The extended mapping class group, M 0d},
of R is the group of isotopy classes of all homeomorphisms of R. Note that the
isotopy classes of orientation reversing homeomorphisms are also included in M 0d},
and so M odR is a subgroup of M 0d}z of index two. The pure mapping class group
PM odR of R is defined as the group of isotopy classes of all orientation-preserving

homeomorphisms R -—> R preserving all boundary components of R setwise.

The extended mapping class group, M 0d}, can also be defined as the quotient
H omeo(R) by the normal subgroup H omeoo(R) where H omeo(R) is the group of
homeomorphisms of R and H omeoo(R) is the group of homeomorphisms of R which

are isotopic to the identity.

A mapping class, g E M 0d}, is called pseudo-Anosov if A is nonempty and if
g"(a) 79 a, for all a in A and any n 75 0. g is called reducible if there is a nonempty
subset B Q A such that a set of disjoint representatives can be chosen for B and
9(8) = B. In this case, B is called a reduction system for 9. Each element of B

is called a reduction class for g. A reduction class, a, for g, is called an essential

reduction class for g, if for each 5 E A such that i(a,B) 75 O and for each integer
m 54$ 0, gm(5) 75 S. The set, 3g, of all essential reduction classes for g is called
the canonical reduction system for g. The correspondence 9 —> [39 is canonical. In

particular, it satisfies g(B;,) = thg—l for all g, h in M 0d}.

The following theorem gives a classification of the elements of the mapping class

groups for orientable surfaces of negative Euler characteristic.

Theorem (Thurston) [FLP] : If Y is an orientable surface of negative Euler
characteristic, then a mapping class [h] of M ods must be either periodic (i.e., [h"] =

1 for some n 75 O), or reducible, or pseudo-Anosou.

The isotopy class of a Dehn twist about a circle a, is denoted by to, , where [a] : a.
Dehn twists are the key elements in the mapping class groups. If T is a compact
surface of genus at least one, then PM odT is generated by Dehn twists about non-

separating circles.

It is a well known fact that fta f ‘1 = tfij) for all a in A, f E M 0d}, where
e( f) = 1 if f has an orientation preserving representative and e( f) = —1 if f has
an orientation reversing representative. The following are well known results about

Dehn twists.

Lemma 1.1 Let a, B E A and i, j be nonzero integers. Then, t; = t}, 4: a = B and

i = j.

PROOF. If a = fl and i = j, then obviously t; = tie: To see the other implication,
we consider the reduction systems. Since 6 is the canonical reduction system for tie
and a is the canonical reduction system for t; and t; = t2, their canonical reduction

systems must be equal. So, a = ,8. Then we have t; 2 t{,. Then, i = j, since to, is

an infinite order element in M ad’s. I

Lemma 1.2 Let 01,6 be distinct elements in A. Let i, j be two nonzero integers.

Then, tgtf, = tjfit; 4: i(a,fl) = 0.

PROOF. tgtf, = tfgt; 4:) tgtgtg‘ = t’fl 4:) t’

it.

41) i(a,fi) = 0 (“<:” : Clear. “:>”: i(a,[3) # 0 => tg‘m) 7é B for all m E 2" since

(6) : ti? ‘1’ tidfl) = 5 (by Lemma 1.1)

a is an essential reduction class for to, ..) I

Every mapping class in the extended mapping class group induces an automor-
phism of the complex of curves in the following way: Let g E M ad}, a E A. Assume
that G E g and a E a. If H : R —> R is a homeomorphism which is isotopic to G ,
then G (a) and H (a) are isotopic circles. If b is a circle on R which is isotopic to a,
then G(a) and G(b) are isotopic circles on R. Hence, by defining g.(a) to be the
isotopy class of the circle G (a) on R, we get a well defined map g... : A —> A. Let
01,6 E A such that i(a,fi) = 0. Choose a E a and b E 6 such that aflb = Q. If
G E 9, then G(a) and G(b) are disjoint circles. Therefore, i(g(a),g(fi)) = O. This
implies that g. extends to a simplicial map on C (S) It is easy to see that g... is

injective and surjective, and hence, an automorphism of C (S)

1.4 Superinjective Simplicial Maps of Complexes
of Curves

In the rest of the thesis, we assume that S is a closed, connected, orientable surface

of genus g 2 3.

Definition 1.3 A simplicial map A : C(S) —> C(S) is called superinjective if the
following condition holds: if (1,6 are two vertices in C (S) such that i(a, fl) # 0, then

i(A(a),A(fi)) # 0-

Lemma 1.4 A superinjective simplicial map, A : C (S) —> C (S) , is injective.

PROOF. Let a and B be two distinct vertices in C(S). If i(a,B) 75 0, then
i(A(a),A(fi)) # 0, since A preserves nondisjointness. So, A(a) sé A(B). If i(a,fi) =
O, we choose a vertex 7 of C(S) such that i(7,oz) = O and i(7,fl) 3f 0. Then,
i(A(7),A(a)) = 0, i(A(7),A(5)) 75 0. So, A(a) 79 A(B). Hence, A is injective. I

Lemma 1.5 Let and be two distinct vertices of C(S), and let A : C(S) —> C(S) be
a superinjective simplicial map. Then, a and ,8 are connected by an edge in C (S) if

and only if A(a) and A(fi) are connected by an edge in C(S).

PROOF. Let a, 6 be two distinct vertices of C (S) By Lemma 1.4, A is injective.
So, A(a) 75 A(fl). Then we have, a and 6 are connected by an edge 4:) i(a,fl) = 0
{it i(A(a),A(fl)) = 0 (by superinjectivity) (i) A(a) and A(B) are connected by an

edge. I

Let P be a set of pairwise disjoint circles on S. P is called a pair of pants decom-
position of S, if S p is a disjoint union of genus zero surfaces with three boundary
components, pairs of pants. A pair of pants of a pants decomposition is the image
of one of these connected components under the quotient map q : Sp —> S together
with the image of the boundary components of this component. The image of the
boundary of this component is called the boundary of the pair of pants. A pair of
pants is called embedded if the restriction of q to the corresponding component of Sp

is an embedding.

Lemma 1.6 Let A : C(S) —> C (S) be a superinjective simplicial map. Let P be a
pair of pants decomposition of S. Then, A maps the set of isotopy classes of elements

of P to the set of isotopy classes of elements of a pair of pants decomposition, P’, of

S.

PROOF. The set of isotopy classes of elements of P forms a top dimensional
simplex, A, in C (S) Since A is injective, it maps A to a top dimensional simplex
A’ in C(S). Pairwise disjoint representatives of the vertices of A’ give a pair of pants

decomposition P’ of S. I

By Euler characteristic arguments it can be seen that there exist exactly 3g — 3
circles and 29 — 2 pairs of pants in a pair of pants decomposition of S. An ordered set
(a1, ..., a39__3) is called an ordered pair of pants decomposition of S, if {a1, ...,a39_3}
is a pair of pants decomposition of S. Let P be a pair of pants decomposition of S.
Let a and b be two distinct elements in P. Then, a is called adjacent to b w.r.t. P

iff there exists a pair of pants in P which has a and b on its boundary.

Remark: Let P be a pair of pants decomposition of S. Let [P] be the set of
isotopy classes of elements of P. Let 01,6 E [P]. We say that a is adjacent to
B w.r.t. [P] if the representatives of a and d in P are adjacent w.r.t. P. By
Lemma 1.6, A gives a correspondence on the isotopy classes of elements of pair of
pants decompositions of S. A([P]) is the set of isotopy classes of a pair of pants

decomposition which corresponds to P, under this correspondence.

Lemma 1.7 Let A : C (S) —> C (S) be a superinjective simplicial map. Let P be a pair
of pants decomposition of S. Then, A preserves the adjacency relation for two circles,
i.e. if a, b E P are two circles which are adjacent w.r.t. P and [a] = a, [b] 2 B, then

A(a),A(fi) are adjacent w.r.t. A([P]).

PROOF. Let P be a pair of pants decomposition of S. Let a, b be two adjacent
circles in P and [a] = a, [b] = ,8. Let Po be a pair of pants of P, having a and b on
its boundary. By Lemma 1.6, we can choose a pair of pants decomposition, P’ , such

that A([P]) = [P’].

10

Either P0 is embedded or nonembedded. In the case Po is embedded, either a
and b are the boundary components of another pair of pants or not. In the case Po
is nonembedded, either a or b is a separating curve on S. So, there are four possible
cases for P0. For each of these cases, in Figure 1.2, we show how to choose a circle c

which essentially intersects a and b and does not intersect any other circle in P.

 

 

 

 

 

Figure 1.2. Four possible cases for P0

Let 7 = [c]. Assume that A(a) and A(fl) do not have adjacent representatives.
Since i(ry, a) aé 0 7e i(7,fi), we have i(A(7), A(a)) sé 0 aé i(A('y), A(fl)) by superinjec-
tivity. Since i(y, [e]) = 0 Ve E P\ {a, b}, we have i(A(7),A([e])) = O Ve E P\ {a, b}.
But this is not possible because A('y) has to intersect geometrically essentially with
some isotopy class other than A(a) and A(B) in the image pair of pants decomposition

to be able to make essential intersections with A(a) and A(B). This gives a contra-

11

diction to the assumption that A(a) and A(fi) do not have adjacent representatives.

Let P be a pair of pants decomposition of S. A curve x E P is called a 4—curve

in P, if there exist four distinct circles in P, which are adjacent to x w.r.t. P.

A nontrivial circle a on S is called k—separating (or a genus k circle) if the surface
S, is disconnected and one of its components is a genus k surface where 1 S k < 9.
If S, is connected, then a is called nonseparating. Two circles on S are topologically
equivalent if and only if they are either both nonseparating or k-separating for some

k.

Lemma 1.8 Let A : C (S) —) C (S) be a superinjective simplicial map. Then, A sends
the isotopy class of a k-separating circle to the isotopy class of a k-separating circle,

wherelSkSg—l.

PROOF. Let a = [a] where a is a k-separating circle where 1 S k S g — 1. Since
the genus of S is at least 3, a is a separating curve of genus at least 2. So, it is

enough to consider the cases when k 2 2.

   

(i) (ii)

Figure 1.3. Pants decompositions with separating circles

12

 

Figure 1.4. Adjacent circles

Case I: Assume that k = 2. Let 32 be a subsurface of S of genus 2 having a as
its boundary. We can choose a pair of pants decomposition Q = {a1,a2,a3,a4} of
S2 as shown in Figure 1.3, (i). Then, we can complete Q U {a} to a pair of pants
decomposition P of S in any way we like. By Lemma 1.6, we can choose a pair of

pants decomposition, P’, of S such that A([P]) = [P’].

Let a; be the representative of A([a,-]) which is in P’ for i = 1, ..,4 and a’ be the

representative of A([a]) which is in P’.

