combinatorial topology and group theorybje1/gt_groups.pdf · combinatorial complexes of course, one...

Brent Everitt

Combinatorial Topology and GroupTheory

– Monograph –

October 15, 2008

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Contents

1 Combinatorial Complexes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 1-Complexes (ie: Graphs) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 1

1.1.1 The category of graphs . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 11.1.2 Quotients and subgraphs . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 31.1.3 Balls, spheres, paths and homotopies . . . . . . . . . . . . . .. . . . . . . . . . . . 41.1.4 Forests and Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 6

1.2 The category of2-complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2.1 2-complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 81.2.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 91.2.3 Maps of2-complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.4 Homotopies and homeomorphisms . . . . . . . . . . . . . . . . . . .. . . . . . . . 12

1.3 Aside: comparision with “proper” topology . . . . . . . . . . . . . . . . . . .. . . . . . . . 141.3.1 The category of CW complexes . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 141.3.2 A functor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 14

1.4 Quotients of2-complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 141.4.1 Quotients in general . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 141.4.2 Group actions and their quotients . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 151.4.3 Quotients by a subcomplex . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 161.4.4 Pushouts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 17

1.5 Pullbacks and Higman composition . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 211.5.1 Pullbacks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 211.5.2 Higman composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 24

1.6 Notes on Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 24

2 Topological Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.1 Products of paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 272.2 The fundamental group . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 282.3 Maps induce homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 292.4 π1 and quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 302.5 Cores . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 322.6 π1 is a topological invariant . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 332.7 Homomorphisms (almost) induce maps . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 332.8 Interlude on Abelian groups . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 342.9 Homology of graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 352.10 The rank of graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 362.11 Homology of 2-complexes . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 372.12 *Cohomology of 2-complexes . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 392.13 Notes on Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 39

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VI Contents

3 Coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 413.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 41

3.1.1 Coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 413.1.2 Lifting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 433.1.3 Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 46

3.2 Actions, intermediate and universal covers . . . . . . . . . .. . . . . . . . . . . . . . . . . 483.2.1 Group actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 483.2.2 Intermediate covers . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 493.2.3 Universal covers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 503.2.4 Monodromy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . 53

3.3 Operations on coverings . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 543.3.1 Pushouts of covers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 543.3.2 Pullbacks of covers . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 553.3.3 Higman compositions of covers . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 56

3.4 Lattices of covers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 563.4.1 Aside: posets and lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 563.4.2 The poset of intermediate covers . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 593.4.3 The lattice of intermediate covers . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 60

3.5 Notes on Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 62

4 Galois theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 634.1 Galois groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 63

4.1.1 Automorphisms and Galois groups . . . . . . . . . . . . . . . . . .. . . . . . . . . 634.1.2 Constructing automorphisms. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 644.1.3 From covers to subgroups . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 664.1.4 From subgroups to covers . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 66

4.2 Galois covers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 664.2.1 Galois covers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 664.2.2 Galois covers and the fundamental group . . . . . . . . . . . .. . . . . . . . . . 684.2.3 An inverse Galois theorem . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 68

4.3 Galois correspondences . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 694.3.1 The Galois correspondence . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 694.3.2 Galois correspondence for the universal cover . . . . . .. . . . . . . . . . . . 704.3.3 Lattice excision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 70

4.4 Notes on Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 71

5 Generators and Relations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 735.1 Generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 735.2 Schreier generators forπ1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 735.3 Presentations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 735.4 The substitution theorem . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 745.5 Presentation forπ1in terms of Schreier generators . . . . . . . . . . . . . . . . . . . . . 755.6 The presentation 2-complex . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 775.7 Handles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 785.8 Notes on Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 79

6 Graphs and free groups. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 816.1 Free groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 816.2 Rank of a free group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 826.3 Graphs of finite rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 836.4 Free presentations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 846.5 Interlude: categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 846.6 Free objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 846.7 Nielsen transformations: finite rank versus finite generation . . . . . . . . . . . . . 85

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Contents VII

6.8 Four definitions of free group . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 866.9 Interlude: thep-adic numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 876.10 Interlude: Lie groups. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 876.11 Ihara’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 876.12 Notes on Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 88

7 Coverings and groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 897.1 Regular trees and the Cayley complex of a free group . . . . .. . . . . . . . . . . . . 897.2 Nielsen-Schreier and the ranks of subgroups of free groups. . . . . . . . . . . . . . 907.3 Normal forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 927.4 Local properties and Marshall Hall’s theorem . . . . . . . . .. . . . . . . . . . . . . . . . 927.5 Some theorems of Greenberg . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 937.6 Howsen’s theorem and the Hanna Neumann conjecture . . . . .. . . . . . . . . . . . 947.7 The Reidemeister-Schreier theorem . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 967.8 Cayley complexes of quotients . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 967.9 Monodromy representations and Miller’s theorem . . . . . .. . . . . . . . . . . . . . . 977.10 Notes on Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 103

8 Amalgams. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 1058.1 Type I amalgams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 1058.2 Type II amalgams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 1088.3 Coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . 1108.4 The Kurosh theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 1118.5 title? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 1118.6 Graphs of groups and their amalgams . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 1158.7 Virtually free groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 1178.8 Notes on Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 123

9 Serre’s Arboreal Dictionary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1259.1 Representation varieties . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . 1279.2 Notes on Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 127

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1

Combinatorial Complexes

Of course, one has to face the question, what is the good categoryof spaces in which to do homotopy theory? –John Frank Adams.

For us the question is: “what is the good category in which to do combinatorial grouptheory?”. This chapter introduces it, and studies some of the important constructions onecan do in it.

1.1 1-Complexes (ie: Graphs)

1.1.1 The category of graphs

Definition 1.1 (1-complex: first go).A 1-complexor graph is a non-empty setX with aninvolutary map−1 : X → X and an idempotents : X → X0, whereX0 is the set of fixedpoints of−1.

Thus a graph has0-cells orverticesX0 and1-cells oredgesX1 = X \X0. One says thatthe edgee ∈ X1 hasstart vertexs(e) andterminal vertexs(e−1) and thinks of theinverseedgee−1 as juste, but traversed in the reverse direction (or with the reverseorientation).The edgee is incidentwith the vertexv if e ∈ s−1(v). We draw pictures like Figure 1.1, but

s(e)

s(e−1)

e

Fig. 1.1.edge of a graph

it is important to keep in mind that they are purely for illustrative purposes. For instance,if the vertex set has cardinality that of the power set of the continuum, then there are notenough points on a piece of paper for a picture to fit. Definition 1.1 is a little terse, andsometimes it is useful to spell it out a little more:

Definition 1.2 (1-complex: second go).A 1-complexor graphX is composed of two dis-joint non-empty setsX0 andX1, together with two incidence maps and an inverse map,

s, t : X → X0 and −1 : X → X,

such that, (i).s(v) = v for all v ∈ X0; (ii). v−1 = v for all v ∈ X0, e−1 ∈ X1 ande−1 6= e = (e−1)−1 for all e ∈ X1; (iii). t(x) = s(x−1) for all x ∈ X .

More terminology: anarc is an edge/inverse edge pair, and anorientationO for X , is asetO of edges containing exactly one edge from each arc. Writee for the arc containing the


2 1 Combinatorial Complexes

edgee (so thate−1 = e). The graphX is finitewhenX0 is finite andlocally finitewhen thesets−1(v) is finite for everyv ∈ X0. Thus a finite graph may have infinitely many edges, asituation that possibly differs from that in combinatorics. The cardinality of the sets−1(v)is thevalencyof the vertexv. A pointed graphis a pairXv := (X, v) for v ∈ X a vertex.

Exercise 1.3 (1-complex: third go). Here is one more, recursive definition of graph thatmakes clearer the connection with the2-complexes of§1.2. Firstly, a0-complex is a setX ;a map of0-complexes is a mapf : X → Y of sets; a pointed0-complex is a pair(X, v)consisting of a0-complexX andv ∈ X , and a map of pointed0-complexesf : (X, v)→(Y, u) is a set mapf : X → Y with f(v) = u; finally, the0-sphereS0 is the set withtwoelements.

With these preliminaries out of the way, a graphX is a graded setX = X0, X1 withXk 6= ∅, such that

(C1). there is an involutory map−1 : X → X with fixed point setX0;(C2). X0 is 0-complex;(C3). eachα ∈ X1 hasboundary∂α = (eα, fα) with eα the0-sphereS0 andfα : eα

v →

X(0)u a pointed mapping of0-complexes.

Here is the exercise: show that definitions 1.1, 1.2 and this version are all equivalent.

The trivial graph has a single vertex and no edges. Figure 1.2 shows some more exam-ples of graphs, including some with countably many edges. These will tend to be moreinteresting than finite graphs.

. . .

Fig. 1.2.Examples of graphs: theS1 graph at left and two graphs with a countably infinite numberof vertices and edges.

We want graphs to form the objects of a category, so amap of graphs is a set mapf : X → Y with f(X0) ⊆ Y 0, such that the diagram on the left of Figure 1.3 commutes,whereσX is one of the mapssX or −1 for X , andσY similarly, ie: fsX(x) = sY f(x)andf(x−1) = f(x)−1. Notice that a map is allowed to squash edges down to vertices: we

X Y

X Y

σX σY

f

f sX(e)

tX(e)

ef(e)

Fig. 1.3.graph mappingf : X → Y

call a mapdimension preservingif in addition to the above we havef(X1) ⊆ Y 1. A mapf : Xv → Yu of pointed graphs is a graph mapf : X → Y with f(v) = u.

The commuting off with s and−1 is meant to be a combinatorial version of continuity.Note in particular that ifx is a vertex thenfsX(x) = sY f(x) is vacuous, but ifx is an edgemapped byf to a vertex as on the right in Figure 1.3, then the condition becomesfsX = f ,


1.1 1-Complexes (ie: Graphs) 3

so thatf(e) is the image ofsX(e); the condition also ensures thattX(e) is mapped tof(e),and not left “hanging”.

For a fixed vertexv ∈ X0, ands−1X (v) ∈ X1 the edges with start vertexv, it is easy to

see that a mapf : X → Y induces a maps−1X (v)→ s−1

Y (f(v)) to the set of edges startingatf(v). We call this thelocal continuityof f at the vertexv.

A mapf : X → Y is an isomorphismif it is dimension preserving and a bijection onthe vertex and edge sets. WriteX ∼= Y .

Exercise 1.4.Show that iff : X → Y is an isomorphism then the mapf−1 : Y → X ,which is the inverse off on Y 0 andY 1, is also a graph isomorphism. Show that the setAut(X) of graph isomorphismsX → X forms a group under composition.

A group action is a homomorphismGϕ→ Aut(X), and we abbreviateϕ(g)(x) to g(x).

It preserves orientationif there is an orientationO for X with g(O) = O for all g ∈ G.

Exercise 1.5.An inversionof a graphX is a mapf : X → X which interchanges theedges of some arc, ie:f(e) = e−1 for somee ∈ X1. A group action is said to bewithoutinversionsiff no g ∈ G acts as an inversion onX . Show that an action by a groupGpreserves orientation if and only if it acts without inversions.

A groupG actsfreely iff the action is free on the vertices, ie: for anyg ∈ G andv avertex,g(v) = v implies thatg is the identity element.

Exercise 1.6.If G acts freely and orientation preservingly on a graph, then show that theaction is free on the edges too.

Graph isomorphisms are pretty rigid, and it is sometimes useful to have a relation witha bit more slack. Thus, asubdivisionof an edgee in a graphX replaces it by two newedges and a new vertex as in Figure 1.4, or is the reverse of this process. WriteX ↔ X ′

when two graphs differ by the subdivision of a single edge. Two graphsX andY are then

v1

v2

e

v1

v2

v′

e1

e2

Fig. 1.4.Subdividing an edge

homeomorphic, writtenX ≈ Y , when there is a finite sequenceX = X0 ↔ X1 ↔ · · · ↔Xk = Y of subdivisions connecting them. It is easy to see that homeomorphism is anequivalence relation for graphs. A property that is invariant under homeomorphism, ie: ifX ≈ Y are homeomorphic, thenX has the property if and only ifY does, is called atopological invariant.

1.1.2 Quotients and subgraphs

The most useful construction in the category of graphs is thequotient:

Definition 1.7 (quotient relations and quotient graphs).If X is a graph, then aquotientrelation is an equivalence relation∼ onX such that

(i). x ∼ y ⇒ s(x) ∼ s(y) andx−1 ∼ y−1 (ii) . x ∼ x−1 ⇒ [x] ∩X0 6= ∅,

where[x] is the equivalence class under∼ of x. If ∼ is a quotient relation on a graphXthen definesX/∼ and−1 on the set of equivalence classesX/∼ by

(i). sX/∼[x] = [sX(x)] (ii) . [x]−1 = [x−1].



Notice that edges can be equivalent to vertices, but if an edge is equivalent to its inversethen it must also be equivalent to a vertex. This ensures that[e] 6= [e]−1 in the quotient.

Proposition 1.8.If ∼ is a quotient relation thenX/∼ with the mapssX/∼ and−1 definedabove is a graph, and the quotient mapq : X → X/∼ given byq(x) = [x] is a map ofgraphs.

The proof is a straight forward exercise. In particular, thefixed points inX/∼ of the new−1 map are precisely those equivalence classes[x] wherex ∼ v for somev ∈ X0. Thusthe quotient has vertices the[v] for v ∈ X0 (and these classes may well include some ofthe edges of the old graphX) and edges those[e] with [e] ∩X0 = ∅.

The two main examples of graph quotients arise by factoring out the action of a group,or by squashing a subgraph down to a vertex. For the first we have the following,

Proposition 1.9.Let ∼ be the equivalence relation onX given by the orbits of a groupaction. Then∼ is a quotient relation if and only if the group action is orientation preserving,and we writeX/G := X/∼ for the quotient of the graph by the action of the group.

A subgraphis a graph mappingX → Y that is an isomorphism onto its image. Equiv-altently, it is a subsetX ⊂ Y , such that the mapss and−1 give a graph when restricted toX .

Let X → Y be a subgraph and define∼ on Y by x ∼ y iff x = y or bothx andylie in Y . Then this is a quotient relation and we writeY/X for Y/∼, the quotient ofYby the subgraphX . It is what results by squashingX to a vertex. Extending this a little, ifXi, (i ∈ I) is a family of disjoint subgraphs inY then define∼ by x ∼ y iff x = y or x andy lie in thesameXi, and writeY/Xi for the corresponding quotient. Note the differencebetween this, where eachXi has been squashed to a distinct vertexvi, andY/(

∐Xi),

where the whole union is squashed to just the one vertex.ThecoboundaryδX of a subgraphX → Y consists of those edgese ∈ Y with s(e) ∈ X

andt(e) 6∈ X .

Exercise 1.10.Let X1, X2 andY be graphs andfi : Y → Xi dimension preserving mapsof graphs. Let∼ on the disjoint unionX1

∐X2 be the equivalence relationgenerated by

thex ∼ y iff there is az ∈ Y with x = f1(z) andy = f2(z). Thus,x ∼ y iff there arex0, x1, . . . , xk with x0 = x andxk = y, and for eachj, there isz ∈ Y with xj = fi(z),xj+1 = fi+1 mod2(z). Show that∼ is a quotient relation if and only if there are orientationsO, Oi for Y, Xi with fi(O) ⊆ Oi.

