simplicial objects and homotopy groups j. wupeople.brandeis.edu/~syzygy/jiewusimplicial.pdf5....

Simplicial Objects and Homotopy Groups

J. Wu†

Department of Mathematics, National University of Singapore,Singapore

E-mail address: [email protected]: www.math.nus.edu.sg/~matwujie

†Partially supported by a grant from the National University of Singapore.

Contents

Chapter 1. ∆-Objects and Homology 51. ∆-sets 52. Geometric Simplicial Complexes 73. Abstract Simplicial Complexes and ∆-sets 254. ∆-complexes and the geometric realization of ∆-sets 285. Homology of ∆-sets 306. Simplicial Homology and Singular Homology 33

Chapter 2. Simplicial Sets and Homotopy 351. Simplicial Sets 352. Geometric Realization of Simplicial Sets 503. Homotopy and Fibrant Simplicial Sets 534. The Relations between Spaces and Simplicial Sets 585. Homotopy Groups 636. Simplicial Fibration 88

Chapter 3. Simplicial Group Theory 1051. Moore Chains and Homotopy Groups 1062. Simplicial Abelian Groups and the Hurewicz Theorem 1103. Free Group Constructions 1154. Free Products with Amalgamation 1265. Simplicial and ∆-Structure on Configuration Spaces (Homotopy Groups

and Braids) 132

Bibliography 157

3

CHAPTER 1

∆-Objects and Homology

1. ∆-sets

Definition 1.1. A ∆-set means a sequence of sets X = {Xn}n≥0 with facesdi : Xn → Xn−1, 0 ≤ i ≤ n, such that(1.1) didj = djdi+1

for i ≥ j, which is called the ∆-identity.

Remark 1.2. One can use coordinate projections for catching ∆-identity:

di : (x0, . . . , xn) −→ (x0, . . . , xi−1, xi+1, . . . , xn).

Let O+ be the category whose objects are finite ordered sets and whose mor-phisms are functions f : X → Y such that f(x) < f(y) if x < y. Note that theobjects in O+ are given by [n] = {0, 1, . . . , n} for n ≥ 0 and the morphisms in O+are generated by di : [n− 1] −→ [n] with

di(j) ={

j if j < ij + 1 if j ≥ i

for 0 ≤ i ≤ n, that is di is the ordered embedding missing i. We may write thefunction di in matrix form:

di =(

0 1 · · · i− 1 i i+ 1 · · · n− 10 1 · · · i− 1 i+ 1 i+ 2 · · · n.

).

The morphisms di satisfy the following identity:

djdi = di+1dj

for i ≥ j.

Remark 1.3. For seeing that morphisms in O+ are generated by di, observethat any morphism in O+ means an ordered embedding, which can be written asthe compositions of di’s.

Let S denote the category of sets.

Proposition 1.4. ∆-sets are one-to-one correspondent to contravariant func-tors from O+ to S.

Proof. Let F : O+ → S be a contravariant functor. Define Xn = F ([n]) and

di = F (di) : Xn = F ([n])→ Xn−1 = F ([n− 1]).Then X is a ∆-set.

Conversely suppose that X is a ∆-set. Define the F : O+ → S by settingF ([n]) = Xn and F (di) = di. Then F is a contravariant functor. �

5

6 1. ∆-OBJECTS AND HOMOLOGY

A ∆-set G = {Gn}n≥0 is called a ∆-group if each Gn is a group, and each facedi is a group homomorphism. In other words, a ∆-group means a contravariantfunctor from O+ to the category of groups. More abstractly, for any category C,a ∆-object over C means a contravariant functor from O+ to C. In other words,a ∆-object over C means a sequence of objects over C, X = {Xn}n≥0 with facesdi : Xn → Xn−1 as morphisms in C.

Example 1.1 (n-simplex). The n-simplex ∆+[n], as a ∆-set, is as follows:

∆+[n]k = {(i0, i1, . . . , ik) | 0 ≤ i0 < i1 < · · · < ik ≤ n}for k ≤ n and ∆+[n]k = ∅ for k > n. The face dj : ∆+[n]k → ∆+[n]k−1 is given by

dj(i0, i1, . . . , dk) = (i0, i1, . . . , îj , . . . , ik),

that is deleting ij . Let σn = (0, 1, . . . , n). Then

(i0, i1, . . . , ik) = dj1dj2 · · · djn−kσn,where j1 < j2 < · · · < jn−k with {j1, . . . , jk} = {0, 1, . . . , n} r {i0, i1, . . . , ik}. Inother words, any elements in ∆[n] can be written an iterated face of σn.

Definition 1.5. A ∆-map f : X → Y means a sequence of functionsf : Xn → Yn

for each n ≥ 0 such that f ◦ di = di ◦ f , that is the diagram

Xnf - Yn

Xn−1

di

? f - Yn−1

di

?

commutes. A ∆-subset A of a ∆-set X means a sequence of subsets An ⊆ Xn suchthat

di(An) ⊆ An−1for all 0 ≤ i ≤ n

2. GEOMETRIC SIMPLICIAL COMPLEXES 7

Definition 1.7. Let X be a ∆-set and let S ⊆⋃∞

n=0Xn. The ∆-subsetgenerated by S is defined by

〈S〉∆ =⋂{A ⊆ X |S ⊆

∞⋃n=0

An A = {An} is a ∆− subset of X}.

For x ∈ Xn, 〈{x}〉∆ is simply denoted by 〈x〉∆. A ∆-set X is called monogenic ifit is generated by a single element.

Proposition 1.8. Let X be a ∆-set and let S ⊆⋃∞

n=0Xn. Then

〈S〉∆n = (S ∩Xn) ∪⋃

0 ≤ j1 < j2 < · · · < jk ≤ n+ k1 ≤ k


Exercise 2.1. (1). The barycentric coordinates ti(x) of x with respectto a0, . . . , an are continuous on x.

(2). x ∈ ∂σ ⇐⇒ ti(x) = 0 for some 0 ≤ i ≤ n. Thus x ∈ Intσ ⇐⇒ ti(x) > 0for all 0 ≤ i ≤ n.

(3). σ is compact and convex in Rm, where a subset A of Rm is called convexif for each pairs x, y ∈ A the line segment joining x and y lies in A.

(4). Given a simplex, there is one and only one geometrically independent setof points spanning σ.

(5). σ equals to the union of all line segments joining a0 to points of thesimplex τ spanned by a1, . . . , an. Intσ is convex and is open in planespanned by the points a0, a1, . . . , an; its closure is σ. Furthermore Intσis the union of all open line segments joining a0 to the points of Intτ .

(6). Recall that in Rm, the norm of a point x = (x1, . . . , xm) is defined to be

‖x‖ =

√√√√ m∑i=1

x2i .

The m-dimensional unit ball Dm is defined by Dn = {x ∈ Rm | ‖x‖ ≤ 1}and the (m− 1)-dimensional unit sphere Sm−1 is defined to be

Sm−1 = {x ∈ Rm | ‖x‖ = 1}.

There is a homeomorphism of σ with the unit ball Dn that carries ∂σonto the unit sphere Sn−1.

2.2. Geometric Simplicial Complex. For having (geometric) simplicial com-plexes with arbitrary many simplices and arbitrary dimension of its simplices, weneed to extend our m-dimensional Euclidean space to an infinite dimensional vectorspace as follows:

Let J be an arbitrary index set, and let RJ be the J-fold product of R withitself. An element in RJ is a function from J to R denoted by x = (xα)α∈J .The product RJ is vector space with addition given by the usual component-wiseaddition and multiplication by scalars, that is, (x+ y)α = xα + yα and (cx)α = cxαfor α ∈ J .

Let EJ be the subset of RJ consisting of all points (xα)α∈J such that xα = 0for all but finitely many values of α. Then EJ is a vector subspace of RJ with abasis given by eα for α ∈ J , where eα is the map from J to R with eα(α) = 1and eα(β) = 0 for β 6= α. (Note. {eα}α∈J does not form a basis for RJ if J isan infinite set. Since {eαα∈J} is a basis, we can also consider that EJ is the vectorsubspace of RJ generated by eα for α ∈ J . ) The space EJ has a metric defined bysetting

|x− y| = max{|xα − yα|α∈J .

Definition 2.2. A geometric simplicial complex K is a collection of simplices,all contained in some Euclidean space EJ for some index set J such that

(1). if σn is a simplex in K and τp is a face of σn, then τp is in K; and(2). if σn and τp are simplices of K, then σn∩τp is either empty, or a common

face of σn and τp.The dimension of K is defined to be

dimK = sup{dimσ | σ a simplex ofK}.


So an n-dimensional simplicial complex means a simplicial complex without sim-plices of dimension higher than n. If L is a sub-collection of K that contains allfaces of its elements, then L is a simplicial complex in its own right, called a sub-complex of K. One special subcomplex of K is the collection of all simplices of Kdimension at most n, called the n-skeleton of K denoted by sknK. The points ofthe collection sk0K are called vertices of K.

Exercise 2.2. A collection K of simplices is a simplicial complex if and onlyif the following hold:

(1). Every face of a simplex of K is in K;(2). Every pair of distinct simplices of K have disjoint interior.

Definition 2.3. Let|K| =

⋃σn∈K

σn ⊆ RJ

the union of the simplices of K. Giving each simplex its natural topology as sub-space of RN for some large N , we then define the topology on |K| by requiringthat: a subset A of |K| is closed in |K| if and only if A ∩ σ is closed in σ for eachsimplex σ ∈ K. It is easy to see that this defines a topology on |K|. The space |K|is called polyhedron of K.

Note. In some references, a polyhedron means |K| for a finite simplicial simplexK, that is, K only has finitely many simplices. In such a case, since we can choosethe index set J to be a finite set, |K| ⊆ Rm for some large m and our re-definedtopology on |K| coincides with the subspace topology of |K| in Rm. If K hasinfinitely many simplices, then our redefined topology could be different from thesubspace topology. For instance, let K be the collection of all line segments fromthe origin to the points in the unit circle S1. The underline set |K| is the sameas the unit disk D2, but one can check that the topology on |K| is different fromD2. In fact, one can show that the redefined topology on |K| has more open sets(and so more closed sets) than the subspace topology, that is, the topology of |K|is a finer (larger) than the topology |K| inherits as a subspace of EJ , because if Ais closed in |K| in subspace topology then A ∩ σ is closed in σ for every σ ∈ Kand so A is closed in |K|. The redefined topology on |K| is important for makingsimplicial maps, which will be defined later, to be continuous.

Exercise 2.3. Prove the following statements:(1). If L is a subcomplex of K, then |L| is a closed subspace of |K|. In

particular, if σ ∈ K, then σ is a closed subspace of |K|.(2). A map f : |K| → X is continuous if and only if f |σ is continuous for every

σ ∈ K.(3). |K| is Hausdorff.(4). If K is finite, then |K| is compact. Conversely if a subset A of |K| is

compact, then A ⊆ |L| for some finite subcomplex L of K.

Exercise 2.4. Prove the following statements:(1). If K is a simplicial complex, then the intersection of any collection of

subcomplex of K is a subcomplex of K.(2). If {Kα} is a collection of simplicial complexes in EJ , and if the intersection

of every pair |Kα| ∩ |Kβ | is the polyhedron of a simplicial complex which


is a subcomplex of both Kα and Kβ , then the union⋃αKα is a simplicial

complex.

Exercise 2.5. Let K be a simplicial complex. Show that for any point x ∈ |K|there is a unique simplex σ of K such that x ∈ Intσ.

Definition 2.4. Given simplicial complexes K and L, a function

f : |K| → |L|

is called a simplicial map if it satisfies the following conditions:

(1). If a is a vertex of K, then f(a) is a vertex of L.(2). If a0a1 . . . an) is a simplex of K, then f(a0), f(a1), . . . , f(an) span a sim-

plex of L (possibly with repeats).(3). If x =

∑ni=0 tia

i is a point in a simplex a0a1 . . . an) of K, then

f(x) =n∑

i=0

tif(ai).

That is f is linear on each simplex.

From the definition, for having a simplicial map, one needs:

(1). a function f which sends the vertices of K to vertices of L such that(2). Whenever a0, a1, · · · , an span a simplex of K, f(a0), f(a1), · · · , f(an)

spans a simplex of L.

