Simplicial Objects and Homotopy Groups. J. Wu

Transcription

1 Simplicial Objects and Homotopy Groups J. Wu Department of Mathematics, National University of Singapore, Singapore address: URL:

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

3 Contents Chapter 1. -Objects and Homology sets 5 2. Geometric Simplicial Complexes 7 3. Abstract Simplicial Complexes and -sets complexes and the geometric realization of -sets Homology of -sets Simplicial Homology and Singular Homology 33 Chapter 2. Simplicial Sets and Homotopy Simplicial Sets Geometric Realization of Simplicial Sets Homotopy and Fibrant Simplicial Sets The Relations between Spaces and Simplicial Sets Homotopy Groups Simplicial Fibration 88 Chapter 3. Simplicial Group Theory Moore Chains and Homotopy Groups Simplicial Abelian Groups and the Hurewicz Theorem Free Group Constructions Free Products with Amalgamation Simplicial and -Structure on Configuration Spaces (Homotopy Groups and Braids) 132 Bibliography 157 3

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5 CHAPTER 1 -Objects and Homology 1. -sets Definition 1.1. A -set means a sequence of sets X = {X n } n 0 with faces d i : X n X n 1, 0 i n, such that (1.1) d i d j = d j d i+1 for i j, which is called the -identity. Remark 1.2. One can use coordinate projections for catching -identity: d i : (x 0,..., x n ) (x 0,..., x i 1, x i+1,..., x n ). Let O + be the category whose objects are finite ordered sets and whose morphisms are functions f : X Y such that f(x) < f(y) if x < y. Note that the objects in O + are given by [n] = {0, 1,..., n} for n 0 and the morphisms in O + are generated by d i : [n 1] [n] with { d i j if j < i (j) = j + 1 if j i for 0 i n, that is d i is the ordered embedding missing i. We may write the function d i in matrix form: ( ) d i 0 1 i 1 i i + 1 n 1 =. 0 1 i 1 i + 1 i + 2 n. The morphisms d i satisfy the following identity: for i j. d j d i = d i+1 d j Remark 1.3. For seeing that morphisms in O + are generated by d i, observe that any morphism in O + means an ordered embedding, which can be written as the compositions of d i s. Let S denote the category of sets. Proposition sets are one-to-one correspondent to contravariant functors from O + to S. Proof. Let F : O + S be a contravariant functor. Define X n = F ([n]) and d i = F (d i ): X n = F ([n]) X n 1 = F ([n 1]). Then X is a -set. Conversely suppose that X is a -set. Define the F : O + S by setting F ([n]) = X n and F (d i ) = d i. Then F is a contravariant functor. 5

6 6 1. -OBJECTS AND HOMOLOGY A -set G = {G n } n 0 is called a -group if each G n is a group, and each face d i is a group homomorphism. In other words, a -group means a contravariant functor 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 = {X n } n 0 with faces d i : X n X n 1 as morphisms in C. Example 1.1 (n-simplex). The n-simplex + [n], as a -set, is as follows: + [n] k = {(i 0, i 1,..., i k ) 0 i 0 < i 1 < < i k n} for k n and + [n] k = for k > n. The face d j : + [n] k + [n] k 1 is given by d j (i 0, i 1,..., d k ) = (i 0, i 1,..., î j,..., i k ), that is deleting i j. Let σ n = (0, 1,..., n). Then (i 0, i 1,..., i k ) = d j1 d j2 d jn k σ n, where j 1 < j 2 < < j n k with {j 1,..., j k } = {0, 1,..., n} {i 0, i 1,..., i k }. In other words, any elements in [n] can be written an iterated face of σ n. Definition 1.5. A -map f : X Y means a sequence of functions f : X n Y n for each n 0 such that f d i = d i f, that is the diagram X n f Yn d i d i f X n 1 Y n 1 commutes. A -subset A of a -set X means a sequence of subsets A n X n such that d i (A n ) A n 1 for all 0 i n <. A -set X is called to be isomorphic to a -set Y, denoted by X = Y, if there is a bijective -map f : X Y. Let X be a -set and let A be a -subset. Clearly the inclusion A X, that is, A n X n for each n 0, is a -map. Proposition 1.6. Let X be a -set and let x X n be an element. Then there exists a unique -map f x : + [n] X such that f x (σ n ) = x. Proof. From the assumption f x (σ n ) = x, we have f x (i 0, i 1,..., i k ) = f x (d j1 d j2 d jn k σ n ) = d j1 d j2 d jn k f x (σ n ) = d j1 d j2 d jn k x. This defines a -map f x such that f x (σ n ) = x. The simplicial map f x : + [n] X is called representing map of x.

7 2. GEOMETRIC SIMPLICIAL COMPLEXES 7 Definition 1.7. Let X be a -set and let S n=0 X n. The -subset generated by S is defined by S = {A X S A n A = {A n } is a subset of X}. n=0 For x X n, {x} is simply denoted by x. A -set X is called monogenic if it is generated by a single element. Proposition 1.8. Let X be a -set and let S n=0 X n. Then S n = (S X n ) d j1 d j2 d jk (S X n+k ) for each n 0. Proof. Let K n = (S X n ) 0 j 1 < j 2 < < j k n + k 1 k < 0 j 1 < j 2 < < j k n + k 1 k < d j1 d j2 d jk (S X n+k ) Then d i (K n ) K n 1 by the -identity (1.1) for 0 i n. Thus K = {K n } n 0 is a -subset of X. If A is a -subset of X with S n=0 A n, then clearly K n A n for each n 0. Thus K = S and hence the assertion. Problem 1.1. Classify all monogenic -sets. 2. Geometric Simplicial Complexes In this section, we give a review on (geometric) simplicial complex Simplices. Definition 2.1. A set {a 0,..., a n } of (n + 1) points in R m is called geometrically independent if the vectors a 1 a 0, a 2 a 0,..., a n a 0 are linearly independent. A geometric n-simplex { } n n σ n = x = t i a i t i 0 and t i = 1 R m i=0 with subspace topology, where {a 0,..., a n } linearly independent. Sometimes we write σ = a 0 a 1 a n for meaning that σ is spanned by the vertices a 0, a 1,..., a n. The numbers t i are uniquely determined by the point x, which are called barycentric coordinates of x of σ with respect to a 0,..., a n. The points a i are called the vertices of σ n. If {a i0, a i1,..., a ir } is a subset of {a 0,..., a n } with 0 i 0 < < i r n, then the subspace r n t r = t j a ij λ j 0 and t j = 1 j=0 is call a face of σ n. The number n is called the dimension of σ. A face of σ different from σ itself is called a proper face. The union of all proper faces of σ is called the boundary of σ, denoted by σ. The interior Int(σ) of σ is defined by Intσ = σ σ, sometimes called open simplex. i=0 i=0

8 8 1. -OBJECTS AND HOMOLOGY Exercise 2.1. (1). The barycentric coordinates t i (x) of x with respect to a 0,..., a n are continuous on x. (2). x σ t i (x) = 0 for some 0 i n. Thus x Intσ t i (x) > 0 for all 0 i n. (3). σ is compact and convex in R m, where a subset A of R m is called convex if 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 set of points spanning σ. (5). σ equals to the union of all line segments joining a 0 to points of the simplex τ spanned by a 1,..., a n. Intσ is convex and is open in plane spanned by the points a 0, a 1,..., a n ; its closure is σ. Furthermore Intσ is the union of all open line segments joining a 0 to the points of Intτ. (6). Recall that in R m, the norm of a point x = (x 1,..., x m ) is defined to be x = m x 2 i. i=1 The m-dimensional unit ball D m is defined by D n = {x R m x 1} and the (m 1)-dimensional unit sphere S m 1 is defined to be S m 1 = {x R m x = 1}. There is a homeomorphism of σ with the unit ball D n that carries σ onto the unit sphere S n Geometric Simplicial Complex. For having (geometric) simplicial complexes with arbitrary many simplices and arbitrary dimension of its simplices, we need to extend our m-dimensional Euclidean space to an infinite dimensional vector space as follows: Let J be an arbitrary index set, and let R J be the J-fold product of R with itself. An element in R J is a function from J to R denoted by x = (x α ) α J. The product R J is vector space with addition given by the usual component-wise addition and multiplication by scalars, that is, (x + y) α = x α + y α and (cx) α = cx α for α J. Let E J be the subset of R J consisting of all points (x α ) α J such that x α = 0 for all but finitely many values of α. Then E J is a vector subspace of R J with a basis given by e α for α J, where e α is the map from J to R with e α (α) = 1 and e α (β) = 0 for β α. (Note. {e α } α J does not form a basis for R J if J is an infinite set. Since {e α α J } is a basis, we can also consider that EJ is the vector subspace of R J generated by e α for α J. ) The space E J has a metric defined by setting x y = max{ x α y α α J. Definition 2.2. A geometric simplicial complex K is a collection of simplices, all contained in some Euclidean space E J for some index set J such that (1). if σ n is a simplex in K and τ p is a face of σ n, then τ p is in K; and (2). if σ n and τ p are simplices of K, then σ n τ p is either empty, or a common face of σ n and τ p. The dimension of K is defined to be dim K = sup{dim σ σ a simplex ofk}.