1.1.3 Balls, spheres, paths and homotopies

Two particular graphs are theI1-graph andS1-graph as shown in Figure 1.5. A graphX is

I1 S1

Fig. 1.5.theI1 andS1-graphs.

then a1-ball either if it is the trivial graph or is homeomorphic toI1, and a1-sphereif it ishomeomorphic toS1.

It is easy to see that the vertices of a1-ball can be labelled asv0, . . . , vn and the edgesase±1 . . . , e±n with s(ei) = vi−1 ands(e−1

i ) = vi. Thus a1-ball hasend verticesv0, vn inan obvious sense. The following is also easily proved by induction:


1.1 1-Complexes (ie: Graphs) 5

Lemma 1.11.A graphX is a1-sphere if and only if eitherX = S1, or there are non-trivial1-ballsBi, (i = 1, 2) with end verticesvi1, vi2, such thatX = B1

∐B2/∼,

B1

B2X =

where the equivalence classes of∼ arev11, v21, v12, v22 and thex for all other cellsx ∈ X1

∐X2.

A path in X is a graph mappingB → X with B a 1-ball. It is convenient not to insistthat the map preserve dimension, but by Exercise 1.12 below,we can always replace the1-ball by another so that the map is dimension preserving if necessary. In any case, the imagein X is a sequence of edgese1 . . . eℓ, that are consecutively incident in the obvious way:s(e−1

i ) = s(ei+1), and there is no harm in thinking about paths in terms of theirimages.We will use both points of view interchangably as convenient. A path joins the verticess(e1), s(e

−1ℓ ) that are the images of the end vertices of the1-ball, and isclosedif these end

vertices have the same image.

Exercise 1.12.let f : B → X be a path with edges labellede±1 . . . , e±n as in the commentsbefore Lemma 1.11, and image edgese′1 . . . e′ℓ in X . Show there are1 ≤ i1 ≤ i1 ≤ · · · ≤iℓ ≤ n with f(eij

) = e′j and all other edges mapped to vertices. Thus,B can be replacedby a1-ball B′ and dimension preserving mapf ′ : B′ → X having the same image path.

Exercise 1.13.Show that a closed pathB → X is the same thing as a mappingS → Xfor someS ≈ S1.

If X → Y is a graph map andB → X a path, then there is an induced path inY givenby the compositionB → X → Y . Thus, a graph mapping sends paths to paths.

The graphX is connectedif any two vertices can be joined by a path. Theconnectedcomponentof X containing the vertexv consists of those vertices that can be joined tov,together with all their incident edges. A connected graph has finitely many edges if andonly if it is finite and locally finite.

A path e1 . . . eℓ contains aspur if ei+1 = e−1i for somei, ie: the path consecutively

traverses an edge and its inverse. Anelementary homotopyof a path,e1 . . . eiei+1 . . . eℓ ↔e1 . . . ei(ee

−1)ei+1 . . . eℓ inserts or deletes aspur as in Figure 1.6. Two paths arefreely

insert/delete

e1 . . . eiei+1 . . . eℓ e1 . . . ei(ee−1)ei+1 . . . eℓ

ei

ei+1

ei

ei+1

e

Fig. 1.6.elementary homotopy

homotopiciff there is a finite sequence of elementary homotopies taking one to the other.Paths homotopic to a trivial path consisting of a single vertex and no edges are said tobe homotopically trivial. We leave it as an exercise to formulate these notions for a pathB → X thought of as a mapping.

Exercise 1.14.Show that homotopic paths have the same start and end vertices, and thushomotopically trivial paths are necessarily closed. Show that homotopy is an equivalencerelation on the paths with common fixed endpoints.



1.1.4 Forests and Trees

A path in a graphX is reducedwhen it contains no spurs. By removing spurs we can ensurethat if there exists a path between two vertices, then there must exist a reduced path; indeedfor any two vertices, there is a reduced path between them if and only if they lie in the samecomponent.

Proposition 1.15.The following are equivalent for a graphX :

1. There is at most one reduced path joining any two vertices;2. any closed path is homotopically trivial;3. any non-trivial closed path contains a spur.

A graph satisfying any of the conditions of Proposition 1.15is called aforest, and aconnected forest is atree.

Proof. 1 ⇒ 2: a closed path is necessarily contained in a component of thegraph, thusby assumption there is a unique reduced path connecting any two of it’s vertices. The pathcannot contain just a single vertexu and edgee (which it circumnavigates some numberof times), for if so, then the edgee and the trivial path atu are distinct reduced paths fromu to u. Thus any closed path contains at least two distinct vertices. We show the path ishomotopic to the trivial path based at one of them, sayu. Let v be another vertex of thepath andw the reduced path running fromu to v. Then the path decomposes into two parts,w1 running fromu to v andw2 running fromv to u. If w1 6= w then it cannot be reduced,hence must contain a spur.Removing it and continuing, we have a series of homotopies thatreduces the pathw1 to the pathw as in the left of Figure 1.7. Similarly forw2 andw−1

u

vw1

whomotopies

u

v

w2

whomotopies

u

Fig. 1.7.homotoping a closed path to the trivial path

(which is the unique reduced path fromv to u) as on the right of Figure 1.7. Thus our pathis homotopic toww−1, which in turn can be reduced by the removal of spurs to the trivialpath based atu.

2⇒ 1: if u andv lie in different components then there are obviously no reduced pathsconnecting them. Otherwise, ifw1 andw2 are reduced paths running fromu to v then thepathw1w

−12 is homotopically trivial, hence contains a spur. As thewi are reduced, the spur

must be at the beginning or the end of the closed path, ie: involve the first edges ofw1

andw2, or the last edges as in Figure 1.8. Shifting attention to thereduced subpaths not

or

u

v

u

v

Fig. 1.8.uniqueness of reduced paths

involving this spur and continuing, we getw1 = w2. We leave the equivalence of parts 2and 3 as an Exercise.2


1.2 The category of2-complexes 7

Exercise 1.16.LetX be a finite graph, remembering that this only means that the vertex setX0 is finite. If each vertex has valency at least two, show thatX contains a homotopicallynon-trivial closed path. Deduce that ifT is a finite tree, then|T 1| = 2(|T 0| − 1).

We’ll have more to say about trees later. We finish this section by considering how toapproximate a graph by a tree: ifX is a connected graph, then aspanning treeis a subgraphT → X that is a tree and contains all the vertices ofX (ie: T 0 = X0). The followingexercise shows that under some mild set-theoretic assumptions, spanning trees always exist.

Exercise 1.17.Recall the well-ordering principle from set theory: any setX can be givenan order≤ (see Definition 3.40) so that for any elementsx, y ∈ X , eitherx ≤ y or y ≤ x,and for anyS ⊂ X there is as ∈ S with s ≤ x for all x ∈ X . In particular, the edgesetX1 of a graph can be well-ordered. Choose a basepoint vertexv0, and consider thosevertices at distance one fromv0, ie: thev 6= v0 with s(e) = v0, s(e−1) = v for some edgee. For each such, choose an edgeev that is minimal in the well-ordering amongst the edgesjoining v0 to v. Let T1 be the subgraph consisting ofv0, its distance1 neighbours and thechosen edges.

1. Show thatT1 is a tree. Continue the construction inductively: at stepk, take the treeTk−1 constructed at the end of stepk− 1, and for each vertexv of X a distance1 froma vertex ofTk−1, choose a minimal edge as above. LetTk be the subgraph consistingof Tk−1 together with the distance1 vertices and minimal edges. Show thatTk is atree.

2. Show thatT , the union of all theTk, is the required spanning tree.

In a graph the vertex and edge sets are arbitrary, so the edge set can have a wildly differentcardinality from the vertex set, causing difficulties with some arguments. This shortcomingis avoided by spanning trees, which have a number of edges that is roughly the same as thenumber of vertices of the graph they span:

Proposition 1.18.LetX be a connected graph andT → X a spanning tree. Then

|T 1| =

2|X0| − 1, if X is finite,|X0|, if X is infinite.

Proof. The result for finite graphs is the content of Exercise 1.16. If X is an infinite graphwith spanning treeT , then the edge set ofT must be infinite, as a finite edge set only spans|T 1| + 1 vertices by the first part. ThenX0 = T 0 =

⋃e∈T 1sT (e), sT (e−1) has the

cardinality that ofT 1. 2

Exercise 1.19.Let Ti → X be a family of mutually disjoint trees in a connected graphX .Show there is a spanning treeT → X containing theTi as subgraphs, and such thatq(T )is a spanning tree forX/Ti, whereq : X → X/Ti is the quotient map.

A spanning forestis a subgraphΦ → X that is a forest and contains all the vertices ofX . By consideringq−1(T ′) for some spanning treeT ′ of the (connected) graphX/Φ, showthat any spanning forest can be extended to a spanning tree.

1.2 The category of2-complexes

Now to the category of combinatorial2-complexes, where we mimic the constructions inthe previous section as much as possible.



1.2.1 2-complexes

Definition 1.20 (2-complex).A combinatorial2-complexX is a graded setX = X0, X1,X2 with Xk 6= ∅, and such that

(C1). if X(1) := X0∐

X1, there is an involutory map−1 : X(1) → X(1) with fixed pointsX0 and an idempotents : X(1) → X0, makingX(1) a graph;

(C2). eachσ ∈ X2 hasboundary∂σ = (Xσ, aσ) with Xσ ≈ S1 andaσ : Xσx → X

(1)v a

pointed dimension preserving map of graphs.

The elements of the setX2 are the2-cells orfacesof the complex with the pair(Xσ, aσ)the boundary of the faceσ ∈ X2. The mapaσ is theattaching mapof the face (see Figure1.9). The underlying graphX(1) is called the1-skeleton. The edges of a2-complex aredistinguished by the map−1, whereas the vertices and faces are not. Partly this helps withthe accounting later on, but it also reflects the motivating topology:1-spheres are connectedwhereas the0-sphere is not, so the−1 map allows us to choose a connected boundary foredges in some sense.

aσ

Xσ σ

v

x

γ

Fig. 1.9. Boundary of a face consisting of a1-sphere and an attaching map, which may wrap thesphere around the face several times. The vertexv appears in the boundary ofσ and the red pathwrapping twice around is a boundary path ofσ starting atv.

One thinks of a face as a disc attached to the1-skeleton with boundary a closed pathas in Figure 1.9, and again, although these pictures providetopological motivation, theydo not cary as much information as the definition. For example, the imageaσ(Xσ) ofa face boundary can be a path that wraps around the same set of edges several times.A example of this phenomenon is the “torus” complex below, and we will encounter animportant example later where a face has boundary withXσ having many edges, but theimageaσ(Xσ) a single loop.

The pointing of the attaching maps will prove to be essentiallater on, allowing us inparticular to distinguish faces using the1-skeleton. Callv = aσ(x) thedistinguished vertexof the faceσ.

We say that the vertexv appears in the boundaryof the faceσ whenever(aσ)−1(v) 6= ∅.Thus there is a vertexx of Xσ mapping tov via the attaching mapaσ, and indeed theremay be several of them. We will call the vertices in(aσ)−1(v) ⊂ Xσ the appearences ofvin the boundary of the faceσ. For such a vertexx, if we take a pathγ circumnavigating the1-sphereXσ in some direction as shown in Figure 1.9, then we call its image aboundarypath ofσ starting atv. The vertexv appears a total of|(aσ)−1(v)| times in the boundaryof σ, and each appearence gives rise to a pair of boundary paths starting atv.

Exercise 1.21.Formulate an equivalent version of definition 1.20 so that instead of the faceattaching maps, the boundary of a face is a closed pathe1 . . . eℓ with a distinguished vertexin the underlying graph.

Many of the concepts and adjectives of graphs can be applied directly to2-complexes byapplying them to the1-skeleton. Thus, we have the obvious notions of a finite2-complex,connected2-complexes, paths in2-complexes, etc.



1.2.2 Examples

Figure 1.10 gives three different versions of a2-complex that we will call the2-sphere.The first version (top left) is a straight pictorial version of definition 1.20 in this case: the1-skeleton in the middle is a graph with two vertices and two edges; meanwhile there aretwo faces with the attaching maps shown, and each face has distinguished vertex one of thevertices inX(1). This version is both the most accurate and the most cumbersome.

In the second version (bottom left) we have adopted the convention that parts of thecomplex with the same label give the same cell and drawn the complex “face-centrically”with the faces thought of as discs sewn onto the1-skeleton. Carrying out the identificationssuggested by this picture gives the third version on the right.

Xσ1

X(1)

Xσ2aσ1 aσ2

e2

e1

e−11

e−12

v1 v2

e1

e2

v1 v1v2 v2

e1

e2

σ1

e2

e1

σ2

σ1

σ2

v1

v2

e1

e2

Fig. 1.10.2-sphere: pictorial version of definition 1.20 (top left); face-centric version with faces sewnon (bottom left) and topologically suggestive version (right).

Xσ1

X(1)

Xσ2aσ1 aσ2

e2

e1

e−11

e−12

v1 v2

e1

e2

Fig. 1.11.Not the2-sphere

Figure 1.11 gives a very similar complex, differing only in that both faces now havethe same distinguished vertex. For reasons that are a littleobscure at the moment, we verymuch prefer the complex of Figure 1.10 to that of Figure 1.11.For now we will contentourselves with the following comparision: both complexes have the same1-skeleton, andonce maps of2-complexes are defined in§1.2.3, we will see that the only isomorphismof Figure 1.10 that restricts to the identity on the1-skeleton is the identity isomorphism.In Figure 1.11 on the other hand, there will be a non-trivial isomorphism restricting to theidentity on the1-skeleton. This distinction will have ramifications in Chapter 4.

We call Figure 1.12 the (real) projective planeRP2, a combinatorial model for the disc

with antipodal points on the boundary identified. We have again drawn the complex bothface-centrically andala Definition 1.20. Similarly, Figure 1.13 shows various versions ofthe torus complex.

1.2.3 Maps of2-complexes

To complete the definition of the category of2-complexes we need to define maps betweenthem. The principle is the same as that for maps of graphs in§1.1.1: continuity is captured



σ

v

v

e

e

X(1)

Xσaσ

ve

e

e

Fig. 1.12.Projective plane complex

Xσe1 e−11

e2

e−12

ve1 e2

σ

v

v

v

v

e1 e1

e2

e2

Fig. 1.13.Various versions of the torus (distinguished vertices leftoff)

by making the presumptive map commute with the various attaching maps of the cells. Thedefinition is complicated by the fact that a map may squash a face down to a path in the1-skeleton.

Definition 1.22 (maps of2-complexes).A mapf : X → Y of 2-complexes is a mapf : X(1) → Y (1) of the underlying graphs, and for each faceσ ∈ X2 we have eitherf(σ)is a faceτ ∈ Y 2 or f(σ) is a closed path in the graphY (1) such that,

1. if ∂σ = (Xσ, aσ) with aσ : Xσu → X

(1)v , andf(σ) = τ a face with∂τ = (Y τ , aτ )

andaτ : Y τx → Y

(1)w , then there is anisomorphismεσ : Xσ

u → Y τx making the diagram

below left commute;

Xσu Y τ

x

X(1)v Y

(1)w

aσ aτ

∼=

f

Xσu Sx

X(1)v Y

(1)w

aσ

f

2. if f(σ) is the closed pathS → Y with S ≈ S1, then there is a mapXσu → Sx making

the diagram above right commute, and moreover, the pathS → X is homotopicallytrivial.