Proposition 2.5. A simplicial map f : |K| → |L| is continuous.

Proof. The assertion follows from that f restricted to each simplex is contin-uous. �

Exercise 2.6. Suppose that f : sk0K → sk0 L is a bijective correspondencesuch that the vertices a0, a1, . . . , an spanned a simplex in K if and only

f(a0), f(a1), . . . , f(an)

spanned a simplex in L. Then the induced simplicial map f : |K| → |L| is ahomeomorphism, called linear isomorphism or simplicial homeomorphism ofK withL.

An important concept is the star of a vertex in a simplicial complex.

Definition 2.6. Let K be a simplicial complex and let v be a vertex of K.The star of v in K, denoted by St v or St(v,K), is the union of the interior ofthose simplices of K that have v as vertex. Its closure, denoted by S̄tv, is calledthe closed star of v in K. Note that S̄tv is the union of all simplices of K having vas a vertex. The set S̄tv r St v is called the link of v in K, denoted by Lk v.

A picture for the link of v is as follows:


v

Lk v

The star can be measured by continuous functions defined as follows: Let vbe a vertex in K and let x be a point in |K|. Then x is interior to precisely onesimplex of K, whose vertices are (say) a0, a1, . . . , an. Then

x =n∑

i=0

tiai

with ti > 0 andn∑

i=0

ti = 1. We define the barycentric coordinate tv(x) of x with

respect to v by setting

tv(x) ={ti if v = ai for some 0 ≤ i ≤ n0 otherwise.

Proposition 2.7. Let K be a simplicial complex and let v be a vertex. Thentv : |K| → R is continuous.

Proof. Given any simplex σ ofK, tv(x)|σ is either identically 0 or the barycen-tric coordinate of x with respect to the vertex v of σ. Thus tv(x)|σ is continuous.By the definition of the topology on |K|, tv(x) is continuous on |K|. �

Proposition 2.8. Let K be a simplicial complex and let v be a vertex. Then

St v = {x ∈ |K| | tv(x) > 0}.

Thus St v is an open neighborhood of v in |K|.

Proof. Exercise. �

Proposition 2.9 (Star Covering). Let K be a simplicial complex. Then

|K| =⋃

v∈sk0 K

St v.

Proof. For any x ∈ |K|, there exists a unique simplicial σ such that x ∈ Intσ.Let v be a vertex of σ. Then x ∈ St v and hence the result. �

Proposition 2.10. Let x ∈ S̄tv. Then the line segment xv lies in S̄tv. More-over the line segment starting from v meets Lk v exactly one point.



2.3. Simplicial Approximation.

Definition 2.11. Let K and L be simplicial complexes and let f : |K| →|L| be a continuous map. A simplicial map g : |K| → |L| is called a simplicialapproximation to f if, for each vertex v of K,

f(St(v,K)) ⊆ St(g(v), L).

If f is a simplicial map, then f is a simplicial approximation to itself becausef sends each simplex of K to a simplex of L and hence f(St(v,K)) ⊆ St(f(v), L).

Proposition 2.12. Let h : |K| → |L| and k : |L| → |M | have simplicial ap-proximation f : K → L and g : L → M , respectively. Then g ◦ f is a simplicialapproximation to k ◦ h.

Proof. We know that g ◦ f is a simplicial map. If v is a vertex of K, then

h(St(v,K)) ⊆ St(f(v), L)

because f is a simplicial approximation to h. Thus

k(h(St(v,K))) ⊆ k(St(f(v), L)) ⊆ St(g(f(v)),M)

because g is a simplicial approximation to k. �

Proposition 2.13. Let K and L be simplicial complexes, and let f : |K| → |L|be a continuous map. Suppose that, for each vertex a of K, there exists a vertex bof L such that f(St(a,K)) ⊆ St(b, L). Then there exists a simplicial approximationg to f , such that g(a) = b for each vertex a of K.

Proof. It suffices to check that g(a0), . . . , g(an) span a simplex of L whenevera0, a1, . . . , an span a simplex of K.

Let σ = a0a1 · · · an be the simplex of K spanned by a0, a1, . . . , an. Let x ∈ Intσbe a point in the interior. Then

x ∈n⋂

i=0

St ai

It follows that

f(x) ∈n⋂

i=0

f(St ai) ⊆n⋂

i=0

St(g(ai)),

that is tg(ai)(f(x)) > 0 for each 0 ≤ i ≤ n. By the definition of the function tv, theunique simplex that contains f(x) in its interior must have each g(ai) as a vertex,and hence has a face spanned by g(a0), g(a1), . . . , g(an). �

Definition 2.14. Let X and Y be topological spaces. Let A be a subspaceof X. Let f, g : X → Y be continuous maps such that f |A = g|A. We call f ishomotopic to g relative to A if there exists a continuous map F : X × [0, 1] → Ysuch that F (x, 0) = f(x), F (x, 1) = g(x) for every x ∈ X and F (a, t) = f(a) forevery a ∈ A and 0 ≤ t ≤ 1.

Theorem 2.15. Let K and L be simplicial complexes, and let f : |K| → |L|be a continuous map. Then any simplicial approximation g to f is homotopic to frelative to the subspace of K of those points x such that f(x) = g(x).


Proof. Let x be a point of K. Then there is a unique simplex σ = a0a1 · · · ansuch that x ∈ Intσ. From the proof of the above proposition, f(x) lies in theinterior of a simplex τ of L that contains a face spanned by g(a0), . . . , g(an). Thusτ contains the point g(x), and so the line segment f(x)g(x) lies in τ . Since thelinear homotopy

F (x, s) = (1− s)f(x) + sg(x)can be defined, the assertion follows. �

To show the existence of simplicial approximation to any continuous maps, weneed the concept of subdivision discussed in next subsection.

2.4. Barycentric Subdivision.

Definition 2.16. Let K be a geometric simplicial complex in EJ . A complexK ′ is called to be a subdivision of K if

(1). Each simplex of K ′ is contained in a simplex of K.(2). Each simplex of K equals to the union of finitely many simplices of K ′.

These conditions imply that the union of the simplices of K ′ is equal to theunion of the simplices of K, that is, |K| = |K ′| as sets. The finiteness part ofcondition (2) guarantee that |K ′| and |K| are equal as topological spaces.

Definition 2.17. Suppose that K is a simplicial complex in EJ , and w is apoint in EJ such that each ray emanating from w intersects |K| in at most onepoint. We define the cone on K with vertex w to be the collection of all simplicesof the form a0a1 · · · apw, where a0a1 · · · ap is a simplex in K, along all faces of suchsimplices. Denote the cone by K ∗ w.

The cone K ∗ w is pictured as follows:w

K

Definition 2.18. Let K be a simplicial complex. Suppose that Lp is a subdi-vision of skpK. Let σ be a (p + 1)-simplex of K. Note that |∂σ| is a polyhedronof a subcomplex of skpK and so it is a polyhedron of a subcomplex, denoted byLσ, of Lp. If wσ is an interior point of σ, then the cone Lσ ∗ wσ is a simplicialcomplex whose underlying space is σ. We define Lp+1 to be the union of Lp andthe simplicial complexes Lσ ∗ wσ as σ runs over all (p + 1)-simplices of K. ThenLp+1 is a simplicial complex. (Check this!), called subdivision of skp+1K obtainedby starring Lp from the points wσ.


Definition 2.19. Let σ = v0v1 · · · vn be an n-simplex. The barycenter of σ isthe point

σ̂ =n∑

i=0

1n+ 1

vi,

that is σ̂ is the point of Intσ all of those barycentric coordinates with respect tothe vertices are equal.

If σ is 1-simplex, then σ̂ is the midpoint. If σ is a 0-simplex, then σ̂ = σ. Ingeneral, σ̂ is the centroid of σ.

Definition 2.20. Let K be a simplicial complex. We define a sequence ofsubdivisions of the skeletons of K as follows: Let L0 = sk0K. Assume that Lp isdefined as a subdivision of skpK. Let Lp+1 be the subdivision of skp+1K obtained

by starring Lp from the barycenter of the (p+ 1)-simplices of K. The union∞⋃

p=0Lp

is a subdivision of K, called barycentric subdivision of K, denoted by sdK. Definethe iterated barycentric subdivision recursively by sdnK = sdn−1(sdK) for n > 1.

Let σ be a 2-simplex. Then sdσ is shown in the picture below:

The simplices in sdK can be described as follows. Define a partial order onthe simplices of K by setting σ1 < σ2 if σ1 is a proper face of σ2.

Proposition 2.21. The simplicial complex sdK equals to the collection of allsimplices of the form

σ̂1σ̂2 · · · σ̂n,where σ1 < σ2 < · · · < σn in K.

Proof. The proof is given by induction on p that the assertion holds for skpKfor each p ≥ 0. The assertion holds for sk0K as sd sk0K = sk0K. Suppose thatthe assertion holds for skpK. By definition of barycentric subdivision, the assertionholds for skp+1K. Since sdK =

⋃p

sd skpK, the assertion follows. �

An important property of barycentric subdivision is as follows, which plays akey role for proving the simplicial approximation theorem.

Theorem 2.22. Given a finite simplicial complex K, and given any � > 0,there is a positive integer N such that each simplex of sdN K has diameter lessthan �.


Proof. 1. If σ = a0a1 · · · an is a simplex, then the diameter diamσ = max{|ai −aj | | 0 ≤ i ≤ j ≤ n}, the maximum distance between vertices.

Let l = max{|ai − aj | | 0 ≤ i < j ≤ n}. For each vertex ai, let D(ai, l) ={x | |x − ai| ≤ l} be the ball of radius l centered at ai. Then D(ai, l) is convex.Since E(ai, l) contains all vertices of σ, σ ⊆ D(ai, l). Thus

|x− ai| ≤ l

for any x ∈ σ and each 0 ≤ i ≤ n.Now given any x ∈ σ, consider the ball D(x, l) = {y | |y − x| ≤ l}. Then

σ ⊆ D(x, l) because each ai ∈ D(x, l) as |ai − x| ≤ l. Thus, for any z ∈ σ,|x− z| ≤ l. Hence diamσ = l.2. If σ = a0a1 · · · an is a simplex, then for any x ∈ σ

|σ̂ − x| ≤ nn+ 1

diamσ.

Note that for each at

|at − σ̂| =∣∣∣∣at − n∑

i=0

1n+1a

i

∣∣∣∣

=

∣∣∣∣∣∣∣∣∣∣∑

0 ≤ i ≤ ni 6= t

1n+1 (a

t − ai)

∣∣∣∣∣∣∣∣∣∣≤

∑0 ≤ i ≤ ni 6= t

1n+1 |a

t − ai|

≤ nn+1diamσ.

Let l′ = max{|at − σ̂| | 0 ≤ t ≤ n}. Then l′ ≤ nn+1diamσ. The ball

D(σ̂, l′) = {x | |x− σ̂| ≤ l′}

contains all vertices of σ and so σ ⊆ D(σ̂, l′). It follows that

|x− σ̂| ≤ l′ ≤ nn+ 1

diamσ

for any x ∈ σ.3. For any simplicial complex L, let

meshL = sup{diamσ | σ is a simplex of L}.

Let K be an n-dimensional finite simplicial complex. Then

mesh sdK ≤ nn+ 1

meshK.

The proof is given by induction on the skeleton skpK. For sk0K, we havemesh sd sk0K = mesh sk0K = 0. Suppose that

mesh sd skpK ≤p

p+ 1mesh skpK.


Consider skp+1K. By the definition, sd skp+1K is the union of sd skpK and thesimplices of the form τ ∗ σ̂ for (p + 1)-simplices σ of K, where τ is a simplex ofsd ∂σ. If σ′ is a simplex of sd skpK, then

diamσ′ ≤ pp+ 1

mesh skpK ≤p

p+ 1mesh skp+1K ≤

p+ 1p+ 2

mesh skp+1K.

If σ′ = τ ∗ σ̂ with τ a simplex of sd ∂σ, then, by Step 2,

diamσ′ ≤ p+ 1p+ 2

diamσ ≤ p+ 1p+ 2

mesh skp+1K.

The induction is finished and hence the statement.4. Since K is a finite simplicial complex, let dimK = n and let d = meshK. Then

mesh sdN K ≤(

n

n+ 1

)Nd.