Since a1 and a3 are 4—curves in P, by Lemma 1.7 and Lemma 1.4, a’l and at, are
4-curves in P’. Notice that a curve in a pair of pants decomposition can be adjacent

to at most four curves w.r.t. the pair of pants decomposition.

a3 is a 4-curve. So, there are two pairs of pants, A and B, of P’ such that each
of them has a], on its boundary. Let x, y, z,t be as shown in Figure 1.4. Since a3 is a
4—curve and x, y, z,t are the only curves which are adjacent to at, w.r.t. P’, x, y, z,t

are four distinct curves.

a3 is adjacent to al w.r.t. P implies that a; is adjacent to a’1 w.r.t. P’. Then,

W.L.O.G. we can assume that x = a’l. So, a’1 is a boundary component of A. Since

13

a’1 is a 4—curve in P’, a’1 is also a boundary component of a pair of pants different
from A. Since a’1 75 z, a’1 7E t and a’1 96 a3, a’1 is not a boundary component of
B. So, there is a new pair of pants, C, which has a’1 on its boundary. Let v,w
be as shown in Figure 1.4. Since a’l is a 4—curve and y, u, w, a3 are the only curves
which are adjacent to a’1 w.r.t. P’, y,v,w,ag are four distinct curves. Since a1
is adjacent to each of a,a2,a3,a4 w.r.t. P, a’l is adjacent to each of a’,a’2,a§,af,
w.r.t. P’. Then, {y,v,w,ag} = {a’,a’2,a§,af,} and, so, {y,v,w} = {a’,a’2,af,}.
Similarly, since a3 is adjacent to each of a, a1, a2, a4 w.r.t. P, as is adjacent to each
of a’,a’1,a’2,af, w.r.t. P’. Then, since each of x,y,z,t is adjacent to a3, we have
{x,y,z,t} = {a’,a’1,a’2,af,}. Since at = a’l, {y,z,t} = {a’,a§,a’4} = {y,v,w}. Then
{2,t} = {v,w}. Hence, A U B U C is a genus 2 subsurface of S having y as its

boundary.

Since {y,v,w} = {a’,a’2,af,}, y is a’ or a’2 or a. Assume that y = a’2. Then, a’2
can’t be adjacent to a; w.r.t. P’ as it can be seen from Figure 1.4. But this gives
a contradiction since a2 is adjacent to a4 w.r.t. P, a’2 must also be adjacent to a1,
w.r.t. P’. Similarly, if y = a2, we get a contradiction since of, wouldn’t be adjacent
to a; w.r.t. P’, but it should be as a4 is adjacent to a2 w.r.t. P. Hence, y = a’.

This proves that a’ is a genus 2 curve.

Case II: Assume that k 2 3. Let S], be a subsurface of S of genus k having a as
its boundary. We can choose a pair of pants decomposition Q = {a1, a2, ..., a3k_2} of
S1,, and then complete Q U {a} to a pair of pants decomposition P of S such that
each of a,- is a 4—curve in P for i = 1,2,...,3k — 2 and a, a1,a3 are the boundary
components of a pair of pants of Q. In Figure 1.3, (ii), we show how to choose Q
when k = 4. In the other cases, when k = 3 or k 2 5, a similar pair of pants

decomposition of Sk can be chosen.

I4

Let P’ be a pair of pants decomposition of S such that A([P]) = [P’]. Let a:-
be the representative of A([a,-]) which is in P’, for i = 1, ...,3k — 2, and a’ be the

representative of A([a]) which is in P’. Let Q’ = [a],a’2, ...,agk_2}.

For every 4-curve x in P’, there exist two pairs of pants A(x) and B(x) of P’
having x as one of their boundary components. Let 0(2)) 2 A(x) U B(x). The
boundary of C(x) consists of four distinct curves, which are adjacent to x w.r.t.
P’. Since all the curves a1, ...,a3k_2 in Q are 4-curves in P, all the corresponding
curves a’1,...,a§,k_2 are 4-curves in P’ by Lemma 1.7 and Lemma 1.4. Let N’ =

C(a’1)U C(ag) U u C(agk_2).
Claim: N’ is a genus k subsurface of 5 having a’ as its boundary.

Proof: Each boundary component of C(afi) is either a’ or a;- for some j = 1, ..., 3k—
2. Since a: is in the interior of C(afi), a; is in the interior of N’ for i = 1,2, ...,3k—2.
So, all the boundary components of C (a2) which are different from a’ are in the
interior of N’. Hence, N’ has at most one boundary component, which could be
a’. We know that a’ is adjacent to two distinct curves, a’1,ag, w.r.t. P’ since a is
adjacent to a1,a3 w.r.t. P. Suppose that a’ is in the interior of N’. Then a’ is only
adjacent to curves w.r.t. P’ which are in Q’. On the other hand, being adjacent to
a’1,ag, it has to be adjacent to a third curve, a;- in Q’, w.r.t. P’. But each of such
curves has four distinct adjacent curves w.r.t. P’ which are different from a’ already
since for j aé 1,3, each of a,- has four distinct adjacent curves w.r.t. P, which are
different from a by our choice of P. So, a’ cannot be adjacent to a;- w.r.t. P’ when
j # 1, 3. Hence, a’ is not in the interior of N’, it is on the boundary of N’. So, N’
is a subsurface of S having a’ as its boundary. To see that N’ has genus h, it is

enough to realize that {a’1,...,a§k_2} is a pair of pants decomposition of N’. Hence,

a’ is a k-separating circle. I

15

Lemma 1.9 Let A : C(S) —> C(S) be a superinjective simplicial map. Let t be a
k-separating circle on S, where 1 g k g g — 1. Let SI,S2 be the distinct subsurfaces
of S of genus k and g—k respectively which have t as their boundary. Let t’ E A([t]).
Then t’ is a k-separating circle and there exist subsurfaces S[,S§ of S of genus is
and g — h respectively which have t’ as their boundary such that A(C(S1)) Q C(51)
and ms») c C(85).

PROOF. Let t be a k-separating circle where 1 _<_ k _<_ g — 1. Since the genus of S
is at least 3, t is a separating curve of genus at least 2. So, it is enough to consider

the cases when k 2 2.

Let 51,82 be the distinct subsurfaces of S of genus k and g— h respectively which
have t as their boundary. Let t’ E A([t]). By Lemma 1.8, t’ is a k-separating circle.
As we showed in the proof of Lemma 1.8, there is a pair of pants decomposition P1
of SI, and P1 U {t} can be completed to a pair of pants decomposition P of S such
that a set of curves, Pl’, corresponding (via A) to the curves in P1 can be chosen such
that PI’ is a pair of pants decomposition of a subsurface that has t’ as its boundary.
Let S] be this subsurface. Let S], be the other subsurface of S which has t’ as its
boundary. A pairwise disjoint representative set, P’, of A([P]) containing Pl’ U {t’}
can be chosen. Then, by Lemma 1.6, P’ is a pair of pants decomposition of S. Let
P2 = P\(P1Ut) and P5 = P’ \(Pl’ Ut’). Then P2, P2’ are pair of pants decompositions

of S2, 35 respectively as P1, P1’ are pair of pants decompositions of 51,5] respectively.

Now, let a be a vertex in C (SI). Then, either a E [P1] or a has a nonzero geomet-
ric intersection with an element of [P1]. In the first case, clearly A(a) E C (Si) since
elements of [P1] correspond to elements of [Pi] S C (Si) In the second case, since A
preserves zero and nonzero geometric intersection (since A is superinjective) and a
has zero geometric intersection with the elements of [P2] and [t] and nonzero inter-

section with an element of [P1], A(a) has zero geometric intersection with elements of

16

[P2’] and [t’] and nonzero intersection with an element of [Pf]. Then, A(a) E C (S[)

Hence, A(C(Sl)) Q C(Si). The proof of A(C(S2)) g C(Sé) is similar. I

Lemma 1.10 Let A : C(S) —> C(S) be a superinjective simplicial map. Then A
preserves topological equivalence of ordered pairs of pants decompositions of S, (i.e.
for a given ordered pair of pants decomposition P = (c1,c2, ...,c3g_3) of S, and a
corresponding ordered pair of pants decomposition P’ = (c’1,c’2, ...,c§g_3) of S, where
[c[] = A([c,-]) V i = 1,2, ...39—3, there exists a homeomorphism F : S —> S such that
F(c,) = c: V i=1,2,...3g—3).

PROOF. Let P be a pair of pants decomposition of S and A be a nonembedded
pair of pants in P. The boundary of A consists of the circles x,y where x is
a 1-separating circle on S and y is a nonseparating circle on S. Let R be the
subsurface of S of genus g — 1 which is bounded by x and let T be the subsurface
of S of genus 1 which is bounded by 2:. Let P1 be the set of elements of P \ {x}
which are on R and P2 be the set of elements of P \ {x} which are on T. Then,
P1,P2 are pair of pants decompositions of R,T respectively. So, P2 = {y} is a
pair of pants decomposition of T. By Lemma 1.9, there exists a 1—separating circle
30’ E A([x]) and subsurfaces T’, R’, of S, of genus 1 and g — 1 respectively such that
A(C(R)) Q C(R’) and A(C(T)) Q C(T’). Since [P1] Q C(R), we have A([P1]) g C(R’).
Since [P2] 9 C (T), we have A([P2]) Q C (T’ ) Since A preserves disjointness, we can
see that a set, Pl’, of pairwise disjoint representatives of A([P1]) disjoint from x’ can
be chosen. By counting the number of curves in Pl’, we can see that PI’ is a pair of
pants decomposition of R’. Similarly, a set, P], of pairwise disjoint representatives of
A([P2]) disjoint from x’ can be chosen. By counting the number of curves in P2’, we
can see that P; is a pair of pants decomposition of T’ . Since P2 has one element, y,
P; has one element. Let y’ E P2’. Since x’,y’ correspond to x, y respectively and y

and y’ give pair of pants decompositions on T and T’ (which are both nonembedded

17

pairs of pants) and x and x’ are the boundaries of R and R’, we see that A “sends”

a nonembedded pair of pants to a nonembedded pair of pants.

Let B be an embedded pair of pants of P. Let x,y,z E P be the boundary
components of B. We consider two cases:
(i) At least one of x, y or z is a separating circle.

(ii) All of x, y, z are nonseparating circles.

In the first case, W.L.O.G assume that x is a k—separating curve for 1 S k < 9.
Let 81,52 be the distinct subsurfaces of S of genus k and g - h respectively which
have x as their boundary. W.L.O.G. assume that y,z are on S2. Let x’ E A([x]). By
Lemma 1.9, there exist subsurfaces, S], 55, of S of genus k and g - h respectively
which have x’ as their boundary such that A(C(S1)) Q C(S[) and A(C(Sg)) Q C(Sé).
Then, since y U 2 Q Sg, A({[y], [z]}) Q C(Sé). Let y’ E A([y]),z’ E A([z]) such that
{x’, y’, z’} is pairwise disjoint. Let P’ be a set of pairwise disjoint representatives of
A([P]) which contains 113’, y’, z’ . P’ is a pair of pants decomposition of S. Then, since
x is adjacent to y and z w.r.t. P, x’ is adjacent to y’ and z’ w.r.t. P’ by Lemma
1.7. Then, since x’ Uy’ Uz’ Q SQ, and x’ is the boundary of SQ, there is an embedded
pair of pants in S; which has x’ ,y’ , z’ on its boundary. So, A “sends” an embedded
pair of pants bounded by x, y, z to an embedded pair of pants bounded by x’, y’, 2’

in this case.

In the second case, we can find a nonseparating circle w and a 2—separating circle
t on S such that {x,y,z,w} is pairwise disjoint and x,y,z,w are on a genus 2
subsurface, SI, that t bounds as shown in Figure 1.5. Let P1 = {x,y, z,w}. P1 is a
pair of pants decomposition of S 1 . We can complete P1U{t} to a pants decomposition

PofS.