The first part is just saying that when a faceσ is mapped to a faceτ , then the attach-ing mapaτ of τ is the compositionY τ → Xσ → X(1) → Y (1), going anti-clockwisearound the diagram. Alternatively, replacingY τ by the isomorphic copyXσ, τ has bound-ary ∂f(σ) = (Xσ, faσ), with faσ : Xσ

u → X(1)v → Y

(1)f(v) a pointed map of graphs as in

Figure 1.14. Iff(σ) a path, then the edges forming the boundaryaσ(Xσ) of σ are mappedby f to edges of this path.

With Exercise 1.21 in mind, it is an easy exercise to reformulate Definition 1.22 as fol-lows: if f(σ) is a faceτ , then the closed path forming the boundary ofσ is mapped viafto the closed path forming the boundary ofτ .



Xσ σaσ f

f(σ)

Fig. 1.14.Composing an attaching map with a map of complexes gives the attaching map for theimage of a face.

Thus, the boundaries of faces are mapped to the boundaries offaces, and the distin-guished vertex of a face is mapped to the distinguished vertex of the image face. If a face ismapped to a path in the1-skeleton, then this path must be the image of the boundary pathof the face (hence closed) and is homotopically trivial. Themotivating example is squash-ing a face flat: the boundary gets squashed too, into a path of the forme1 . . . eℓe

−1ℓ . . . e−1

1 .In this example the result is clearly homotopically trivial, but we will really require thiscondition in Chapter 2 for the fundamental groupπ1 to be a functor.

We now establish a number of conventions regarding maps thatwill hold throughout thisbook. They all involve the face isomorphisms that appear in part 1 of Definition 1.22. Thefirst relates to the question, “when are two maps the same?”. Suppose thatX andY are2-complexes and

f, g : X → Y,

maps of2-complexes. Thenf andg are identical whenf(x) = g(x) for all cells x ∈X (obviously),and if f(σ) = g(σ) = τ for facesσ ∈ X andτ ∈ Y , the associatedisomorphismsεσ andε′σ : Xσ → Y τ of Definition 1.22 part 1 areidentical.

Our next convention concerns the formation of compositions. Suppose that

Xf−→ Y

g−→ Z,

are maps of2-complexes. Then the compositiongf is formed in the usual way, with theadditional proviso that ifσ ∈ X is a face such thatgf(σ) is a face ofZ, the isomorphismXσ → Zgf(σ) is the composition of the isomorphisms,

Xσ ∼=−→ Y f(σ) ∼=

−→ Zgf(σ).

These two conventions have ramifications for commuting diagrams of complexes andmaps, which are after all, just statements about two maps being the same. For example,when we say that the diagram of maps and complexes on the left commutes,

X

Y1

Y2

Z

f1

f2

g1

g2

Xσ

Yf1(σ)1

Yf2(σ)2

Zgifi(σ)

∼=

∼=

∼=

∼=

then fori = 1, 2, the compositionsgifi are the same as the dotted map across the middle.If σ is a face ofX that maps to a facegifi(σ) of Z, then the diagram of isomorphisms onthe right must also commute.

Let f : X → Y be a map of2-complexes,u, v vertices withf(u) = v, andτ a facein Y . For a faceσ of X that maps toτ let εσ : Xσ → Xτ be the isomorphism makingthe diagram in Definition 1.22 part 1 commute. This isomorphism then induces a mapεσ : (aσ)−1(u) → (aτ )−1(v) between the appearences ofu in the boundary ofσ and theappearences ofv in the boundary ofτ . Since this is true for all theσ mapping toτ , we canformulate the,



Definition 1.23 (local continuity). If f : X → Y is a map of2-complexes,u, v verticeswith f(u) = v, andτ a face inY , then thelocal continuitymap associated to this triple is

∐εσ :

∐

f(σ)=τ

(aσ)−1(u)→ (aτ )−1(v),

where the disjoint union is over the facesσ mapping toτ , and∐

εσ is the disjoint union ofthe isomorphismsεσ.

An example is given in Figure 1.15 with the mapping of the “plane” complex to the torus.

Exercise 1.24.Show that if the right hand side of the set in Definition 1.23 isempty, thenso is the left hand side.

σ1

Xσ1

σ2

Xσ2

σ3

Xσ3

σ4

Xσ4

u

Y τ

f

Fig. 1.15. local continuity for a map of2-complexes: the inifinite plane complex on the left mapsvia f to the torus on the right;τ is the single face of the torus and for the single vertexv, the set(aτ )−1(v) ∈ Y τ consists of the four vertices marked with little red and bluecircles and squares. Afixed vertexu mapping tov appears in the boundary of the four facesσi with (fσi)−1(u) consistingof a single vertex in each case. For all other facesσ we have(aσ)−1(u) = ∅. The red and blue circlesand squares on the left map via the local continuity map of Definition 1.23 to the corresponding oneson the right.

A map isdimension preservingif the map of the underlying graphs preserves dimensionandf(X2) ⊂ Y 2. It is anisomorphismif it is dimension preserving, and a bijection on thevertex, edge and face sets. In this case one easily sees that,as for graphs, the inverse mapf−1 is also an isomorphismf−1 : Y → X (just reverse the horizontal arrows in the leftcommuting diagram of Definition 1.22(1)!) so that the set of automorphismsf : X → X

forms a group Aut(X) under composition. A group actionGϕ→ Aut(X) then preserves

orientation, is without inversion and free, when the underlaying map of graphs has theseproperties.

A subcomplexis a mappingX → Y of 2-complexes that is an isomorphism onto itsimage.

Exercise 1.25.Formulate an equivalent version of the definition of a subcomplexX of Y ,with the sets ofk-cellsXk of X as subsets of the sets ofk-cellsY k of Y .

1.2.4 Homotopies and homeomorphisms

As with graphs in§1.1.3 we can deform paths in a combinatorial manner, mimicing thecontinuous homotopies of paths in topology.

Let e1 . . . eℓ be a path in the2-complexX . An elementary homotopyeither inserts ordeletes a spur as in§1.1.3 or inserts or deletes the boundary of a face in the followingsense. Ifσ ∈ X2 is a face with boundary∂σ = (Xσ, aσ), then by Exercise 1.13,aσ(Xσ)



insert/delete

e1 . . . eiei+1 . . . eℓ e1 . . . ei(e′

j . . . e′ke′1 . . . e′j−1)ei+1 . . . eℓ

ei

ei+1

ei

ei+1

σ

Fig. 1.16.elementary homotopy–inserting/deleting the boundary of aface.

is a closed path, saye′1 . . . e′k. The homotopy then inserts into (or deletes from) the paththe result of completely traversing the closed path, starting at one of its vertices, so thatall the incidences match up in the obvious way, ie: so thats(e′j) = t(e′j−1) is the vertext(ei) = s(ei+1).

We make two remarks before proceeding. The first is that the pointing of the face attach-ing maps plays no role: ifaσ : Xσ

u → X(1)v is the attaching map then the vertexv lies in the

closed pathaσ(Xσ), but we do not insist that the traversal of the boundary startand finishat this vertex. The other is that pictures such as the right hand side of Figure 1.16 shouldbe approached with care. Theentireboundary path ofσ must be traversed, including anyrepetitions. Later we will have faces with boundary a closedpath that travelsn times, say,around an edge loop. Any homotopy involving this boundary must then travel the fullntime around the loop.

σ σ σ

Fig. 1.17.homotoping a path across a face

Two paths arehomotopicif and only if there is a finite sequence of elementary homo-topies, taking one to the other. A path homotopic to the trivial path based at one of thevertices it passes through is said to behomotopically trivial. For example, two paths run-ning different ways around a face are homotopic as shown in Figure 1.17. To get from thefirst picture to the second, insert the boundary of the face; to get from the second to thethird, remove the obvious spurs.

Exercise 1.26.Show that homotopic paths have the same start and end vertices, and thushomotopically trivial paths are necessarily closed. Show that homotopy is an equivalencerelation on the paths with common fixed endpoints.

We can also subdivide2-complexes to get homeomorphic ones, although this will playless of a role than it does with graphs (graphs homeomorphic to S1 were essential to thedefinition of2-complex). What we want to do is summarized by Figure 1.18: replace anexisting faceσ by two new facesσ1, σ2 by creating a new arc running between vertices ofσ. We are then free to choose new distinguished vertices for theσi.

Exercise 1.27.Formulate the definition of subdividing a face in terms of Definition 1.20by using the description of1-spheres given by Lemma 1.11.



σ

σ1

σ2

e

Fig. 1.18.subdividing a face

Write X ↔ X ′ when the two complexes differ by the subdivision of an edge orface,so thatX andY are thenhomeomorphic, writtenX ≈ Y , when there is a finite sequenceX = X0 ↔ X1 ↔ · · · ↔ Xk = Y of subdivisions (of either tyoe) connecting them.It is easy to see that homeomorphism is an equivalence relation and we have topologicalinvariants for2-complexes as well as graphs.

Figure 1.19 shows a series of subdivisions of the2-sphere.

σ1

σ2

v1

v2

e1

e2

σ1

σ2

v1

v2

e1

e2

e3

σ1

σ2

v1

v2

e1

e2

e3

Fig. 1.19.subdividing a sphere

1.3 Aside: comparision with “proper” topology

1.3.1 The category of CW complexes

1.3.2 A functor

1.4 Quotients of2-complexes

Often we want to squash unwanted parts of a complex away, gluecomplexes together, factorout the action of a group, and so on. In otherwords, we want to be able to take quotients ofcomplexes. We do this by defining an equivalence relation on the cells of the complex, andthen defining a new complex whose cells are the equivalence classes of the relation. It turnsout that there are a number of subtleties complicating the exposition, arising if we want tobe able to identify cells of different dimensions.

1.4.1 Quotients in general

All quotients start with an equivalence relation:

Definition 1.28 (quotient relation on a2-complex).If X is a2-complex, then aquotientrelation is an equivalence relation on the vertices, edges, paths andfaces ofX such that

1.∼ restricted to the1-skeletonX(1) is a graph quotient relation as in Definition 1.7,with quotient map of graphsq : X(1) → X(1)/∼;


1.4 Quotients of2-complexes 15

2. if Ω is an equivalence class of∼ with Ω ⊂ X2, then there is a(XΩ, aΩ) with XΩ ≈S1 andaΩ : XΩ

u → (X(1)/∼)v dimension preserving, such that for all facesσ ∈ Ωwith ∂σ = (Xσ, aσ), there is an isomorphismXσ → XΩ making the diagram belowleft commute:

Xσ XΩ

X(1) X(1)/∼

aσaΩ

∼=

q

Xσ S

X(1) X(1)/∼

aσ

q

3. if Ω is an equivalence class containing a face but withΩ 6⊂ X2, then there is a ho-motopically trivial pathS → X(1)/∼ such that for all facesσ ∈ Ω there is a mapXσ → S making the diagram above right commute.

Part 2 is just saying that equivalent faces have boundaries that fold up in the quotientgraph to give the same thing. Similarly, if a face is to be identified with a closed path thenpart 3 forces the boundary of the face to be identified with it as well. The homotopicallytrivial condition is a little obscure at the moment: it’s role will become clearer in Chapter2.

Definition 1.29 (quotient2-complex).If ∼ is a quotient relation on the2-complexX thendefine the quotientX/∼ as follows

1. the1-skeleton(X/∼)(1) is the quotient graphX(1)/∼ with quotient map of graphsq : X(1) → X(1)/∼;

2. the faces(X/∼)2 consist of those equivalence classesΩ with Ω ⊂ X2. Such a facehas boundary∂Ω = (XΩ, aΩ) as given by Definition 1.28 part 2.

Proposition 1.30.If ∼ is a quotient relation thenX/∼ is a2-complex and the quotient mapq : X → X/∼ given byq(x) = [x] is a map of2-complexes.

Proof. That the quotient is a2-complex is immediate from definition 1.29, and the com-muting diagrams given there are precisely the definition ofq being a map of2-complexes.2

1.4.2 Group actions and their quotients

When a groupG acts on a2-complexX we can replaceX by a complex on which theaction of G is trivial, ie: every element ofG acts as the identity. Thus we may factorout group actions, and we do this by forming a quotient. Recall from §1.2.3 the groupAut(X) of automorphisms of the2-complexX , and a group action is a homomorphismG

ϕ→ Aut(X).Let∼ be the equivalence relation onX given by the orbits of the action, so thatx ∼ y

iff y = g(x) for someg ∈ G (and then necessarilyx andy have the same dimension).

Proposition 1.31 (quotient by a group action).Let∼ be the equivalence relation on the2-complexX given by the orbits of a group action. Then∼ is a quotient relation if and onlyif the group action is orientation preserving.

We write X/G for the quotient complexX/∼. Thus, although one may in principleconsider group actions that don’t preserve orienatation, as the primary purpose of suchactions is to form quotients, we will only consider orientation preserving actions. Comparethis with Proposition 1.9, noting how the2-complex structure imposes no new conditionsfor ∼ to be a quotient relation.



Proof. We have a quotient relation on the1-skeletonX(1) if and only if the action pre-serves orientation by Proposition 1.9. This leaves us with the second part of Definition 1.28to worry about. IfΩ ⊂ X2 is an equivalence class of faces, fix a faceσ ∈ Ω, and let(XΩ, aΩ) = (Xσ, qaσ), whereq is the graph quotient map. Ifτ ∼ σ thenσ = g(τ) forsomeg ∈ G, so Definition 1.22 part 1 gives an isomorphismXτ → Xσ with the diagrambelow left commuting:

Xτ Xσ

X(1) X(1)

aτ aσ

∼=

g

X(1) X(1)

X(1)/∼

g

q q

The triangular diagram above right commutes by the nature ofthe quotient map: ify =g(x) then [x] = [y] so thatq(x) = q(y). Now stitch the two diagrams together, so that(XΩ, aΩ) = (Xσ, qaσ) does the job. 2

Consider as an example Figure 1.20, where a Euclidean plane complex is rolled into aninfinite tube by the action of the integersZ.

X =

v

u

e

[v]

[u]

[e]

X/Z =

Fig. 1.20.Let Z act on theX (left) by mapping1 ∈ Z to the automorphism that translatesX onestep to the right, as shown by the red arrow: The quotientX/Z is an infinite rolled up tube. Ifm ∈ Z,then its effect onX is to translatem steps to the right, whereas its effect onX/Z is to rotate a cellmtimes around the tube, bringing it back to itself.

Exercise 1.32.In the definition of the quotientX/G we took(XΩ, aΩ) = (Xσ, qaσ) forsome fixedσ ∈ Ω. Show that we are free to choose instead a different face fromΩ: if τ ∼ σand we take(XΩ, aΩ) = (Xτ , qaτ ) instead, then this new version ofX/G is isomorphicto the old one by the identity map.

1.4.3 Quotients by a subcomplex

Now for a quotient that involves some serious squashing: ifY → X is a subcomplex wedefine a new complex whereY has been compacted down to a single vertex.

Define∼ onX to be the equivalence relation with the following equivalence classes: (i).all the cells inY (of whatever dimension) form one class; (ii). every other class has theform [x] = x. Thus, we havex ∼ y if and only if eitherx = y or bothx andy lie in thesubcomplexY .

Proposition 1.33.The relation∼ defined above is a quotient relation.

Write X/Y for the corresponding quotient, thequotient ofX by the subcomplexY : itis what results from collapsingY to a vertex and propagating the effects of this on theincidence of cells throughoutX , but otherwise leaving the cells ofX unaffected.