9 2. GEOMETRIC SIMPLICIAL COMPLEXES 9 So an n-dimensional simplicial complex means a simplicial complex without simplices of dimension higher than n. If L is a sub-collection of K that contains all faces of its elements, then L is a simplicial complex in its own right, called a subcomplex of K. One special subcomplex of K is the collection of all simplices of K dimension at most n, called the n-skeleton of K denoted by sk n K. The points of the collection sk 0 K are called vertices of K. Exercise 2.2. A collection K of simplices is a simplicial complex if and only if 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 = σ n K σ n R J the union of the simplices of K. Giving each simplex its natural topology as subspace of R N for some large N, we then define the topology on K by requiring that: a subset A of K is closed in K if and only if A σ is closed in σ for each simplex 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 simplex K, that is, K only has finitely many simplices. In such a case, since we can choose the index set J to be a finite set, K R m for some large m and our re-defined topology on K coincides with the subspace topology of K in R m. If K has infinitely many simplices, then our redefined topology could be different from the subspace topology. For instance, let K be the collection of all line segments from the origin to the points in the unit circle S 1. The underline set K is the same as the unit disk D 2, but one can check that the topology on K is different from D 2. 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 E J, because if A is closed in K in subspace topology then A σ is closed in σ for every σ K and so A is closed in K. The redefined topology on K is important for making simplicial 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 {K α } is a collection of simplicial complexes in E J, and if the intersection of every pair K α K β is the polyhedron of a simplicial complex which

10 OBJECTS AND HOMOLOGY is a subcomplex of both K α and K β, then the union α K α 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 Int σ. 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 a 0 a 1... a n ) is a simplex of K, then f(a 0 ), f(a 1 ),..., f(a n ) span a simplex of L (possibly with repeats). (3). If x = n i=0 t ia i is a point in a simplex a 0 a 1... a n ) of K, then f(x) = n t i f(a i ). i=0 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 a 0, a 1,, a n span a simplex of K, f(a 0 ), f(a 1 ),, f(a n ) 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 continuous. Exercise 2.6. Suppose that f : sk 0 K sk 0 L is a bijective correspondence such that the vertices a 0, a 1,..., a n spanned a simplex in K if and only f(a 0 ), f(a 1 ),..., f(a n ) spanned a simplex in L. Then the induced simplicial map f : K L is a homeomorphism, called linear isomorphism or simplicial homeomorphism of K with L. 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 of those simplices of K that have v as vertex. Its closure, denoted by Stv, is called the closed star of v in K. Note that Stv is the union of all simplices of K having v as a vertex. The set Stv St v is called the link of v in K, denoted by Lk v. A picture for the link of v is as follows:

11 2. GEOMETRIC SIMPLICIAL COMPLEXES 11 Lk v v The star can be measured by continuous functions defined as follows: Let v be a vertex in K and let x be a point in K. Then x is interior to precisely one simplex of K, whose vertices are (say) a 0, a 1,..., a n. Then n x = t i a i with t i > 0 and n i=0 i=0 t i = 1. We define the barycentric coordinate t v (x) of x with respect to v by setting { ti if v = a t v (x) = i for some 0 i n 0 otherwise. Proposition 2.7. Let K be a simplicial complex and let v be a vertex. Then t v : K R is continuous. Proof. Given any simplex σ of K, t v (x) σ is either identically 0 or the barycentric coordinate of x with respect to the vertex v of σ. Thus t v (x) σ is continuous. By the definition of the topology on K, t v (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 t v (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 = St v. v sk 0 K Proof. For any x K, there exists a unique simplicial σ such that x Int σ. Let v be a vertex of σ. Then x St v and hence the result. Proposition Let x Stv. Then the line segment xv lies in Stv. Moreover the line segment starting from v meets Lk v exactly one point. Proof. Exercise.

12 OBJECTS AND HOMOLOGY 2.3. Simplicial Approximation. Definition 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 simplicial approximation 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 because f sends each simplex of K to a simplex of L and hence f(st(v, K)) St(f(v), L). Proposition Let h: K L and k : L M have simplicial approximation f : K L and g : L M, respectively. Then g f is a simplicial approximation 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 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 b of L such that f(st(a, K)) St(b, L). Then there exists a simplicial approximation g to f, such that g(a) = b for each vertex a of K. Proof. It suffices to check that g(a 0 ),..., g(a n ) span a simplex of L whenever a 0, a 1,..., a n span a simplex of K. Let σ = a 0 a 1 a n be the simplex of K spanned by a 0, a 1,..., a n. Let x Int σ be a point in the interior. Then n x St a i It follows that f(x) i=0 n f(st a i ) i=0 n St(g(a i )), that is t g(a i )(f(x)) > 0 for each 0 i n. By the definition of the function t v, the unique simplex that contains f(x) in its interior must have each g(a i ) as a vertex, and hence has a face spanned by g(a 0 ), g(a 1 ),..., g(a n ). Definition Let X and Y be topological spaces. Let A be a subspace of X. Let f, g : X Y be continuous maps such that f A = g A. We call f is homotopic to g relative to A if there exists a continuous map F : X [0, 1] Y such that F (x, 0) = f(x), F (x, 1) = g(x) for every x X and F (a, t) = f(a) for every a A and 0 t 1. Theorem 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 f relative to the subspace of K of those points x such that f(x) = g(x). i=0

13 2. GEOMETRIC SIMPLICIAL COMPLEXES 13 Proof. Let x be a point of K. Then there is a unique simplex σ = a 0 a 1 a n such that x Int σ. From the proof of the above proposition, f(x) lies in the interior of a simplex τ of L that contains a face spanned by g(a 0 ),..., g(a n ). Thus τ contains the point g(x), and so the line segment f(x)g(x) lies in τ. Since the linear 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, we need the concept of subdivision discussed in next subsection Barycentric Subdivision. Definition Let K be a geometric simplicial complex in E J. A complex K 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 the union of the simplices of K, that is, K = K as sets. The finiteness part of condition (2) guarantee that K and K are equal as topological spaces. Definition Suppose that K is a simplicial complex in E J, and w is a point in E J such that each ray emanating from w intersects K in at most one point. We define the cone on K with vertex w to be the collection of all simplices of the form a 0 a 1 a p w, where a 0 a 1 a p is a simplex in K, along all faces of such simplices. Denote the cone by K w. The cone K w is pictured as follows: w K Definition Let K be a simplicial complex. Suppose that L p is a subdivision of sk p K. Let σ be a (p + 1)-simplex of K. Note that σ is a polyhedron of a subcomplex of sk p K and so it is a polyhedron of a subcomplex, denoted by L σ, of L p. If w σ is an interior point of σ, then the cone L σ w σ is a simplicial complex whose underlying space is σ. We define L p+1 to be the union of L p and the simplicial complexes L σ w σ as σ runs over all (p + 1)-simplices of K. Then L p+1 is a simplicial complex. (Check this!), called subdivision of sk p+1 K obtained by starring L p from the points w σ.

14 OBJECTS AND HOMOLOGY Definition Let σ = v 0 v 1 v n be an n-simplex. The barycenter of σ is the point n 1 ˆσ = n + 1 vi, i=0 that is ˆσ is the point of Intσ all of those barycentric coordinates with respect to the vertices are equal. If σ is 1-simplex, then ˆσ is the midpoint. If σ is a 0-simplex, then ˆσ = σ. In general, ˆσ is the centroid of σ. Definition Let K be a simplicial complex. We define a sequence of subdivisions of the skeletons of K as follows: Let L 0 = sk 0 K. Assume that L p is defined as a subdivision of sk p K. Let L p+1 be the subdivision of sk p+1 K obtained by starring L p from the barycenter of the (p + 1)-simplices of K. The union L p p=0 is a subdivision of K, called barycentric subdivision of K, denoted by sd K. Define the iterated barycentric subdivision recursively by sd n K = sd n 1 (sd K) for n > 1. Let σ be a 2-simplex. Then sd σ is shown in the picture below: The simplices in sd K can be described as follows. Define a partial order on the simplices of K by setting σ 1 < σ 2 if σ 1 is a proper face of σ 2. Proposition The simplicial complex sd K equals to the collection of all simplices of the form ˆσ 1ˆσ 2 ˆσ n, where σ 1 < σ 2 < < σ n in K. Proof. The proof is given by induction on p that the assertion holds for sk p K for each p 0. The assertion holds for sk 0 K as sd sk 0 K = sk 0 K. Suppose that the assertion holds for sk p K. By definition of barycentric subdivision, the assertion holds for sk p+1 K. Since sd K = sd sk p K, the assertion follows. p An important property of barycentric subdivision is as follows, which plays a key role for proving the simplicial approximation theorem. Theorem Given a finite simplicial complex K, and given any ɛ > 0, there is a positive integer N such that each simplex of sd N K has diameter less than ɛ.