18

 

Figure 1.5. Nonseparating circles, 23, y, z, bounding a pair of pants

Let 52 be the subsurface of S of genus g — 2 which is not equal to 31 and that
is bounded by t. By Lemma 1.9, there exist a 2-separating circle t’ E A([t]) and
subsurfaces, S[,S§, of S of genus 2 and g — 2 respectively such that A(C(S1)) Q
C(81) and A(C(Sg)) Q C(Sé). Since P1 Q 51, A([P1]) Q C(Si). We can choose
a pairwise disjoint representative set, P’, of A([P]) containing Pl’. P’ is a pair of
pants decomposition of S. Let x’ ,y’ ,z’ ,w’ E Pl’ be the representatives of x,y,z,w
respectively. Then, since t is adjacent to z and w w.r.t. P, t’ is adjacent to z’ and
w’ w.r.t. P’ by Lemma 1.7. Then, since t’ U 2’ U w’ Q S] and t’ is the boundary
of S], there is an embedded pair of pants in S] which has t’, z’, w’ on its boundary.
Since 2 is a 4-curve in P, z’ is a 4—curve in P’. Since 2 is adjacent to x,y w.r.t.

’ is on the boundary of a pair of pants

P, z’ is adjacent to x’,y’ w.r.t. P’. Since 2
which has w’, t’ on its boundary, and z’ is adjacent to x’, y’, there is a pair of pants
having x’, y’, 2’ on its boundary. So, A “sends” an embedded pair of pants bounded

by x, y, z to an embedded pair of pants bounded by x’, y’, z’ in this case too.

Assume that P = (c1,c2, ..., c39-3) is an ordered pair of pants decomposition of
S. Let c; E A([c,]) such that the elements of {c’1,c’2,...,cgg_3} are pairwise dis-
joint. Then, P’ : (c’1,c’2,...,cgg_3) is an ordered pair of pants decomposition of

S. Let (B1,B2, ..., B2942) be an ordered set containing pairs of pants of P. By

19

the arguments given above, there are corresponding, “image”, ordered pairs of pants
( [, Bg, ..., B594). Nonembedded pairs of pants correspond to nonembedded pairs of
pants and embedded pairs of pants correspond to embedded pairs of pants. Then, by
the classification of surfaces, there exists an orientation preserving homeomorphism
h,- : B,- —> B: , Vi = 1, ..., 2g—2. We can compose each h,- with an orientation preserv-
ing homeomorphism r, which switches the boundary components, if necessary, to get
h; = r,- oh, to agree with the correspondence given by A on the boundary components,
(i.e. For each boundary component a of B,- for i = 1, ..., 2g—2, A([q(a)]) = [q’(h;(a))]
where q : S p —-> S and q’ : S]; —> S are the natural quotient maps). Then for two
pair of pants with common boundary component, we can glue the homeomorphisms
by isotoping the homeomorphism of the one pair of pants so that it agrees with the
homeomorphism of the other pair of pants on the common boundary component. By
adjusting these homeomorphisms on the boundary components and gluing them we

get a homeomorphism F: S —> S such that F(c,) = c: Vi = 1,2, ...,3g — 3. I

Remark: Let 8 be an ordered set of vertices of C (S) having a pairwise disjoint
representative set E. Then, E can be completed to an ordered pair of pants decom-
position, P, of S. We can choose an ordered pairwise disjoint representative set, P’,
of A([P]) by Lemma 1.6. Let E’ be the elements of P’ which correspond to the ele-
ments of E. By Lemma 1.10, P and P’ are topologically equivalent as ordered pants
decompositions. Hence, the set E and E’ are topologically equivalent. So, A gives a
correspondence which preserves topological equivalence on a set which has pairwise
disjoint representatives. In particular, A “sends” the isotopy class of a nonseparating

circle to the isotopy class of a nonseparating circle.

We use the following lemma to understand some more properties of superinjective

simplicial maps.

20

 

 

 

Figure 1.6. Circles intersecting once

Lemma 1.11 (Ivanov) [Iv1] Let 011 and (12 be two vertices in C (S) Then,
i(al,a2) = 1 if and only if there exist isotopy classes 0:3,a4, (15 such that
(i) i(a,,a,) = 0 if and only if the i”’ and j’” circles on Figure 1.6 are disjoint.
{ ii ) if (14 is the isotopy class of a circle C4, then C4 divides S into two pieces, and
one of these is a torus with one hole containing some representatives of the isotopy

classes (11, 02 .

Lemma 1.12 Let A : C (S) —> C (S) be a superinjective simplicial map. Let a, B be
two vertices of C(S). If i(a,B) = 1, then i(A(a),A(B)) = 1.

PROOF. Let a, B be two vertices of C (S) such that i(a, B) = 1. Then, by Ivanov’s
Lemma, there exist isotopy classes 013,014,015 such that i(a,,a,~) = 0 if and only if
i’”, j’h circles on Figure 1.6 are disjoint and if 04 is the isotopy class of a circle C4, then
C4 divides S into two pieces, and one of these is a torus with one hole containing
some representative of the isotopy classes 01,012. Then, since A is superinjective
i(A(a,-),A(a,-)) = 0 if and only if i’h,j”‘ circles on Figure 1.6 are disjoint, and by
Lemma 1.9, if A(a4) is the isotopy class of a circle C7,, then Cf, divides S into two
pieces, and one of these is a torus with one hole containing some representative of the

isotopy classes A(al), A(ag). Then, by Ivanov’s Lemma, i(A(a),A(fi)) = 1. I

21

CHAPTER 2

Induced Map On Complex Of Arcs

2. 1 Introduction

In this chapter, we first define the complexes of arcs of an orientable surface, R, with
nonempty boundary. Then, for a closed, connected, orientable surface, S, of genus
at least 3, we describe how to “code” properly embedded arcs on Sc (c is nonsepa-
rating) with circles on S. Using this “coding”, and the properties of superinjective
simplicial maps of the complex of curves that we studied in chapter 1, we show that
a superinjective simplicial map of the complex of curves of S induces an injective
simplicial map of the complex of arcs of Sc. We also prove that a simplicial map,
A : C (S) ——> C (S) is superinjective if and only if it is induced by a homeomorphism of

the surface S .

Throughout this chapter, we assume that S is a closed, connected, orientable
surface, S, of genus at least 3, A : C (S) —> C (S) is a superinjective simplicial map,

c, d are nonseparating simple closed curves on S, and [d] = A([c]).

22

2.2 The Complex B(R)

For an orientable surface R with nonempty boundary, an arc i on R is called pr0perly
embedded if (9i Q BR and i is transversal to OR. i is called nontrivial (or essential)
if i cannot be deformed into OR in such a way that the endpoints of i stay in 6R
during the deformation. The complex of arcs, B(R), on R is an abstract simplicial
complex. Its vertices are the isotopy classes of nontrivial properly embedded arcs i in
R. A set of vertices forms a simplex if these vertices can be represented by pairwise

disjoint arcs.

Let T be a compact orientable surface with a nonempty set of punctures. Assume
that T has negative Euler Characteristic. An ideal triangulation of T is a collection
of embedded arcs with disjoint interiors connecting the punctures of T such that
if we cut T along all these arcs, we will get a collection of ideal triangles. For such
punctured surfaces we can define the complex of arcs similar to the previous definition
by taking the set of vertices to be the isotopy classes of non-trivial arcs connecting
punctures of the surface. Then we can see that, considering up to isotopy, the ideal

triangulations are exactly the top dimensional Simplices of this arc complex.

B(R) is a finite dimensional complex. It is known that every two top dimen—
sional Simplices A,A’ of B(R) can be connected by a chain of Simplices A =
A0, A1, ..., Am = A’ such that any two consecutive Simplices A” A,“ have exactly
one common face of codimension 1, [Mos]. This chain connectedness property of B ( R)

implies the following lemma.

Lemma 2.1 If an injection )4 : B(Sc) —> B(Sd) agrees with an element h... on a top
dimensional simplex, where h. is induced by a homeomorphism h : Sc —> Sd, then u
agrees with h... on B(SC).

PROOF. Suppose that it agrees with h, on a top dimensional simplex, A. Then,

23

h:1 o ,u fixes A pointwise. Let w be a vertex in B(Sc). If w E A, then w is fixed by
h:1 o p. Suppose that w is not in A. Take a top dimensional simplex containing w.
Call it A’. There exists a chain A = A0, A1, ..., Am = A’ of top dimensional Simplices
in B(SC), connecting A to A’ such that any two consecutive Simplices A,,A,-+1
have exactly one common face of codimension 1. This follows from the Connectivity
Theorem for Elementary Moves of Mosher, [Mos], appropriately restated for surfaces
with boundaries. Since h:1 o u is injective, A1 must be sent to a top dimensional
simplex by h:1 o u. Let wl be the vertex of A1 which is not in A. Since A is fixed
by h:1 o p, the common face of A and A1 is fixed. A1 is the unique top dimensional
simplex containing the common codimension 1 face of A and A1 other than A. Since
h: 1 o ,u(A1) is a top dimensional simplex having this common face which is different
from A, A1 must be sent onto itself by h:1 o p. This implies that wl is fixed. By an
inductive argument, we can prove that all the top dimensional Simplices in the chain
are fixed. This shows that w is fixed. Hence, h:1 o u is the identity on B(Sc) and u

agrees with h... on B(SC). I

Let V(Sc) and V(Sd) be the sets of vertices of B(SC) and B(Sd) respectively. First,
we will prove that A induces a map A... : V(Sc) —> V(Sd) with certain properties. Then

we will prove that A... extends to an injective simplicial map A. : B(SC) —-> B(Sd).

2.3 Pairs of Disjoint Arcs and Neighborhoods

Lemma 2.2 Let a and b be two disjoint arcs on SC connecting the two boundary
components, 61, 02, of Sc. Let N be a regular neighborhood of a U b U 81 U 02 in Sc.

Then, (N ,a, b) 9—.” (Sf, a0, be) where S,2 is a standard sphere with four holes and a0, be

are arcs as shown in Figure 2.1.

PROOF. Let F = (91 U 02 U a U b. Let N be a regular neighborhood of F in SC.

24

 

Figure 2.1. Unlinked disjoint arcs of type 2

Figure 2.2. The graph of a, b and the boundary components

N deformation retracts onto I‘. So, N and I‘ have the same Euler characteristics.
Let m be the genus of N and n be the number of boundary components of N. It is

easy to see that x(F) = —2. Then, —2 = x(N) =2—2m—n. So, 2m+n =4.

There are three possibilities for m and n:

(i)m=0,n=4, (ii)m=1,n=2, (iii)m=2,n=0.

Since there are 2 boundary components corresponding to 81, (92, none of which is
the boundary component on a given side of the arc a, n is at least 3. So, only (i)

can hold. Therefore, N is homeomorphic to S}, a sphere with four holes.

By using Euler characteristic arguments we can see that a regular neighborhood

of 81 U ('32 U a in N is homeomorphic to a pair of pants, P. Let w be the boundary

25

component of P different from (91,82.