Proof. ∼ is clearly a graph quotient relation on the1-skeleton. IfΩ is an equivalence classcontaining faces then it either contains a single faceσ (and so let(XΩ, aΩ) = (Xσ, aσ))or is all of Y (in which case Definition 1.28 part 3 kicks in, so letS → (X/Y )(1) consistof a single vertex forS mapping to the vertex that isY ). 2

A typical quotient by a subcomplex arises whenT → X is a spanning tree for the1-skeleton and we formX/T as in Figure 1.21.

. . .

T → X

q

Fig. 1.21.squashing a spanning tree down to a vertex

1.4.4 Pushouts

The pushout is a rather general construction which arises whenever a pair of complexes areglued together across a common subcomplex. Before giving the definition we need somepreliminary notions about equivalence relations which mayor may not be well known tothe reader. We place them in an exercise:

Exercise 1.34.Recall that the formal definition of an equivalence relationon a setX is asubsetS ⊂ X such that, (i).S contains the diagonal,(x, x) ∈ S for all x ∈ X , (ii). S issymmetric,(x, y) ∈ S ⇒ (y, x) ∈ S, and (iii). (x, y), (y, z) ∈ S ⇒ (x, z) ∈ S. Show thatif S1, S2 are equivalence relations onX then so isS1 ∩ S2, and hence ifY is any subsetof X we may define theequivalence relation generated byY to be the intersection of allequivalence relationsS with Y ⊂ S.

Definition 1.35 (pushout).Let X1, X2 andY be2-complexes andgi : Y → Xi maps of2-complexes. Let∼ on the disjoint unionX1

∐X2 be the equivalence relation generated

by thex ∼ y iff there is az ∈ Y with x = g1(z) andy = g2(z). If ∼ is a quotient relationthen call the quotientX1

∐X2/∼ thepushoutof the datagi : Y → Xi, and denote it by

X1

∐Y Y2.

Figure 1.22 illustrates the schematic setup:

Exercise 1.36.If ∼ is the relation described in Definition 1.35, show thatx ∼ z iff therearex0, x1, . . . , xk ∈ X1

∐X2 with x0 = x andxk = z, andy1, . . . , yk ∈ Y , such that

g1(y1) = x0, g2(y1) = x1, g2(y2) = x1, g1(y2) = x2, . . . and so on.

Defineti : Xi → X1

∐Y X2 to be the compositionsXi → X1

∐X2 → X1

∐X2/∼

of the inclusion ofXi in the disjoint union and the quotient map.

Proposition 1.37.If Y 6= ∅, theXi are connected and the∼ of Definition 1.35 is a quotientrelation, then the pushout is connected, and the mapsti make the diagram on the leftcommute.

Y

X1

X2

X1

∐Y X2

g1

g2

t1

t2

Y

X1

X2

X1

∐Y X2

g1

g2

t1

t2

Zt′1

t′2



Y

X1

X2

g1

g2

g1(Y )

g2(Y ) X1

‘Y

X2

Fig. 1.22.the pushoutX1

‘Y

X2 of the datagi : Y → Xi

Moreover if thegi are dimension preserving, the pushout is universal in that if Z, t′1, t′2 are a2-complex and maps making such a square commute, then there isa maph : X1

∐Y X2 →

Z making the diagram above right commute.

The proposition gives a clue as to why the name pushout: the datagi : Y → Xi formingthe input to the construction gives the two sides of the commutative square top left, andX1

∐Y X2 “pushes out” these two sides to complete the square. The proof of the proposi-

tion, while routine, is a good test of all the definitions so far, so we go through it in somedetail.

Proof. Let u be a vertex ofY andvi = gi(u) ∈ Xi. By the connectedness of theXi,vertices in either one of them can be joined by a path to the appropriatevi. Suppose that[xi] are vertices in the pushout withxi ∈ Xi. Then the image undert1 of a path fromx1 to v1 finishes at the vertex[v1] and joins up with the image of a path undert2 fromv2 to x2, since[v1] = [v2], and so the pushout is connected. Ify ∈ Y then its imageunder the two mapsY → Xi → X1

∐Y X2 are equivalent by the definition of∼. Thus

t1g1(y) = t2g2(y) and the square commutes.Suppose now that we have complexes and mapsZ, t′i making the outside square on

the right hand side commute. After a moments thought it is obvious what the maph :X1

∐Y X2 → Z ought to be: every cell of the pushout has the form[x] for x some cell of

one of theXi. Defineh[x] = t′i(x) whenx ∈ Xi. We leave it to the reader to show that wehave a well defined map, and also a map of graphs.

To finish the verification thath is a map, it is generally easier, rather than showing Defi-nition 1.22 directly, to use the reformulation where boundaries of faces are closed paths inthe1-skeleton and a map must send boundaries to boundaries. IfΩ = [σ] is a face of thepushout withσ ∈ X1 say, thent1 mapsσ to Ω. But then the boundary ofσ must map to theboundary oft′1(σ), and as the bottom triangle in the diagram above commutes, this forcesthe boundary ofΩ to be mapped to the boundary oft′1(σ) as well.

But we also showh is a map via Definition 1.22, to satisfy ourselves that such things arepossible (and as a consequence, to see why we won’t tend to in the future!). Because thegi

are dimension preserving, the image underh of a faceΩ = [σ] is always a face, so we justhave part 1 of Definition 1.22 to check. Suppose for specificness thatσ is a face ofX1, letτ = t′1(σ), and consider the following diagram:

Xτ Xσ XΩ

Z(1) X(1)1

(X1

∐Y X2)

(1)

∼=∼=

The lefthand square commutes and is given to us courtesy of the mapt′1 and Definition 1.22part 1; the righthand square commutes and is a product of Definition 1.28 part 2 applied to



the quotient that is the pushout. The maph, goes from the bottom right to the bottom leftcomplex, and it is, by defintion, the map making the triangle formed with the maps alongthe bottom edge, commute. Define a mapε from the top right to the top left complexes tobe the composition of one of the isomorphisms with the inverse of the other. We leave it asan exercise in diagram chasing to see that the large square around the outside commutes,ie: thataτε = hfΩ. This is exactly what we need forh to be a map. 2

Exercise 1.38 (pointed pushout).Suppose thegi : Yy → (Xi)xiare maps of pointed

complexes. Ifq : X1

∐X2 → (X1

∐X2)/∼ is the quotient map, letx = q(x1) = q(x2),

and call(X1

∐Y X2)x thepointed pushout. Show that we have a Proposition analogous to

Proposition 1.37 with all complexes and maps pointed.

We now come to the slightly delicate matter of the actualexistenceof pushouts. Giventhe setup of Definition 1.35 the pushout cannot always be formed, precisely because thequotient cannot always be formed. This is true even for graphs, as Figure 1.23 illustrates.

Y X

g1

g2

Fig. 1.23.Datagi : Y → Xi for which the pushout does not exist. HereX1 = X2 = X andY areall graphs with a single edge joining two vertices and the maps g1 andg2 send the edge ofY to theedgee and its inversee−1 respectively. The resulting equivalence relation onX hase ∼ e−1 and sois not a quotient relation.

A complete solution to the existence of pushouts whenY is a graph is given in Exercise1.39.

Exercise 1.39.Show that ifY is a graph and thegi are dimension preserving, then thepushout exists if and only if there are orientationsO, Oi for Y, Xi with gi(O) ⊆ Oi. Thusin particular, if the graphsg1(Y ) andg2(Y ) are disjoint, then the pushout can always beformed.

A simple example of the situation of Example 1.39 is theStallings fold, shown in Figure1.24. Here the graphY is a single edge joining two vertices. An even simpler one, shownin Figure 1.25, is thewedgeof a pair of complexes:Y is now the trivial complex consistingof just a single vertex.

Y X

g1

g2

Fig. 1.24.Stallings fold: hereY is a single edgee joining two distinct vertices, andX1 = X2 = X,gi : Y → X, (i = 1, 2) with g1(s(e)) = g2(s(e)) andg1(e) 6= g2(e)

−1

Things are less simple when theXi are2-complexes. Nevertheless, pushouts exist mostof the time:

Proposition 1.40.If X1, X2 and Y are 2-complexes andgi : Y → Xi dimension pre-serving maps, with orientationsO, Oi for Y, Xi such thatgi(O) ⊆ Oi, then the pushoutexists.



•Y

X1

X2

X1

‘Y

X2

g1

g2

t1

t2

Fig. 1.25.Wedge of spheres:Y is the trivial vertex.

Proof. The relation∼ of Definition 1.35 is a quotient relation on the1-skeleton by Ex-ercise 1.39. As thegi are dimension preserving, all the cells in an equivalence class havethe same dimension, and we are thus left with part 2 of Definition 1.28 to do. LetΩ bean equivalence class of faces, fix aτ ∈ Ω, and take(XΩ, fΩ) = (Xτ , qaτ ), with qaτ

dimension preserving as thegi are. Suppose thatσ ∈ Ω and that we are in the special casethatσ = g1(x), τ = g2(x) for some facex in Y . Then we get a commutative diagram verysimilar to the one in the proof of Proposition 1.37,

Xσ1 Y x Xτ

2

(X1

∐Y X2)

(1)Y (1) (X1

∐Y X2)

(1)

∼=∼=

by splicing together the diagrams supplied by the mapsg1, g2 and a diagram for the mapsg′i : Y → Xi → X1

∐X2 whereqg′1 = qg′2. The bottom left and righthand complexes

both map to the pushout, making a commuting square that attaches to the bottom of thediagram, and we take the composition of the two isomorphismsto give an isomorphismXσ

1 → Xτ2 . The reader can then check that the outside circuit of this large diagram is what

we need to verify part 2 of Definition 1.28.

X1 Y X2

Xσ1 Xσ

2

g1 g2

Fig. 1.26. Initial data for a typical pushout of Chapter 3:Y has six vertices, edges and faces,X1

has three of everything andX2 has two of everything. The boundaries of all faces are hexagonal, butwe’ve only shown one in each case; the others are identical, with the distinguished vertices exhaustingthe six, three and two possibilities respectively. The attaching maps wrap around the1-skeletons asshown.

In general, whenσ ∼ τ we haveσ = σ0 = g1(x1), σ1 = g2(x1), . . . , σk−1 =g1(xk), τ = σk = g2(xk), and the requirements for a quotient relation can be verifiedby repeatedly applying the process of the previous paragraph. 2


1.5 Pullbacks and Higman composition 21

Pushouts will really prove their mettle in Chapters 3-4, where thegi will be coverings.Figure 1.26 illustrates the kind of initial set-up we will have, and Figure 1.27 the resultingpushout.

X1

‘Y X2

Fig. 1.27.Pushout resulting from the set-up in Figure 1.26. There is a single vertex, edge and face inthe quotient, and the face has a hexagonal boundary with the attaching map wrapping it around the1-skeleton six times as shown.

Our final example of a pushout arises when one of the mapsgi is just an inclusion, sothat the initial data consists of two complexesX1 andX2, and a mapg from asubcomplexof X1 to X2. The resulting pushout (when it exists) is the result of glueing X1 andX2

together via the attaching mapg. See Figure 1.28.

g1(Y )

X1

X2

X1

‘Y

X2

g2

Fig. 1.28.Glueing complexes together via an attaching map.

1.5 Pullbacks and Higman composition

We now come to a pair of constructions which both start with the roughly the same kind ofinput data: a complexY , a family of complexesXi, and a family of mapsgi : Xi → Y .The first of these, the pullback, is dual to the pushout, or, touse categorical terminology,is a “co”-pushout. It is essentially what we get if we reversethe directions of all the mapsin the description of the pushout. Pullbacks, like pushouts, will play a crucial role in thetheory of coverings of complexes in Chapters 3-4: pullbackswill act like a kind of “union”of complexes and pushouts like a kind of “intersection”. In§3.4.3 we will be able to bemuch more precise about what we mean by this.

The other construction, Higman composition, is less well known, and can be performedonly in very special circumstances. Nevertheless, when possible, it will prove extremelypowerful, and this makes its inclusion worthwhile.

1.5.1 Pullbacks

It is easier to do graphs first, then extend to2-complexes proper:



Definition 1.41 (pullbacks of graphs).Let X1, X2 andY be graphs andgi : Xi → Ymaps of graphs. ThepullbackX1

∏Y X2 has vertices (respectively edges) thex1 × x2,

xi ∈ X0i (resp.xi ∈ X1

i ), such thatg1(x1) = g2(x2). The incidence maps ares(e1×e2) =s(e1)× s(e2), and(e1 × e2)

−1 = e−11 × e−1

2 . See Figure 1.29.

g1

g2

u1 × u2

v1 × v2

e1 × e2

X1

QY

X2

u1

v1

e1

X1

u2

v2

e2X2

u

v

eY

Fig. 1.29.Construction of the pullback for graphs: the verticesui andvi map via thegi to u andv,and the edgesei map via thegi to e. Thus in the pullback we get verticesu1 ×u2 andv1 × v2 joinedby an edgee1 × e2.

We now set up for the pullback of2-complexes by seeing how the boundaries of faces intheXi behave when we pullback the1-skeletons using this recipe. Suppose that thegi aredimension preserving, and thatσ is a face ofY andσi faces of theXi mapping toσ via thegi. We get a by now familar commuting diagram:

Xσ11 Y σ Xσ2

2

X(1)1 Y (1) X

(2)2

∼=∼=ε2ε1

aσ1 aσ2

Suppose also that as we move clockwise around the sphereY σ the edges aree1, e2, . . . , ek.Then the edges of theXσi

i can be labelledei1, ei2, . . . , eik, with ej = εi(eij) and the closedpathaσiei1, a

σiei2, . . . , aσieik the set of boundary edges forσi with aσieij mapping via

gi to aσej in Y .The upshot is that the pullback contains a path of edges

aσ1e11 × aσ2e21, aσ1e12 × aσ2e22, . . . , a

σ1e1k × aσ2e2k, (1.1)

and sincesaσ1e11 = taσ1e1k andsaσ2e21 = taσ2e2k, this path in the pullback is closed.The idea is to “sew a face” into the1-skeleton having boundary this closed path, by takingY σ and using the attaching mapaσ1ε−1

1 × aσ2ε−12 that sendsej to aσ1e1j × aσ2e2j . More-

over, all the pointings are respected, so that if the pointedmapsaσi send verticesui to vi,andaσ sendsu to v, then the attaching map in the pullback sendsu to v1 × v2.

Definition 1.42 (pullbacks of2-complexes).Let X1, X2 andY be2-complexes andgi :Xi → Y dimension preserving maps of2-complexes. Thepullback X1

∏Y X2 hasℓ-

dimensional cells thex1 × x2, for xi ∈ Xℓi , such thatg1(x1) = g2(x2). The incidence

maps ares(e1 × e2) = s(e1)× s(e2), (e1 × e2)−1 = e−1

1 × e−12 and

∂(σ1 × σ2) = (Y σ, aσ1ε−11 × aσ2ε−1

2 ),

whereg1(σ1) = σ = g2(σ2), and theεi are the isomorphisms in the diagram above.


1.5 Pullbacks and Higman composition 23

σ1 × σ2

σ1

σ2

⇐ and

∈ X1

∈ X2v1 × v2 v1

v2

∂(σ1 × σ2) = ∂σ1 × ∂σ2

∂σ1

∂σ2

Fig. 1.30.pullback of faces

We get a schematic like Figure 1.30 wheneverg1(σ1) = g2(σ2). For i = 1, 2 we nowdefine maps

ti : X1

∏

Y

X2 → Xi,

as follows. Each cell of the pullback has the formx1 × x2 with thexi ∈ Xi, so letti(x1 ×x2) = xi. If σ1 × σ2 is a face of the pullback, then writingX = X1

∏Y X2, we have

Xσ1×σ2 = Y σ for the boundary of this face, whereg1(σ1) = σ = g2(σ2) ∈ Y . For theisomorphism

Xσ1×σ2∼=−→ Xσi

i ,

we takeε−1i : Y σ → Xσi

i . We leave it as an exercise for the reader to show that the diagramof Definition 1.22 part 1 commutes, so that theti are indeed mappings of2-complexes.