15 2. GEOMETRIC SIMPLICIAL COMPLEXES 15 Proof. 1. If σ = a 0 a 1 a n is a simplex, then the diameter diam σ = max{ a i a j 0 i j n}, the maximum distance between vertices. Let l = max{ a i a j 0 i < j n}. For each vertex a i, let D(a i, l) = {x x a i l} be the ball of radius l centered at a i. Then D(a i, l) is convex. Since E(a i, l) contains all vertices of σ, σ D(a i, l). Thus x a i 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 a i D(x, l) as a i x l. Thus, for any z σ, x z l. Hence diam σ = l. 2. If σ = a 0 a 1 a n is a simplex, then for any x σ ˆσ x n diam σ. n + 1 Note that for each a t a t ˆσ = = at n i=0 0 i n i t 0 i n i t n n+1diam σ. Let l = max{ a t ˆσ 0 t n}. Then l 1 n+1 ai 1 n+1 (at a i ) 1 n+1 at a i D(ˆσ, l ) = {x x ˆσ l } n n+1diam σ. The ball contains all vertices of σ and so σ D(ˆσ, l ). It follows that x ˆσ l n n + 1 diam σ for any x σ. 3. For any simplicial complex L, let mesh L = sup{diam σ σ is a simplex of L}. Let K be an n-dimensional finite simplicial complex. Then mesh sd K n mesh K. n + 1 The proof is given by induction on the skeleton sk p K. mesh sd sk 0 K = mesh sk 0 K = 0. Suppose that mesh sd sk p K p p + 1 mesh sk p K. For sk 0 K, we have

16 OBJECTS AND HOMOLOGY Consider sk p+1 K. By the definition, sd sk p+1 K is the union of sd sk p K and the simplices of the form τ ˆσ for (p + 1)-simplices σ of K, where τ is a simplex of sd σ. If σ is a simplex of sd sk p K, then diam σ p p + 1 mesh sk p K p p + 1 mesh sk p+1 K p + 1 p + 2 mesh sk p+1 K. If σ = τ ˆσ with τ a simplex of sd σ, then, by Step 2, diam σ p + 1 p + 2 diam σ p + 1 p + 2 mesh sk p+1 K. The induction is finished and hence the statement. 4. Since K is a finite simplicial complex, let dim K = n and let d = mesh K. Then mesh sd N K ( ) N n d. n + 1 Thus mesh sd N K 0 as N and hence the result. Remark By inspecting the proof, the assertion also holds if K is a finite dimensional simplicial complex such that mesh K < +. In practice it may be necessary to subdivide only part of a simplicial complex K, 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 sk A 1 K = A and let sk A n K be the union of A and the simplices σ of K with dim σ n. Definition Let K be a geometric simplicial complex in E J and let A be a 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 get subdivision. Definition Let K be a simplicial complex and let A be a subcomplex of K. We define a sequence of subdivisions of the relative skeletons sk A n K of K as follows: Let L 1 = A and let L 0 = sk A 0 K. Assume that L p is defined as a subdivision of sk A p K. Let L p+1 be the subdivision of sk p+1 K obtained by starring L p from the barycenter of the (p + 1)-simplices of K NOT in A. The union L p p=0 is a subdivision of K, called barycentric subdivision of K relative to A, denoted by sd(k, A). Define the iterated barycentric subdivision recursively by sd n (K, A) = sd n 1 (sd(k, A)) for n > 1. The barycentric subdivision relative to a subcomplex is shown by the picture:

17 2. GEOMETRIC SIMPLICIAL COMPLEXES 17 A A Recall that τ < σ if and only if τ is a proper face of σ. Proposition 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 K A, together with the vertices of A. Distinct points a 1, a 2,..., a q, ˆσ 1, ˆσ 2,..., ˆσ m (with dim σ i dim σ i+1 for each i) span a simplex of sd(k, A) if and only if a 1, a 2,..., a q span a simplex σ of A, σ j is a simplex of K A for 1 j m with σ < σ 1 < σ 2 < < σ m. Note. If m = 0, then it just requires a 1,..., a q that span a simplex of A. If q = 0, it just requires σ m > σ m 1 > > σ 1 and in this case ˆσ 1, ˆσ 2,..., ˆσ m forms a simplex disjoint from A. Proof. The proof follows by induction on the relative skeleton sk A n K. Observe that the identity map sd(k, A) K is not a simplicial map if K A because the vertices ˆσ of sd(k, A) are not the vertices of K. Proposition 2.27 (Simplicial Approximation to the Identity Map). Let K be a simplicial complex and let A be a subcomplex of K. Let f be any function assigning simplices σ of K A to one of its vertices. Then there exists a simplicial approximation h to the identity map sd(k, A) K such that h A = id A and h(ˆσ) = f(σ) for each simplex σ of K A. Proof. First we need to check that h is a simplicial map. Let τ be a simplex of sd(k, A). Then τ = a 1 a 2 a pˆσ 1ˆσ 2 ˆσ q, where a 1, a 2,..., a q span a simplex σ 0 of A, σ j is a simplex of K A and σ 0 < σ 1 < σ 2 < < σ q. We check that the vertices (2.1) {a 1, a 2,..., a p, f(ˆσ 1 ), f(ˆσ 2 ),..., f(ˆσ q )} span a simplex in K. Since σ 0 < σ 1 < σ 2 < < σ q, each σ j is a face of σ q and so any vertex of σ j is a vertex of σ q. Thus the elements in Equation (2.1) are vertices of σ q 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:

18 OBJECTS AND HOMOLOGY There exists a simplex µ of K such that µ has v as a vertex and the interior Int τ Int µ. 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 = a 1 a 2 a pˆσ 1ˆσ 2 ˆσ q, where a 1, a 2,..., a q span a simplex σ 0 of A, σ j is a simplex of K A and σ 0 < σ 1 < σ 2 < < σ q. 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 = σ q. From the previous paragraph, h(v) is a vertex of µ and so we only need to check that Int τ Int µ. Let x Int(τ). Then p q x = s i a i + t j ˆσ i i=1 with s i, t j > 0 and p i=1 s i + q j=1 t j = 1. Let b 0, b 1,..., b m be the vertices of σ q. Let m a i = s i,k b k ˆσ j = k=0 j=1 m t i,k b k k=0 with s i,k, t i,k 0, m k=0 s i,k = 1 and m k=0 t j,k = 1. Since ˆσ q is the barycenter of σ q, t q,k = 1 m+1 for 0 k m. Now m p q x = ( s i s i,k + t j t j,k )b k. with k=0 k=0 For each k, we have p s i s i,k + i=1 i=1 m p s i s i,k + i=1 j=1 Thus x Int(σ q ) and hence the result. j=1 q t j t j,k = 1. j=1 q t j t j,k t q t q,k = t q m + 1 > 0. Corollary Let K be a simplicial complex and let A be a subcomplex of K. 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 of K such that B A =, then for a vertex a of sd(k, A) not in B, there exists a vertex b of K such that b B and St(a, sd(k, A)) St(b, K).