Let I" be the graph that we get when we contract 81 and 82 to two points. If a is
isotopic to b in N, the two arcs corresponding to a and b on I" should be homotopic
to each other relative to these two points. But this gives a contradiction since the
arcs intersect only at these end points and the union of the arcs is 7’ which is a
circle and such two arcs cannot be homotopic relative to their end points on a circle.
So, a is not isotopic to b in N. Since b connects the two boundary components
01 and 62 and 01,62 E P, b O P is nonempty. W.L.O.G. assume that b intersects
the boundary components of P transversely and doesn’t intersect a. Since N is
a regular neighborhood of a U b U 01 U 62 and b is a properly embedded essential
are which is nonisotopic to a in N, b O P contains exactly two essential properly
embedded arcs, let’s call them b1, b2. One of them starts on 01 and ends on w and
the other one starts on 62 and ends on w. Let P’ be a regular neighborhood of
61 U 62 U a in P such that P’ O b = b1 U b2. Let x be the boundary component of P’,
which is different from 61, 62. We choose this neighborhood to get rid of the possible
inessential arcs of b in P. P’ is a pair of pants. Since N is a sphere with four holes,
the complement of P’ in N is a pair of pants, R. Let y and z be the boundary
components of R which are different from x. Then, we have a homeomorphism
(b : (N, 81,62,x,y, z) ——> (S42,ro,to,xo,yo,zo) where ro,to,xo,yo,zo are as shown in

Figure 2.3.

Let Po be the pair of pants bounded by r0, to, x0. Let x1, x2 be two parallel copies
of 9:0 in P0 as shown in Figure 2.3 such that each of them intersects ¢(b) transversely
at exactly 2 points and none of x1, x2 intersects <b(a). Let A0 be the annulus which
is bounded by x2 and x1, and Bo be the annulus which is bounded by x1,xo. Let

Q, be the pair of pants bounded by re, to, 1:2.

By the classification of properly embedded essential arcs on a pair of pants, there

26

 

 

Figure 2.3. Disjoint arcs on pairs of pants

exists an isotopy on Q0 carrying ¢(a) and Qofl¢(b) to Qoflao and Qoflbo respectively,
where a0, be are as shown in the figure. Let It : Q, X I —> Q, be such an isotopy. We

can extend h: to Ft : (Q0 U A0) x I —-> (Q0 U A0) so that it, is id on x1 for all t E I.

Let R0 be the pair of pants bounded by xmyo, zo. Suppose that R0 (1 ¢(b) is an
inessential arc in R0. Then it can be deformed into the interior of Po and then we get
¢(a) is isotopic to ¢(b) in P0 which implies that a is isotopic to b in N. This gives
a contradiction. So, b 0 R0 is an essential arc in R0. Then, by the classification of
properly embedded essential arcs on a pair of pants there exists an isotopy carrying
R0 0 ¢(b) to R0 0 b0. Let r : R, x I —> R, be such an isotopy. We can extend r to
i" : (ROUBO) x I —+ (ROUBO) so that 1"“, is id on x1 for all t E I. Then, by gluing the

extensions 17: and i" we get an isotopy 29 on S,2 x I which fixes each of re, to, re, ya, 20.

By the classification of isotopy classes of arcs relative to the boundary on an
annulus, 191(¢(b)) 0 (A, U Bo) can be isotoped to tfio (b0) 0 (A, U Bo) for some 19 E Z.
Let’s call this isotopy u. Let [4 denote the extension by id to N. Then we have,
t;0k(fl1('l91(¢(b)))) = b0. Clearly, tgo" o [11 o 291 fixes each of re, to, x0, ya, 20. Hence, we

get a homeomorphism, tgok o .111 o 191 o 45 : (N,a, b) —> (S4240, b0). I

27

 

(i) (ii) (iii)

Figure 2.4. Disjoint arcs and neighborhoods

Let a and b be two disjoint arcs connecting a boundary component of Sc to
itself. Then, a and b are called linked if their end points alternate on the boundary

component. Otherwise, they are called unlinked.

The proofs of Lemma 2.3, Lemma 2.4 and Lemma 2.5 are similar to the proof of

Lemma 2.2. So, we do not prove these lemmas here. We only state them.

Lemma 2.3 Let a and b be two disjoint arcs which are unlinked, connecting one
boundary component (9,, of Sc to itself for k = 1,2. Let N be a regular neighborhood
of a U bUB;c on Sc. Then, (N, a, b) E“ (Siambo) where ambo are the arcs drawn on

a standard sphere with four holes, S} , as shown in Figure 2.4, (i).

Lemma 2.4 Let a and b be two disjoint arcs on Sc such that a connects one bound-
ary component (9,, of Sc to itself for some k = 1,2 and b connects the boundary
components 61 and 62 of Sc. Let N be a regular neighborhood of a U b U 61 U 62.
Then, (N, a, b) E“ (S42,ao,bo) where ambo are the arcs drawn on a standard sphere

with four holes, S}, as shown in Figure 2.4, (ii).

28

Lemma 2.5 Let a and b be two disjoint, linked arcs connecting one boundary com-
ponent 8;, of SC to itselffor k = 1, 2. Let N be a regular neighborhood of a U bU 81,.
Then, (N ,a,b) E“ (Siambo) where E? is a standard surface of genus one with two

boundary components, and a0, b0 are as shown in Figure 2.4, (iii).

2.4 Properties of Superinjective Simplicial Maps

Let AI be a sphere with k holes and k 2 5. A circle a on M is called an n-circle
if a bounds a disk with n holes on M where n 2 2. If a is a 2-circle on M, then
there exists up to isotopy a unique nontrivial embedded arc a’ on the two-holed disk
component of Ma joining the two holes in this disc. If a and b are two 2-circles
on M such that the corresponding arcs a’, b’ can be chosen to meet exactly at
one common end point, and a = [a],6 = [b], then {a,fl} is called a simple pair.
A pentagon in C (M) is an ordered 5—tuple (a1,ag,a3,a4,as), defined up to cyclic
permutations, of vertices of C(M) such that i(aj,aj+1) = O for j = 1,2,...,5 and
i(aj, 01k) 75 0 otherwise, where as = a]. A vertex in C (M) is called an n-vertex if it
has a representative which is an n—circle on M. Let M’ be the interior of M. There
is a natural isomorphism X : C (M ’ ) —> C (M) which respects the above notions and
the corresponding notions in [K]. Using this isomorphism, we can restate a theorem

of Korkmaz as follows:

Theorem 2.6 (Korkmaz) [K] Let M be a sphere with n holes and n 2 5. Let
a, B be two 2-vertices of C (M) Then {a, ,8} is a simple pair ifi’ there exist vertices
71, 72, ...,7n_2 of C (M) satisfying the following conditions:

(i) (unmanafi) 2'8 a Pentagon in C(M),

(ii) '71 and 7,,_2 are 2-vertices, 72 is a 3-vertex and ’71: and 7,,4, are k-vertices

for3gksg.

29

 

 

Figure 2. 5. “Horizontal” and “vertical” circles

(22“) {0,73,74,757U'37n—2}r {0,72,74,75a"'37n—2}: {18:73:74,751"')7n—2}7 and

{71,72,74,75, ..., 7,,_2} are codimension-zero simplices.

By using the following lemmas, we will see some more properties of A.

Lemma 2.7 Let c,x,y,z,h,v be essential circles on S. Suppose that there ex-
ists a subsurface N of S and a homeomorphism (p : (N, c,x,y,z,h,v) —>
(No,co,xo,yo,zo,ho,vo) where N, is a standard sphere with four holes having
co, x0, ya, 20 on its boundary and ho, v0 (horizontal, vertical ) are two circles which have
geometric intersection 2 and algebraic intersection 0 as indicated in Figure 2. 5. Then,
there exist 0’ E A([c]),x’ E A([x]),y’ E A([y]),z’ E A([z]),h’ E A([h]),v’ E A([v]),N’ C

S and a homeomorphism X : (N’,c’,x’,y’,z’,h’,v’) —+ (No,co,xo,yo,zo,ho,vo).

PROOF. Let A: {c,x,y,z,v}. Any two elements in A which are isotopic in S
bound an annulus on S. Let B be a set consisting of a core from each annulus which
is bounded by elements in A, circles in A which are not isotopic to any other circle in
A, and v. We can extend B to a pants decomposition P of S. Since the genus of S is
at least 3, there are at least four pairs of pants of P. Note that {v} is a pair of pants

decomposition of N . Each pair of pants of this pants decomposition of N is contained

30

 

Figure 2.6. Sphere with five holes

in exactly one pair of pants in P. Hence, there is a pair of pants R of P whose interior
is disjoint from N and has at least one of c, x, y, 2 as one of its boundary components.
W.L.O.G. assume that R has y on its boundary. Let T be a regular neighborhood,
in R, of the boundary components of R other than y. Let t,w be the boundary
components of T which are in the interior of R. Then, y, t, w bound an embedded
pair of pants Q in R. Let N = N U Q. Then, we can extend No to NO and extend
cp to a homeomorphism (,7) : (N,a,x,y,z,h,v,t,w) —-> (No,co,xo,yo,zo,ho,vo,to,wo),

where No is as shown in Figure 2.6.

Using Lemma 1.10, we can choose pairwise disjoint representatives c’,x’,y’,z’,
v'.t'.w' of A([c]).A([x]).i<[y1>.i([z1). A([v]),A([tl),/\([wl) respectively et. there
exists a subsurface N’ of S and a homeomorphism x : (N ’,c’ ,x’ ,y’ ,z’ ,v’ ,t’ ,w’ )
-+ (No,co,xo,yo,zo,vo,to,wo)- Clearly, we have, z'(lhl),[0l) = z'(lhlilml) = 0,
i([h],[y]) = i([h],[z]) = 0. Since c,x,y,z are all essential circles on S, we have
i([h],[v]) 75 0. Then, since A is superinjective, we have, i(A([h]),A([c])) = 0,
i(A([h]),A([l‘lD = i(A([h]),A([i/ll) = i(A([h]),A(lzm = 0 and z(A([h]),Mlle r4 0-

31

Then, a representative h’ of A([h]) can be chosen such that h’ is transverse to
v’, h’ doesn’t intersect any of c’,x’,y’,z’, and i(A([h]),A([v]) = lh’ fl v’|. Since
i(A([h]), A([v])) ¢ 0, h’ intersects v’. Hence, h’ is in the 4—holed sphere bounded by

c’,x’,y’,z’ in N’.

N and N ’ are spheres with five holes in S. Since c,x,y,z are essential circles
in S, the essential circles on N are essential in S. Similarly, since c’,x’,y’,z’ are
essential circles in S , the essential circles on N’ are essential in S. Phrthermore,
we can identify C(N) and C(N’) with two subcomplexes of C(S) in such a way
that the isotopy class of an essential circle in N or in N’ is identified with the
isotopy class of that circle in S. Now, suppose that a is a vertex in C (N ) Then,
with this identification, (1 is a vertex in C(S) and 0: has a representative in N .
Then, i(a, [e]) = i(a, [23]) = i(a, [t]) = i(a, [w]) = i(a, [2]) = 0. Then there are two
possibilities: (i) i(a, [v]) = O and i(a,[y]) = O (in this case or = [v] or a = [y]),
(ii) i(a, [v]) # O or i(a, [y]) 76 0. Since A is injective, A(a) is not equal to any of
[c’], [x’], [t’], [w’], [2’]. Since A is superinjective, we have, i(A(a), [c’]) = i(A(a), [x’]) =
i(A(a), [t’]) = i(A(a), [w’]) = i(A(a), [z’]) = 0. Then, there are two possibilities: (i)
i(A(a),[v’]) = 0 and i(A(a),[y’]) = O (in this case A(a) = [v’] or A(a) = [y’]), (ii)
i(A(a), [v’]) 76 O or i(A(a), [y’]) 5% 0. Then, a representative of A(a) can be chosen
in N’. Hence, A maps the vertices of C (S) that have essential representatives in

N to the vertices of C(S) that have essential representatives in N’, (i.e. A maps

C(N) Q C(S) to C(N’) Q C(S)). Similarly, A maps C(N) Q C(S) to C(N’) Q C(S).