Proposition 1.43.The ti are dimension preserving maps making the diagram below leftcommute,

X1

∏Y X2

X1

X2

Y

t1

t2

g1

g2

X1

∏Y X2

X1

X2

Y

t1

t2

g1

g2

Z

t′1

t′2

Moreover, the pullback is universal in that ifZ, t′1, t′2 are a2-complex and maps making

such a square commute, then there is a mapZ → X1

∏Y X2 making the diagram above

right commute.

Proof. We haveg1t1(x1 × x2) = g2t2(x1 × x2), and writingX = X1

∏Y X2, the iso-

morphismXσ1×σ2 → Y σ for bothgiti is the composition,

Xσ1×σ2ε−1

i−→ Xσi

iεi−→ Y σ,

hence identical (to the identity map). Ifz is a cell ofZ, then the commuting of the largesquare on the outside of the righthand diagram gives thatg1t

′1(z) = g2t

′2(z), so thatt′1(z)×

t′2(z) is a cell of the pullback. DefineZ → X1

∏Y X2 by z 7→ t′1(z) × t′2(z), and for the

isomorphismZσ → Xt′1(σ)×t′2(σ) = Y git′

i(σ), take the composition

Zσ ∼=−→ X

t′i(σ)i

∼=−→ Y git

′

i(σ),

(necessarily identical fori = 1, 2 as the square commutes) provided by the mapst′i andgi. We leave it to the reader to check that this is a map of2-complexes with the requiredproperties. 2

Comparison with the pushout reveals good news and bad news. The good news is thatthe construction of the pullback is built into the definition, so there is never any questionabout the existence of pullbacks. Pullbacksalwaysexist. The bad news is that pullbacksare not in general connected. The following goes some way to alleviating this:



Exercise 1.44 (pointed pullbacks).Suppose thegi(Xi)xi→ Yy are maps of pointed com-

plexes. Thenx = x1×x2 is a vertex of the pullback. Write(X1

∏Y X2)x for the connected

component of the pullback containingx1×x2. Show that we have a Proposition analogousto Proposition 1.43 for the pointed pullback, with all complexes and maps pointed.

X1 Y X2

Xσ1 Xσ

2

g1 g2

Fig. 1.31.Initial data for a typical pullback of Chapters 3-4:Y has six vertices, edges and faces,X1

has three of everything andX2 has two of everything. The boundaries of all faces are hexagonal, butwe’ve only shown one in each case; the others are identical, with the distinguished vertices exhaustingthe six, three and two possibilities respectively. The attaching maps wrap around the1-skeletons asshown.

X1

‘Y

X2

Fig. 1.32.Pullback resulting from the set-up in Figure 1.26. There is asingle vertex, edge and face inthe quotient, and the face has a hexagonal boundary with the attaching map wrapping it around the1-skeleton six times as shown.

If the reader needed any more convincing about the duality between pullbacks andpushouts then we can run the example of§1.4.4 backwards and use the pullback to getback to where we started. Thus in Figure 1.31 we have the same two complexesX1, X2 asin the pushout example, but this timeY is the end result of that example and the mapsgi arethe ti from Proposition 1.37. The resulting pullback, shown in Figure 1.32 is the startingpoint of the pushout example, and the mapsti given by Proposition 1.43 are the startinggi

from that example. Notice how the number of faces proliferates (rather than declines as itdoes in the pushout):X1 has three, each of which can be paired up with the two inX2, giv-ing six faces in the pullback. As the distinguished verticescycle through the three verticesof X1 (resp. two inX2), their products cycle through the six vertices ofX1

∏Y X2.

1.5.2 Higman composition

1.6 Notes on Chapter 1


1.6 Notes on Chapter 1 25

e2

e1

fi1

fi2fi3

Fig. 1.33.handle


3

Coverings

We return to the combinatorial topology of2-complexes and develop their covering spacetheory.

3.1 Basics

3.1.1 Coverings

Definition 3.1 (covering of2-complexes).A mapf : Y → X of 2-complexes is acover-ing if and only if

1. f preserves dimension;2. for every pair of verticesu andv, with f(u) = v, the local continuity (see§1.1.1) off

atu,s−1

Y (u)→ s−1X (v),

is a bijection.3. for every pair of verticesu andv, with f(u) = v, and faceτ of X , the local continuity

(see Definition 1.23) off atv,∐

εσ :∐

f(σ)=τ

(aσ)−1(u)→ (aτ )−1(v)

is a bijection.

The terminologycoverandlift is used for images and pre-images of a covering map. Iff(y) = x, then one says thaty coversx, or thatx lifts to y. The set of all lifts ofx, oralternatively the setf−1(x) of all cells coveringx, is itsfiber.

Note that each of theεσ is an isomorphism, so the local continuity map is a disjointunion of a set of maps, each of which is therestriction of a bijection. In anycase, each isindividually an injection.

The last part of each definition express “local isomorphism”properties of covering maps:if u is a vertex coveringv, thenY looks the same nearu asX does nearv. Specifically, theconfiguration of edges around a vertex looks the same both upstairs and downstairs (Figure3.1 left): for every vertexu of Y , the coveringf is a bijection from the set of edges inYwith start vertexu to the set of edges inX with start vertexf(u).

Given a vertexv and a faceτ downstairs containing the vertex in its boundary, this facelooks the same nearv as its lifts do near any vertexu coveringv: if τ containsv in itsboundaryk times, so there arek “wedge-shaped” pieces ofτ fitting together aroundv, thenthere arek wedge-shaped pieces fitting together aroundu, where these wedges belong tofacesσ that coverτ (see Figure 3.1 right). Note that the wedges upstairs don’t necessarilybelong to distinct faces.


42 3 Coverings

u

v

f

u

v

f

Fig. 3.1.The local continuity maps are bijections for coverings

Part 3 of the definition gives in particular that∑

f(σ)=τ

|(aσ)−1(u)| = |(aτ )−1(v)|,

so thatv appears the same number of times in the boundary ofτ asu does in the boundariesof all the facesσ in the fiber ofτ .

One commonly sees the assumption that in a covering, both thecovering complexY andthe covered complexX are connected, but we won’t assume this at the moment. Indeed,we will find it positively useful in some situations tonot assume that a covering complexbe connected.

Exercise 3.2.Let f : Y → X be a covering andY a connected component ofY . Showthat restrictingf to Y gives a covering. Show that we may not restrict a covering to anarbitrary subcomplex and still get a covering.

Y Xf

Fig. 3.2.A simple graph covering: the two vertices ofY cover the single vertex ofX and the twoarcs ofY similarly.

σ1

σ2

u1

u2

e1

e2

σ

v

v

e

e

fY X

Fig. 3.3. The graph covering of Figure 3.2 extended to a covering of2-complexes:Y is now the2-sphere with the faceσi havingui as its distinguished vertex. Theσi cover the single face ofX,which, unlike in Figure 3.2, has been drawn face-centrically.

Figure 3.2 shows a very simple graph covering with the red path on the left covering thered path on the right. Figure 3.3 extends this to a covering of2-complexes.


3.1 Basics 43

Exercise 3.3.Call a mapf : Y → X of graphs animmersionwhen it preserves dimensionand the local continuity maps areinjections; call a map of2-complexes an immersion if therestriction to the1-skeletons is an immersion and the local continuity maps in Definition3.1 part 3 are inejctions. Give examples of immersions that are not coverings.

Exercise 3.4.Show that for anyX the identity mapX → X is a covering.

3.1.2 Lifting

When we have a coveringf : Y → X , the complexesX andY look the same as longas we restrict our attention to small pieces. If two complexes look the same as each otherthen we should be able to pull back parts ofX to find parts ofY mapping to them. Puttingthese together, when we have a covering we should be able to pull back small pieces ofXand getsmallpieces ofY covering them. This process is called is calledlifting. The smallpieces turn out to be paths and faces.

Proposition 3.5 (path and spur lifting). Let f : Y → X be a covering withf(u) = vvertices.

1. If γ is a path inX starting atv then there is a pathγ in Y starting atu and coveringγ. Moreover, ifγ1, γ2 are paths inY starting atu and covering the same path inX ,thenγ1 = γ2.

2. A path inY covering a spur is itself a spur. Consequently, two paths inY coveringfreely homotopic paths are freely homotopic.

Part 1 is calledpath lifting and part 2 isspur lifting. Call γ the lift of γ at u. Thus apath can always be lifted to one starting at any vertex covering its initial vertex and thislift is unique. As with so many such results, it is the uniqueness of the lift, rather than theexistence, that turns out to be most useful.

Proof. The existence ofγ is easily seen, as in Figure 3.4, since ifγ = e1 . . . en, there is

v

u

e1

e′1

e2

e′2

en

e′n

X

Y

Fig. 3.4.Lifting paths in a covering: individual edges can be lifted via the bijection between the edgesstarting at a vertex upstairs and the edges starting at a vertex it covers downstairs. Paths are then liftedby repeated edge lifts.

an edgee′1 coveringe1 under the bijection induced byf of the edges with start vertexuand those with start vertexv. This edgee′1 must end at a vertex that covers the end vertexof edgee1, as coverings (being maps of complexes) preserve vertex-edge incidences. Theprocess can be repeated starting at this new vertex to giveγ. For the uniqueness, the firstedges of theγi both have initial vertexu and map toe1, hence must be the same edge.Continuing in this manner along the two paths gives their equality. For the second part, thepathγ ∈ Y must have the forme1e2, where the “middle vertex” is the start of the edgese−11 ande2. Use the injectivity of the cover on the edges starting at this vertex to deduce

thate−11 = e2. 2


44 3 Coverings

Exercise 3.6.Let f : Y → X be a covering andγ = α1α2 a path inX (hencet(α1) =s(α2)). Show that the lift at a vertexu ∈ Y of γ is the pathα1α2 consisting of the liftα1

of α1 atu followed by the lift ofα2 at t(α1).

Exercise 3.7.Show that paths cannot necessarily be lifted by an immersion, but when theycan, they are unique. Show that spur lifting is a property also enjoyed by immersions.

The first spin-off of path lifting justifies the usage of the word “cover”, and is nota prioriobvious from the definition:

Proposition 3.8 (surjectivity of coverings).If f : Y → X is a covering withX connectedthenf is a surjective map of2-complexes, ie: every cell ofX is the image underf of somecell ofY .

Proof. Path lifting gives the surjectivity on the vertices and the local continuity maps givesit on the edges and faces: letu be a vertex ofY , and by connectedness, we can join anyvertexv′ of X to f(u) by a path. Lift this path tou, so that its terminal vertex inY mapsvia f to v′. For any edge or facex of X , take a vertexv lying in its boundary (which canjust be the start vertex of the edge if we have an edge), so thatthe setss−1

X (v) or (ax)−1(v)are non-empty, and letu be a vertex mapping tov. The bijectivity of the local continuitymaps then gives an edge or face mapping viaf to the one we started with.2

Exercise 3.9.Illustrate by an example why the connectedness ofX is necessary for Propo-sition 3.8.

Proposition 3.10 (face lifting).Let f : Y → X be a covering,τ a face ofX , v a vertexthat appears in its boundary andγ a boundary path ofτ starting atv. Let u be a vertexcoveringv and γ the lift ofγ at u. Then there is a faceσ of Y that coversτ , containsu inits boundary and hasγ as a boundary path starting atu.

Thus the boundaries of faces lift to the boundaries of faces as shown schemeatically inFigure 3.5.

τ

γ

eγ

v

u

X

Y

τ

σ

γ

eγ

v

u

X

Ylift

Fig. 3.5.Face lifting

Proof. Suppose thatτ has boundary∂τ = (Xτ , aτ ), so that(aτ )−1(v) 6= ∅ asv lies inits boundary. Hence there is a vertexx of the1-sphereXτ that maps via the attaching mapaτ to v. Definition 3.1 part 3 gives a vertexy of Y lying in

∐f(σ)=τ (aσ)−1(u) mapping

to x via the local continuity off , ie: there is a faceσ of Y coveringτ , containingu in itsboundary, and with the diagram,

Y σy Xτ

x

Y(1)u X

(1)v

aσ aτ

∼=

f

commuting. In particular there is a boundary path ofσ starting atu that coversγ, and bythe uniqueness of lifts, this must be the pathγ. 2


3.1 Basics 45

The result of being able to find pre-images of paths and faces by a covering is thathomotopies of paths can also be “pulled back” through a covering:

Corollary 3.11 (homotopy lifting). LetY → X be a covering. Then two paths that coverhomotopic paths are themselves homotopic.

Proof. The homotopy between the covered paths is realised by a finitesequence of inser-tions or deletions of spurs and face boundaries. By spur and face lifting, those sections ofthe covering paths mapping to these spurs and face boundaries are themselves spurs andface boundaries, while by uniqueness of path lifting, the remaining pieces are identical.Thus the same sequence of elementary homotopies can be realised between the coveringpaths as between the covered ones.2

Exercise 3.12.Let f : Y → X be a covering andγ1, γ2 paths inX related by an elemen-tary homotopy, ie:γ2 is what results from inserting a spur or a face boundary intoγ1. Letγ1 be the lift ofγ1 at some vertexu of Y andγ the result of lifting to the appropriate vertexthe elementary homotopy and performing it onγ1. Show thatγ is the lift γ2 of γ2 atu.

Another spin-off of homotopy lifting (and Exercise 3.12) isthe following characterisa-tion of the image of the induced homomorphism between fundamental groups:

Corollary 3.13. Letf : Y → X be a covering withf(u) = v and

f∗ : π1(Y, u)→ π1(X, v),

the induced homomorphism. Thenf∗ is injective and a closed pathγ at v represents anelement off∗π1(Y, u) if and only if the liftγ of γ at u is closed.

The injectivity of the induced homomorphism is probably thesingle most importantproperty of coverings: it means that the fundamental group of the covering space can beindentified with a subgroup of the fundamental group of the space that is being covered.The appropriate context in which to develop this idea properly will be the Galois theory ofcoverings in Chapter 4, so we will make a bigger deal of it then.

Proof. If two elements ofπ1(Y, u) map to the same element ofπ1(X, v) then they are rep-resented by closed paths atu covering homotopic paths inX . Homotopy lifting thus givesthat the closed paths themselves are homotopic, and so the two elements of the fundamentalgroup coincide, thus establishing the injectivity of the homomorphism. For the second part,if γ is closed then its homotopy class maps viaf∗ to the homotopy class ofγ. Conversely,if γ represents an element in the image of the homomorphism then there is a closed pathγ1

atv, homotopic toγ, and withf(γ1) = γ1 for γ1 closed atu. By Exercise 3.12, the lift ofγatu is what results by lifting these homotopies; in particular,γ1 and the lift are homotopic,so that the lift is closed. 2

Exercise 3.14.Let f : Y → X be a covering withf(u) = v andu′ the terminal vertexof a pathµ ∈ Y starting atu. Show thatf∗π1(Y, u) = hf∗π1(Y, u′)h−1, whereh is thehomotopy class off(µ).

This thread of ideas culminates in the following general lifting result.