19 2. GEOMETRIC SIMPLICIAL COMPLEXES 19 Proof. Let h be a simplicial approximation to the identity. vertex 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). Let a be any Since h A = id A, 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 by requiring 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 Simplicial Approximation Theorem. Theorem 2.29 (Finite Simplicial Approximation Theorem). Let K and L be simplicial complexes. Suppose that K is finite. Let f : K L be a continuous map. Then there exists N such that the map f has a simplicial approximation g : sd N K L. Proof. Since {St w w is a vertex of L} is an open cover of L, the space K is covered by the open sets A = {f 1 (St w) w is a vertex of L}. Since K is a compact metric space, there exists a Lebesgue number λ such that any subset of K with diameter less than λ lies in an element of A. (If there were no such λ, there is a sequence C n of subsets of K such that diam C n < 1/n but C n does not lie in any element of A. Choose x n C n. By compactness, there is a subsequence x ni convergent to a point x K. Then x lies in an element f 1 (St w) of A, and so C ni f 1 (St w) when i is sufficiently large. This gives a contradiction.) Choose N such that each simplex of sd N K has diameter less than λ/2. Let a be a vertex of sd N K. Then the diameter of St v is less than λ because, for x, y St v, there exist simplices σ and τ of sd N K such that both σ and τ have v as a vertex with x σ and y τ and so x y x a + y a < λ 2 + λ 2 = λ. Thus St v f 1 (St w) for a vertex w of L. The assertion follows from Proposition 2.13 now. Now we consider the relative case. Definition Let K be a simplicial complex and let A be a subcomplex of 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 the same as the subcomplex of sd K of simplices having no vertices in sd A. The next lemma states that the maximum diameter of the stars of sd N (K, A) of vertices in the supplement Ā tends to 0. Lemma Let K be a simplicial complex with a subcomplex A. Suppose that there are finitely many simplices of K A. Given any ɛ > 0, there exists N such that sup{diam St(v, sd N (K, A)) v Ā } < ɛ.

20 OBJECTS AND HOMOLOGY Proof. Let τ be a 1-simplex of sd 2 (K, A). Suppose that τ has a vertex a in A. Then another vertex of τ is either in A or the barycenter ˆσ for a simplex σ of sd(k, A) having a as a vertex. Thus σ Ā and so ˆσ Ā. In other words, NO 1-simplices of sd 2 (K, A) that can have vertices in both A and Ā. It follows that NO n-simplices of sd 2 (K, A) that can have vertices in both A and Ā. (If an n-simplex had vertices a in A and b in Ā, then its face from a to b is a 1-simplex having its vertex a in A and b in Ā which contradicts to that no 1-simplex can have vertices in both A and Ā.) Let  denote the supplement of A in sd(k, A), that is,  is the set of simplices of sd 2 (K, A) that have no vertices in A. Then, if τ is a simplex of sd 2 (K, A) with a vertices in Ā, then τ Â. Note that the subdivision sdn (K, A) includes the norelative subdivision sd N 2  of Â. Thus sd N 2  is a subcomplex of sd N (K, A). Now we show by induction that: For each N 2 if τ is a simplex of sd N (K, A) having a vertex in Ā, then τ sd N 2 Â. This statement has been proved when N = 2. Suppose that it holds for N 1 with N 3. Let τ be a simplex of sd N (K, A) with a vertex in Ā. Then τ has the expression a 1 a 2 a pˆσ 1 ˆσ q, where a 1,..., a p span a simplex σ 0, each σ j is a simplex of sd N 1 (K, A) with σ 0 < σ 1 < < σ q. Since each a i is not in Ā, we may assume that ˆσ j Ā for some 1 j q. It follows that σ j has a vertex in Ā. Since σ j is a face of σ q as σ j < σ q, σ q has a vertex in Ā. By induction, σ q sd N 3  and σ 0 must be empty as it is a face of the simplex σ q of sd N 3 Â. It follows that τ sd N 2 Â. The induction is finished and hence the statement. Thus for any vertex v of sd N (K, A) such that v Ā St(v, sd N (K, A)) = {Int τ τ is a simplex of sd N (K, A) having v as a vertex} {Int τ τ is a simplex of sd N 2  having v as a vertex} = St(v, sd N 2 Â). The assertion follows from Theorem 2.22 now. Now we construct a new simplicial complex K + = sd(sd(k, A), A Ā). In other words, K + is obtained by doing barycentric subdivision on those simplices of sd(k, A) that are not in A and Ā. Thus K+ is a subdivision between sd(k, A) and sd 2 (K, A). A picture is as follows: bar(a) A The vertices v of K + are one of the following three cases:

21 2. GEOMETRIC SIMPLICIAL COMPLEXES 21 Type I. v is a vertex of A; Type II. v is a vertex of Ā; Type III. v is the barycenter ˆσ for a simplex σ of sd(k, A) that has vertices in both A 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+ is given 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 = ˆσ 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 = id A Ā. The last rule forces h to push down the vertices of the form ˆσ into A. Lemma 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 = a 1 a 2 a pˆσ 1 ˆσ q, where a 1, a 2,, a p span a simplex σ 0 in A Ā, σ 1,..., σ q are simplices of sd(k, A) that have vertices in both A and Ā, and σ 0 < σ 1 < < σ q. Since v A is a vertex of τ, p 1 and v {a 1,..., a p }. It follows that a i Ā for each 1 i q because if not, then σ 0 is a simplex having vertices in both A and Ā which contradicts to that σ A Ā. Thus τ only have Type I and Type II vertices, namely τ has NO vertices in Ā. By the definition of the map h, we have h(a i ), h(ˆσ j ) A for 1 i p and 1 j q. Hence h(τ) is a simplex of A having v as vertex because h A = id A. The assertion follows by taking the union of the interior of τ having v as a vertex. Proposition Let K and L be a simplicial complexes and let A be a subcomplex 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 of K A. 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 approximation g : sd N (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  be the supplement of A in sd(k, A) as discussed in the proof of Lemma Then there exists N 2 such that sup{diam St(v, sd N (K, A)) v  } < λ. (Note. Here we replace  as Ā in Lemma 2.31 by considering sd(k, A) as K.)

22 OBJECTS AND HOMOLOGY By Proposition 2.13, it suffices to show that, for every vertex v of sd N (K, A), there exists a vertex w of L such that Case I. v Â. Then St(v, sd N (K, A)) (f h) 1 (w). diam St(v, sd N (K, A)) < λ and so there exists a w such that St(v, sd N (K, A)) (f h) 1 (w). Case II. v Â. By Corollary 2.28, St(v, sd N (K, A)) St(b, sd 2 (K, A)) for some vertex b of sd 2 (K, A) NOT in Â. Then b A. (This is because, by definition, Â is the subcomplex of sd 2 (K, A) of the simplices having no vertices in A. In particular, Â contains all of vertices, as 0-simplices, that are not in A.) We claim that St(b, sd 2 (K, A)) St(b, K + ). Let τ be a simplex of sd 2 (K, A) having a vertex b A. Then τ has the expression τ = a 1 a 2 a pˆσ 1ˆσ 2 ˆσ 2, where a 1,..., a p span a simplex σ 0 of A, σ j is a simplex of sd(k, A) for 1 j q, with σ 0 < σ 1 < < σ q. Since b is a vertex of τ, b {a 1,..., a p } with p 1. Since σ 0 is a face of each σ j for 1 j q, b is a vertex of σ j for 1 j q. It follows that τ K + = sd(sd(k, A), A Ā). This proves that Now by Lemma 2.32, we have Since f A is a simplicial map, we have St(b, sd 2 (K, A)) St(b, K + ). h(st(b, K + )) St(b, A). f(st(b, A)) St(f(b), L). By combining the previous equations, we have St(v, sd N (K, A)) St(b, sd 2 (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 : sd N (K, A) L. For checking that, we can make a choice of g such that g A = f A. Let v be a vertex of A. In the argument of Case II above, we can choose b = v by Corollary Thus we have St(v, St N (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 that g A = f A. The proof is finished now.