It is easy to see that {[h], [v]} is a simple pair in N. Then, by Theorem 2.6,
there exist vertices 71,72,73 of C(N) such that (71,72, [h],’73, [v]) is a pentagon in

C(N), '71 and 73 are 2-vertices, '72 is a 3-vertex, and {[h],73}, {[h],72}, {[v],'y3}

and {71, 72} are codimension—zero Simplices of C (N)

32

 

(i) (ii)

Figure 2. 7. Circles on torus with two boundary components

Since A is superinjective and c,x,y,z are essential circles, we can see that
(A(71),A(’72), A([h]),A(73), A([v])) is a pentagon in C(N’). By Lemma 1.10, A(71)
and A(73) are 2-vertices, and A(72) is a 3-vertex in C(N’). Since A is an injective
simplicial map {A([h]),A(73)}, {A([h]),A(72)}, {A([vlL A(73)} and {A(71)w\(72)}
are codimension-zero Simplices of C (N ’). Then, by Theorem 2.6, {A([h]),A([v])}
is a simple pair in N’. Since A([h]) has a representative, h’, in N’, such that
i(A([h]),A([v]) = [h’ H v’l and {A([h]),A([v])} is a simple pair in N’, there exists

a homeomorphism X 2 (Ni, CI) IL", y’) zit h’) v’) _f (NO, Co, x01 yo, 20) how v0) ‘ I

Lemma 2.8 Let c,x,y,z,m,n be essential circles on S. Suppose that there ex-
ists a subsurface N of S and a homeomorphism cp : (N, c,x,y,z,m, n) ——),p
(No,co,xo,yo,zo,mo,no) where N, is a standard torus with two boundary compo-
nents, cmxo, and ya, 20, mmno are circles as shown in Figure 2.7. Then, there exist
0’ E A([c]),x’ E A([x]),y’ E A([y]),z’ E A([z]),m’ E A([m]),n’ E A([n]),N’ Q S and a

homeomorphism X such that (N’,c’,x’,y’,z’,m’,n’) _’x (cho, xo,yo,zo,mo,no).

33

PROOF. Since the genus of S is at least 3, c cannot be isotopic to x in S. So,
we can complete {c, x, y, z} to a pair of pants decomposition, P, of S. Since {y, 2}
gives a pair of pants decomposition on N, by Lemma 1.10, there exists a subsurface
N’ Q S which is homeomorphic to No and there are pairwise disjoint representa-
tives c’,x’,y’,z’ of A([c]),A([x]),A([y]),A([z]) respectively and a homeomorphism 45
such that (N’,c’,x’,y’, 2’) —>¢ (No,co,xo,yo,zo). Then by Lemma 1.12, we have the

following:

) ( ( ) A([Zl))=1. Z'(l77e],[2l)=1=>i(/\([nl)./\([Zl))
2' [midi/l):1=>i(>\([ml)v\([vl))=1, z'(lnHyl) =1=>i(/\([nl),x\([yl))

) ( ( ) A([ClD = 0, z'(lnl, [6]) = 0 => z'(/\([nl),/\([Cl)) = 0,
([w])) = 0. Z'(lnl, [w]) = 0 => z(A([n]). A([313] )= 0,
([v])) = 0-

1,
1,

There are representatives m1 E A([m]),n’ E A([n]) such that [mlfly’l = [mlflz’] =
1,]m1flc’|=]mlflx’|= O, [n’r‘iy’] = |n’flz’| =1,|n’flc’| = |n’flx’] = [mlfln’l = 0

with all transverse intersections.

Since p is a homeomorphism, we have [¢(m1) fl <b(n’)| = O,](b(n’) 0 yo] = 1,
WW 0 Zol = 1,|¢(n') nCol = |¢(n') onl = 0 = WW) 0 Col = |¢(m1)fl $0] = 0,

[45(7’11) O yo] = |¢(m1)O zol = 1.

Let’s choose parallel copies y1, y2 of ya and 21, 22 of 20 as shown in Figure 2.7 so
that each of them has transverse intersection one with §b(m’) and ¢(n’). Let P1, P2 be
the pair of pants with boundary components co, yo, zo, and $0,312, 22 respectively. Let
Q1, Q2, R1, R2 be the annulus with boundary components {yo,y1},{y1,y2}, {zo, 2:1},
{21, 22} respectively. By the classification of isotopy classes of families of properly
embedded disjoint arcs in pair of pants ¢(m’) 0 P1, ¢(m’) 0 P2, ¢(n’) (1 P1 and

¢(n’) 0 P2 can be isotoped to the arcs mo 0 P1, mo 0 P2, no 0 P1, no 0 P2 respectively.

34

Let K, : P1 x I -+ P1, r : P2 x I ——> P2 be such isotopies. By a tapering argument,
we can extend it and r and get it : (P1 U Q1 U R1) x I —> (P1 U Q1 U R1) and
f : (PgUQ2UR2) x I-—> (P2UQ2UR2) so that Fit is id on y1U21 and it is id on

yl U 21 for all t E I . Then, by gluing these extensions we get an isotopy 19 on No x I .

By the classification of isotopy classes of arcs (relative to the boundary) on an
annulus, 191(gb(n’)) (1 (R1 U R2) can be isotoped to t’;0(no) 0 (R1 U R2) for some
k E Z. Let’s call this isotopy u. Let [i denote the extension by id to No. Simi-
larly, 291(qb(n’)) 0 (Q1 U Q2) can be isotoped to t;0(no) 0 (Q1 U Q2) for some I E Z.
Let’s call this isotopy u. Let 17 denote the extension by id to No. Then, “glu-
ing” the two isotopies [2 and 17, we get a new isotopy, e, on No. Then we have,
t;0’(t;"(61(191(qb(n’))))) = no. Clearly, t; o t"c o 61 o 291 fixes co,xo,yo,zo. So,

30

we get tgo’ o t;" o 61 o 191 o d) : (N’,c’,x’,y’,z’,n’) —> (No,co,xo,yo,zo,no). Let
X = tgo’otg'okoel 0291 Oct. Then, we also get, X(m1) is isotopic to ma because of the inter-
section information. Let m’ = X”1(mo). Then we get, X : (N’,c’,x’,y’,z’,m’,n’) —->

(Narc01$03y01Z0)moano)- I

2.5 Arcs and Encoding Simplices

Let i be an essential properly embedded are on Sc. Let A be a boundary component
of Sc which has one end point of i and B be the boundary component of Sc which
has the other end point of i. Let N be a regular neighborhood of i U A U B in SC.
By Euler characteristic arguments, N is a pair of pants. The boundary components
of N are called encoding circles of i on SC. These circles correspond to nontrivial
circles on S, which are called encoding circles of i on S. The set of isotopy classes
of encoding circles on S is called the encoding simplex, A,, of i (and of [i]). Note

that [c] is a vertex in the encoding simplex of i.

35

An essential properly embedded are i on SC is called type 1.1 if it joins one bound-
ary component 6,, of SC to itself for k = 1, 2 and if 0,, Ui has a regular neighborhood
N in SC, which has only one circle on its boundary which is inessential w.r.t. SC. If
N has two circles on its boundary which are inessential w.r.t. Sc, then i is called
type 1.2. We call i to be type 2, if it joins the two boundary components ('31 and ('32
of Sc to each other. An element [i] E V(Sc) is called type 1.1 (1.2, 2) if it has a type

1.1 (1.2, 2) representative.

Let 6’, 82 be the boundary components of Sd. We prove the following lemmas in

order to show that A induces a map A... : V(SC) —> V(Sd) with certain properties.

Lemma 2.9 Let 8;, E OS, for some k E {1, 2}. Then, there exists a unique 6’ E 850;
for some I E {1, 2} such that if i is a properly embedded essential are on Sc connecting

8,, to itself, then there exists a properly embedded arc j on 5,; connecting 8’ to itself

such that A(A,) = A,.

PROOF. Assume that each of 81 and 82 satisfies the hypothesis. Let i be a
properly embedded, essential, type 1.1 arc connecting 8,, to itself. Then, there exist
properly embedded arcs, jl, connecting 61 to itself, and j2, connecting 02 to itself,
such that A(A,) = A3, and A(A,) = A”. Then, we have A], = A1“:- Note that a
properly embedded essential arc i is type 1.1 iff A,- has exactly 3 elements. Otherwise
A,- has 2 elements. Since i is type 1.1 and A(A,) = A], and A(A,-) = A], and A is
injective, jl and jg are type 1.1. We can choose a pairwise disjoint representative set
{a, b, d} of A], on S. Since A], 2 A12, {a, b, d} is a pairwise disjoint representative
set for A,, on S. Then, the curves (1,13 on S; which correspond to a,b on S and
81 bound a pair of pants, P, on 5,; containing an arc, j], isotopic to jl. Similarly,
EL, b,02 bound a pair of pants, Q, on 5,; containing an arc, jg, isotopic to jg. Let’s

cut Sd along a and b. Then, P is the connected component of SdUan containing 81

36

and Q is the connected component of SdUaUg, containing 62. P 75 Q since 62 is not
in P and 02 is in Q. Then P and Q are distinct connected components meeting
along a and b. Hence, 5,; is P U Q, a twice holed torus. This implies that S is a
genus 2 surface which gives a contradiction since the genus of S is at least 3. So,

only one boundary component of Sd can satisfy the hypothesis.

Since i is type 1.1, A,- contains [c] and two other isotopy classes of nontrivial
circles which are not isotopic to c in S. Let P’ be a pairwise disjoint representative
set of A([A,-]), containing d. By the proof of Lemma 1.10, P’ bounds a pair of pants
on S. Since the genus of S is at least 3, P’ bounds a unique pair of pants on S,
which corresponds to a unique pair of pants, Q , in 5,1 which has only one inessential
boundary component. Let 0’“) be this inessential boundary component. Let j be
an essential properly embedded arc connecting 6’“) to itself in Q. Then, we have

A(Ai) = AJ' .

Now, to see that 0’“) is independent of the type 1.1 are i connecting 0,, to itself,

we prove the following claim:

Claim 1: If we start with two type 1.1 arcs i and j starting and ending on (9],,

then 6’“) = 8’”).

Proof of Claim 1: Let [i], [j] be type 1.1 and i, j connect 6,, to itself. W.L.O.G.
we can assume that i and j have minimal intersection. First, we prove that there is
a sequence j = r0 ——> r1 —+ —> rn+1 = i of essential properly embedded arcs joining
(9;, to itself so that each consecutive pair is disjoint, i.e. the isotopy classes of these

arcs define a path in B(Sc), between i and j.

If lifljl = 0, then take r0 = j, r1 = i. We are done. Assume that lifljl = m > 0.
We orient i and j arbitrarily. Then, we define two arcs in the following way: Start

on the boundary component 81,, on one side of the beginning point of j and continue

37

 

 

Figure 2.8. Splitting the arc j along the end of i

along j without intersecting j, till the last intersection point along i. Then we would
like to follow i, without intersecting j, until we reach 0],. So, if we are on the correct
side of j we do this; if not, we change our starting side from the beginning and follow
the construction. This gives us an arc, say jl. We define jg, another arc, by changing
the orientation of j and following the same construction. It is easy to see that jl, jg
are disjoint properly embedded arcs connecting 0;, to itself as i and j do. One can
see that jl, jg are essential arcs since i, j intersect minimally. In Figure 2.8, we show
the beginning and the end points of i, the essential intersections of i, j , and jl, jg

near the end point of i.