Proposition 3.15 (map lifting). If f : Yu → Xv is a (pointed) covering andg : Zx → Xv

a map withZ connected, then there is a mapg : Zx → Yu making the diagram

Yu

Zx Xv

fg

g

commute if and onlyg∗π1(Z, x) ⊂ f∗π1(Y, u). If g exists then it is unique.


46 3 Coverings

One way to think of this result is as a generalisation of path lifting: if Z is a1-ball thenthe mapg : Z → X is a path inX starting atv. As the fundamental group of a1-ball istrivial, the conditiong∗π1(Z, x) ⊂ g∗π1(Y, u) is trivially satisfied, as the left hand side isthe identity subgroup. The resulting mapg : Z → Y is a path inY starting atu, and thecommuting of the diagram just says that this new path covers the old one.

Proof. The “only if” part can be dispensed with quickly: it follows as π1 is a functor. Asf g = g we getf∗g∗ = g∗ so thatg∗π1(Z, x) = f∗g∗π1(Z, x) ⊂ g∗π1(Y, u).

Suppose we have the condition on the images of the fundamental groups. We proceed todefine a mapg having the required properties: ifz is a vertex ofZ, then by connectednessthere is a path joining it tox. Take the image of this path byg and then lift the result viathe coveringf to a path atu. Defineg(z) to be the end vertex of the resulting path inY .Edges and faces are similar: take a path inZ from x to a vertex in the boundary of theedge/face; map the grouping of path and edge/face toX via g, lift via f (using face liftingif necessary) and obtain an edge/face inY that will be our image.

We show that the result is well defined on the vertices. The edges and faces are similar.Suppose then thatγ1, γ2 are paths inZ from x to the vertexz. The diagram then commutesby the definition ofg and we leave it to the reader to show that it is a map of complexes.2

The process used in the construction ofg in the proof of Proposition 3.15.

Exercise 3.16 (lifting trees).Let f : X → Y be a covering andT ⊂ X a tree. Show that

1. f−1(T ) ⊂ Y is a forest.2. If Ti ⊂ Y, (i ∈ I) are the connected components of the forestf−1(T ), thenf maps

eachTi isomorphically ontoT .3. If Y/Ti andX/T are the quotients then there is an induced coveringY/Ti → X/T

making the diagram,Y Y/Ti

X X/T

commute, where the horizontal arrows are the quotient maps.4. This procedure is independent of the treeT : if T ′ ⊂ X is another tree such that there

is a homeomorphismα : X/T → X/T ′ with αq = q′ for q, q′ : X → X/T, X/T ′ thequotient maps, then there is an isomorphismY/Ti → Y/T ′

i .

3.1.3 Degree

We now come to an important numerical invariant that can be attached to a covering. Look-ing back at the example of Figure 3.2, we have a graph coveringof X where both the fiberof the single vertex and the fiber of the single edge contain two cells. Extending this cov-ering to one of2-complexes as in Figure 3.3, the fiber of the single face also contains twocells. This is no coincidence,

Proposition 3.17 (covering degree).If f : Y → X is a covering withX connected, thenany two fibers have the same cardinality.

This common cardinality of the fibers is called thedegreeof the covering, written

deg(Y → X) or deg(Y/X).

The connectedness ofX is essential, for ifX has componentsX1 andX2 say, andfi :Yi → Xi are coverings of different degree then we can cobble together a new coveringf :Y = Y1

∐Y2 → X1

∐X2 with f |Yi

= fi. The cardinality of the fibers now depends onwhich component ofXi they lie above. Anyway, the connectedness ofX is used explicitlyin the proof:


3.1 Basics 47

Proof. If v, u are vertices ofX andγ a path fromv to u, then liftingγ to a pathγ at anyvertex of the fiber ofv and taking its end vertext(γ), gives a (set) mapping from the fiberof v to the fiber ofu. By the uniqueness of the lift of the pathγ−1 at any vertex in the fiberof u, this mapping is injective. Interchanging the roles ofv andu gives an injective map inthe reverse direction, hence the fibers ofv andu have the same cardinality. Similarly, bythe second part of the definiton of covering and path lifting,if e is an edge ofX then thereis a bijection between the fiber ofe and the fiber of its start vertexs(e). This establishes thedegree for graphs.

Now to the fiber of a face. Letτ be a face ofX and letv be a vertex in its boundary with|(aτ )−1(v)| = m > 0. We will show that this face and vertex have fibers with the samecardinality, ie:|f−1(τ)| = |f−1(v)|. Thesem vertices are marked by little red circles inFigure 3.6 (form = 2). Let u1, . . . , uk be the vertices in the fiber ofv and consider those

• • •. . .. . . . . .

. . .. . . . . .

•

Xσi

Xτ

u1 u2 uk v9>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>;

f−1(v)

f

∼=

Fig. 3.6.Showing that the cardinality of the fiber of the faceτ is the same as the cardinality of thefiber of a vertexv in its boundary. The two fibers are shown on the left, and if|(aτ )−1(v)| consistsof them red vertices, then there arem|f−1(τ )| blue vertices mapped in anm-to-1 fashion onto thefiber ofv.

facesσi in the fiber ofτ containing one of the verticesuj in their boundary. Then in fact,everyface in the fiber ofτ must be in this collection of faces, by the incidence preservingproperty Definition 1.22 part 1, of the mapf . Thus, this set of facesσi is the fiber ofτ .Now, eachσi in this fiber hasXσi isomorphic toXτ , and soXσi hasm vertices thatcorrespond to them red vertices under this isomorphism. Consider this set ofm|f−1(τ)|vertices, marked in blue in Figure 3.6. As eachσi in the fiber maps toτ , each of these bluevertices is sent by the attaching maps to one of theuj in the fiber ofv. Thus, the variousattaching mapsaσi map this set of blue vertices to the fiber ofv.

We haven’t used the covering yet! Here is where we do: fix a vertexuj in the fiber ofv,so that the definition of covering applied to the tripleuj, v andτ givesm blue vertices map-ping via the various attaching maps touj. Thus, them|f−1(τ)| blue vertices are mappedin anm-to-1 fashionontothe fiber ofv, ie:

m|f−1(τ)| = m|f−1(v)|,

and we are done.2

Degree plays a similar role for coverings as does dimension for vector spaces or indexfor groups. In particular, “ifH is a subgroup of index one in a groupG thenH = G”, is anargument whose combinatorial topology version is,

Corollary 3.18. A degree one covering of a connected complex is an isomorphism.

The Corollary follows immediately from the surjectivity ofcoverings Proposition 3.8,and the definition of degree. This simple little result will play a crucial role in the proof ofthe Galois correspondence of§4.3.1. Notice that this is the second time in as many sectionsthat subgroups have made a natural appearence in covering space theory (the other timewas Corollary 3.13).


48 3 Coverings

Exercise 3.19 (degree of tree lifting).Show that in Exercise 3.16 part 3 we have

deg(Y → X) = deg(Y/Ti → X/T ).

3.2 Actions, intermediate and universal covers

3.2.1 Group actions

Recall from§1.2.3 that a group acts freely on a2-complex precisely when it acts freely onthe vertices, ie: for anyg ∈ G and vertexv, if g(v) = v theng is the identity. It turns outthat such group actions give us a plentiful supply of coverings:

Proposition 3.20.If a groupG acts orientation preservingly and freely on a2-complexXthen the quotient mapq : X → X/G is a covering.

Proof. is an exercise in not getting yourself confused! It may help to remember that thecells of X/G are the equivalence classes of the action and the fiber underq of any cellconsists of those cells ofX in the equivalence class. Suppose for definiteness thatG =gi, i ∈ I.

Thatq is a covering of the underlying graphs is straight-forward:if E is an equivalenceclass of edges thenE = gi(e), i ∈ I for some edgee of X , and inX/G it has start vertexthe equivalence classV = gi(v), i ∈ I for v = s(e). Then for the vertexgi(v) coveringV , the quotient maps the edgegi(e) ontoE. Two different edges starting atgi(v) cannotlie in the same class, for then a non-trivial element ofG would fix this vertex, contradictingthe freeness of the action.

SupposeΣ is a face of the quotient withV a vertex lying in its boundary. Thus there arev ∈ V andσ ∈ Σ with V = gi(v), i ∈ I, Σ = gi(σ), i ∈ I andgi(v) lying in theboundary ofgi(σ) as in Figure 3.7 (where for illustrative purposesG is finite of orderk).We are free, by Exercise 1.32, to choose the boundary ofΣ from amongst the faces inΣ,

XσXg1(σ) Xgk(σ). . .

• • •v g1(v) gk(v)

. . .

Fig. 3.7.A face of the quotientX/G and a vertex lying in its boundary

so we choose∂Σ = (Xσ, qaσ). Then the elements of(aΣ)−1(V ) are those vertices ofXσ

that are sent by the attaching mapaσ to a vertex of the equivalence classV . If v ∈ V issome vertex of this class, then the elements of

∐gi(σ)∈Σ(aσ)−1(v) are the vertices of the

variousXgi(σ) that are sent via their attaching mapsagi(σ) to v. Our job is to show thatthe local continuity map

∐εgi(σ) is a bijection between these two sets, whereεgi(σ) is the

isomorphismXgi(σ) → Xσ induced byg−1i .

For injectivity, we know that theεgi(σ) are individually injective; if fori 6= j there areverticesxi ∈ Xgi(σ) andxj ∈ Xgj(σ) sent via their attaching maps tov thengig

−1j is not

the identity but nevertheless fixesv, a contradiction. Thus the local continuity maps mustbe injective.

Let x be a vertex in(aΣ)−1(V ), hence sent by the attaching mapaσ to somegi(v).There is a vertexy of Xgi(σ) sent tox by the isomorphismεgi(σ), and since the attachingmap of the facegi(σ) is the composition


3.2 Actions, intermediate and universal covers 49

Xgi(σ)εgi(σ)

−→ Xσ aσ

−→ X(1) g−1

−→ X(1),

we get thaty is sent via this attaching map tov, and soy ∈∐

gi(σ)∈Σ(aσ)−1(v). The localcontinuity map is thus a surjection.2

3.2.2 Intermediate covers

Lemma 3.21 (intermediate graph coverings).Let Yq−→ X

r−→ Z be dimension pre-

serving maps of graphs and letp = rq. If any two ofp, q andr are coverings, then so is thethird.

Proof. Given three sets and three set maps forming a commutative triangle, then any twoof them a bijection implies that the third is also a bijection. This simple fact, applied to thesets of edges starting at a triple of verticesx, y, z with x = q(y) andz = r(x), gives theresult. 2

Lemma 3.22 (intermediate coverings of2-complexes).Let Yq−→ X

r−→ Z be dimen-

sion preserving maps of2-complexes and letp = rq. Then,

1. if q, r are coverings then so isp, and2. if p, q are coverings then so isr.

Proof. We do1 and leave2, which is similar, to the reader. First, we need to know aboutthe fiber underp of a faceτ ∈ Z:

p−1(τ) =∐

σ∈r−1(τ)

q−1(σ).

Let z ∈ Z be a vertex andy ∈ Y a vertex covering it via the (graph) coveringp. Letx = q(y) so thatz = r(x). We need to show that the local continuity map is a bijectionfrom the appearences ofy in the faces ofp−1(τ) and the appearences ofz in τ . Localcontinuity of the coveringr gives a bijection between the latter and the appearences ofx inthe faces ofr−1(τ), which is a disjoint union over the appearences ofx in each individualfaceσ ∈ r−1(τ). For eachσ ∈ r−1(τ), local continuity of the coveringq gives a bijectionbetween the appearences ofx in σ and the appearences ofy in the faces ofq−1(σ). Thedisjoint union of these appearences ofy in the faces ofq−1(σ) asσ varies overr−1(τ)gives the appearences ofy in the faces ofp−1(τ), and since the local continuity associatedto p is the composition of the local continuities ofq andr, we are done. 2

Why do we not have a result that says that ifp andr are coverings then so isq, like wedo for graphs? The problem is that for coverings of2-complexes, the fibers of a cell playa central role: we need to know about the fiber of a face in orderto verify that a map is acovering. Now, in the two cases of Lemma 3.22, we can describethe fiber of a face underthe map we are interested in, in terms of the two maps that we already know about. This isnot so when we know aboutp andr and are interested inq.

If Yq→ X

r→ Z are coverings, so that the compositionY

p→ Z is a covering, then call

q andr coveringsintermediateto p. It turns out that the set of coverings intermediate to afixed coveringY → Z carries a great deal of structure, and we will explore this more fullyin §3.4.

Exercise 3.23.Let Y be a graph andY1, Y2 → Y subgraphs of the form,

Y = Y1 Y2e

(†)


50 3 Coverings

1. If p : Y → X is a covering withX single vertexed, then the real line is a subgraphα : R → Y , with α(e0) = e andpα(ek) = p(e) for all k ∈ Z.

2. If Y1 is a tree,p : Y → X , r : Z → X coverings, andα : Y2 → Z an isomorphismonto its image, then there is an intermediate coveringY

q→ Z

r→ X .

3. If W → Y is a covering andY1 a tree, thenW also has the form(†) for some subgraphsY ′

1 , Y ′2 →W , with Y ′

1 a tree.

3.2.3 Universal covers

Much of the discussion of coverings so far as been in the abstract: we haven’t seen manyactual covers! By Exercise 3.4 we know that at the very least,a complex is a cover of itself,but that is rather trivial. In this section we show that a complex always has another coverwhich is at the other extreme, in that it is as “big” as possible.

Definition 3.24 (universal covers).A coveringf : Y → X is universalif and only if forany coveringq : Z → X there is a coveringp : Y → Z making the diagram,

Y Z

Xf

p

q

commute.

Equivalently,Y → X is universal when any other coveringZ → X of X is intermediateto it.

Exercise 3.25.Show that ifY1, Y2 → X are universal covers then there is an isomorphismY1∼= Y2. Thus, universal covers, if they exist, are unique.

But universal covers do exist! To construct them we mimic a standard construction intopology: letX be connected and fix a reference point vertexv0 of X and letX be thefollowing complex. Its verticesv are the homotopy classes of paths inX starting atv0.Suppose we have two such vertices,v1 andv2, the homotopy classes of the pathsγ1 andγ2. Then there is an edgee of X connectingv1 to v2 if and only if there is an edgee of Xsuch that the pathγ1e is homotopic toγ2. Figure 3.8 shows the kind of schematic set-upwe have.

e

v0

γ

γe

Fig. 3.8.How an edge ofX gives an edge ofeX : there is an edgeee with start vertex the homotopyclass ofγ and finish vertex the homotopy class ofγe.

Exercise 3.26.Let v1 and v2 be vertices ofX corresponding to the homotopy classes ofthe pathsγ1 andγ2. If there is a pathγ = e1 . . . ek in X from v1 to v2, then there is a pathγ = e1 . . . ek in X with the edgeei arising from the edgeei as above, andγ1γ homotopicto γ2.



Lemma 3.27.The graphX is connected. Definef : X → X(1) by f(v) = the terminalvertex ofγ, wherev is the homotopy class of the pathγ, andf(e) = e, where the edgeearises frome as above. Thenf is a graph covering.