Thus mesh sdN K → 0 as N →∞ and hence the result. �

Remark 2.23. By inspecting the proof, the assertion also holds if K is a finitedimensional simplicial complex such that meshK < +∞.

In practice it may be necessary to subdivide only part of a simplicial complexK, so as to leave alone a given subcomplex A. For doing subdivision in this case,we have the relative skeleton filtration defined as follows: Let skA−1K = A and letskAn K be the union of A and the simplices σ of K with dim σ ≤ n.

Definition 2.24. Let K be a geometric simplicial complex in EJ and let A bea subcomplex of K. A complex K ′ is called to be a subdivision of K relative A if

(1). Each simplex of K ′ is contained in a simplex of K.(2). Each simplex of K equals to the union of finitely many simplices of K ′.(3). A is a subcomplex of K ′.

The last condition requires that the simplices of the subcomplex A do not getsubdivision.

Definition 2.25. Let K be a simplicial complex and let A be a subcomplexof K. We define a sequence of subdivisions of the relative skeletons skAn K ofK as follows: Let L−1 = A and let L0 = skA0 K. Assume that Lp is definedas a subdivision of skAp K. Let Lp+1 be the subdivision of skp+1K obtained bystarring Lp from the barycenter of the (p + 1)-simplices of K NOT in A. The

union∞⋃

p=0Lp is a subdivision of K, called barycentric subdivision of K relative

to A, denoted by sd(K,A). Define the iterated barycentric subdivision recursivelyby sdn(K,A) = sdn−1(sd(K,A)) for n > 1.

The barycentric subdivision relative to a subcomplex is shown by the picture:


A

A

Recall that τ < σ if and only if τ is a proper face of σ.

Proposition 2.26. Let K be a simplicial complex and let A be a subcomplex.Then the vertices of sd(K,A) are the barycenters of the simplices of KrA, togetherwith the vertices of A. Distinct points

a1, a2, . . . , aq, σ̂1, σ̂2, . . . , σ̂m

(with dim σi ≤ dim σi+1 for each i) span a simplex of sd(K,A) if and only ifa1, a2, . . . , aq span a simplex σ of A, σj is a simplex of K r A for 1 ≤ j ≤ m withσ < σ1 < σ2 < · · · < σm.

Note. If m = 0, then it just requires a1, . . . , aq that span a simplex of A. Ifq = 0, it just requires σm > σm−1 > · · · > σ1 and in this case σ̂1, σ̂2, . . . , σ̂m formsa simplex disjoint from A.

Proof. The proof follows by induction on the relative skeleton skAn K. �

Observe that the identity map | sd(K,A)| → |K| is not a simplicial map ifK rA 6= because the vertices σ̂ of sd(K,A) are not the vertices of K.

Proposition 2.27 (Simplicial Approximation to the Identity Map). Let Kbe a simplicial complex and let A be a subcomplex of K. Let f be any functionassigning simplices σ of KrA to one of its vertices. Then there exists a simplicialapproximation h to the identity map | sd(K,A)| → |K| such that h|A = idA andh(σ̂) = f(σ) for each simplex σ of K rA.

Proof. First we need to check that h is a simplicial map. Let τ be a simplexof sd(K,A). Then τ = a1a2 · · · apσ̂1σ̂2 · · · σ̂q, where a1, a2, . . . , aq span a simplexσ0 of A, σj is a simplex of K rA and σ0 < σ1 < σ2 < · · · < σq. We check that thevertices

(2.1) {a1, a2, . . . , ap, f(σ̂1), f(σ̂2), . . . , f(σ̂q)}span a simplex in K. Since σ0 < σ1 < σ2 < · · · < σq, each σj is a face of σq and soany vertex of σj is a vertex of σq. Thus the elements in Equation (2.1) are verticesof σq and therefore they span a simplex in K.

Next we check that h is a simplicial approximation to id: | sd(K,A)) → |K|.Let v be any vertex of sd(K,A). We need to show that

St(v, sd(K,A)) ⊆ St(h(v),K).Let τ be a simplex having v as a vertex. We are going to show that:


There exists a simplex µ of K such that µ has v as a vertex and theinterior Int τ ⊆ Intµ.

If so, then Int τ ⊆ St(h(v),K) for any simplex τ of sd(K,A) having v as a vertex.From the definition of star, we then conclude that St(v, sd(K,A)) ⊆ St(h(v),K).

Now we prove the above statement. By Proposition 2.26,

τ = a1a2 · · · apσ̂1σ̂2 · · · σ̂q,

where a1, a2, . . . , aq span a simplex σ0 of A, σj is a simplex of K rA and

σ0 < σ1 < σ2 < · · · < σq.

If q = 0, then v is a vertex of A and so h(v) = v. In this case, we choose µ = τ .Suppose that q > 0. We choose µ = σq. From the previous paragraph, h(v) is avertex of µ and so we only need to check that Int τ ⊆ Intµ. Let x ∈ Int(τ). Then

x =p∑

i=1

siai +

q∑j=1

tj σ̂i

with si, tj > 0 and∑p

i=1 si +∑q

j=1 tj = 1. Let b0, b1, . . . , bm be the vertices of σq.

Let

ai =m∑

k=0

si,kbk

σ̂j =m∑

k=0

ti,kbk

with si,k, ti,k ≥ 0,∑m

k=0 si,k = 1 and∑m

k=0 tj,k = 1. Since σ̂q is the barycenter ofσq, tq,k = 1m+1 for 0 ≤ k ≤ m. Now

x =m∑

k=0

(p∑

i=1

sisi,k +q∑

j=1

tjtj,k)bk.

withm∑

k=0

p∑i=1

sisi,k +q∑

j=1

tjtj,k

= 1.For each k, we have

p∑i=1

sisi,k +q∑

j=1

tjtj,k ≥ tqtq,k =tq

m+ 1> 0.

Thus x ∈ Int(σq) and hence the result. �

Corollary 2.28. Let K be a simplicial complex and let A be a subcomplex ofK. Let a be any vertex of sd(K,A). Then there exists a vertex b of K such that

St(a, sd(K,A)) ⊆ St(b,K)

such that if a ∈ A, then we can choose b = a. Moreover if B is a subcomplex ofK such that B ∩ A = ∅, then for a vertex a of sd(K,A) not in |B|, there exists avertex b of K such that b 6∈ |B| and

St(a, sd(K,A)) ⊆ St(b,K).


Proof. Let h be a simplicial approximation to the identity. Let a be anyvertex of sd(K,A). By the definition of simplicial approximation,

St(a, sd(K, a)) = id(st(a, sd(K,A))) ⊆ St(h(a),K).

Since h|A = idA, h(a) = a if a ∈ A. If B is a subcomplex of K such that B∩A = ∅,then we can make a choice of the simplicial approximation h to the identity byrequiring that if a is a vertex of sd(K,A) not in |B|, then h(a) is a vertex not in|B| by the above proposition. �

2.5. Simplicial Approximation Theorem.

Theorem 2.29 (Finite Simplicial Approximation Theorem). Let K and L besimplicial complexes. Suppose that K is finite. Let f : |K| → |L| be a continuousmap. Then there exists N such that the map f has a simplicial approximationg : sdN K → L.

Proof. Since {Stw | w is a vertex of L} is an open cover of |L|, the space |K|is covered by the open sets

A = {f−1(Stw) | w is a vertex of L}.

Since |K| is a compact metric space, there exists a Lebesgue number λ such thatany subset of |K| with diameter less than λ lies in an element of A. (If there wereno such λ, there is a sequence Cn of subsets of |K| such that diamCn < 1/n butCn does not lie in any element of A. Choose xn ∈ Cn. By compactness, thereis a subsequence xni convergent to a point x ∈ |K|. Then x lies in an elementf−1(Stw) of A, and so Cni ⊆ f−1(Stw) when i is sufficiently large. This gives acontradiction.)

Choose N such that each simplex of sdN K has diameter less than λ/2. Let a bea vertex of sdN K. Then the diameter of St v is less than λ because, for x, y ∈ St v,there exist simplices σ and τ of sdN K such that both σ and τ have v as a vertexwith x ∈ σ and y ∈ τ and so

|x− y| ≤ |x− a|+ |y − a| < λ2

+λ

2= λ.

Thus St v ⊆ f−1(Stw) for a vertex w of L. The assertion follows from Proposi-tion 2.13 now. �

Now we consider the relative case.

Definition 2.30. Let K be a simplicial complex and let A be a subcomplexof K. The supplement of A in K, denoted by Ā, is the set of simplices of sd(K,A)that have NO vertices in A. Clearly Ā is a subcomplex of sd(K,A), which is thesame as the subcomplex of sdK of simplices having no vertices in sdA.

The next lemma states that the maximum diameter of the stars of sdN (K,A)of vertices in the supplement |Ā| tends to 0.

Lemma 2.31. Let K be a simplicial complex with a subcomplex A. Suppose thatthere are finitely many simplices of K r A. Given any � > 0, there exists N suchthat

sup{diam St(v, sdN (K,A)) | v ∈ |Ā|} < �.


Proof. Let τ be a 1-simplex of sd2(K,A). Suppose that τ has a vertex a inA. Then another vertex of τ is either in A or the barycenter σ̂ for a simplex σof sd(K,A) having a as a vertex. Thus σ 6∈ Ā and so σ̂ 6∈ |Ā|. In other words,NO 1-simplices of sd2(K,A) that can have vertices in both A and |Ā|. It followsthat NO n-simplices of sd2(K,A) that can have vertices in both A and |Ā|. (If ann-simplex had vertices a in A and b in |Ā|, then its face from a to b is a 1-simplexhaving its vertex a in A and b in |Ā| which contradicts to that no 1-simplex canhave vertices in both A and |Ā|.)

Let Â denote the supplement of A in sd(K,A), that is, Â is the set of simplicesof sd2(K,A) that have no vertices in A. Then, if τ is a simplex of sd2(K,A) with avertices in |Ā|, then τ ∈ Â. Note that the subdivision sdN (K,A) includes the no-relative subdivision sdN−2 Â of Â. Thus sdN−2 Â is a subcomplex of sdN (K,A).Now we show by induction that:

For each N ≥ 2 if τ is a simplex of sdN (K,A) having a vertex in |Ā|,then τ ∈ sdN−2 Â.

This statement has been proved when N = 2. Suppose that it holds for N − 1with N ≥ 3. Let τ be a simplex of sdN (K,A) with a vertex in |Ā|. Then τ hasthe expression a1a2 · · · apσ̂1 · · · σ̂q, where a1, . . . , ap span a simplex σ0, each σj isa simplex of sdN−1(K,A) with σ0 < σ1 < · · · < σq. Since each ai is not in |Ā|,we may assume that σ̂j ∈ |Ā| for some 1 ≤ j ≤ q. It follows that σj has a vertexin |Ā|. Since σj is a face of σq as σj < σq, σq has a vertex in |Ā|. By induction,σq ∈ sdN−3 Â and σ0 must be empty as it is a face of the simplex σq of sdN−3 Â.It follows that τ ∈ sdN−2 Â. The induction is finished and hence the statement.

Thus for any vertex v of sdN (K,A) such that v ∈ |Ā|

St(v, sdN (K,A)) =⋃{Int τ | τ is a simplex of sdN (K,A) having v as a vertex}

⊆⋃{Int τ | τ is a simplex of sdN−2 Â having v as a vertex}

= St(v, sdN−2 Â).

The assertion follows from Theorem 2.22 now. �

Now we construct a new simplicial complex K+ = sd(sd(K,A), A ∪ Ā). Inother words, K+ is obtained by doing barycentric subdivision on those simplices ofsd(K,A) that are not in A and Ā. Thus K+ is a subdivision between sd(K,A) andsd2(K,A). A picture is as follows:

A

bar(A)

The vertices v of K+ are one of the following three cases:


Type I. v is a vertex of A;Type II. v is a vertex of Ā;

Type III. v is the barycenter σ̂ for a simplex σ of sd(K,A) that has vertices in bothA and Ā.