[i O j1| < m, [i O jg| < m since we eliminated at least one intersection with i.
We also have [.71 O j] = [jg H j] = 0 since we never intersected j in the construction.
Notice that jl and jg are not oriented, and i is oriented.

Claim 2: jl is not isotopic to jg.

Proof of Claim 2: Suppose that jl and jg are isotopic. Then, they are not linked,
and there is a band B such that 8B Q jl U jg U (9;, and BB \ (jl U jg) is a disjoint

union of two arcs 51, 82 on 61,, where each of 61 and eg starts at an end point of jl

38

and ends at an end point of jg. By the construction, there is an end point of each
of jl and jg near the end point of i on 6],, on the are t shown in the figure, and
there is not any point ofj on t. Then, either t = 51 or t = eg. W.L.O.G. assume
that t = 51. Then, 51 Q BB has the end point of i. Since 51 Q 81,, it has only one
side on Sc, and hence on B. Then, since 51 Q 8B has the end point of i, by the
construction we can see that the last intersection point along i of i and j has to lie
in the band B. Then, since j does not intersect any of j1 and jg, and B has a point
of j in the interior, j has to live in B. Then, since 51 doesn’t contain any point of
j, and j does not intersect any of jl and jg, the end points of j has to lie on eg,
which implies that j is an inessential are on Sc. This gives a contradiction. Hence,

jl and jg cannot be isotopic.
Claim 3: Either jl or jg is of type 1.1.

Proof of Claim 3: If jl and jg are linked, then a regular neighborhood of jl U ng0k
in Sc is a genus one surface with two boundary components by Lemma 2.5. In this
case, both arcs have to be type 1.1 since encoding circles on Sc for jl, jg can be
chosen in this surface and this surface has only one boundary component which is a

boundary component of Sc. So, both arcs have to be type 1.1.

If jl and jg are unlinked, then a regular neighborhood, N, of jl U jg U0,- in Sc is a
sphere with four holes by Lemma 2.3. Encoding circles of jl, jg on S, can be chosen
in N. Assume that jl is type 1.2 and has two encoding circles which are the two
boundary components of the surface Sc. Then, jg would have to be type 1.1 since N
can have at most two boundary components which are the boundary components of
Sc and regular neighborhoods of j1 U (9;, and jg Uolk have only one common boundary

component which is Bk. This proves Claim 3.

Let r1 E {j1,jg} and r1 be type 1.1. By the construction we get, [i (“I r1] < m,

39

| j 0 r1] = 0. Now, using i and r1 in place of i and j we can define a new type
1.1 arc rg with the properties, [i O rg| < [i 0 r1], In (1 rg| = 0. By an inductive
argument, we get a sequence of arcs such that every consecutive pair is disjoint,
i = r,,+1 —> rn —> rn_1 —> —-> r1 —> r0 = j. This gives us a path in B(Sc) between i
and j. By using the results of Lemma 2.7 and 2.8, we can see that each pair of disjoint
type 1.1 arcs give us the same boundary component. Then by using the sequence given
above, we conclude that i and j give us the same boundary component. This proves

Claim 1.

Let i0 be a properly embedded, type 1.1 arc on Sc connecting 8,, to itself. Let
8’ = 8’(’°). If i is a properly embedded, type 1.1 arc on Sc connecting 8,, to itself,
then by the arguments given above we have 8’ = 8’“), and there exists a properly
embedded arc j on 5.; connecting 8’ to itself such that A(A,) = A,. Suppose that
i is a properly embedded, type 1.2 arc on Sc connecting 8,, to itself. Let A,- be
the encoding simplex of [i]. Since i is type 1.2, A,- contains [c] and only one other
isotopy class of a nontrivial circle which is not isotopic to c in S. A pairwise disjoint
representative set, P, of A,- corresponds to a nonembedded pair of pants on S. Let
P’ be a pairwise disjoint representative set of A([A,-]), containing d. By extending
P to a pair of pants decomposition of S, and applying Lemma 1.10, we can see that
P’ corresponds to a unique nonembedded pair of pants of S which corresponds to
a unique pair of pants, Q, in Sd which has two inessential boundary components
containing 8’. Let j be a nontrivial properly embedded arc connecting 8’ to itself in

Q. Then, we have A(A,) = A,. Hence, 8’ is the boundary component that we want.

40

2.6 Induced Map

We define a map a : {81,8g} —> {81, 82} using the correspondence which is given by

Lemma 2.9.
Lemma 2.10 o is a bijection.

PROOF. Let x,y,z,t,h,c be essential circles on S such that x,y,z,t bound a
4-holed sphere and h, c be two circles which intersect geometrically twice and alge-

braically zero times in this subsurface as shown in 2.9, (i).

By using Lemma 2.7, we can see that there exist pairwise disjoint representa-
tives x’ ,y’ , z’ ,t’ , h’ ,d of A([x]), A([y]), A([z]),A([t]), A,([h])A ([c]) respectively such that
x’, y’, z’, t’ bound a 4—holed sphere, and h’, d intersect geometrically twice and alge-

braically zero times in this subsurface as shown in 2.9, (ii).

The curve h corresponds to two arcs, say i1,ig, on Sc which start and end on
different boundary components, 8),, 81, of Sc respectively. W.L.O.G. assume that
{x,z,c} and {y,t,c} are encoding circles of i1 and ig on S. {x,z,c} and {y,t,c}
correspond to {x’,z’,d} and {y’,t’,d} which bound pairs of pants L and R on S
respectively. Assume that L corresponds to a pair of pants, L’, on 54 which has
a boundary component 8’" of S01 and R corresponds to a pair of pants, R’, on 3,;
which has a boundary component 8" of S4. h’ corresponds to two arcs, say jl and
jg on L’ and R’ which start and end on different boundary components, 8’" and 8”
of S1 respectively. By the definition of a we have, 0(8),) = 8’" and 0(8)) = 8” for

k,l,m,n E {1,2}, k 75 l,m 75 n. So, a is onto. Hence, a is a bijection. I

Lemma 2.11 Let [i] E V(S C). If i connects 8,, to 81 on Sc where k, l E {1, 2},
then there exists a unique [j] E V(Sd) such that j connects 0(8k) to 0(81) and
A(Ag) ‘2 AJ’ .

41

 

 

(i) (ii)

Figure 2.9. Sphere with four holes

PROOF. Let [i] E V(SC) and let i connect 8;, to 81 on Sc where k,l E {1,2}.
Let A, be the encoding simplex of i. Then, a pairwise disjoint representative set of
A([A,]) containing d corresponds to a unique pair of pants in 5,; which has boundary
components 0(8),) and 0(8)). By the classification of properly embedded arcs in pair
of pants, there exists a unique isotopy class of nontrivial properly embedded arcs
which start at 0(8k) and end at 0(8)), in this pair of pants. Let j be such an arc,

we have A(A,) = A,.

Let e be an essential properly embedded arc such that e connects 0(8),) to 0(8))
and A(A,) = A3. Then, we have Ae = AJ- = A(A,). Let a, b,d be a pairwise
disjoint representative set of A(A.) on S. Then there are properly embedded arcs
j1,e1 isotopic to j, 6 respectively which are in pair of pants bounded by a, b, d. Since
the genus of S is at least 3, there is at most one pair of pants which has a, b, d on its
boundary. So, j1,e1 are in the same pair of pants. Since they both connect the same
boundary components in this pair of pants, they are isotopic. So, [7'] = [e]. Hence,

[7'] is the unique isotopy class such that j connects 0(8),) to 0(8)) and A(A,) = Aj.

42

A induces a unique map A... : V(Sc) —> V(Sd) such that if [i] E V(Sc) then A.([i])
is the unique isotopy class corresponding to [i] where the correspondence is given by
Lemma 2.11. Using the results of the following lemmas, we will prove that A. extends

to an injective simplicial map on the whole complex, B(Sc).

Lemma 2.12 A, : V(SC) —> V(Sd) extends to a simplicial map A. : B(SC) —+ B(Sd).

PROOF.It is enough to prove that if two distinct isotopy classes of essential properly
embedded arcs in Sc have disjoint representatives, then their images under A... have
disjoint representatives. Let a, b be two disjoint representatives of two distinct classes
in V(Sc). Let 81,8g be the two boundary components of Sc. We consider the

following cases:

 

 

Figure 2.10. Circles associated to type 2 arcs

Case 1: Assume that a and b connect the two boundary components of Sc.
W.L.O.G. assume that the end points of a map to the same point under the quotient
map q : Sc —-> S, and the end points of b map to the same point under q. By

Lemma 2.2, we have, (N, a, b) E (Sf,ao,bo) where N is a regular neighborhood of

43

a U b U 81 U 8g. Then, we have (q(N),q(a),q(b),q(81)) E“ (Mo,m1,mg,m3) where

Mo,m1,mg,m3 are as shown in 2.10.

It is easy to see that (Mo,m1,mg,m3) E“ (No,no,mo,zo), where No,no,rn0, 2,,
are as given in Lemma 2.8. Then, there exists a homeomorphism (p : (q(N),q(a),
q(b),q(61)) —> (Nanammzo). Let U = r"(m4),y = r"(ms),z = 90—10%) where
m4,m5,m6 are as shown in 2.10. Suppose that q(a) is isotopic to q(b) on S. Since
q(a) is disjoint from q(b), there exists an annulus having q(a) U q(b) as its boundary.
Since each of q(a) and q(b) intersects c transversely once, c cuts this annulus into
a band B on S, such that 88 Q a U b U 85c. But this is not possible since a is
not isotopic to b on Sc. Hence, q(a) is not isotopic to q(b) on S. Then, y and z
are essential circles on S. Since q(81) is equal to c on S, q(81) is an essential circle
on S. Since each of q(a) and q(b) intersects c transversely once, q(a) and q(b) are
essential circles on S. Since v intersects q(b) transversely once, u is also essential on
S. Note that d E A([q(81)]). Then, by using Lemma 2.8, we can find a’ E A([q(a)]),
b’ E A([q(b)]), N’ Q S and a homeomorphism X : (N’,a’,b’,d) ——> (N, nmmo, 20). If

we cut N’ along d, we get two disjoint arcs, a,b corresponding to a’, b’ respectively.

Since a connects 81 and 8g, the encoding simplex of [a] is {[c], 7} where 7 is the
class of a l-separating circle on S. There exists x E 7 such that x bounds a genus 1
subsurface Q and q(a) and c intersect transversely once on Q. Since i( [x], [e]) = 0,
i([x], [q(a)]) = O, and A is superinjective, we have i(A([x]), [d]) = O, i(A([x]), A([q(a)]))
= 0. Then we choose a representative x’ of A([x]) which is disjoint from a’ U d. By
Lemma 3.7, x’ is a 1-separating circle bounding a subsurface R containing a’ U (1.
Then, it is easy to see that {[x’], [d]} is the encoding simplex of [d]. Then, since
the encoding simplex of [d] is the image of the encoding simplex of [a] under A, and
both a and [a] are type 2, by the definition of A..., [d] is the image of [a] under A...
Similarly, [b] is the image of [b] under A... This shows that A.([a]) and A.([b]) have

44

disjoint representatives in and b respectively.