23 2. GEOMETRIC SIMPLICIAL COMPLEXES 23 Theorem 2.34 (Relative Simplicial Approximation Theorem). Let K and L be a simplicial complexes and let A be a subcomplex of K. Suppose that there are finitely many simplices of K A. Let f : K L be any continuous map such that f A is a simplicial map. Then there exists N and a simplicial map g : sd N (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 Then g A = f A and g is homotopic to f h relative to A by Theorem Since h is a simplicial approximation to the identity map with h A = id A, h id K relative to A and so f h f relative to A. It follows that g f relative to A. by 2.6. Some Applications. Recall that the homotopy groups π n (X) is defined π n (X) = [S n, X] the set of homotopy classes of the pointed continuous maps from S n to X up to pointed homotopy (namely homotopy relative to the basepoint). A direct consequence of simplicial approximation theorem is as follows: Theorem π r (S n ) = 0 for r < n. Proof. Let f : S r S n be a pointed continuous map. Let K be the simplicial complex such that K = S r and let L be the simplicial complex such that L = S n. (We can choose K and L as the boundary of an (r + 1)-simplex and an (n + 1)- simplex, respectively.) Consider S r and S n as polyhedron of simplicial complexes. By the simplicial approximation theorem, there exists a subdivision K of K and a simplicial map g : K = K = S r L = S n such that g is homotopic to f relative to the basepoint. Since g is a simplicial map, the map g must send K into the skeleton sk r (L) because K is an r-dimensional simplicial complex. Since r < n, sk r (L) L and so g is not onto. Thus there is a point x S n such that g(s r ) S n {x} = R n. Hence g is homotopic to the constant map relative to the basepoint by a linear homotopy. It follows that f is homotopic to the constant map relative to the basepoint. Another direct consequence is to give a computation of the fundamental group. Theorem π 1 (S 1 ) = Z. Proof. By simplicial approximation theorem, we only need to consider the simplicial maps from a subdivision K of the circle K to the circle K, where K is the boundary of a 2-simplex with vertices w 0, w 1, w 2 in the order of counterclockwise along the circle, where w 0 is regarded as the basepoint. Let K has vertices v 0, v 1,..., v n in the order of counter-clockwise along the circle, where v 0 is regarded as the basepoint. Let g : K K be a pointed simplicial map. See the picture:

24 v 0 w 2 w OBJECTS AND HOMOLOGY v n g v 3 v 2 v 1 w 1 Then g maps the circle (v 0, v 1,..., v n, v 0 ) (in order) into the sequence (g(v 0 ), g(v 1 ),..., g(v n ), g(v 0 )), where g(v i ) is one of w j with g(v 0 ) = w 0. If g(v i ) = g(v i+1 ), then g is constant on the 1-simplex [v i, v i+1 ] and so, up to homotopy, we can remove g(v i+1 ). For the subsequence (g(v i ), g(v i+1 ), g(v i+2 )), if g(v i ) = g(v i+2 ), then (g(v i ), g(v i+1 ), g(v i+2 )) is a path from g(v i ) to g(v i+1 ) and backwards to g(v i ) and so we can collapse this subsequence. Assume that v 0,..., v n have minimal number of vertices in the homotopy class of g. Then the only sequence of (g(v 0 ), g(v 1 ),..., g(v n ), g(v 0 )) is given by one of the following: (1). constant map (w 0 ), (2). around the circle n-times positively: (w 0, w 1, w 2, w 0, w 1, w 2, w 0,..., w 0, w 1, w 2, w 0 ), or (3). around the circle m-times negatively: (w 0, w 2, w 1, w 0, w 2, w 1, w 0,..., w 0, w 2, w 1, w 0 ). This shows that any pointed continuous map f : S 1 S 1 is homotopic to one of the maps g n : S 1 S 1 z z n for n Z, relative to the basepoint. In particular, π 1 (S 1 ) is generated by [g 1 ]. On the other hand, one can show that g n is not homotopic to g m relative to the basepoint if n m, as g n is the continuous mapping that goes around S 1 for n times. This gives that π 1 (S 1 ) = S 1. Example 2.1 (Cohomotopy Sets). Let X be a pointed space. Then the n-th cohomotopy set is defined to be π n (X) = [X, S n ], the set of homotopy classes of pointed continuous maps from X to the n-sphere up to pointed homotopy. For instance, π r (S n ) = [S r, S n ] = π n (S r ). The fundamental problem in algebraic topology is to determine π r (S n ) for general r and n. When n = 1, it was known that π r (S 1 ) = 0 for r 1 and π 1 (S 1 ) = Z. When n > 1, we know from the above theorem that π r (S n ) = 0 for r < n. When r = n, π n (S n ) = 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

25 3. ABSTRACT SIMPLICIAL COMPLEXES AND -SETS 25 corresponding homology group. Or one can try to directly compute π n (S n ) using simplicial methods or other methods.) For r > n, π r (S n ) is known up to certain range by very nontrivial works contributed by many topologists, but far unknown for general r even if n = 2. By using simplicial methods, one might try a different approach to study the higher homotopy groups of spheres. Consider S n as the boundary of an (n + 1)-simplex σ n+1. So, as a simplicial complex, S n has (n+2) vertices labeled in order by {0, 1, 2,..., n+1}. Observe that, in σ n+1, any nonempty subset of the vertices {0, 1, 2,..., n + 1} spans a simplex of σ n+1. Since S n is the boundary of σ n+1, any proper subset of {0, 1, 2,..., n + 1} spans a simplex of S n. Let K be a simplicial complex and let sk 0 K be the set of vertices of K. Suppose that f : K S n is a simplicial map. Then f sends vertices of K to the vertices of S n. 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 that v 0, v 1,..., v r span a simplex of K. Then {f(v 0 ), f(v 1 ),..., f(v r )} must span a simplex of S n, equivalently, {f(v 0 ), f(v 1 ),..., f(v r )} is a proper subset of {0, 1,..., n+1}. Namely at least at one color is missing in {f(v 0 ), f(v 1 ),..., f(v r )}. Conversely suppose that there is a coloration on the vertices of K using (n + 1) different colors, that is there is a function from sk 0 K to {0, 1,..., n}, such that if v 0, v 1,..., v r span a simplex of K, then at least one color is missing among the colors of v 0, v 1,..., v r. Then the coloration induces a unique simplicial map from K to S n. This establishes the one-to-one correspondence between the set of simplicial maps from K to S n and the set of colorations on vertices of K that satisfying the above rule. Thus, for studying simplicial maps from K to S n, one can study how to make those colorations on the vertices of K satisfying the above rule. For studying π r (S n ), one needs to understand: (1). The colorations on the vertices of K satisfying the above rule, where K = S r, namely K is a triangulation of S r. This will give simplicial maps K S n. In general, one may study the colorations on the vertices of K, where K is a triangulation of a manifold. (2). The colorations on the vertices of K satisfying the above rule, where K = S r [0, 1]. This will control the homotopy. The above two problems are not well-understood so far. But it will be very interesting if one could make progress on this as it attacks the fundamental open problem in algebraic topology. Exercise 2.7. Determine all possible simplicial maps from K S 1, where K is a simplicial complex such that K = S Abstract Simplicial Complexes and -sets 3.1. Definition and Geometric Realizations. Definition 3.1. An abstract simplicial complex K is a collection of finite nonempty sets, such that if A is an element in K, so is every nonempty subset of A.

26 OBJECTS AND HOMOLOGY The element A of K is called a simplex of K; its dimension is one less than the number of its elements. Each nonempty subset of A is called a face of A. The dimension of K is the supremum of the dimensions of its simplices. The vertex set V (K) is the union of the one-point elements of K; we shall make no distinction between the vertex v and the 0-simplex {v}. A sub collection of K that is itself a complex 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 the vertex set of K to the vertex set of K such that {a 0, a 1,..., a n } K if and only if {f(a 0 ), f(a 1 ),..., f(a n )} K. Definition 3.2. Let K be a geometric simplicial complex. Let V be the vertex set of K. Let K be the collection of all subsets {a 0, a 1,..., a n } of V such that a 0, a 1,..., a n span a simplex of K. The collection K is called vertex scheme of K, or abstraction of K. Theorem 3.3. A relation between abstract simplicial complexes and geometric simplicial complexes is as follows: (1). Every abstract simplicial complex K is isomorphic to the vertex scheme of some geometric simplicial complex. (2). Two geometric simplicial complexes are linearly isomorphic if and only if their 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 E J spanned by finite subsets of the standard basis {e α } α J for E J. It is easy to see that J is a simplicial complex. Moreover if σ and τ are two simplices of J, then their combined 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 index set J = V. We specify a subcomplex K of J by the condition that for each abstract simplex {a 0,..., a n } K, the geometric simplex spanned by e a0, e a1,..., e an is to be in K. It is immediate that K is a geometric simplicial complex and K is isomorphic to the vertex scheme of K Subdivision of Abstract Simplicial Complexes. Let V be a set with a partial order <. Let S(V ) = {{a 0, a 1,..., a n } a i V a 0 < a 1 < < a n }. Then clearly S(V ) is an abstract simplicial complex with V as its vertex set because if {a 0, a 1,..., a n } is well-ordered, then any nonempty subset of {a 0, a 1,..., a n } is well-ordered. The subdivision of abstract simplicial complexes follows the ideas of barycentric subdivision of geometric simplicial complexes. Let K be an abstract simplicial complex. Let A, B K. Let K be partially ordered by if A is a proper face of B. Define A < B sd(k) = S(K) called subdivision of K. If K is the geometric realization of K, then, by definition, the geometric realization of sd(k) is sd(k).