By Proposition 2.27, there exists a simplicial approximation h to the identity of|K+| = | sd(sd(K,A), A ∪ Ā)| → | sd(K,A)| such that h on the vertices of K+ isgiven by the following rule:

(1). If v is a vertex of A, then h(v) = v;(2). If v is a vertex of Ā, then h(v) = v;(3). If v = σ̂ for a simplex σ of sd(K,A) that has vertices in both A and Ā,

then h(v) is a vertex of σ in A.The first two rules requires that h|A∪Ā = idA∪Ā. The last rule forces h to pushdown the vertices of the form σ̂ into A.

Lemma 2.32. If v is a vertex of A, then h(St(v,K+)) ⊆ St(v,A).

Proof. Let τ be a simplex of K+ having v as a vertex. Since

K+ = sd(sd(K,A), A ∪ Ā),τ has the expression

τ = a1a2 · · · apσ̂1 · · · σ̂q,where a1, a2, · · · , ap span a simplex σ0 in A∪Ā, σ1, . . . , σq are simplices of sd(K,A)that have vertices in both A and Ā, and σ0 < σ1 < · · · < σq. Since v ∈ A is a vertexof τ , p ≥ 1 and v ∈ {a1, . . . , ap}. It follows that ai 6∈ Ā for each 1 ≤ i ≤ q becauseif not, then σ0 is a simplex having vertices in both A and Ā which contradicts tothat σ ∈ A ∪ Ā. Thus τ only have Type I and Type II vertices, namely τ has NOvertices in Ā. By the definition of the map h, we have

h(ai), h(σ̂j) ∈ Afor 1 ≤ i ≤ p and 1 ≤ j ≤ q. Hence h(τ) is a simplex of A having v as vertexbecause h|A = idA. The assertion follows by taking the union of the interior of τhaving v as a vertex. �

Proposition 2.33. Let K and L be a simplicial complexes and let A be asubcomplex of K. Let

h : |K+| = | sd(sd(K,A), A ∪ Ā)| = |K| → | sd(K,A)| = |K|be the simplicial map defined above. Suppose that there are finitely many simplices ofKrA. Let f : |K| → |L| be any continuous map such that f ||A| is a simplicial map.Then there exists N such that the composite f ◦ h has a simplicial approximationg : sdN (K,A) with the property that g||A| = f ||A|.

Proof. Let

A = {(f ◦ h)−1(St(w,L)) | w is a vertex of L}be an open covering of |K|. Let λ be a Lebesgue number of A.

Let Â be the supplement ofA in sd(K,A) as discussed in the proof of Lemma 2.31.Then there exists N ≥ 2 such that

sup{diam St(v, sdN (K,A)) | v ∈ |Â|} < λ.

(Note. Here we replace Â as Ā in Lemma 2.31 by considering sd(K,A) as K.)


By Proposition 2.13, it suffices to show that, for every vertex v of sdN (K,A),there exists a vertex w of L such that

St(v, sdN (K,A)) ⊆ (f ◦ h)−1(w).

Case I. v ∈ |Â|.Then

diam St(v, sdN (K,A)) < λ

and so there exists a w such that St(v, sdN (K,A)) ⊆ (f ◦ h)−1(w).Case II. v 6∈ |Â|.

By Corollary 2.28,

St(v, sdN (K,A)) ⊆ St(b, sd2(K,A))

for some vertex b of sd2(K,A) NOT in |Â|. Then b ∈ A. (This is because, bydefinition, Â is the subcomplex of sd2(K,A) of the simplices having no vertices inA. In particular, Â contains all of vertices, as 0-simplices, that are not in A.) Weclaim that

St(b, sd2(K,A)) ⊆ St(b,K+).Let τ be a simplex of sd2(K,A) having a vertex b ∈ A. Then τ has the expression

τ = a1a2 · · · apσ̂1σ̂2 · · · σ̂2,

where a1, . . . , ap span a simplex σ0 of A, σj is a simplex of sd(K,A) for 1 ≤ j ≤ q,with σ0 < σ1 < · · · < σq. Since b is a vertex of τ , b ∈ {a1, . . . , ap} with p ≥ 1.Since σ0 is a face of each σj for 1 ≤ j ≤ q, b is a vertex of σj for 1 ≤ j ≤ q. Itfollows that τ ∈ K+ = sd(sd(K,A), A ∪ Ā). This proves that

St(b, sd2(K,A)) ⊆ St(b,K+).

Now by Lemma 2.32, we have

h(St(b,K+)) ⊆ St(b, A).

Since f ||A| is a simplicial map, we have

f(St(b, A)) ⊆ St(f(b), L).

By combining the previous equations, we have

St(v, sdN (K,A)) ⊆ St(b, sd2(K,A))= St(b,K+)⊆ h−1(St(b, A))⊆ (f ◦ h)−1(St(f(b), L)).

This finishes the proof that f ◦ h has a simplicial approximation g : | sdN (K,A)| →|L|.

For checking that, we can make a choice of g such that g||A| = f ||A|. Letv be a vertex of A. In the argument of Case II above, we can choose b = v byCorollary 2.28. Thus we have

St(v,StN (K,A)) ⊆ (f ◦ h)−1(St(f(v), L)).

Thus the simplicial map g sends v to f(v) for each vertex v ∈ A. It follows thatg||A| = f ||A|. The proof is finished now. �


Theorem 2.34 (Relative Simplicial Approximation Theorem). Let K and Lbe a simplicial complexes and let A be a subcomplex of K. Suppose that thereare finitely many simplices of K r A. Let f : |K| → |L| be any continuous mapsuch that f ||A| is a simplicial map. Then there exists N and a simplicial mapg : sdN (K,A)→ L such that g||A| = f ||A| and g is homotopic to f relative to |A|.

Proof. Let g be the simplicial map in Proposition 2.33. Then g||A| = f ||A|and g is homotopic to f ◦ h relative to A by Theorem 2.15. Since h is a simplicialapproximation to the identity map with h|A = idA, h ' id|K| relative to A and sof ◦ h ' f relative to A. It follows that g ' f relative to A. �

2.6. Some Applications. Recall that the homotopy groups πn(X) is definedby

πn(X) = [Sn, X]

the set of homotopy classes of the pointed continuous maps from Sn to X up topointed homotopy (namely homotopy relative to the basepoint). A direct conse-quence of simplicial approximation theorem is as follows:

Theorem 2.35. πr(Sn) = 0 for r < n.

Proof. Let f : Sr → Sn be a pointed continuous map. Let K be the simplicialcomplex such that |K| ∼= Sr and let L be the simplicial complex such that |L| = Sn.(We can choose K and L as the boundary of an (r + 1)-simplex and an (n + 1)-simplex, respectively.) Consider Sr and Sn as polyhedron of simplicial complexes.By the simplicial approximation theorem, there exists a subdivision K ′ of K anda simplicial map g : |K ′| = |K| ∼= Sr → |L| ∼= Sn such that g is homotopic to frelative to the basepoint. Since g is a simplicial map, the map g must send K ′

into the skeleton skr(L) because K ′ is an r-dimensional simplicial complex. Sincer < n, skr(L) 6= L and so g is not onto. Thus there is a point x ∈ Sn such thatg(Sr) ⊆ Sn r {x} ∼= Rn. Hence g is homotopic to the constant map relative tothe basepoint by a linear homotopy. It follows that f is homotopic to the constantmap relative to the basepoint. �

Another direct consequence is to give a computation of the fundamental group.

Theorem 2.36. π1(S1) = Z.

Proof. By simplicial approximation theorem, we only need to consider thesimplicial maps from a subdivision K ′ of the circle K to the circle K, where Kis the boundary of a 2-simplex with vertices w0, w1, w2 in the order of counter-clockwise along the circle, where w0 is regarded as the basepoint. Let K ′ hasvertices v0, v1, . . . , vn in the order of counter-clockwise along the circle, where v0 isregarded as the basepoint. Let g : K ′ → K be a pointed simplicial map. See thepicture:


g

w1

w0

w2

vn

v0

v2

v1v3

Then g maps the circle (v0, v1, . . . , vn, v0) (in order) into the sequence

(g(v0), g(v1), . . . , g(vn), g(v0)),

where g(vi) is one of wj with g(v0) = w0. If g(vi) = g(vi+1), then g is constant onthe 1-simplex [vi, vi+1] and so, up to homotopy, we can remove g(vi+1). For the sub-sequence (g(vi), g(vi+1), g(vi+2)), if g(vi) = g(vi+2), then (g(vi), g(vi+1), g(vi+2)) isa path from g(vi) to g(vi+1) and backwards to g(vi) and so we can collapse thissubsequence. Assume that v0, . . . , vn have minimal number of vertices in the ho-motopy class of g. Then the only sequence of (g(v0), g(v1), . . . , g(vn), g(v0)) is givenby one of the following:

(1). constant map (w0),(2). around the circle n-times positively:

(w0, w1, w2, w0, w1, w2, w0, . . . , w0, w1, w2, w0),

or(3). around the circle m-times negatively:

(w0, w2, w1, w0, w2, w1, w0, . . . , w0, w2, w1, w0).

This shows that any pointed continuous map f : S1 → S1 is homotopic to one ofthe maps

gn : S1 → S1 z 7→ zn

for n ∈ Z, relative to the basepoint. In particular, π1(S1) is generated by [g1].On the other hand, one can show that gn is not homotopic to gm relative to thebasepoint if n 6= m, as gn is the continuous mapping that goes around S1 for ntimes. This gives that π1(S1) = S1. �

Example 2.1 (Cohomotopy Sets). Let X be a pointed space. Then the n-thcohomotopy set is defined to be

πn(X) = [X,Sn],

the set of homotopy classes of pointed continuous maps from X to the n-sphere upto pointed homotopy. For instance, πr(Sn) = [Sr, Sn] = πn(Sr). The fundamentalproblem in algebraic topology is to determine πr(Sn) for general r and n. Whenn = 1, it was known that πr(S1) = 0 for r 6= 1 and π1(S1) = Z. When n > 1,we know from the above theorem that πr(Sn) = 0 for r < n. When r = n,πn(Sn) = Z. (This is a direct consequence of Hurewicz Theorem which gives that,for simply connected spaces, the first non-trivial homotopy group is the same as the

3. ABSTRACT SIMPLICIAL COMPLEXES AND ∆-SETS 25

corresponding homology group. Or one can try to directly compute πn(Sn) usingsimplicial methods or other methods.) For r > n, πr(Sn) is known up to certainrange by very nontrivial works contributed by many topologists, but far unknownfor general r even if n = 2. By using simplicial methods, one might try a differentapproach to study the higher homotopy groups of spheres.

Consider Sn as the boundary of an (n + 1)-simplex σn+1. So, as a simplicialcomplex, Sn has (n+2) vertices labeled in order by {0, 1, 2, . . . , n+1}. Observe that,in σn+1, any nonempty subset of the vertices {0, 1, 2, . . . , n+ 1} spans a simplex ofσn+1. Since Sn is the boundary of σn+1, any proper subset of {0, 1, 2, . . . , n + 1}spans a simplex of Sn.

LetK be a simplicial complex and let sk0K be the set of vertices ofK. Supposethat f : K → Sn is a simplicial map. Then f sends vertices of K to the vertices ofSn. Thus, for each vertex v of K, there is a color f(v) = i for some 0 ≤ i ≤ n+ 1.This colorizes the vertices of K by using (n + 2) different colors. Suppose thatv0, v1, . . . , vr span a simplex of K. Then

{f(v0), f(v1), . . . , f(vr)}

must span a simplex of Sn, equivalently, {f(v0), f(v1), . . . , f(vr)} is a proper subsetof {0, 1, . . . , n+1}. Namely at least at one color is missing in {f(v0), f(v1), . . . , f(vr)}.

Conversely suppose that there is a coloration on the vertices of K using (n+1)different colors, that is there is a function from sk0K to {0, 1, . . . , n}, such thatif v0, v1, . . . , vr span a simplex of K, then at least one color is missing among thecolors of v0, v1, . . . , vr. Then the coloration induces a unique simplicial map fromK to Sn.

This establishes the one-to-one correspondence between the set of simplicialmaps from K to Sn and the set of colorations on vertices of K that satisfying theabove rule. Thus, for studying simplicial maps from K to Sn, one can study how tomake those colorations on the vertices of K satisfying the above rule. For studyingπr(Sn), one needs to understand:

(1). The colorations on the vertices of K satisfying the above rule, where|K| ∼= Sr, namely K is a triangulation of Sr. This will give simplicialmaps K → Sn. In general, one may study the colorations on the verticesof K, where K is a triangulation of a manifold.