Proof. If v is a vertex ofX corresponding to the pathγ = e1 . . . ek, thene1 . . . ek is a pathconnectingv to the homotopy class of the empty path, and soX is connected. To see thatfis a covering we need that it is a dimension preserving map (which we leave to the reader)and that for every pair of verticesv andv with f(v) = v, it induces a bijection from theedges starting atv to the edges starting atv. Suppose then thatv is a vertex correspondingto the class of the pathγ ande1, e2 are edges connectingv to verticesv1 andv2. Letv be theterminal vertex ofγ (so thatf(v) = v) ande an edge starting atv with f(e1) = f(e2) = e.Then the verticesv1 andv2 both correspond to the homotopy class of the pathγe, and sov1 = v2. The edgesei both arise by applying the construction described above to the pair ofverticesv andv1 = v2, and as only one edge can arise this way, we havee1 = e2. The localcontinuity maps are thus injective. For an edgee starting atv, and a vertexv in its fibercorresponding to the pathγ from v0 to v, there is, by definition, an edgee connectingv andthe vertex corresponding to the pathγe. This gives the surjectivity of the local continuitymaps. 2

X1 X2 X3

Fig. 3.9.Three complexes. Note thatX2 andX3 have the same1-skeleton.

eX1eX2

eX3

Fig. 3.10.The1-skeletons of the complexeX for theX of Figure 3.9.

Figure 3.9 gives three complexes and the graphsX are given in Figure 3.10. IfX is theS1-graph on the left, then there is a 1-1 correspondence between the homotopy classes ofpaths and paths of the forme . . . e (k times) ore−1 . . . e−1 (k times). ThusX has verticesvk for k ∈ Z. There is an edge connecting the vertex of the pathe . . . e (k times) to thevertex of the pathe . . . e (k + 1 times), to give the infinite2-valent tree at left in Figure3.10. Similarly the single-vertexed graph with two edges has X the4-valent infinite tree.

The last of the three complexes is the torus, which has exactly the same1-skeleton asX2, but the presence of a face drastically changes the graphX. In X2 paths of the formγandγe1e2e

−11 e−1

2 give distinct vertices inX2, but thesamevertex inX3.One more thing before proceeding with the definition ofX : the graphX gives a covering

of the graphX , but not necessarily a universal one. We can see this in the examples above


52 3 Coverings

whereX2 andX3 have the same1-skeleton but quite differentX . Indeed, it is not hard toconstruct a graph coveringX2 → X3, soX3 is most definitely not universal.

Now to the faces ofX , which arise out of the faces ofX , much as the edges do. First,we urge the reader to do the following exercise.

Exercise 3.28.In the graph coveringf : X → X , show that the boundaries of faces inXlift to closed paths inX. More precisely, letu be a vertex ofX , σ a face containingu inits boundary andγ a boundary path ofσ starting atu. If v is a vertex coveringu via the(graph) coveringf andγ is the lift of γ atv, thenγ is a closed path inX.

Thus the boundaries of faces inX give rise to closed paths inX , and the idea behind theconstruction of the2-skeleton ofX is to “sew” faces into these closed paths (actually wesaw this phenomenon when we constructed the graphX3 in Figure 3.10).

Fix a triple consisting of a vertexu of X , a vertexv of X covering it, and a faceσ ofX . Suppose∂σ = (Xσ, aσ) and thatu appearsk times in the boundary ofσ. In particular,there arek paths circumnavigatingXσ clockwise, and mapping via the attaching mapaσ

to k boundary paths ofσ starting atu.Let x be one of thek appearences ofu in the boundary ofσ; γ the path circumnavigating

Xσ clockwise starting atx andaσ(γ) the resulting boundary label ofσ starting atu. Liftthe labelaσ(γ) to v (via the graph coveringf : X → X), to get by Exercise 3.28, a closedpath inX. If there is already a face ofX coveringσ and with boundary the lift ofaσ(γ)to v, then do nothing. Otherwise, sew in a new faceσ with this boundary in the followingway: let ∂σ = (Xσ, aeσ) whereaeσ is the composition ofaσ and the lift ofaσ(γ) to X.Pointaeσ using the verticesx ∈ Xσ andv ∈ X.

Repeat the procedure above for all such triplesu, v, σ, rememberingnot to sew in a newface if there is already one coveringσ with the lifted boundary label.

Xσe1 e−11

e2

e−12

ue1 e2

v

Fig. 3.11.Universal cover of the torus: the fourγ circumnavigatingXσ are shown in red, blue, greenand yellow. The lifts of theaσ(γ) to v are shown ineX on the right. As each is a different path, weget four distinct faces coveringσ and containingv in their boundary.

Figure 3.11 shows the result of performing this process withthe complexX3 of Figure3.9, sewing faces onto the1-skeleton of Figure 3.10. Figure 3.12 shows the effect onX ofan extra face inX .

Proposition 3.29.Definef : X → X on the1-skeleton as in Lemma 3.27, and for eachface σ arising as above from a faceσ of X , definef(σ) = σ. Thenf is a covering of2-complexes.

Corollary 3.30. The2-complexX is simply connected.

Proof. Let γ be a closed path inX based at the vertexv corresponding to the empty path atv0. Then by Exercise 3.26 it covers a pathγ at v0 that is homotopically trivial. Homotopylifting via the coveringf of Proposition 3.29 gives thatγ is homotopically trivial too, andhence thatX is simply connected. 2



Xσe1 e−11

e2

e−12

e1

Xτ

ue1 e2

v

Fig. 3.12.ComplexX (left) obtained by sewing another faceτ onto the torus and its universal covereX (right). Now the red and blueγ circumnavigatingXσ lift to the same set of edges ineX (taken ina different order), as do the green and yellowγ. We thus get two faces in the fiber ofσ that containvin their boundary.

This leaves one last piece of remaining business:

Proposition 3.31.The coveringf : X → X is universal.

Proof. Given a coveringq : Y → X we construct a coveringX → Y by “cover and lift”.2

Exercise 3.32.Show, using universal coverings, that ifX is a graph andγ1, γ2 are reducedhomotopic paths inX with the same start vertex, thenγ1 = γ2. [hint: lift the paths toXand use properties of reduced paths in trees to deduce that these lifts are identical.]

3.2.4 Monodromy

Whenf : Y → X is a covering we will eventually get an action of two groups onY , or atleast on parts ofY . The more important of these is the Galois group of the covering, whichacts on the whole complexY , and forms the principle subject of Chapter 4.

If v is a vertex ofX then the lesser of these two actions is that of the fundamental groupπ1(X, v) on the fiberf−1(v) of the vertexv. Thus, the fundamental group is acting asa permutation group of the set of vertices lying overv. Such permutation representationsof the fundamental groups of2-complexes will play a key role in the proof of results likeMiller’s theorem in Chapter 7.

To define the action, see that it makes sense, and is a homomorphism

π1(X, v)→ Sym(f−1(v)),

we require no more than the path and homotopy lifting of§3.1.2. So, the “path-liftingaction” would probably be a sensible name: the action would then do exactly what it sayson the box! However, it is traditional in topology to call this actionmonodromy, and so wewill too.

The definition is illustrated in Figure 3.13: letγ be a closed path atv representing anelement (which, by our custom, we also callγ) of the fundamental groupπ1(X, v). Todefine the image of a vertexu in the fiber ofv, take the liftγ of γ at u, and suppose thatthis lift has end vertexx. Defineγ(u) = x.

We obviously have a well-defined issue to deal with, so that the action does not dependon our choice of path. Ifγ1 is another closed path atv representing the same element of thefundamental group (so in the group we haveγ = γ1), then the pathsγ, γ1 are homotopic.This homotopy can be lifted, via homotopy lifting, to a homotopy between the liftsγ andγ1 atu, and so the two lifts are homotopic inY .

But homotopic paths have the same endpoints! Thusγ1 ends atx as well, and we get thesame image vertex ofu irrespective of the path chosed to representγ.


54 3 Coverings

• • • •. . .. . . . . . γ

eγ

vu xf

9>>>>>>>>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>>>>>>>>;

f−1(v)

Fig. 3.13.Defining the monodromy using path lifting.

By Exercise 3.6 the lift at a vertex of the pathγ1γ2 is the pathγ1γ2 obtained by liftingγ1 and thenγ2. Thus the map sendingγ to the permutation off−1(v) defined above is agroup homomorphismπ1(X, v)→ Sym(f−1(v)), and we have our action. Summarising,

Proposition 3.33.If Y → X is a covering then monodromy gives a homomorphism

π1(X, v)→ Sym(f−1(v)).

In particular, a covering of finite degree gives a homomorphism fromπ1(X, v) to a finitegroup.

Suppose thatX has just the one vertexv, so that the fiber ofv consists of all the verticesof Y . Monodromy thus gives an action on the whole0-skeleton ofY . The next exerciseshows that in general this action cannot be extended any further than this, ie: there areexamples where it cannot be extended from the0-skeleton to the1-skeleton.

Exercise 3.34. 1. Let X be the complex of Figure 3.14. Describe the universal coverX → X , showing that in particular that it is a covering of degree6.

X(1)

e1 e2

Xσ1

e1

e1

Xσ2

e2

e2

Xτ

e1

e2

e1

e2

e1

e2

Fig. 3.14.complexX for Exercise 3.34.

2. Show that inX there exists an edge that joins two vertices, but no edge joining theimages of these two vertices under the monodromy action ofπ1(X, v). Thus it is notpossible to define an automorphism ofX at this edge. Deduce that there can thereforebe no homomorphism fromπ1(X, v) to the automorphism group of the1-skeletonextending the monodromy action on the0-skeleton.

3.3 Operations on coverings

In the first chapter we had three constructions arising from acollection of complexes andmaps between them: the pushout, pullback and Higman composition. In this section weshow that all three are useful ways of creating new coveringsfrom old. Throughout thissection all complexes are connected.

3.3.1 Pushouts of covers

In this subsection, letf : Y → Z be a fixed covering of connected2-complexes, andfor i = 1, 2, let Y

pi→ Xi

qi→ Z be connected coverings intermediate tof . Thus we

have the commuting diagram of Figure 3.15 left, with all the maps in sight coverings.


3.3 Operations on coverings 55

As the pi are dimension preserving, we may, by Proposition 1.40, formthe pushoutX1

∐Y X2, obtaining in the process mapsti : Xi → X1

∐Y X2 as the composition

Xi → X1

∐X2 → X1

∐X2/∼ of the inclusion ofXi in the disjoint union and

the quotient map. The universality of the pushout, Proposition 1.37, applied to the mapsqi : Xi → Z, gives the commuting diagram on the right of Figure 3.15. If[x] is a cell ofthe quotientX1

∐X2/∼, thenx ∈ Xi for somei, and so the new maph : X1

∐Y X2 → Z

sends[x] to qi(x) ∈ Z.

Y

X1

X2

Z

p1

p2

q1

q2f

Y

X1

X2

X1

‘Y

X2

p1

p2

t1

t2

Zq1

q2

Fig. 3.15.

Proposition 3.35 (pushouts of covers).The maps

ti : Xi → X1

∐

Y

X2, for i = 1, 2, andh : X1

∐

Y

X2 → Z,

are coverings. Thus, the pushout of two intermediate coverings is an intermediate covering.

Proof. We show that theti are coverings, so part 2 of Proposition 3.22 gives ush for free.For ease of expression we dot1, with t2 completely analogous. It is clearly dimensionpreserving, leaving us to show that the local continuity maps are bijections. This is similarfor both edges and faces, so suppose that[u] is a vertex of the pushout andv a vertex ofX1

mapping to it viat1. Thusu andv are equivalent under the quotient relation of the pushout,so we may write[v] for the vertex instead. Suppose we have a cell of the pushout incidentwith the vertex[v]: an edge[e′] with start vertex[v] or a face[σ′] with [v] appearing in itsboundary. Thus,e′ andσ′ are cells of the disjoint unionX1

∐X2, and there is a vertex

v′ ∈ X1

∐X2, equivalent tov, that is the start vertex ofe′, or appears in the boundary

of σ′. The equivalence betweenv andv′ is realized by a sequence of lifts and covers (ofvertices) through the coveringspi : Y → Xi. The same sequence, but using path or facelifting instead, yields an edgee ∈ X1 or a faceσ ∈ X1, equivalent to thee′ or σ′, and withv the start vertex ofe or v appearing in the boundary ofσ.

Two objects atv (edges starting atv or appearances ofv in faces) mapping undert1 tothe same object at[v] must then map underh to the same object inZ. As ht1 = q1, thesetwo objects atv map via the coveringq1 to the same object atq(v1), a contradiction, andso the local continuity maps are injective.2

Exercise 3.36.Is the pushout of two immersions an immersion?

3.3.2 Pullbacks of covers

In this subsection, letf : Y → Z be a fixed covering of connected2-complexes, and fori = 1, 2, let Y

pi→ Xi

qi→ Z be connected coverings intermediate tof . Thus we have the

commuting diagram on the left of Figure 3.16 with all the mapsin sight coverings. Asthe coveringsqi are dimension preserving, we may, via Definition 1.42, form the pullbackX1

∏Z X2, obtaining in the process mapsti : X1

∏Z X2 → Xi given byti : x1 × x2 7→

xi. The universality of the pullback, Proposition 1.43, applied to the mapspi : Y →Xi, gives the commuting diagram on the right of Figure 3.16. Thenew maph : Y →X1

∏Z X2 takes a celly ∈ Y and maps it to the cellp1(y)× p2(y) ∈ X1

∏Z X2.


56 3 Coverings

Y

X1

X2

Z

p1

p2

q1

q2f

X1

QZ

X2

X1

X2

Z

t1

t2

q1

q2

Y

p1

p2

Fig. 3.16.

Proposition 3.37 (pullbacks of covers).The maps

ti : X1

∏

Z

X2 → Xi, for i = 1, 2, andh : Y → X1

∏

Z

X2,

are coverings. Thus, the pullback of two intermediate coverings is an intermediate cover-ing.

Proof. It suffices, by part 2 of Proposition 3.22 to show thath is a covering. It is dimensionpreserving as thepi are, and so it remains to show that the various local continuity mapsare bijections. This is similar for both edges and faces: suppose thatv1 × v2 is a vertex ofthe pullback andu a vertex ofY mapping to it viah. If two objects atu (edges starting atu or appearances ofu in faces) map underh to a single object atv1 × v2, then these twomap via thepi to single objects at thevi in theXi. The coveringspi then ensure that theoriginal two objects coincide, hence the injectivity of thelocal continuity maps.

Surjectivity requires a couple more steps: start with an object at the vertexv1× v2 of thepullback. It maps via theti to objects at thevi, and they in turn map via theqi to thesameobject atv = qi(vi). The path and face lifting provided by the coverspi give two objects atu mapping to this single object atu, one viaq1p1 and the other viaq2p2. But then these twoobjects map via the coveringf to this single object atu, and so must be the same object. Bydefinition, the image viah of this single object atu must be the original object atv1 × v2

that we started with. 2

Exercise 3.38.Is the pullback of two immersions an immersion?

Exercise 3.39.Let Y → X ← Z be coverings withZ a forest. Show that the pullbackX

∏X Z is also a forest.

We pause to observe a slight asymmetry to the duality betweenpushouts and pullbacks:given coveringsr1, r2 : Y1, Y2 → X , the t1, t2 : Y1

∏X Y2 → Y1, Y2 are coverings,

whereas coveringsq1, q2 : Z → Y1, Y2 do not necessarily give coveringst1, t2 : Y1, Y2 →Y1

∐Z Y2, unless theqi are intermediateZ → (Y1 or Y2)→ X . Indeed, taking theY1 = Y2

to be two copies of the left hand graph,

Y1 = Y2 = = Z =qi

and the coveringsqi : Z→Y1 or Y2 (described here by drawing the fibers of the vertices),then theti provided by the pushout construction are not coverings of the pushout.