27 3. ABSTRACT SIMPLICIAL COMPLEXES AND -SETS Abstract Simplicial Complexes and -sets. Let K be an abstract simplicial complex. Let K n be the set of n-simplices of K. Define the faces d i : K n K n 1, 0 i n, as follows: Let the vertices of K be ordered. If {a 0, a 1,..., a n } is an n-simplex of K with a 0 < a 1 < < a n, then define d i {a 0, a 1,..., a n } = {a 0, a 1,..., a i 1, a i+1,..., a n }. Proposition 3.4. Let K = {K n } n 0 with faces defined as above. Then K is a -set. Proof. Exercise. Definition 3.5. A -set X is called polyhedral if there exists an abstract simplicial 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 2 X0 be the set of all subsets of X 0. Define φ: X n 2 X0 n 0 by setting φ(x) = {f x (0), f x (1),..., f x (n)} for x X n. Theorem 3.6. Let X be a -set. Then X is polyhedral if and only if the following holds: (1). There exists an order of X 0 such that, for each x X n, f x (0) f x (1) f x (n). (2). The function φ: n 0 X n 2 X0 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 the construction, we have f x (0) < f x (1) < < f x (n). Conversely suppose that X is a set satisfies the two conditions in the statement. First we show that, for x X n, the cardinal of {f x (0), f x (1),..., f x (n)} is n + 1. Otherwise there exists 0 i < j n such that f x (i) = f x (j). Then d i (x) = f x d i (0, 1,..., n) = f x (0, 1,..., i 1, i + 1,..., n) and so φ(d i (x)) = {f x (0), f x (1),..., f x (i 1), f x (i + 1),..., f x (n)} = {f x (0), f x (1),..., f x (i 1), f x (i), f x (i + 1),..., f x (n)} = φ(x). Thus d i (x) = x, which contradicts to that X n X n 1 =. Let the vertices of K be the elements of X 0. Define a subset {a 0, a 1,..., a n } X 0 to be an n-simplex of K if and only if {a 0, a 1,..., a n } Im(φ), where the elements are given in the order that a 0 < a 1 < < a n.

28 OBJECTS AND HOMOLOGY For checking that K is an abstract simplicial complex, let {a 0, a 1,..., a n } be an n-simplex of K such that {a 0, a 1,..., a n } = φ(x). From the above arguments, x X n and so we may assume that a i = f x (i) for 0 i n. Let {a i0, a i1,..., a ip } be a subset of {a 0, a 1,..., a n } with 0 i 0 < i 1 < < i p n. Let {j 0, j 1,..., j n p 1 } = {0, 1,..., n} {i 0, i 1,..., i p } with 0 j 0 < j 1 < < j n p 1. Then Thus d j0 d j1 d jn p 1 x = f x d j0 d j1 d jn p 1 (0, 1,..., n) = f x (i 0, i 1,..., i p ). φ(d j0 d j1 d jn p 1 x) = {a i0, a i1,..., a ip } and so {a i0, a i1,..., a ip } is a p-simplex of K. Finally the function g : X K with g(x) = {f x (0),..., f x (n)} for x X n is bijective -map. This proves that X = K. 4. -complexes and the geometric realization of -sets The standard geometric n-simplex n is defined by n n = {(t 0, t 1,..., t n ) t i 0 and t i = 1}. Define d i : n 1 n by setting i=0 d i (t 0, t 1,..., t n 1 ) = (t 0,..., t i 1, 0, t i,..., t n 1 ). Exercise 4.1. Show that The maps d i satisfy the following identity: for i j. The boundary d j d i = d i+1 d j n = n i=0d i ( [n 1]) is the union of all faces of n. Let Int( n ) = n n be the interior of n, called open simplex. Definition 4.1. A -complex structure on a space X is a collection of maps such that C(X) = {σ α : n X α J n n 0}. (1). σ α Int( n ) : Int( n ) X is injective, and each point of X is in the image of exactly one such restriction σ α Int( n ). (2). For each σ α C(X), each face σ α d i C(X). (3). A set A X is open if and only if σ 1 α (A) is open in n for each σ α C(X).

29 4. -COMPLEXES AND THE GEOMETRIC REALIZATION OF -SETS 29 Define with d i : C n (X) C n 1(X) given by for 0 i n. C n (X) = {σ α : n X α J n } C(X) d i (σ α ) = σ α d i 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 = ( n, x)/ = n K n /, x K n n 0 n=0 where ( n, x) is n labeled by x K n and is generated by (z, d i x) (d i z, x) for any x K n and z n 1 labeled by d i x. For any x K n, let σ x : n = ( n, x) K be the canonical characteristic map. The topology on K is defined by A K is open if and only if the pre-image σx 1 (A) is open in n for any x K n and n 0. Proposition 4.2. Let K be a -set. Then K is -complex. Proof. Let sk n K = {K j } 0 j n be the n-skeleton of K. Then sk n K is a -subset of K. Let + [n] = sk n 1 + [n] be the boundary of + [n]. Observe that + [n] = n + [n] = n. From the push-out diagram there is a push-out diagram x K n + [n] x K n + [n] skn 1 K push skn K, n sk n 1 K x K n x K n n push skn K. Thus sk n K is obtained from sk n 1 K by attaching cells with labels in K n. By induction, it follows that K is a -complex.

30 OBJECTS AND HOMOLOGY 5. Homology of -sets Recall that a chain complex of groups means a sequence C = {C n } of groups with differential n : C n C n 1 such that n n+1 is trivial, that is Im( n+1 ) Ker( n ) and so the homology is defined by H n (C) = Ker( n )/ Im( n+1 ), which is a coset in general. A chain complex C is called normal if Im( n+1 ) is a normal subgroup of Ker( n ) for each n. In this case H n (C) is a group for each n. Proposition 5.1. Let G be a -abelian group. Define n = n ( 1) i d i : G n G n 1. Then n 1 n = 0, that is, G is a chain complex under. Proof. i=0 n 1 n = n 1 ( 1) i d i = = = i=0 = 0. 0 i<j n 0 i<j n 0 i<j n n ( 1) j d j j=0 ( 1) i+j d i d j + ( 1) i+j d i d j + ( 1) i+j d i d j + 0 j i n 1 0 j<i+1 n 0 j<i n ( 1) i+j d i d j ( 1) i+j d j d i+1 ( 1) i+j 1 d j d i Let X be a -set. The homology H (X; G) of X with coefficients in an abelian group G is defined by H (X; G) = H (Z(X) G, ), where Z(X) = {Z(X n )} n 0 and Z(X n ) is the free abelian group generated by X n. Proposition 5.2. Let 1 C i C p C 1 be any short exact sequence of chain complexes of (possibly non-commutative) groups. Then there is a long exact sequence H k+1 (C ) k+1 H k (C ) i H k (C) p H k (C ). Moreover if C and C are normal chain complexes, then k+1 is a group homomorphism for each k.

31 5. HOMOLOGY OF -SETS 31 Proof. Consider the commutative diagram C k+2 i Ck+2 p C k+2 C k+1 i p Ck+1 C k+1 C k i Ck p C k C k 1 i p Ck 1 C k 1. Let x C k+1 with (x) = 1. There exists x C k+1 such that p( x) = x. Since p( ( x)) = (p( x)) = (x) = 1, there exists x C k such that i( x) = ( x). Now i( ( x)) = (i( x)) = ( x) = 1. Thus x is a circle in C and so { x} defines an element in H k (C ). Let ˆx be another element in C k+1 such that p(ˆx) = x. Then p( xˆx 1 ) = 1 and so there exists an element z C k+1 such that i(z) = x 1ˆx. Now i( x (z)) = ( x)( ( x)) 1 (ˆx) = (ˆx). Thus { x} H k (C ) is independent on the choice of the pre-image of x in C k+1. Suppose that x = x (y) with (x) = 1 for some y C k+2. There exists ỹ C k+2 such that p(ỹ) = y. Then with This shows that x = p( x (ỹ)) x = ( x (ỹ)) = ( x) = x. k+1 : H k+1 (C ) H k (C ) {x} { x} is well-defined. Assume that C and C are normal chain complexes. For x, x C k+1 with (x) = (x ) = 1. Then p( x x ) = 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( x) = ( x). Let y C k with (y) = 1 and i (y) is trivial in H k (C). Then there exists ỹ C k+1 such that i(y) = (ỹ).