(2). The colorations on the vertices of K satisfying the above rule, where|K| ∼= Sr × [0, 1]. This will control the homotopy.

The above two problems are not well-understood so far. But it will be very interest-ing if one could make progress on this as it attacks the fundamental open problemin algebraic topology. �

Exercise 2.7. Determine all possible simplicial maps from K → S1, where Kis a simplicial complex such that |K| ∼= S2.

3. Abstract Simplicial Complexes and ∆-sets

3.1. Definition and Geometric Realizations.

Definition 3.1. An abstract simplicial complex K is a collection of finitenonempty sets, such that if A is an element in K, so is every nonempty subsetof A.


The element A of K is called a simplex of K; its dimension is one less thanthe number of its elements. Each nonempty subset of A is called a face of A. Thedimension of K is the supremum of the dimensions of its simplices. The vertexset V (K) is the union of the one-point elements of K; we shall make no distinctionbetween the vertex v and the 0-simplex {v}. A sub collection of K that is itself acomplex is called a subcomplex of K. Two abstract simplicial complexes K and K′are called to be isomorphic if there exists a bijective correspondence f mapping thevertex set of K to the vertex set of K′ such that {a0, a1, . . . , an} ∈ K if and only if{f(a0), f(a1), . . . , f(an)} ∈ K′.

Definition 3.2. Let K be a geometric simplicial complex. Let V be thevertex set of K. Let K be the collection of all subsets {a0, a1, . . . , an} of V suchthat a0, a1, . . . , an span a simplex of K. The collection K is called vertex schemeof K, or abstraction of K.

Theorem 3.3. A relation between abstract simplicial complexes and geometricsimplicial complexes is as follows:

(1). Every abstract simplicial complex K is isomorphic to the vertex schemeof some geometric simplicial complex.

(2). Two geometric simplicial complexes are linearly isomorphic if and only iftheir vertex schemes are isomorphic as abstract simplicial complexes.

Proof. We leave (b) as an exercise. To prove (a), we proceed as follows:Given an index set J , let ∆J be the collection of all simplices in EJ spanned byfinite subsets of the standard basis {eα}α∈J for EJ . It is easy to see that ∆J isa simplicial complex. Moreover if σ and τ are two simplices of ∆J , then theircombined vertex set is geometrically independent and spans a simplex of ∆J .

Now let K be an abstract simplicial complex with vertex set V . Choose the in-dex set J = V . We specify a subcomplex K of ∆J by the condition that for each ab-stract simplex {a0, . . . , an} ∈ K, the geometric simplex spanned by ea0 , ea1 , . . . , eanis to be in K. It is immediate that K is a geometric simplicial complex and K isisomorphic to the vertex scheme of K. �

3.2. Subdivision of Abstract Simplicial Complexes. Let V be a set witha partial order

3. ABSTRACT SIMPLICIAL COMPLEXES AND ∆-SETS 27

3.3. Abstract Simplicial Complexes and ∆-sets. Let K be an abstractsimplicial complex. Let Kn be the set of n-simplices of K. Define the faces

di : Kn → Kn−1, 0 ≤ i ≤ n,as follows:

Let the vertices of K be ordered. If {a0, a1, . . . , an} is an n-simplex of Kwith a0 < a1 < · · · < an, then define

di{a0, a1, . . . , an} = {a0, a1, . . . , ai−1, ai+1, . . . , an}.

Proposition 3.4. Let K∆ = {Kn}n≥0 with faces defined as above. Then K∆is a ∆-set.


Definition 3.5. A ∆-set X is called polyhedral if there exists an abstractsimplicial complex K such that X ∼= K∆.

In general, a ∆-set may not be polyhedral.

Exercise 3.1. Let X = ∆+[1] ∪∆+[1]0 ∆+[1] be the union of two copies of∆+[1] by identifying the vertices. Show that X is not polyhedral.

Now let X be a ∆-set and let 2X0 be the set of all subsets of X0. Define

φ :∐n≥0

Xn −→ 2X0

by setting φ(x) = {fx(0), fx(1), . . . , fx(n)} for x ∈ Xn.

Theorem 3.6. Let X be a ∆-set. Then X is polyhedral if and only if thefollowing holds:

(1). There exists an order of X0 such that, for each x ∈ Xn,fx(0) ≤ fx(1) ≤ · · · ≤ fx(n).

(2). The function φ :⋃

n≥0Xn −→ 2X0 is one-to-one.

Proof. If X ∼= K∆ for some abstract simplicial complex K, then each n-simplex is uniquely determined by its vertices. Thus φ is one-to-one. From theconstruction, we have fx(0) < fx(1) < · · · < fx(n).

Conversely suppose that X is a ∆ set satisfies the two conditions in the state-ment. First we show that, for x ∈ Xn, the cardinal of {fx(0), fx(1), . . . , fx(n)}is n+ 1. Otherwise there exists 0 ≤ i < j ≤ n such that fx(i) = fx(j). Then

di(x) = fxdi(0, 1, . . . , n) = fx(0, 1, . . . , i− 1, i+ 1, . . . , n)and so

φ(di(x)) = {fx(0), fx(1), . . . , fx(i− 1), fx(i+ 1), . . . , fx(n)}= {fx(0), fx(1), . . . , fx(i− 1), fx(i), fx(i+ 1), . . . , fx(n)}= φ(x).

Thus di(x) = x, which contradicts to that Xn ∩Xn−1 = ∅.Let the vertices of K be the elements of X0. Define a subset

{a0, a1, . . . , an} ⊆ X0to be an n-simplex of K if and only if {a0, a1, . . . , an} ∈ Im(φ), where the elementsare given in the order that a0 < a1 < · · · < an.


For checking that K is an abstract simplicial complex, let {a0, a1, . . . , an} bean n-simplex of K such that {a0, a1, . . . , an} = φ(x). From the above arguments,x ∈ Xn and so we may assume that ai = fx(i) for 0 ≤ i ≤ n. Let {ai0 , ai1 , . . . , aip}be a subset of {a0, a1, . . . , an} with 0 ≤ i0 < i1 < · · · < ip ≤ n. Let

{j0, j1, . . . , jn−p−1} = {0, 1, . . . , n}r {i0, i1, . . . , ip}

with 0 ≤ j0 < j1 < · · · < jn−p−1. Then

dj0dj1 · · · djn−p−1x = fxdj0dj1 · · · djn−p−1(0, 1, . . . , n)= fx(i0, i1, . . . , ip).

Thusφ(dj0dj1 · · · djn−p−1x) = {ai0 , ai1 , . . . , aip}

and so {ai0 , ai1 , . . . , aip} is a p-simplex of K.Finally the function

g : X −→ K∆

with g(x) = {fx(0), . . . , fx(n)} for x ∈ Xn is bijective ∆-map. This proves thatX ∼= K∆. �

4. ∆-complexes and the geometric realization of ∆-sets

The standard geometric n-simplex ∆n is defined by

∆n = {(t0, t1, . . . , tn) | ti ≥ 0 andn∑

i=0

ti = 1}.

Define di : ∆n−1 → ∆n by setting

di(t0, t1, . . . , tn−1) = (t0, . . . , ti−1, 0, ti, . . . , tn−1).

Exercise 4.1. Show that The maps di satisfy the following identity:

djdi = di+1dj

for i ≥ j.

The boundary∂∆n = ∪ni=0di(∆[n− 1])

is the union of all faces of ∆n. Let Int(∆n) = ∆n r ∂∆n be the interior of ∆n,called open simplex.

Definition 4.1. A ∆-complex structure on a space X is a collection of maps

C(X) = {σα : ∆n → X | α ∈ Jn n ≥ 0}.

such that(1). σα|Int(∆n) : Int(∆n)→ X is injective, and each point of X is in the image

of exactly one such restriction σα|Int(∆n).(2). For each σα ∈ C(X), each face

σα ◦ di ∈ C(X).

(3). A set A ⊆ X is open if and only if σ−1α (A) is open in ∆n for eachσα ∈ C(X).

4. ∆-COMPLEXES AND THE GEOMETRIC REALIZATION OF ∆-SETS 29

DefineC∆n (X) = {σα : ∆n → X | α ∈ Jn} ⊆ C(X)

with di : C∆n (X)→ C∆n−1(X) given by

di(σα) = σα ◦ di

for 0 ≤ i ≤ n.

Exercise 4.2. C∆(X) = {C∆n (X)}n≥0 is a ∆-set.

Let K be a ∆-set. The geometric realization |K| of K is defined to be

|K| =∐

x ∈ Knn ≥ 0

(∆n, x)/ ∼=∞∐

n=0

∆n ×Kn/ ∼,

where (∆n, x) is ∆n labeled by x ∈ Kn and ∼ is generated by

(z, dix) ∼ (diz, x)

for any x ∈ Kn and z ∈ ∆n−1 labeled by dix. For any x ∈ Kn, let σx : ∆n =(∆n, x) → |K| be the canonical characteristic map. The topology on K is definedby A ⊆ |K| is open if and only if the pre-image σ−1x (A) is open in ∆n for anyx ∈ Kn and n ≥ 0.

Proposition 4.2. Let K be a ∆-set. Then |K| is ∆-complex.

Proof. Let sknK = {Kj}0≤j≤n be the n-skeleton of K. Then sknK is a∆-subset of K. Let ∂∆+[n] = skn−1 ∆+[n] be the boundary of ∆+[n]. Observethat

|∆+[n]| ∼= ∆n |∂∆+[n]| ∼= ∂∆n.From the push-out diagram ⊔

x∈Kn

∂∆+[n] - skn−1K

push

⊔x∈Kn

∆+[n]?

∩

- sknK,?

∩

there is a push-out diagram ⊔x∈Kn

∂∆n - | skn−1K|

push

⊔x∈Kn

∆n?

∩

- | sknK|.?

∩

Thus | sknK| is obtained from skn−1K| by attaching cells with labels in Kn. Byinduction, it follows that |K| is a ∆-complex. �


5. Homology of ∆-sets

Recall that a chain complex of groups means a sequence C = {Cn} of groupswith differential ∂n : Cn → Cn−1 such that ∂n ◦ ∂n+1 is trivial, that is Im(∂n+1) ⊆Ker(∂n) and so the homology is defined by

Hn(C) = Ker(∂n)/ Im(∂n+1),

which is a coset in general. A chain complex C is called normal if Im(∂n+1) is anormal subgroup of Ker(∂n) for each n. In this case Hn(C) is a group for each n.

Proposition 5.1. Let G be a ∆-abelian group. Define

∂n =n∑

i=0

(−1)idi : Gn → Gn−1.

Then ∂n−1 ◦ ∂n = 0, that is, G is a chain complex under ∂∗.

Proof.

∂n−1 ◦ ∂n =n−1∑i=0

(−1)idin∑

j=0

(−1)jdj

=∑

0≤i

5. HOMOLOGY OF ∆-SETS 31

Proof. Consider the commutative diagram

C ′k+2⊂

i - Ck+2p-- C ′′k+2

C ′k+1

∂′

?⊂

i - Ck+1

∂

? p-- C ′′k+1

∂′′

?

C ′k

∂′

?⊂

i - Ck

∂

? p -- C ′′k

∂′′

?

C ′k−1

∂′

?⊂

i - Ck−1

∂

? p-- C ′′k−1.

∂′′

?

Let x ∈ C ′′k+1 with ∂′′(x) = 1. There exists x̃ ∈ Ck+1 such that p(x̃) = x. Since

p(∂(x̃)) = ∂′′(p(x̃)) = ∂′′(x) = 1,

there exists x̄ ∈ C ′k such that i(x̄) = ∂(x̃). Now

i(∂′(x̄)) = ∂(i(x̄)) = ∂ ◦ ∂(x̃) = 1.

Thus x̄ is a circle in C ′ and so {x̄} defines an element in Hk(C ′).Let x̂ be another element in Ck+1 such that p(x̂) = x. Then

p(x̃x̂−1) = 1

and so there exists an element z ∈ C ′k+1 such that i(z) = x̃−1x̂. Now

i(x̄∂′(z)) = ∂(x̃)(∂(x̃))−1∂(x̂) = ∂(x̂).