Case 2: Assume that a connects one boundary component of SC to itself and b
connects the other boundary component of Sc to itself. Then, since a and b are
disjoint, they are both type 1.1 arcs. W.L.O.G. we can assume that a connects 81
to itself and b connects 8g itself in such a way that q(8a) = q(8b) where q is the
quotient map q : Sc —> S. Then, since a U 81 is disjoint from b U 8g, we can find
disjoint regular neighborhoods, N1, of aU81 and Ng of bU8g on Sc. Then, Q(N1UN2)
is a 4—holed sphere. Let x, y, z,t be the boundary components of this subsurface s.t.
x,z,c are encoding circles for a on S and y,t,c are encoding circles for b on S.
Let h = q(a U b). Then, h and c intersect geometrically twice and algebraically zero

times in this subsurface as shown in 2.9, (i).

By using Lemma 2.7, we can see that there exist pairwise disjoint representa-
tives x’,y’, z’,t’, h’,d of A([x]), A([y]), A([z]), A([t]), A([h]), A([c]) respectively such that
x’, y’, z’, t’ bound a 4—holed sphere, and h’, d intersect geometrically twice and alge-

braically zero times in this subsurface as shown in 2.9, (ii).

The curve h’ corresponds to two essential disjoint arcs when we cut this four
holed sphere along d. As in case 1, by the definition of A..., these two disjoint arcs

are representatives for A...([a]) and (A.([b]) which proves the lemma for Case 2.

Case 3: Assume that a,b are unlinked, connecting 8,- to itself for some i = 1, 2.
By Lemma 2.3, there is a homeomorphism gb such that (Sf, a0, b0) ’-_‘-’¢, (N, a, b) where

N is a regular neighborhood of a U b U 8,- in Sc and a0, b0 are as shown in 2.11.

Since ho, x0, co and v0, yo, co are the boundary components of regular neighbor-
hoods of co U b0 and co U a0 on S} respectively, ¢(ho), ¢(xo) and ¢(’Uo),¢(yo)
are encoding circles for b and a on Sc respectively. Then, q(¢(vo)),q(¢(ho))

are essential circles on S where q : SC —> S is the quotient map. We have,

45

 

 

Figure 2.11. Arcs and their encoding circles, I

(S§,co,xo,yo,zo,ho,vo) g (N.¢(co),¢(xo),¢(yo),¢(zo),¢(ho),¢(vo)). Since a and b

are essential arcs on Sc, q(¢(xo)) and q(¢(yo)) are essential circles on S. q(¢(co))
is equal to c on S. So, q(q’)(co)) is an essential circle on S. Since a is not iso-
topic to b on Sc, q(¢(zo)) is an essential circle on S. So, we have that all of
Q(¢(vo)),Q(¢(ho)), Q(¢(Co)),(J(¢($o)),q(¢(yo)),¢1(¢(zo)), are essential circles on 5-
Note that d E A([qb(co)]). Then, by using Lemma 2.7, we can find x’ E A([¢(xo)]),y’ E
A([¢(yo)]), 2’ E A([¢(zo)]),h’ E A([<b(ho)]),v’ E A([¢(vo)]), N’ Q S and a homeomor-
phism X : (S},co,xo,yo,zo,ho,vo) ——> (N’,d,x’,y’,z’,h’,v’). Then, h’,x’ are encoding
circles for A.([b]) on S and v’, y’ are encoding circles for A.([a]) on S, and X(bo), x(ao)

are disjoint representatives for A,([b]) and A..([a]) respectively.

Case 4: Assume that a connects one boundary component of Sc to itself and b
connects the two boundary components of SC to each other. The proof is similar to

the proof of Case 3.

Case 5: Assume that a, b are linked, connecting 8,- to itself for some i = 1, 2. By
Lemma 2.5, there is a homeomorphism qb : (No, a0, be) —> (N, a, b) where N is a regu-
lar neighborhood of 8,-Uan in Sc and NO, a0, b0 are as in Figure 2.12. Since yo, zo, co

and mo, no, co are the boundary components of regular neighborhoods of a0 U co and

46

 

Figure 2.12. Arcs and their encoding circles, ll

be U co on No respectively, ¢(yo),(b(zo) and ¢(mo),¢(no) are encoding circles for a
and t on s. respectively. We have (News.res.¢<y.).¢(z.).¢(m.).¢(n.)) ext—r
(No,co,xo,yo,zo,mo,no). q(¢(co)) is equal to c on S. So, q(¢(co)) is an essential
circle on S. Since c is a nonseparating circle on S, q(¢(xo)) is an essential circle on
S. Since q(¢(yo)),q(¢(zo)) and q(¢(mo)),q(¢(no)) are encoding circles for essential
arcs a and b on S respectively, q(¢(yo)), q(¢(zo)), q(¢(mo)) and q(¢(no)) are essential
circles on S. Note that d E A([q(¢(co))]). Then, by using Lemma 2.8, we can choose
$2 6 )‘(lQ(¢(xo))l)1 y; E A([Q(¢(yo))l)r23 E A([Q(¢(62))l)rm2 E A([Q(¢(mo))l)tn2 E
A([q(¢(no))]), N’ Q S and a homeomorphism X : (No,co,xo,yo,zo,mo, no) —>
(N’,d.x2.y2,z;rmérn2)- Since q(¢(yo)).4(¢(zo)) and Q(¢(mo)),Q(¢(no)) are encod-
ing circles for a and b on S respectively, y;, z; and m’o,n:, are encoding circles for
A..([a]), A..([b]) on S respectively. Existence of X shows that A.([a]),A.([b]) have dis-
joint representatives. X(ao) and X(bo) are disjoint representatives for A.([a]),A.([b])

respectively.

We have shown in all the cases that if two vertices have disjoint representatives,

then A... sends them to two vertices which have disjoint representatives. Hence, A.

47

extends to a simplicial map A... : B(SC) —> B(Sd). I

Lemma 2.13 Let A : C(S) —> C(S) be a superinjective simplicial map. Then, A. :
B(SC) —+ B(Sd) is injective.

PROOF. It is enough to prove that A... is injective on the vertex set, V(Sc).
Let [i],[j] E V(Sc) such that A,([i]) = A..([j]) = [k]. Then, by the defini-
tion of A.., the type of [i] and [j] are the same. Assume they are both type
1.1. Let [x], [y] and [z], [t] be the essential encoding circles for [i] and [7] re-
spectively. Then, A([x]),A([y]) and A([z]),A([t]) are essential encoding vertices
of [k]. So, {A([x]),A([y])} = {A([z]),A([t])}. Then, since A is injective we get
{[x], [y]} = {[z], [t]}. This implies [i] = [j]. The cases, where [i], [j] are both type 1.2

or type 2 can be proven similar to the first case by using the injectivity of A. I

We have proven that A is an injective simplicial map which preserves the geometric
intersection O and 1 properties. Using these properties and following N.V.Ivanov’s
proof of his Theorem 1.1 [Ivl], it can be seen that A. agrees with a map, h., induced
by a homeomorphism h : Sc —> S; on a top dimensional simplex in B(Sc). Then,
by Lemma 2.1, it agrees with h. on B(Sc). Then, again by Ivanov’s proof, A agrees
with a map, g..., which is induced by a homeomorphism g : S —> S on C (S) This

proves Theorem A.

Remark: Note that at the end of the proof of Theorem 1.4 we appealed to Ivanov’s
proof. In his proof of Theorem 1.1 [Ivl], by using Lemma 1 in his paper and using some
homotopy theoretic results about the complex of curves on surfaces with boundary,
he shows that an automorphism of the complex of curves preserves the geometric
intersection 1 property if the surface has genus at least 2. He uses this property to
induce automorphisms on the complex of arcs using automorphisms of the complex

of curves and gets an element of the extended mapping class group which agrees with

48

the automorphism of the complex of curves. In this thesis, for closed surfaces of genus
at least 3, we prove that a superinjective simplicial map A preserves the geometric
intersection 1 property by using Ivanov’s Lemma 1 and using some elementary surface
topology arguments avoiding any homotopy theoretic results about the complex of
curves. These elementary surface topology arguments can be used to replace the
usage of the homotopy theoretic results about the complex of curves in Ivanov’s
proof to show that an automorphism of the complex curves preserves the geometric
intersection 1 property. Then we follow Ivanov’s ideas and use this property to get
an injective simplicial map of the complex of arcs of Sc (c is nonseparating) using a
superinjective simplicial map of the complex of curves of S and follow Ivanov to get

an element of the extended mapping class group that we want.

49

CHAPTER 3

Injective Homomorphisms of
Subgroups of Mapping Class

Groups

3.1 Introduction

In this chapter we assume that S is a closed, connected, orientable surface of genus at
least 3, and I" = ker(<p), where p : M odg —> Aut(H1(S, Z3)) is the homomorphism

defined by the action of homeomorphisms on the homology.

We study the injective homomorphisms of finite index subgroups of extended map-
ping class groups. First, we study the center of the centralizers of images of Dehn
twists under such homomorphisms. We use the classification of elements of the ex-
tended mapping class groups given by W.Thurston, and the properties of such injec-
tive homomorphisms to make some arguments about the rank of the center of these
centralizers. This helps us to see that an injective homomorphism from a finite index
subgroup of M odg to the whole group induces a superinjective simplicial map on the

complex of curves of the surface S. Then, by using the main result of chapter 2, we

50

prove that this injective homomorphism is induced by a homeomorphism of S and it

has a unique extension to an automorphism of M odg.

3.2 Injective Homomorphisms and Centralizers

For a group G and for subsets A, H Q G, where A Q H, the centralizer of A in H,

CH(A), and center of G, C (G), are defined as follows;

CH(A)={hEH:ha=ahVaEA}, C(G) ={zEGzzg=ng9EG}.
Lemma 3.1 Let H be a subgroup of a group G and let A Q H. Then
HflC(CG(A)) Q C(CH(A))-

PROOF. Let h E H O C(CG(A)). Then, h E H O CG(A). It is clear that
H O Cg(A) = CH(A). So, h E CH(A). Let k E CH(A). Then, k E 00(A) O H Q
Cg(A). Then, since h E C(CG(A)), h commutes with k. Therefore, h E C(CH(A)).

Lemma 3.2 Let K be a finite index subgroup of M odg. Let f : K —) M 0de be an
injective homomorphism. Let F = f‘1(I") O I". Let G be a free abelian subgroup of
F of rank 39 — 3. Ifa E G, then

rank C(Cpr(f(a))) 3 rank C(Cp(a)).

PROOF. Let A = f(G)flC(Cpr(f(a))) and B = (f(G),C(Cpr(f(a)))) be the group
generated by f(G) and C(Cpr(f(a))). Since a E G and G is abelian, f(a) E f(G)
and f(G) is abelian. Then, f(G) Q Cpr(f(a)) since f(G) Q I". Then, B is an abelian
group. Since rank G = 3g - 3 and f is injective on G, rank f (G) = 39 — 3. The
maximal rank of an abelian subgroup of M 0d}; is 3g — 3, [BLM]. So, since f (G) Q B

and B is an abelian group and rank f(G) = 39 — 3 we have rank B = 39 — 3.