32 OBJECTS AND HOMOLOGY By the construction of k+1, k+1 (p(ỹ)) = y. This shows that H k+1 (C ) k+1 Hk (C ) i Hk (C) is exact. Now we show that H k+1 (C) p H k+1 (C ) k+1 H k (C ) is exact. Let y C k+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+1 with (x) = 1 and x = k+1 (x) is trivial in H k (C ). There exists an element z C k+1 such that Let ˆx = i(z) 1 x. Then with (z) = x. p(ˆx) = p(i(z) 1 x) = p( x) = x (ˆx) = (i(z) 1 x) = i( (z) 1 x) = 1. Thus ˆx defines an elements in H k+1 (C) with p ({ˆx}) = {x}. Finally we show that H k (C ) i Hk (C) p Hk (C ) is exact. Since p i is trivial, so is p i. Let y C k with (y) = 1 and p (y) is trivial in H k (C ). There exists an element z C k+1 such that Let z C k+1 such that p( z) = z. Then p(y) = (z). p(y ( z 1 )) = (z)p( ( z 1 )) = (z) (p( z) 1 ) = (z) (z 1 ) = 1. Thus there exists w C k such that i(w) = y ( z 1 ) with i( (w)) = (i(w)) = (y ( z 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 sequence Hk+1 (X, X ; G) k+1 Hk (X ; G) for abelian group G. i Hk (X; G) p Hk (X, X ; G)

33 6. SIMPLICIAL HOMOLOGY AND SINGULAR HOMOLOGY Simplicial Homology and Singular Homology Definition 6.1. Let X be a -complex. Then simplicial homology of X with coefficients in an abelian group G is defined by For any space X, define H (X; G) = H (C (X); G). S n (X) = Map( n, X) be the set of all continuous maps from n to X with d i = d i : S n (X) = Map( n, X) Sn 1 (X) = Map( n 1, X). for 0 i n. Then S (X) = {S n (X)} n 0 is a -set. This allows us to define singular 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 See Hatcher s book [17] for details. H (X, A; G) = H (S (X), S (A); G).

34

35 CHAPTER 2 Simplicial Sets and Homotopy The notion of simplicial sets was arisen from establishing combinatorial models for spaces. Recall that, given any topological space X, there is a singular simplicial set S (X), where S n (X) is the set of all continuous maps from the n-simplex to X. Let C n (X) be the free abelian group generated by S n (X). Then one gets the chain complex 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 by S (X). One may ask whether the homotopy groups π (X) can be obtained in a similar way. The answer is actually yes. In fact π (X) can be obtained directly from S (X) Definitions. 1. Simplicial Sets Definition 1.1. A simplicial set means a -set X together with a collection of degeneracies s i : X n X n+1, 0 i n, such that (1.1) d j d i = d i 1 d j for j < i, (1.2) s j s i = s i+1 s j for j i and s i 1 d j j < i (1.3) d j s i = id j = i, i + 1 s i d j 1 j > i + 1. The three identities for d i d j, s j s i and d i s j are called the simplicial identities. Remark 1.2. One can use deleting-doubling for catching simplicial identities: d i : (x 0,..., x n ) (x 0,..., x i 1, x i+1,..., x n ). s i : (x 0,..., x n ) (x 0,..., x i 1, x i, x i, x i+1,..., x n ). Let O be the category whose objects are finite ordered sets and whose morphisms are functions f : X Y such that f(x) f(y) if x < y. The objects in O are given by [n] = {0,..., n} for n 0, which are the same as the objects in O +. The morphisms in O are generated by d i, which is defined in O +, and the following morphism ( ) s i : [n + 1] [n] s i 0 1 i 1 i i + 1 i + 2 n + 1 = 0 1 i 1 i i i + 1 n for 0 i n, that is, s i hits i twice. 35

36 36 2. SIMPLICIAL SETS AND HOMOTOPY Exercise 1.1. Show that simplicial sets are one-to-one correspondent to contravariant 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 a contravariant functor from O to C. Equivalently a simplicial object C means a sequence of objects X = {X n } n 0 with face morphisms d i : X n X n 1 and degeneracy morphisms s i : X n X n+1, 0 i n, such that the three simplicial identities (1.1), (1.2) and (1.3) hold. A simplicial group means a simplicial object over the category of groups. In other words, a simplicial set X is a simplicial group if and only if (1) each X n is a group and (2) all faces and degeneracies are group homomorphisms. Definition 1.4. A simplicial map (simplicial morphism) f : X Y means a sequence of functions (morphisms) f : X n Y n for each n 0 such that f d i = d i f and f s i = s i f, that is the diagram X n+1 s i X n d i Xn 1 f Y n+1 s i f Y n f di Y n 1 commutes. Moreover if each X n is a subset of Y n such that the inclusions X n Y n form a simplicial map, then X is called a simplicial subset of Y. A simplicial set X is called to be isomorphic to a simplicial set Y, denoted by X = Y, if there is a bijective simplicial map f : X Y Basic Constructions. Example 1.1 (n-simplex). The n-simplex [n], as a simplicial set, is as follows: [n] k = {(i 0, i 1,..., i k ) 0 i 0 i 1 i k n} for k n. The face d j : [n] k [n] k 1 is given by d j (i 0, i 1,..., d k ) = (i 0, i 1,..., î j,..., i k ), that is deleting i j. The degeneracy s j : [n] k [n] k+1 is defined by s j (i 0, i 1,..., d k ) = (i 0, i 1,..., i j, i j,..., i k ), that is doubling i j. Let σ n = (0, 1,..., n) [n] n. Then any elements in [n] can be written as iterated compositions of faces and degeneracies of σ n. Proposition 1.5. Let X be a simplicial set and let x X n be an element. Then there exists a unique simplicial map such that f x (σ n ) = x. f x : [n] X Proof. We leave the proof as an exercise to the reader.

37 1. SIMPLICIAL SETS 37 Definition 1.6. Let X be a simplicial set and A = {A n } n 0 with A n X n. 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 compositions of faces and degeneracies of the elements in A. Example 1.2 (n-sphere). The simplicial n-sphere S n is defined by S n = [n]/ ( [n]), where ( [n]) is the simplicial subset of [n] generated by [n] k for k < n. Let s write explicitly for the elements in the simplicial circle S 1. Now that [1] k = {(i 0,..., i k ) 0 i 0 i k 1} has k + 2 elements. Now By definition, S 1 = [1]/ ( [1]). Thus i {}}{ = {( 0,..., 0, 1,..., 1 0 i k + 1} ( [1]) k = {(0,..., 0), (1,..., 1)}. S 1 k = {, (i 0,..., i k ) 0 i 0 i k 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 S n, we have Sk n = { } for k < n and Sk n = {, (i 0,..., i k ) 0 i 0 i k n with {i 0,..., i k } = {0, 1,..., n}} for k n. Example 1.3 (Cartesian Product). Let X and Y be simplicial sets. Define X Y by setting (X Y ) n = X n Y n with d X Y i = (d X i, dy i ) and sx Y i = (s X i, sy i ). Show that X Y is a simplicial set. Example 1.4. Let f : X Y be a simplicial map. Then is a simplicial subset of Y. Im(f : X Y ) = {Im(f : X n Y n )} n 0 For a sequence of nonnegative integers I = (i 1, i 2,..., i k ) of length k = l(i), denote d I = d i1 d ik and s I = s i1 s ik. Example 1.5 (Basepoint). Let X be a simplicial set and let x 0 X 0. Then the image of the representing map f x0 : [0] X is a simplicial subset of X consisting of only one element f x0 (0,..., 0) = s I (x 0 ) in each dimension. Thus a basepoint of X means a sequence of elements n+1 {}}{ {f x0 ( 0,..., 0)} n 0

38 38 2. SIMPLICIAL SETS AND HOMOTOPY corresponds to the elements x 0 X 0. 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 X 0 has only one element. For a reduced simplicial set, there is a unique choice of basepoints. Moreover any simplicial map between reduced simplicial sets is pointed. Example 1.6 (Push-out). Let C g A f B be simplicial maps. Define X n to be the push-out in the diagram A n f B n g C n push Xn, that is X n is the quotient of the disjoint union B n Cn subject to the equivalence relation generated by f(a) g(a) for a A n. Then X = {X n } n 0 with faces and degeneracies induced from that in B and C forms a simplicial set with a push-out diagram of simplicial sets A f B g C push X. For instance, let [n] = d i (0, 1,..., n) 0 i n [n] be the simplicial subset of [n] generated by all of the faces of the n-simplex σ n = (0, 1,..., n). Then [n] [n] push is a push-out diagram. S n 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 the basepoint of X with the basepoint of Y. In other words, X Y is defined by the push-out diagram X push Y X Y.