Thus {x̄} ∈ Hk(C ′) is independent on the choice of the pre-image of x in Ck+1.Suppose that x′ = x∂′′(y) with ∂′′(x) = 1 for some y ∈ C ′′k+2. There exists

ỹ ∈ Ck+2 such that p(ỹ) = y. Then

x′ = p(x̃∂(ỹ))

withx̄′ = ∂(x̃∂(ỹ)) = ∂(x̃) = x̄.

This shows that∂k+1 : Hk+1(C ′′)→ Hk(C ′) {x} 7→ {x̄}

is well-defined. Assume that C ′ and C ′′ are normal chain complexes. For x, x′ ∈C ′′k+1 with ∂

′′(x) = ∂′′(x′) = 1. Then p(x̃x̃′) = xx′ and so

∂k+1({x}{x′}) = ∂k+1({x})∂k+1({x′})

provided that C ′ and C ′′ are normal.The composite i∗ ◦ ∂k+1 is trivial because i(x̄) = ∂(x̃). Let y ∈ C ′k with

∂′(y) = 1 and i∗(y) is trivial in Hk(C). Then there exists ỹ ∈ Ck+1 such that

i(y) = ∂(ỹ).


By the construction of ∂k+1, ∂k+1(p(ỹ)) = y. This shows that

Hk+1(C ′′)∂k+1- Hk(C ′)

i∗- Hk(C)

is exact.Now we show that

Hk+1(C)p∗- Hk+1(C ′′)

∂k+1- Hk(C ′)

is exact. Let y ∈ Ck+1 such that ∂(y) = 1. Then by the construction of ∂k+1,∂k+1(p(y)) = 1. Thus the composite ∂k+1 ◦ p∗ is trivial. Suppose that x ∈ C ′′k+1with ∂′′(x) = 1 and x̄ = ∂k+1(x) is trivial in Hk(C ′). There exists an elementz ∈ C ′k+1 such that

∂′(z) = x̄.

Let x̂ = i(z)−1x̃. Then

p(x̂) = p(i(z)−1x̃) = p(x̃) = x

with

∂(x̂) = ∂(i(z)−1x̃) = i(∂′(z)−1x̄) = 1.

Thus x̂ defines an elements in Hk+1(C) with p∗({x̂}) = {x}.Finally we show that

Hk(C ′)i∗- Hk(C)

p∗- Hk(C ′′)

is exact. Since p ◦ i is trivial, so is p∗ ◦ i∗. Let y ∈ Ck with ∂(y) = 1 and p∗(y) istrivial in Hk(C ′′). There exists an element z ∈ C ′′k+1 such that

p(y) = ∂′′(z).

Let z̃ ∈ Ck+1 such that p(z̃) = z. Then

p(y∂(z̃−1)) = ∂′′(z)p(∂(z̃−1))= ∂′′(z)∂′′(p(z̃)−1)= ∂′′(z)∂′′(z−1)= 1.

Thus there exists w ∈ C ′k such that i(w) = y∂(z̃−1) with

i(∂′(w)) = ∂(i(w)) = ∂(y∂(z̃−1)) = 1.

and so ∂′(w) = 1. Hence i∗({w}) = {y}. The proof is finished now. �

Let X ′ be a ∆-subset of X. The relative homology H∗(X,X ′;G) is defined by

H∗(X,X ′;G) = H∗((Z(X)/Z(X ′)⊗G, ∂∗).

Corollary 5.3. Let X ′ be a ∆-subset of X. Then there is a long exact se-quence

· · · - Hk+1(X,X ′;G)∂k+1- Hk(X ′;G)

i∗- Hk(X;G)p∗- Hk(X,X ′;G) - · · ·

for abelian group G. �

6. SIMPLICIAL HOMOLOGY AND SINGULAR HOMOLOGY 33

6. Simplicial Homology and Singular Homology

Definition 6.1. Let X be a ∆-complex. Then simplicial homology of X withcoefficients in an abelian group G is defined by

H∆∗ (X;G) = H∗(C∆∗ (X);G).

For any space X, define

Sn(X) = Map(∆n, X)

be the set of all continuous maps from ∆n to X with

di = di∗ : Sn(X) = Map(∆n, X) - Sn−1(X) = Map(∆n−1, X).

for 0 ≤ i ≤ n. Then S∗(X) = {Sn(X)}n≥0 is a ∆-set. This allows us to definesingular homology:

Definition 6.2. For a pair of spaces (X,A), the singular homology H∗(X,A;G)with coefficients in an abelian group G is defined by

H∗(X,A;G) = H∗(S∗(X), S∗(A);G).

See Hatcher’s book [17] for details.

CHAPTER 2

Simplicial Sets and Homotopy

The notion of simplicial sets was arisen from establishing combinatorial modelsfor spaces. Recall that, given any topological space X, there is a singular simplicialset S∗(X), where Sn(X) is the set of all continuous maps from the n-simplex to X.Let Cn(X) be the free abelian group generated by Sn(X). Then one gets the chaincomplex C∗(X) with its homology the singular homology H∗(X; Z). In other words,the singular homology of X is obtained from the free abelian groups generated byS∗(X). One may ask whether the homotopy groups π∗(X) can be obtained in asimilar way. The answer is actually yes. In fact π∗(X) can be obtained directlyfrom S∗(X).

1. Simplicial Sets

1.1. Definitions.

Definition 1.1. A simplicial set means a ∆-set X together with a collectionof degeneracies si : Xn → Xn+1, 0 ≤ i ≤ n, such that

(1.1) djdi = di−1dj

for j < i,

(1.2) sjsi = si+1sj

for j ≤ i and

(1.3) djsi =

si−1dj j < iid j = i, i+ 1sidj−1 j > i+ 1.

The three identities for didj , sjsi and disj are called the simplicial identities.

Remark 1.2. One can use deleting-doubling for catching simplicial identities:

di : (x0, . . . , xn) −→ (x0, . . . , xi−1, xi+1, . . . , xn).

si : (x0, . . . , xn) −→ (x0, . . . , xi−1, xi, xi, xi+1, . . . , xn).

Let O be the category whose objects are finite ordered sets and whose mor-phisms are functions f : X → Y such that f(x) ≤ f(y) if x < y. The objects in Oare given by [n] = {0, . . . , n} for n ≥ 0, which are the same as the objects in O+.The morphisms in O are generated by di, which is defined in O+, and the followingmorphism

si : [n+ 1]→ [n] si =(

0 1 · · · i− 1 i i+ 1 i+ 2 · · · n+ 10 1 · · · i− 1 i i i+ 1 · · · n

)for 0 ≤ i ≤ n, that is, si hits i twice.

35

36 2. SIMPLICIAL SETS AND HOMOTOPY

Exercise 1.1. Show that simplicial sets are one-to-one correspondent to con-travariant functors from O to S.

More abstractly we have the definition of simplicial objects over any category.

Definition 1.3. Let C be a category. A simplicial object over C means acontravariant functor from O to C. Equivalently a simplicial object C means asequence of objects X = {Xn}n≥0 with face morphisms di : Xn → Xn−1 and de-generacy morphisms si : Xn → Xn+1, 0 ≤ i ≤ n, such that the three simplicialidentities (1.1), (1.2) and (1.3) hold. A simplicial group means a simplicial objectover the category of groups. In other words, a simplicial set X is a simplicial groupif and only if (1) each Xn is a group and (2) all faces and degeneracies are grouphomomorphisms.

Definition 1.4. A simplicial map (simplicial morphism) f : X → Y means asequence of functions (morphisms) f : Xn → Yn for each n ≥ 0 such that f ◦ di =di ◦ f and f ◦ si = si ◦ f , that is the diagram

Xn+1 �si

Xndi- Xn−1

Yn+1

f

?� si Yn

f

? di- Yn−1

f

?

commutes. Moreover if each Xn is a subset of Yn such that the inclusions Xn ⊆ Ynform a simplicial map, then X is called a simplicial subset of Y . A simplicial setX is called to be isomorphic to a simplicial set Y , denoted by X ∼= Y , if there is abijective simplicial map f : X → Y .

1.2. Basic Constructions.

Example 1.1 (n-simplex). The n-simplex ∆[n], as a simplicial set, is as follows:

∆[n]k = {(i0, i1, . . . , ik) | 0 ≤ i0 ≤ i1 ≤ · · · ≤ ik ≤ n}

for k ≤ n. The face dj : ∆[n]k → ∆[n]k−1 is given by

dj(i0, i1, . . . , dk) = (i0, i1, . . . , îj , . . . , ik),

that is deleting ij . The degeneracy sj : ∆[n]k → ∆[n]k+1 is defined by

sj(i0, i1, . . . , dk) = (i0, i1, . . . , ij , ij , . . . , ik),

that is doubling ij . Let σn = (0, 1, . . . , n) ∈ ∆[n]n. Then any elements in ∆[n] canbe written as iterated compositions of faces and degeneracies of σn.

Proposition 1.5. Let X be a simplicial set and let x ∈ Xn be an element.Then there exists a unique simplicial map

fx : ∆[n]→ X

such that fx(σn) = x.

Proof. We leave the proof as an exercise to the reader. �

1. SIMPLICIAL SETS 37

Definition 1.6. Let X be a simplicial set and A = {An}n≥0 with An ⊆ Xn.The simplicial subset of X generated by A is defined by

〈A〉 =⋂{A ⊆ Y ⊆ X | Y is a simplicial subset of X},

namely 〈A〉 consists of elements in X that can be written as iterated compositionsof faces and degeneracies of the elements in A.

Example 1.2 (n-sphere). The simplicial n-sphere Sn is defined by

Sn = ∆[n]/∂(∆[n]),

where ∂(∆[n]) is the simplicial subset of ∆[n] generated by ∆[n]k for k < n. Let’swrite explicitly for the elements in the simplicial circle S1.

Now that∆[1]k = {(i0, . . . , ik) | 0 ≤ i0 ≤ · · · ≤ ik ≤ 1}

= {(i︷︸︸︷

0, . . . , 0, 1, . . . , 1 | 0 ≤ i ≤ k + 1}has k + 2 elements. Now

∂(∆[1])k = {(0, . . . , 0), (1, . . . , 1)}.By definition, S1 = ∆[1]/∂(∆[1]). Thus

S1k = {∗, (i0, . . . , ik) | 0 ≤ i0 ≤ · · · ≤ ik ≤ 1}

= {(i︷︸︸︷

0, . . . , 0, 1, . . . , 1 | 1 ≤ i ≤ k}has k + 1 elements including the basepoint ∗ = (0, . . . , 0) ∼ (1, . . . , 1).

For general simplicial n-sphere Sn, we have Snk = {∗} for k < n andSnk = {∗, (i0, . . . , ik) | 0 ≤ i0 ≤ · · · ≤ ik ≤ n with {i0, . . . , ik} = {0, 1, . . . , n}}

for k ≥ n. �

Example 1.3 (Cartesian Product). Let X and Y be simplicial sets. DefineX × Y by setting

(X × Y )n = Xn × Ynwith dX×Yi = (d

Xi , d

Yi ) and s

X×Yi = (s

Xi , s

Yi ). Show that X × Y is a simplicial set.

�

Example 1.4. Let f : X → Y be a simplicial map. ThenIm(f : X → Y ) = {Im(f : Xn → Yn)}n≥0

is a simplicial subset of Y . �

For a sequence of nonnegative integers I = (i1, i2, . . . , ik) of length k = l(I),denote dI = di1 · · · dik and sI = si1 · · · sik .

Example 1.5 (Basepoint). Let X be a simplicial set and let x0 ∈ X0. Thenthe image of the representing map

fx0 : ∆[0] - X

is a simplicial subset of X consisting of only one element fx0(0, . . . , 0) = sI(x0) ineach dimension. Thus a basepoint ∗ of X means a sequence of elements

{fx0(n+1︷︸︸︷

0, . . . , 0)}n≥0


corresponds to the elements x0 ∈ X0.A pointed simplicial set means a simplicial set with a given basepoint. A pointed

simplicial map means a simplicial map that preserves the basepoints.A simplicial set is called reduced if X0 has only one element. For a reduced

simplicial set, there is a unique choice of basepoints. Moreover any simplicial mapbetween reduced simplicial sets is pointed. �

Example 1.6 (Push-out). Let C �g

Af- B be simplicial maps. Define

Xn to be the push-out in the diagram

Anf - Bn

push

Cn

g

?- Xn,

?

that is Xn is the quotient of the disjoint union Bn∐Cn subject to the equivalence

relation generated by f(a) ∼ g(a) for a ∈ An. Then X = {Xn}n≥0 with faces anddegeneracies induced from that in B and C forms a simplicial set with a push-outdiagram of simplicial sets

Af - B

push

C

g

?- X.