3.3.3 Higman compositions of covers

3.4 Lattices of covers

3.4.1 Aside: posets and lattices

At this stage we pause and take a brief look at the general theory of posets and lattices. Thiswill not be comprehensive: we will just familiarise ourselves with the basic terminology


3.4 Lattices of covers 57

and the first of two important examples. There are many books on this subject. We havefollowed [14, Chapter 3].

Partially ordered sets (orposets) formalise the idea of ordering:

Definition 3.40 (poset).A poset is a setP and a binary relation≤ such that

1.≤ is reflexive: x ≤ x for all x ∈ P .2.≤ is antisymmetric: if x ≤ y andy ≤ x thenx = y.3.≤ is transitive: if x ≤ y andy ≤ z thenx ≤ z.

The motivating example is meant to be the integersZ with their usual ordering≤, andthe usual notational conventions from there are used in general: we writex < y to meanx ≤ y butx 6= y. Elementsx, y with x ≤ y or y ≤ x arecomparable, otherwise they areincomparable, a possbility that obviously doesn’t arise with the primordial exampleZ. Wesay thaty coversx, written x ≺ y, whenx < y and if x ≤ z ≤ y then eitherz = x orz = y.

A morphism(or justmap) of posetsf : P → Q is an order-preserving map of sets: ifx ≤ y in P thenf(x) ≤ f(y) in Q. Notice that this is a one way business: comparableelements are sent to comparable elements, but incomparableelements are allowed to be-come comparable. Similarly ananti-morphismis an order-reversing map: ifx ≤ y in Pthenf(y) ≤ f(x) in Q.

Bijective morphisms have inverses, although they may not bemorphisms. A bijectivemorphism with order-preserving inverse is anisomorphism: x ≤ y in P if and only iff(x) ≤ f(y) in Q. Similarly a bijective anti-morphism with order-reversing inverse is ananti-isomorphism.

Posets are often illustrated using theirHasse diagram: a graph whose vertices are theelements ofP and whose edges are the covering relations. Thus, ifx ≺ y then the vertexy is drawn above the vertexx with an edge connecting them. Two examples of Hassediagrams (and posets) are given in Figure 3.17.

∅

1 2 3

1, 2 1, 3 2, 3

1, 2, 3

x y

x ∨ y

Fig. 3.17.Hasse diagram for the poset of subsets of the set1, 2, 3, ordered by inclusion (left) andfor a poset with four elements (right) that is not a lattice.

A special place is reserved for those posets which have supremums and infimums. Ifx, y ∈ P thenz is anupper boundfor them when bothx ≤ z andy ≤ z. It is a least upperboundor supremumor join when it is an upper bound such that for any other upper boundw we havez ≤ w. Similarly, z is a lower boundfor x andy when bothz ≤ x andz ≤ y.It is agreatest lower boundor infimumor meetwhen it is an lower bound such that for anyother lower boundw we havew ≤ z.

It is easy to show that ifx andy have a join then it is unique (hint: any two joins mustbe≤ each other) and similarly for the meet. Writex∨ y for the join andx∧ y for the meetof x andy.

Definition 3.41 (lattice).A lattice is a poset in which every pair of elements has a join anda meet.


58 3 Coverings

The poset on the left of Figure 3.17 is a lattice, as can be checked directly from the Hassediagram, but the example on the right is not: ifx andy are the two minimal elements, thenthey have a join, but no meet.

Exercise 3.42.A 1 in P is a unique maximal element: for allx ∈ P we havex ≤ 1.Similarly a 0 in P is a unique minimal element: for allx ∈ P we have0 ≤ x. Show that afinite lattice has a0 and a1.

Exercise 3.43.A poset is ameet-semilatticeif any two elements have a meet. Dually wehave the notion of ajoin-semilattice. Show that ifP is a finite meet-semilattice with a1thenP is a lattice (dually, ifP is a finite join-semilattice with a0 thenP is a lattice).

Exercise 3.44.Let P andQ be lattices andf : P → Q a lattice isomorphism (respectivelyanti-isomorphism). Show thatf sends joins to joins and meets to meets (resp. joins to meetsand meets to joins), ie:f(x ∨ y) = f(x) ∨ f(y) andf(x ∧ y) = f(x) ∧ f(y).

The most commonly occuring lattice in nature is theBoolean latticeon a setX : itselements are the subsets ofX andA ≤ B iff A ⊂ B. Meets and joins are just intersectionsand unions:A ∧B = A ∩B andA ∨B = A ∪B.

Exercise 3.45.Let X be a finite set,P the Boolean lattice onX andV the real vectorspace with basisX . If v =

∑X λxx ∈ V define|v|2 =

∑X λ2

x, and let2n := v ∈V : |v|2 ≤ 1, then-dimensional cube. Embed the underlying set ofP in V via the mapsendingA ⊂ X to

∑x∈A x, and show that the image ofP is the set of vertices of2n,

while the vertices and edges of2n give the Hasse diagram forP .

A slightly less trivial example is the latticeLn(F) of all subpacesof then-dimensionalvector space over the fieldF, with the ordering given by inclusion of one subspace inanother. The meet of two subspaces is again their intersection, but this time the union istoo small to be their join: the union of two subspaces is not a subspace! Instead we takeU ∨ V = U + V , their sum, consisting of all vectors of the formu + v for u ∈ U andv ∈ V . We leave it to the reader to verify that these are indeed infimums and supremums.

Here is one we are particularly interested in,

Definition 3.46 (lattice of subgroups).Let G be a group. The lattice of subgroupsL(G)has as elements the subgroups ofG ordered by inclusion,H ∧K = H ∩K andH ∨K =〈H, K〉, the subgroup generated byH andK.

S3

id, σ, σ2

id, τ id, στ id, σ2τ

id

Fig. 3.18.Subgroup lattice for the symmetric groupS3: let σ = (1, 2, 3) andτ = (2, 3) so thatστ = (1, 2) andσ2τ = (1, 3).

Exercise 3.47.Show that the set offinite indexsubgroups of a groupG also forms a lattice,with the same meet and join asL(G).

Exercise 3.48.Show that the set offinitely generatedsubgroups of a groupG also forms alattice, with the same meet and join asL(G).



3.4.2 The poset of intermediate covers

In the previous section we said that a lattice of particular interest to us was the latticeof subgroups of a group. In this section and the next, we construct another lattice whoseelements are, more or less, the coverings intermediate to a particular fixed coveringf . Inthe next chapter we’ll see that if we look at this lattice sideways and squint our eyes a little,then it looks the same as the lattice of subgroups of the “group of automorphisms” of thecoveringf .

Throughout this sectionf : Y → Z is a fixed covering of connected2-complexes. Fori = 1, 2, let Y

pi→ (Xi)xi

qi→ Z be connected coverings intermediate tof .

We call these two intermediate coveringsequivalentif and only if there is a isomorphismX1 → X2 making the diagram below right commute:

Y

X1

X2

Z

p1

p2

q1

q2f

Y

X1

X2

Z

p1

p2

q1

q2∼=equivalent ⇒ commutes

This is an equivalence relation on the set of coverings intermediate tof , and we write

L(Yf→ Z) or justL(Y, Z) for the set of equivalence classes of intermediate coverings. The

notation here can become very cumbersome very quickly, so where possible we will writeX ∈ L(Y, Z) to mean that this equivalence class is represented by coveringsY → X → Zintermediate tof .

It is possible to do everything in this section, and the next,in terms of intermediate cov-erings themselves, and not worry about equivalence at all. Nevertheless, it will be essentiallater to make sure all the accounting comes out in the wash. Figure 3.19 shows a pair ofequivalent graph coverings. The graphsXi are necessarily the same, but the coverings aredifferent.

p1

p2

q1 = q2

Fig. 3.19.Equivalent intermediate graph coverings:Z is a bouquet of two loops andY is its universalcover, an infinite4-valent tree. The intermediateX1 = X2, coverZ (as it is so simple) by the samecoveringsq1 = q2. The coverings differ in howY covers theXi via the pi: the first sends thegreen vertex ofY to the red vertex of theXi and the second sends it to the blue vertex. There is anisomorphismX1 → X2 interchanging the two vertices.

We now turnL(Y, Z) into a poset, for which the ordering is this: one (equivalence classof an) intermediate covering is “bigger” than another if thefirst covers the second. Specif-ically, if X1, X2 ∈ L(Y, Z) then defineX1 ≤ X2 precisely when there is a coveringX2 → X1 making the diagram on the left of Figure 3.20 commute.

Presupposing for a minute that this definition makes sense and gives a partial order, wehave


60 3 Coverings

Y

X1

X2

Z

p1

p2

q1

q2cover

Y

X2 X1X ′

2 X ′

1

Z

Fig. 3.20.

Definition 3.49 (poset of intermediate covers).For a fixed coveringf : Y → Z ofconected complexes, the setL(Y, Z) of equivalence classes of connected intermediate cov-erings, together with the partial order≤ defined above is called theposet of intermediatecoverings(to f ).

It is not hard to check that this is well defined: suppose that for i = 1, 2, we have inter-mediate coveringsX ′

i, equivalent to theXi via isomorphismsXi ↔ X ′i. As isomorphisms

are nothing other than degree one coverings, the red map across the middle of the diagramon the right of Figure 3.20 is a covering making the big outside square commmute. ThusX ′

1 ≤ X ′2, and the order doesn’t depend on which representative for the equivalence class

we choose.

Lemma 3.50.The setL(Y, Z) of equivalence classes of coverings intermediate tof : Y →Z, together with the≤ defined above, is a poset.

Proof. Reflexivity and transitivity are immediate, as the identitymap is a covering andthe composition of coverings is a covering. Anti-symmetry requires a moments thought:suppose we have intermediate coveringsX1, X2 ∈ L(Y, Z) with X1 ≤ X2 andX2 ≤ X1,ie: there are coveringsX1 X2 making the appropriate diagrams commute. Letg bethe compositionX1 → X2 → X1 of these two. Then consideration of these commutingdiagrams gives thatp1 = gp1, werep1 : Y → X1 is the covering, and so by the surjectivityof p1, g is the identity map onX1. But then the coveringX1 → X2 must be injective, ie:of degree1, and so an isomorphism. ThusX1 = X2 in L(Y, Z). 2

3.4.3 The lattice of intermediate covers

In the last section we introduced the posetL(Y, Z) of equivalence classes of coveringsintermediate to a fixed coveringY → Z. In this section we show that we have in fact alattice, with join a pullback and meet a pushout.

Throughout,f : Yu → Zv is a fixed pointed covering of connected2-complexes. Allintermediate coveringsY

p→ Xx

q→ Z are connected and pointed, andL(Yu, Zv) is the

poset of equivalence classes of pointed connected intermediate coverings.We start by showing that we have a meet. LetYu→(X1)x1→Zv andYu→(X2)x2→Zv

be intermediate tof , andX1

∐Y X2 the pushout of the coveringsp1 : Yu→(X1)x1 and

p2 : Yu→(X2)x2 . Let x = [x1] = [x2], where· 7→ [·] is the quotient map arising from theconstruction of the pushout, and(X1

∐Y X2)x the resulting pointed pushout.

By Proposition 3.35 we have a new element ofL(Yu, Zv) given by the equivalence classof the intermediate covering,

Yu→(X1

∐

Y

X2)x→Zv.

Our next result shows that if theXi are replaced by equivalent coverings then the newpushout that results is equivalent to the old one:



Proposition 3.51.Let(V1)y1 , (V2)y2 ∈ L(Yu, Zv) be equivalent to(X1)x1 , (X2)x2 via theisomorphisms,

g1 : (X1)x1 → (V1)y1 andg2 : (X2)x2 → (V2)y2 .

Define a mapg1∐g2 : X1∐X2 → V1∐V2 between the disjoint unions byg1∐g2|X1 = g1

andg1 ∐ g2|X2 = g2. Then the map

g : (X1

∐

Y

X2)x → (V1

∐

Y

V2)y, (y = [y1] = [y2]),

defined bygr = r′(g1 ∐ g2), wherer, r′ are the quotient maps arising in the pushouts, isan isomorphism making these pointed pushouts equivalent.

Thus, the pushout can be extended in a well defined way to equivalence classes of inter-mediate coverings, and so for(X1)x1 , (X2)x2 ∈ L(Yu, Zv) we write

(X1

∐

Y

X2)x ∈ L(Yu, Zv),

for the pushout of these two equivalence classes.

Proof (of Proposition 3.51).is a tedious but routine diagram chase.2

Now, Proposition 3.35 gives coveringsti : Xi → X1

∐Y X2 so that(X1

∐Y X2)x ≤

(Xi)xiis a lower bound inL(Yu, Zv) for eachi. If Vy is any other lower bound then we get

coveringsYu → (Xi)xi→ Vy , and by the universality of the pushout, Proposition 1.37,

we have a map(X1

∐Y X2)x → (V )y, which by Proposition 3.22 is a covering. Thus

Vy ≤ (X1

∐Y X2)x, and the pushout is the meet of the two equivalence classes(Xi)xi

.Now to joins, which are similar. LetYu→(X1)x1→Zv and Yu→(X2)x2→Zv be in-

termediate tof , andX1

∏Z X2 the pushout of the coveringsq1 : (X1)x1 → Zv and

q2 : (X2)x2 → Zv. Let x = x1 × x2, a vertex of the pullback, and(X1

∏Z X2)x the

pointed pullback consisting of the connected component containingx.We have a well-definedness result analogous to Proposition 3.51

Proposition 3.52.Let(V1)y1 , (V2)y2 ∈ L(Yu, Zv) be equivalent to(X1)x1 , (X2)x2 via theisomorphisms,

g1 : (X1)x1 → (V1)y1 andg2 : (X2)x2 → (V2)y2 .

Then the mapg : (X1

∏

Z

X2)x → (V1

∏

Z

V2)y, (y = y1 × y2),

defined byg(x1×x2) = g1(x1)×g2(x2) is an isomorphism making these pointed pullbacksequivalent.

Proposition 3.37 gives coveringsti : (X1

∏Z X2)x → (Xi)xi

so that the(Xi)xi≤

(X1

∐Y X2)x and the pullback is an upper bound inL(Yu, Zv) for eachi. If Vy is any

other upper bound we get coveringsVy → (Xi)xi→ Zv, and the by universality of

the pullback, Proposition 1.43, we have a map(V )y → (X1

∏Z X2)x. Proposition 3.22

again, this time applied to the commuting triangle formed by(Y )u → (X1

∏Y X2)Z and

(Y )u → (V )y → (X1

∏Z X2)x, gives that the mapVy → (X1

∏Z X2)x is a covering.

Thus (X1

∐Y X2)x ≤ Vy , and the pullback is the join of the two equivalence classes

(Xi)xi.

Finally, recall from Exercise 3.42 that a1 in a poset is a unique maximal element, anddually, a0 is a unique minimal element. Here is the principle result of§3.4:

Theorem 3.53 (lattice of intermediate coverings).The posetL(Yu, Zv) of pointed con-nected covers intermediate to a fixed coveringf : Yu → Zv is a lattice with join(X1)x1 ∨ (X2)x2 the pullback(X1

∏Z X2)x1×x2 , meet(X1)x1 ∧ (X2)x2 the pushout

(X1

∐Y X2)[xi], unique minimal element0 = Zv and unique maximal element1 = Yu.


62 3 Coverings

The pointing of the covers in this section is essential if onewishes to work withcon-nectedintermediate coverings and also have a lattice structure (both of which we do). Theproblem is the pullback: because it is not in general connected, we need the pointing to tellus which component to choose.

3.5 Notes on Chapter 3


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