51

We have the following exact sequence,
1—+ A —-> f(G) EB C(Cpr(f(a))) —> B —> 1
This gives us
rank f(G)+ rank C(Cpr(f(a))) 2 rank A+ rank B.
Since rank f(G) 2 rank B, we get
rank C(Cpr(f(a))) = rank A.

Then, using Lemma 3.1, we get A = f(G)flC(Cpr(f(a))) Q f(I‘)flC(Cpr(f(a))) Q
C(Cf([‘)(f(a))). Since f is injective on K, C(Cf([‘)(f(a))) is isomorphic to C(Cp(a)).

So, we get rank A 3 rank C (Cp(a)). Then, by the above equation we have

rank C(Cpr(f(a))) S rank C(Cp(a)).

Lemma 3.3 Let F be a finite index subgroup of M 0d} and P Q I". Then, 3N E Z"

s.t. Va E A, C(Cp(t£,")) is an infinite cyclic subgroup of (ta).

PROOF. Let N E Z" such that tf," E I‘ for all a E A. Let 01,8 E A such that
i(a,fi) = 0. Let f E C(Cp(tfi,")). Then, f(a) = a. We want to show that f(fi) = B.
Since i(a, B) = O, to, commutes with t5. Then, ti," commutes with tfg" . Then, since
ti," E F, tfg" E Cp(t£’). Since f E C(Cp(t,’,")), f commutes with tfa”. Then, f(5) = [3.
Since f fixes the isotopy class of every circle which has geometric intersection zero
with a, the reduction, f, of f along a ([BLM]) fixes the isotopy class of every circle
on 3,, where a E [(1]. Since P Q I", f E I". Hence, by Lemma 1.6 [Iv2], f restricts
to each component Q of Sc, and the restriction of f to Q is either trivial or infinite

order. By Lemma 5.1 [IMc] and Lemma 5.2 [IMc], this restriction is finite order.

52

Hence, it is trivial. Then, f = t; for some r E Z, [BLM]. So, C(Cp(tf,")) Q (ta).
Since ti," E C (Cp(t£’ )), C (Cp(tfi," )) is a nontrivial subgroup of an infinite cyclic group.

Hence, it is infinite cyclic. I

Lemma 3.4 Let K be a finite index subgroup of Modg and f : K —> Mod; be an

injective homomorphism. Let a E A. Then there exists N E Z" such that
rank C(Cpr(f(t,’,"))) = 1.

PROOF. Let I‘ = f“1(I") OI". Let a E A, N E Z" such that ti," E I‘. Since

(f(tf» Q C(Cpr(f(tf,"))), and (f(tN)) is an infinite cyclic group we have

a

rank C(Cp,(f(tfi,"))) Z 1-

By Lemma 3.3, rank C (Cp(t,’,” )) = 1. Since ti," is in a free abelian subgroup of I‘
of rank 39 — 3, by Lemma 3.2, rank C(Cpr(f(t,’,"))) 5 rank C(Cp(t£,")). So, we get

rank C(Cpr(f(t,’,"))) S 1.
The inequalities given above imply that rank C (Cpr( f (ti,V ))) = 1. I

Lemma 3.5 Let K be a finite index subgroup of M odg. Let f : K —> M odg be an
injective homomorphism. Then there exists N E Z‘ such that f (t5 ) is a reducible

element of infinite order for all a E A.

PROOF. Let N E Z" such that ti," E K for all a E A. Since f is an injective
homomorphism on K and ti," is an infinite order element of K, f (t5 ) is an infinite
order element of M odg. So, f (t,’,” ) is either a reducible element or p—Anosov. Suppose
it is p—Anosov. Let J be a maximal system of circles containing a where [a] = a
and J’ be the set of isotopy classes of these circles. Let TJ' be. the subgroup of

M odg generated by tg’ VB E J’. T J! is a free abelian subgroup of K and it has

53

rank 39 — 3. Since f is an injection f (T Jr) is also a free abelian subgroup of rank
39 — 3. It contains f (if,” ) By [Mc], CMod§( f (if," )) is a virtually infinite cyclic group
and f(TJI) Q CMod§(f(t£,")). Then 39 — 3 S 1. This gives us 39 _<_ 4. But this is a
contradiction to the assumption that 9 2 3. Hence, f (t5 ) is a reducible element of

infinite order. I

3.3 Injective Homomorphisms of Finite Index
Subgroups

Lemma 3.6 Let K be a finite index subgroup of Modg, and f : K —> Modg be
an injective homomorphism. Then Va E A, f(tg) = tiles) for some M,N E Z“,

8(a) E A.

PROOF. Let I’ = f ”(B ) O I". Since F is a finite index subgroup we can choose
N E 2" such that ti," E I‘ for all a E A. By Lemma 3.5, f(t,’,") is a reducible
element of infinite order in M 0d}. Let C be a realization of the canonical reduction
system of f (if,V ) Let c be the number of components of C and p be the number of
p—Anosov components of f (t: ) Since t,’," E P, f (ti,V ) E I". By Theorem 5.9 [IMc],
C(Cpr(f(tf,"))) is a free abelian group of rank c+p. By Lemma 3.4, c+p = 1. Then,
either c = 1, p = O or c = 0, p = 1. Since there is at least one curve in the canonical

M

reduction system we have c = 1, p = 0. Hence, since f (ti,V ) E I", f (ti,v ) = tfi(a) for

some M E Z", B(a) E A, [BLM], [Iv2]. I

Remark: Suppose that f(tg”) = t; for some 5 E A and M,P E Z" and
f(té") = t? for some 7 E A and N,Q E Z“. Since f(tf’”) = f(tg'M), tg’" = t2)“,
P,Q, M, N E Z". Then, 8 = 7 by Lemma 1.1. Therefore, by Lemma 3.6, f gives a

correspondence between isotopy classes of circles and f induces a map, f... : A —> A,

54

where f,(a) = B(a).

Lemma 3.7 Let K be a finite index subgroup of Modg. Let f : K —> Modfg be
an injective homomorphism. Assume that there exists N E Z“ such that Va E A,

3Q E Z" such that t,’," = t2. Then, f is the identity on K.

PROOF. We use Ivanov’s trick to see that f (ktN k") = f (t;(( a) N): tfijf ’ and
f(ktNk‘1)= f(k)f(tN)f(k)“ = f(k)t3f(k)-1 _. —t;(({j,’°2,9 Va 6 A, Vk e K Then
we have taggk’— — t;(([,()’:”)Q => k(a) = f(k)(a) Va E A, Vk E K by Lemma 1.1. Then,
k"f(k)(a) = a Va E A, Vk E K. Then, k‘1f(k) commutes with to, Va E A,
Vk E K. Since Mod‘S is generated by Dehn twists, k‘1f(k) E C(Modg) Vk E K.
Since the genus of S is at least 3, C(Modg) is trivial. So, k = f (k) Vk E K. Hence,

f=idK. I

Corollary 3.8 Let g : M odg ——> M 0de be an isomorphism and h : M 0d; —) M odg
be an injective homomorphism. Assume that there exists N E Z" such that Va E A,

3Q E Z“ such that h(t,’,") = 9(t2). Then 9 = h.

PROOF. Apply Lemma 3.7 to g'lh with K = Modg. Since for all a in A,

94’1“?) = t3, we have 9’1h 2 Mg. Hence, 9 = h. I

3.4 Induced Map On Complex Of Curves

By the remark after Lemma 3.6, we have that f : K ——> M 0d; induces a map
f... : A —> A. We prove the following lemma which shows that f. is a superinjective

simplicial map on C (S).

Lemma 3.9 Let f : K —+ Mod} be an injection. Let a, 5 E A. Then,

55

liar/3) = 0 Er i(ft(a)rfr(i3)) = 0-

PROOF. There exists N E Z" such that t,’," E K and t’fi” E K. Then
we have the following: i(a,fi) = 0 (it tftg’ = tfgvté," (by Theorem 1.2) 4:)
f(tf,")f(tg’) = f(tg’)f(tfi,") (since f is injective on K) (I) t£(,,)t?,(m = ti(fl)tf:(a)
where P = M(a1,N),Q = M(ag,N) E Z" (by Lemma 3.6) 4:) i(f.(a),f.(fl)) = 0
(by Theorem 1.2). I

Now, we prove the second main theorem of the thesis.

Theorem 3.10 Let f be an injective homomorphism, f : K —-> Modg, then f is
induced by a homeomorphism of the surface S and f has a unique extension to an

automorphism of M odg.

PROOF. By Lemma 3.9, f. is a superinjective simplicial map on C (S) Then, by
Theorem A, f. is induced by a homeomorphism h : S —-> S, i.e. f..(a) = H (a) for
all a in A, where H = [h]. Let XH : M 0d,; —+ M 0d} be the isomorphism defined by
the rule XH(K) = HKH‘1 for all K in Modg. Then for all a in A, we have the

following:

-1 _ -1 _ H-1 M _ _ _ M-c(H"1) __ M-e(H’1)
XH °f(’°=N)—XH (flied—X “Heal—H ltiir’(or)17’—fir-Iona»— 0‘ '

Then, since X”’—1 o f is injective, X”_1 o f = idK by Lemma 3.7. So, XHlK = f.
Hence, f is the restriction of an isomorphism which is conjugation by H, (i.e. f is

induced by h).

Suppose that there exists an automorphism r : M 0d; —> M od'g such that TIK = f.
Let N E Z“ such that ti," E K for all a in A. Since XHIK = f = T]K and ti," E K,
r(t’") = X”(t,’,") for all a in A. Then, by Corollary 3.8, r = XH- Hence, the

0:

extension of f is unique. I

56

BIBLIOGRAPHY

57

BIBLIOGRAPHY

[BLM] J .S.Birman, A.Lubotzky, J.D.McCarthy, Abelian and Solvable Subgroups of

the Mapping Class Group, Duke Mathematical Journal, V.50, No.4 (1983).

[FLP] A.Fathi, F.Laudenbach, V.Poenaru, Travaux de Thurston sur les surfaces,

[H]

[Ivl]

[Iv2]

[IMc]

[Sp]

Seminaire Orsay, Asterisque, Vol.66-67, Soc. Math. de France, 1979.

W.J.Harvey, Geometric structures of surface mapping class groups, Homolog-
ical Group Theory (C.T. Wall, ed), London Math. Soc. Lecture Notes, No.36,
Cambridge Univ. Press, London, 1979, 255-269.

N .V.Ivanov, Automorphisms of Complexes of Curves and of Teichmuller
Spaces, International Mathematics Research Notices, No:14 (1997), 651-666.

N.V.Ivanov, Subgroups of Teichmuller Modular Groups, Translations of Math-
ematical Monographs, Vol. 115, AMS, 1992.

N.V.Ivanov, J .D.McCarthy, On Injective Homomorphisms between Teich-
muller Modular Groups I, Inventiones Mathematics Research Notices 135,
1999, 425-486.

M.Korkmaz, Automorphisms of complexes of curves on punctured spheres and

on punctured tori. Topology and its Applications 95 no. 2 (1999), 85-111.

J .D.McCarthy, Normalizers and centralizers of pseudo-Anosov mapping
classes. The manuscript is available for informal distribution, on request.

L.Mosher, Tiling the Projective Foliation Space of a Punctured Surface, Trans-
actions of the American Mathematical Society Volume 306, Number 1, March
1988.

E.Spanier, Algebraic Topology, 2nd ed., Springer Verlag, New York, 1990.

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