?

For instance, let ∂∆[n] = 〈di(0, 1, . . . , n) | 0 ≤ i ≤ n〉 ⊆ ∆[n] be the simplicialsubset of ∆[n] generated by all of the faces of the n-simplex σn = (0, 1, . . . , n).Then

∂∆[n] ⊂ - ∆[n]

push

∗?

- Sn?

is a push-out diagram. �

Example 1.7 (Wedge and Smash). Let X and Y be pointed simplicial sets.The wedge X ∨ Y of X and Y is the simplicial set obtained by identifying thebasepoint of X with the basepoint of Y . In other words, X ∨ Y is defined by thepush-out diagram

∗ ⊂ - X

push

Y?

∩

- X ∨ Y.?


Note that X ∨ Y can be described as the simplicial subset of X × Y consisting of(x, ∗) for x ∈ X and (∗, y) for y ∈ Y . The smash product X ∧ Y is defined to bethe simplicial quotient X × Y/(X ∨ Y ).

An element x ∈ Xn is called nondegenerate if x 6= si(y) for any y ∈ Xn−1.

Exercise 1.2. Determine nondegenerate elements in S1 ∧ S1. In general, de-termine nondegenerate elements in Sn ∧ Sm.

Remark 1.7. As spaces, we know that Sm ∧ Sn ∼= Sn+m. It should be pointedthat, as simplicial sets, Sm ∧ Sn 6∼= Sm+n although they have the same “homotopytype”.

Example 1.8 (Skeleton of Simplicial Sets). Let X be a simplicial set. Letskn(X) = 〈Xj | j ≤ n〉 is the simplicial subset of X generated by Xj for j ≤ n.Then there is a filtration

sk0X ⊆ sk1X ⊆ · · · sknX ⊆ · · ·with X =

⋃n sknX, called skeleton filtration of X. There are push-out diagrams∐

x∈Xn nondegenerate

∂∆[n]

∐fx|∂∆[n]- skn−1X

∐x∈Xn nondegenerate

∆[n]?

∩

∐fx - sknX

?

∩

for each n ≥ 1. �

Example 1.9 (Cone and suspension). Let X be a pointed simplicial set. Thereduced cone CX is defined by setting

(CX)n = {(x, q) | x ∈ Xn−q, 0 ≤ q ≤ n} with (∗, q) all identified to ∗,

(1.4)di(x, q) =

{(x, q − 1) for 0 ≤ i < q,(di−qx, q) for q ≤ i ≤ n,

si(x, q) ={

(x, q + 1) for 0 ≤ i < q,(si−qx, q) for q ≤ i ≤ n,

where for x ∈ X0, d1(x, 1) = ∗. By identifying x with (x, 0), X is a simplicial subsetof CX. The reduced suspension ΣX is defined by ΣX = CX/X.

We give another description for the construction CX. Let X be any simplicialset without choosing basepoint. Let x̃0 be a new point not in X. Let

(C̃X)n = Xn⊔s−1(Xn−1)

⊔· · ·⊔sn−1(X0)

⊔{sn0 x̃0} = {sn0 x̃0}

n∐k=0

sk−1(Xn−k)

be the disjoint union as a set, where s−1(Xj) = Xj . Regard s−1 as (−1)st degen-eracy operation. From the simplicial identities, we have

(1.5)dis−1 =

{id if i = 0

s−1di−1 if i ≥ 1

sis−1 ={

s−1s−1 if i = 0s−1si−1 if i ≥ 1,


which induces the operations di and si on C̃X exactly given as in Formula (1.4)by identifying sq−1x with (x, q), where d1(s−1x) = x̃0 for x ∈ X0. The resultingsimplicial set C̃X is called unreduced cone of X. The reduced cone CX is then thequotient given by requiring x0 ∼ x̃0 and (x0, 1) = s−1x0 ∼ (s0x0, 0) = s0x0 ∼ s0x̃0for the basepoint vertex x0 ∈ X0. �

Proposition 1.8 (Contractible Criterion). Let X be a pointed simplicial set.Then the inclusion j : X ↪→ CX admits a simplicial retraction if and only if thereexists a function s−1 : Xn → Xn+1 for each n ≥ 0 such that s−1(∗) = ∗ andIdentities (1.5) hold.

Proof. Suppose that there is a simplicial map r : CX → X such that thecomposite r ◦ j = idX . Define s−1(x) = r(x, 1) for x ∈ Xn. Then

dis−1(x) = dir(x, 1) = rdi(x, 1) = r(dis−1(x, 0))sis−1(x) = sir(x, 1) = rsi(x, 1) = r(sis−1(x, 0))

and so Identities (1.5) hold.Conversely, if s−1 : Xn → Xn+1 is defined with Identities (1.5), define the

function r : CX → X by setting r(x, 0) = x and r(x, q) = sq−1x for q > 0. Then ris a simplicial map by Identities (1.5) with r|X = idX and hence the result. �

Theorem 1.9 (Null-homotopy Criterion). Let f : X → Y be a pointed simpli-cial map. Then the extension problem

CXf̃ - Y

X∪

6

f

-

has a solution of a simplicial map f̃ if and only if there exists F : Xn → Yn+1 foreach n ≥ 0 such that the following conditions hold:

(1). d0F = f and diF = Fdi−1 for i > 0;(2). siF = Fsi−1 for i > 0.

Proof. Suppose that the solution f̃ exists. Define F (x) = f̃(x, 1) for x ∈ Xn.Then F satisfies the conditions.

Conversely suppose that there exists F : Xn → Yn+1 satisfying the conditions.Define f̃ : CX → Y by setting

f̃(x, q) =

f(x) if q = 0F (x) if q = 1sq−10 F (x) if q ≥ 2.

Note that f̃ |X = f . We show that f̃ is a simplicial map, that is, dj f̃(x, q) =f̃dj(x, q) and sj f̃(x, q) = f̃sj(x, q). Since f̃ |X = f , these identities hold for q = 0.Consider the case q > 0. Note that

dj f̃(x, q) = djsq−10 F (x) =

d0F (x) if j = 0, q = 1sq−20 F (x) if 0 ≤ j < q, q ≥ 2sq−10 dj−q+1F (x) if j ≥ q,


f̃(dj(x, q)) =

f(x) = if j = 0, q = 1f̃(x, q − 1) = sq−20 F (x) if 0 ≤ j < q, q ≥ 2f̃((dj−qx, q)) = s

q−10 F (dj−qx) if j ≥ q.

Since d0F (x) = f(x) and diF (x) = F (di−1(x)) for i > 0, dj f̃(x, q) = f̃dj(x, q) forany x and q. Now

sj f̃(x, q) = sjsq−10 F (x) =

{sq0F (x) if 0 ≤ j < qsq−10 sj−q+1F (x) if j ≥ q,

f̃sj(x, q) ={f̃(x, q + 1) = sq0F (x) if 0 ≤ j < qf̃(sj−q, q) = s

q−10 F (sj−qx) if j ≥ q.

Since siF (x) = F (si−1(x)) for i > 0, sj f̃(x, q) = f̃sj(x, q) for any x and q. Thisfinishes the proof. �

Exercise 1.3. Show that CS0 ∼= ∆[1].

Proposition 1.10. ΣSn ∼= Sn+1 for n ≥ 0.

Proof. Let σn be the nondegenerate element in Snn . Then, from the definitionof CSn, the nondegenerate element in CSn are given by ∗ ∈ (CSn)0, (σn, 0) ∈(CSn)n and (σn, 1) ∈ (CSn)n+1 with d0(σn, 1) = σn. Thus the nondegenerateelements in ΣSn+1 are given by ∗ ∈ (ΣSn)0 and (σn, 1) ∈ (ΣSn)n+1. It followsthat the representing map

f(σn,1) : ∆[n] - ΣSn

is onto and it factors through the quotient Sn+1 because di(σn, 1) = ∗ in ΣSn forall 0 ≤ i ≤ n. The resulting simplicial map

f̄(σn,1) : Sn+1 - ΣSn

is bijective by counting the elements in each dimension, and hence the result. �

Exercise 1.4. Show that C̃∆[n] ∼= ∆[n+ 1].

Let X and Y be simplicial sets. Write HomS(X,Y ) for the set of simplicialmaps from X to Y . The following example gives a way to construct the mappingspace (as a simplicial set) from X to Y .

Example 1.10 (Mapping Spaces). Let σn = (0, 1, . . . , n) ∈ ∆[n] be the nonde-generate element.

di = fdiσn : ∆[n− 1] - ∆[n]

si = fsiσn : ∆[n+ 1] - ∆[n]

be the representing maps of diσn and siσn for 0 ≤ i ≤ n. Observe that thecomposites

∆[n− 2]dj- ∆[n− 1]

di- ∆[n],

∆[n+ 2]sj- ∆[n+ 1]

di- ∆[n],

∆[n]dj- ∆[n+ 1]

si - ∆[n]


are representation maps of the elements djdiσn, sjsiσn and djsiσn, respectively.From the simplicial identities, the sequences of simplicial sets {∆[n]}n≥0 with diand si is a cosimplicial simplicial set, namely the identities

(1.6)

djdi = di+1dj for i ≥ jsisj = sjsi+1 for i ≥ j

sidj =

djsi−1 j < iid j = i, i+ 1

dj−1si j > i+ 1.

Let Map(X,Y )n = HomS(X ×∆[n], Y ) and let

di = (idX ×di)∗ : Map(X,Y )n = HomS(X ×∆[n], Y )- HomS(X ×∆[n− 1], Y )

= Map(X,Y )n−1,si = (idX ×si)∗ : Map(X,Y )n = HomS(X ×∆[n], Y )

- HomS(X ×∆[n+ 1], Y )= Map(X,Y )n+1

for 0 ≤ i ≤ n. From Identities (1.6), Map(X,Y ) = {Map(X,Y )n}n≥0 with di andsi is a simplicial set, which is called the mapping space from X to Y .

1.3. Polyhedral Simplicial Sets.

Example 1.11 (Simplicial sets from simplicial complex). Let K be an abstractsimplicial complex, and let the vertices of K be ordered. Then we obtain a simplicialset KS = {KSn}n≥0:

KSn = {(v0, v1, . . . , vn) | vi are vertices of a simplex of K with v0 ≤ v1 ≤ · · · ≤ vn},

di(v0, v1, . . . , vn) = (v0, v1, . . . , v̂i, . . . , vn) deleting vi,si(v0, v1, . . . , vn) = (v0, v1, . . . , vi, vi, . . . , vn) doubling vi

with a ∆-map K∆ ⊂ - KS . Observe that KS has the nondegenerate elementsgiven by K∆. �

Definition 1.11. A simplicial set X is called polyhedral if there exists anabstract simplicial complex K such that X ∼= KS .

For a simplicial set X, let XDn denote the set of nondegenerate elements in Xn.

Lemma 1.12. Let X and Y be simplicial sets and let f : X → Y be simplicialmap such that

(1). If x is a nondegenerate element of X, then f(x) is a nondegenerate ele-ment of Y and

(2). The induced function f : XDn → Y Dn is one-to-one for each n ≥ 0.Then f is one-to-one.

Proof. We show by induction that f : Xn → Yn is one-to-one for each n ≥ 0.When n = 0, we are given that f : X0 → Y0 is one-to-one by the second assumption.Suppose that f : Xn−1 → Yn−1 is one-to-one. Let x, y ∈ Xn such that f(x) = f(y).If both x and y are nondegenerate elements, then x = y by the assumption. Supposethat one of x, y is a degenerate element. We may assume that y = sjz for somez ∈ Xn−1 and some 0 ≤ j ≤ n − 1. Then f(x) = f(y) = f(sjz) = sjf(z). By thefirst assumption, x must be a degenerate element. Let x = siw.


If i = j, then

f(w) = disif(w) = dif(siw) = dif(x) = dif(y) = dif(siz) = disif(z) = f(z).

By induction, w = z and so x = siw = siz = y.If i 6= j, we may assume that i < j. Then

f(w) = dif(siw) = dif(x) = dif(y) = dif(sjz) = f(disjz).

By induction, w = disjz and so

x = siw = sidisjz = sisj−1diz = sjsidiz.

It follows that

f(sidiz) = djf(sjsidiz) = djf(x) = djf(y) = djf(sjz) = djsjf(z) = f(z)

and so sidiz = z by induction. Thus

x = sjsidiz = sjz = y

and hence the result. �

Now let X be a simplicial set and let 2X0 be the set of all subsets of X0. Define

φ :∐n≥0

XDn −→ 2X0

by setting φ(x) = {fx(0), fx(1), . . . , fx(n)} for x ∈ Xn.

Theorem 1.13. Let X be a simplicial set. Then X is polyhedral if and only ifthe following holds:

(1). There exists an order of X0 such that, for each x ∈ Xn,

fx(0) ≤ fx(1) ≤ · · · ≤ fx(n).

(2). The function φ :⋃

n≥0XDn −→ 2X0 is one-to-one.

Proof. =⇒ follows from the construction.⇐=. By the proof of Theorem 3.6 of Chapter 1, for each x ∈ XDn , we have

fx(0) < fx(1) < · · · < fx(n) and the abstract simplicial complex K with vertex setX0 is defined by requiring that: {a0, . . . , an} is a simplex of K if and only if theset {a0, . . . , an} ∈ Imφ. Moreover by the proof of Theorem 3.6 of Chapter 1, themap φ induces a ∆-map

g : K∆ ⊂ - Xgiven by

g({fx(0), fx(1), . . . , fx(n)}) = xfor each x ∈ XDn . The ∆-map g extends to a simplicial map

g̃ : KS - X.

Since g maps onto nondegenerate elements of X, the simplicial map g̃ is onto. Sinceg̃ sends each nondegenerate element of KS (that are given by the elements in K∆)to a nondegenerate element of X and g : K∆n → XDn is one-to-one, the map g̃ isone-to-one by the above lemma. Thus g̃ is a simplicial isomorphism and hence theresult. �


1.4. Relations Between ∆-Sets and Simplicial Sets. Let X be a simpli-cial set. By forgetting degeneracy, we can regard X as a ∆-set. On the other hand,given a ∆-set X, we can construct a simplicial set XS by “adding degeneracies”.

Let σn = (0, 1, . . . , n) ∈ ∆[n] be the nondegenerate element. Recall that wehave simplicial maps

di = fdiσn : ∆[n− 1] - ∆[n]si = fsiσn : ∆[n+ 1] - ∆[n]

be the representing maps of diσn and siσn for 0 ≤ i ≤ n. Given a ∆-set X, wedefine a sequence of sets by setting

XSn =∐k≥0

∆[k]n ×Xk/ ∼,

where ∼ is the equivalence relation generated by(a, dix) ∼ (di(a), x)

for any a ∈ ∆[k − 1]n, x ∈ Xk, k ≥ 1 and 0 ≤ i ≤ k. Define the facesdj : XSn → XSn−1

and the degeneraciessj : XSn → XSn+1

by settingdj(a, x) = (dja, x) sj(a, x) = (sja, x)

for 0 ≤ j ≤ n, that is, taking the faces and degeneracies on the first coordinate.We check that dj and sj preserve the equivalence relation:

dj(a, dix) = (dja, dix)

dj(di(a), x) = (djdi(a), x)= (di(dja), x) because di is a simplicial map

sj(a, dix) = (sja, dix)

sj(di(a), x) = (sjdi(a), x)= (di(sja), x) because di is a simplicial map.

Thus dj and sj preserve the equivalence relation and so dj : XSn → XSn−1 andsj : XSn → XSn+1 are well-defined. The simplicial identities automatically hold asthey are induced from the simplicial structure of ∆[k]’s.

The construction looks complicated. But the output simplicial set is not so bigas the equivalence relation identifies many elements. Let

φn : Xn → XSnbe the function defined by φn(x) to be the equivalence class [σn, x] in XSn . Then

dj(φn(x)) = dj [σn, x]= [djσn, x]= [dj(σn−1, x]= [σn−1, djx]= φ(dj(x)).

Thus φ = {φn} : X → XS is a ∆-map.


Lemma 1.14. The ∆-map φ : X → XS is one-to-one.

Proof. Observe that the equivalence relation is generated by connecting someelements in ∆[k]n×Xk and ∆[k− 1]n×Xk−1 for each k. Thus [σn, x] 6= [σn, y] forx 6= y ∈ Xn, and so φ is one-to-one. �

Theorem 1.15. The construction X 7→ XS is a functor from the category of∆-sets and ∆-maps to the category of simplicial sets and simplicial maps. Moreoversuppose that X is a ∆-set such that one of the faces

di : Xn → Xn−1is onto for each n ≥ 1. Then the simplicial set XS is generated by φ(X) with allnondegenerate elements given in φ(X).

Proof. For showing the first assertion, let f : X → Y be a ∆-map. Then themaps

id∆[k]×f : ∆[k]n ×Xk - ∆[k]n × Ykinduces a function fSn : X

Sn → Y Sn because from

id∆[k]×f(dia, x) = (dia, f(x))id∆[k−1]×f(a, dix) = (a, f(dix))

= (a, di(f(x))

the functions {id∆[k]×f} preserves the equivalence relations. Moreover fS ={fSn } → XS → Y S is a simplicial map because the functions {id∆[k]×f} com-mutes with faces and degeneracies. Clear (g ◦ f)S = gS ◦ fS for ∆-maps f : X → Yand g : Y → Z. Thus X 7→ XS is a functor.

For the second assertion, we first show that every equivalence class in XSn canbe represented by an element in ∆[n]n ×Xn. Let (a, x) ∈ ∆[k]n ×Xk.

If k < n, since there exists a face di : Xk+1 → Xk which is onto, x = diy forsome y ∈ Xk+1. Thus (a, x) = (a, diy) ∼ (dia, y) with (di, y) ∈ ∆[k + 1]n ×Xk+1.By iterating this procedure, (a, x) ∼ (b, z) for some (b, z) ∈ ∆[n]×Xn.

If k > n, then all elements in ∆[k]n are given by iterated faces σk. Thusa = di1di2 · · · dik−nσk for some 0 ≤ i1 < i2 < · · · < ik−n ≤ k. It follows that

dik−n ◦ · · · ◦ di2di1(σn) = di1di2 · · · dik−nσk = a

and so(a, x) = (dik−n ◦ · · · ◦ di2di1(σn), x)

∼ (dik−n−1 ◦ · · · ◦ di2di1 , dik−nx)· · ·∼ (σn, di1di2 · · · dik−nx).

Thus every equivalence class in XSn can be represented by an element in ∆[n]n×Xn. Now we let (a, x) ∈ ∆[n]n ×Xn. Note that σn is only nondegenerate elementin

∆[n]n = {(i0, i1, . . . , in) | 0 ≤ i0 ≤ i1 ≤ · · · ≤ in ≤ n}.If a = σn, then (a, x) ∈ φ(X). Otherwise

a = sl1sl2 · · · slkdj1dj2 · · · djkσnfor some 0 ≤ lk < · · · < l2 < l1 ≤ n− 1 and 0 ≤ j1 < j2 < · · · < jk ≤ n with k ≥ 1.(Recall that dj is given by deleting the coordinates in the sequences (i0, . . . , in) and


sj is given by doubling the coordinates. Since a = (i0, . . . , in) 6= (0, 1, . . . , n), wecan apply dj operations and then apply degeneracies backwards to a.) Then

[a, x] = [sl1sl2 · · · slkdj1dj2 · · · djkσn, x]= sl1sl2 · · · slk [dj1dj2 · · · djkσn, x]= sl1sl2 · · · slk [djk ◦ · · · ◦ dj1(σn−k), x]= sl1sl2 · · · slk [σn−k, dj1dj2 · · · djkx]

with [σn−k, dj1dj2 · · · djkx] ∈ φ(X) and hence the result. �

Let f : X → Y be a ∆-map. Then there is a commutative diagram

Xφ - XS

Y

f

? φ - Y S

fS

?

because, for any x ∈ Xn,

fS ◦ φ(x) = fS [σn, x] = [σn, f(x)] = φ ◦ f(x).

Thus φ : X → XS is a natural transformation. More precisely φ is a naturaltransformation from the identity functor of the category of ∆-sets to the functor(−)S composing with the forgetful functor from the category of simplicial sets tothe category of ∆-sets.

Proposition 1.16. The ∆-map φ : X → XS has the following properties:(1). Let Z be any simplicial set and let g : X → Z be a ∆-map. Then there

exists a simplicial map g̃ : XS → Z such that g = g̃ ◦ φ.(2). Suppose that X satisfies the property that one of the faces di : Xn → Xn−1

is onto for each n ≥ 1. Then g̃ is uniquely determined by g.

Proof. The second assertion follows from the first assertion because XS isgenerated by X in this case.

For proving assertion (1), let

Z̄n =∐k≥0

∆[k]n × Zk/ ≈

where ≈ is the equivalence relation generated by

(a, diz) ≈ (di(a), z) (b, siz) ≈ (sib, z)

for any a ∈ ∆[k − 1]n, b ∈ ∆[k + 1]n, z ∈ Zk and 0 ≤ i ≤ k. Then Z̄ = {Z̄n}n≥0 isa simplicial set with the faces

dj : Z̄n → Z̄n−1and the degeneracies

sj : Z̄n → Z̄n+1defined by

dj(a, z) = (dja, z) sj(a, z) = (sja, z)

for 0 ≤ j ≤ n. (Similar to the construction XS , one can check that faces anddegeneracies defined above preserve the equivalence relation.)


Let ψ : Zn → Z̄n be the function defined by ψ(z) = [σn, z]. Then the sequenceof functions ψ = {ψn} : Z → Z̄ is a simplicial map.

Let ḡn : XSn → Z̄n be induced by the functionsid∆[k]×g : ∆[k]n ×Xk - ∆[k]n × Zk.

From the lines in the proof of the above theorem, it follows that the functionḡn is well-defined and ḡ = {ḡn} : XS → Z̄ is a simplicial map. Now there is acommutative diagram

Xφ - XS

Z

g

? ψ - Z̄

ḡ

?

because for any x ∈ Xnḡ ◦ φ(x) = ḡ([σn, x]) = [σn, g(x)] = ψ ◦ g(x).

The assertion will follow by proving that ψ : Z → Z̄ is an isomorphism because ifso, the composite g̃ = ψ−1 ◦ ḡ is the desired map.

To show that ψ : Z → Z̄ is onto, from the proof of the above theorem, everyequivalence class in Z̄n can be represented by an element in ∆[n]n × Zn. Let(a, x) ∈ ∆[n]n × Zn. If a 6= σn, then a = si1si2 · · · sikdj1dj2 · · · djkσn for some0 ≤ lk < · · · < l2 < l1 ≤ n− 1 and 0 ≤ j1 < j2 < · · · < jk ≤ n with k ≥ 1. Then

djk · · · dj2dj1sik · · · si2si1(σn) = djk · · · dj2dj1sik · · · si2(si1σn−1)= djk · · · dj2dj1sik · · · si3(si1si2(σn−1))= djk · · · dj2dj1sik · · · si3(si1si2σn−1)· · ·= djk · · · dj2dj1(si1si2 · · · sikσn−k)= djk · · · dj2(si1si2 · · · sikdi1(σn−k))= djk · · · dj2(si1si2 · · · sikdi1σn−k+1)= djk · · · dj3(si1si2 · · · sikdi1dj2(σn−k+1))= si1si2 · · · sikdj1dj2 · · · djkσn= a.

Thus

(a, x) = (djk · · · dj2dj1sik · · · si2si1(σn), x) ≈ (σn, si1si2 · · · sikdj1dj2 · · · djkx)and so ψ is onto.

To see that ψ is one-to-one, similar to the above argument, every element inset

∐∞k=0 ∆[k]n is the image of the iterated composites of the functions d

simplicial objects and homotopy groups j. wupeople.brandeis.edu/~syzygy/jiewusimplicial.pdf5....

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