39 1. SIMPLICIAL SETS 39 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 be the simplicial quotient X Y/(X Y ). An element x X n is called nondegenerate if x s i (y) for any y X n 1. Exercise 1.2. Determine nondegenerate elements in S 1 S 1. In general, determine nondegenerate elements in S n S m. Remark 1.7. As spaces, we know that S m S n = S n+m. It should be pointed that, as simplicial sets, S m S n = S m+n although they have the same homotopy type. Example 1.8 (Skeleton of Simplicial Sets). Let X be a simplicial set. Let sk n (X) = X j j n is the simplicial subset of X generated by X j for j n. Then there is a filtration sk 0 X sk 1 X sk n X with X = n sk n X, called skeleton filtration of X. There are push-out diagrams fx [n] [n] sk n 1 X for each n 1. x X n nondegenerate x X n nondegenerate [n] fx sk n X Example 1.9 (Cone and suspension). Let X be a pointed simplicial set. The reduced cone CX is defined by setting (1.4) (CX) n = {(x, q) x X n q, 0 q n} with (, q) all identified to, { (x, q 1) for 0 i < q, d i (x, q) = { (d i q x, q) for q i n, (x, q + 1) for 0 i < q, s i (x, q) = (s i q x, q) for q i n, where for x X 0, d 1 (x, 1) =. By identifying x with (x, 0), X is a simplicial subset of CX. The reduced suspension ΣX is defined by ΣX = CX/X. We give another description for the construction CX. Let X be any simplicial set without choosing basepoint. Let x 0 be a new point not in X. Let ( CX) n = X n s 1 (X n 1 ) s n 1(X 0 ) n {s n 0 x 0 } = {s n 0 x 0 } s k 1(X n k ) be the disjoint union as a set, where s 1 (X j ) = X j. Regard s 1 as ( 1)st degeneracy operation. From the simplicial identities, we have { id if i = 0 d i s 1 = s (1.5) { 1 d i 1 if i 1 s 1 s s i s 1 = 1 if i = 0 s 1 s i 1 if i 1, k=0

40 40 2. SIMPLICIAL SETS AND HOMOTOPY which induces the operations d i and s i on CX exactly given as in Formula (1.4) by identifying s q 1 x with (x, q), where d 1(s 1 x) = x 0 for x X 0. The resulting simplicial set CX is called unreduced cone of X. The reduced cone CX is then the quotient given by requiring x 0 x 0 and (x 0, 1) = s 1 x 0 (s 0 x 0, 0) = s 0 x 0 s 0 x 0 for the basepoint vertex x 0 X 0. 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 there exists a function s 1 : X n X n+1 for each n 0 such that s 1 ( ) = and Identities (1.5) hold. Proof. Suppose that there is a simplicial map r : CX X such that the composite r j = id X. Define s 1 (x) = r(x, 1) for x X n. Then d i s 1 (x) = d i r(x, 1) = rd i (x, 1) = r(d i s 1 (x, 0)) s i s 1 (x) = s i r(x, 1) = rs i (x, 1) = r(s i s 1 (x, 0)) and so Identities (1.5) hold. Conversely, if s 1 : X n X n+1 is defined with Identities (1.5), define the function r : CX X by setting r(x, 0) = x and r(x, q) = s q 1x for q > 0. Then r is a simplicial map by Identities (1.5) with r X = id X and hence the result. Theorem 1.9 (Null-homotopy Criterion). Let f : X Y be a pointed simplicial map. Then the extension problem CX f f Y X has a solution of a simplicial map f if and only if there exists F : X n Y n+1 for each n 0 such that the following conditions hold: (1). d 0 F = f and d i F = F d i 1 for i > 0; (2). s i F = F s i 1 for i > 0. Proof. Suppose that the solution f exists. Define F (x) = f(x, 1) for x X n. Then F satisfies the conditions. Conversely suppose that there exists F : X n Y n+1 satisfying the conditions. Define f : CX Y by setting f(x, q) = f(x) if q = 0 F (x) if q = 1 s q 1 0 F (x) if q 2. Note that f X = f. We show that f is a simplicial map, that is, d j f(x, q) = fd j (x, q) and s j f(x, q) = fsj (x, q). Since f X = f, these identities hold for q = 0. Consider the case q > 0. Note that d j f(x, q) = dj s q 1 0 F (x) = d 0 F (x) if j = 0, q = 1 s q 2 0 F (x) if 0 j < q, q 2 s q 1 0 d j q+1 F (x) if j q,

41 1. SIMPLICIAL SETS 41 f(x) = if j = 0, q = 1 f(d j (x, q)) = f(x, q 1) = s q 2 0 F (x) if 0 j < q, q 2 f((d j q x, q)) = s q 1 0 F (d j q x) if j q. Since d 0 F (x) = f(x) and d i F (x) = F (d i 1 (x)) for i > 0, d j f(x, q) = fdj (x, q) for any x and q. Now { s j f(x, q) = sj s q 1 s q 0F (x) if 0 j < q 0 F (x) = s q 1 0 s j q+1 F (x) if j q, { f(x, q + 1) = s q fs j (x, q) = 0F (x) if 0 j < q f(s j q, q) = s q 1 0 F (s j q x) if j q. Since s i F (x) = F (s i 1 (x)) for i > 0, s j f(x, q) = fsj (x, q) for any x and q. This finishes the proof. Exercise 1.3. Show that CS 0 = [1]. Proposition ΣS n = S n+1 for n 0. Proof. Let σ n be the nondegenerate element in Sn. n Then, from the definition of CS n, the nondegenerate element in CS n are given by (CS n ) 0, (σ n, 0) (CS n ) n and (σ n, 1) (CS n ) n+1 with d 0 (σn, 1) = σ n. Thus the nondegenerate elements in ΣS n+1 are given by (ΣS n ) 0 and (σ n, 1) (ΣS n ) n+1. It follows that the representing map f (σn,1) : [n] ΣS n is onto and it factors through the quotient S n+1 because d i (σ n, 1) = in ΣS n for all 0 i n. The resulting simplicial map f (σn,1) : S n+1 ΣS n is bijective by counting the elements in each dimension, and hence the result. Exercise 1.4. Show that C [n] = [n + 1]. Let X and Y be simplicial sets. Write Hom S (X, Y ) for the set of simplicial maps from X to Y. The following example gives a way to construct the mapping space (as a simplicial set) from X to Y. Example 1.10 (Mapping Spaces). Let σ n = (0, 1,..., n) [n] be the nondegenerate element. d i = f diσ n : [n 1] [n] s i = f siσ n : [n + 1] [n] be the representing maps of d i σ n and s i σ n for 0 i n. Observe that the composites [n 2] d j [n 1] d i [n], [n + 2] s j [n + 1] d i [n], [n] d j [n + 1] s i [n]

42 42 2. SIMPLICIAL SETS AND HOMOTOPY are representation maps of the elements d j d i σ n, s j s i σ n and d j s i σ n, respectively. From the simplicial identities, the sequences of simplicial sets { [n]} n 0 with d i and s i is a cosimplicial simplicial set, namely the identities (1.6) d j d i = d i+1 d j for i j s i s j = s j s i+1 for i j d j s i 1 j < i s i d j = id j = i, i + 1 d j 1 s i j > i + 1. Let Map(X, Y ) n = Hom S (X [n], Y ) and let d i = (id X d i ) : Map(X, Y ) n = Hom S (X [n], Y ) Hom S (X [n 1], Y ) = Map(X, Y ) n 1, s i = (id X s i ) : Map(X, Y ) n = Hom S (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 d i and s i is a simplicial set, which is called the mapping space from X to Y Polyhedral Simplicial Sets. Example 1.11 (Simplicial sets from simplicial complex). Let K be an abstract simplicial complex, and let the vertices of K be ordered. Then we obtain a simplicial set K S = {K S n} n 0 : K S n = {(v 0, v 1,..., v n ) v i are vertices of a simplex of K with v 0 v 1 v n }, d i (v 0, v 1,..., v n ) = (v 0, v 1,..., ˆv i,..., v n ) deleting v i, s i (v 0, v 1,..., v n ) = (v 0, v 1,..., v i, v i,..., v n ) doubling v i with a -map K K S. Observe that K S has the nondegenerate elements given by K. Definition A simplicial set X is called polyhedral if there exists an abstract simplicial complex K such that X = K S. For a simplicial set X, let X D n denote the set of nondegenerate elements in X n. Lemma Let X and Y be simplicial sets and let f : X Y be simplicial map such that (1). If x is a nondegenerate element of X, then f(x) is a nondegenerate element of Y and (2). The induced function f : X D n Y D n is one-to-one for each n 0. Then f is one-to-one. Proof. We show by induction that f : X n Y n is one-to-one for each n 0. When n = 0, we are given that f : X 0 Y 0 is one-to-one by the second assumption. Suppose that f : X n 1 Y n 1 is one-to-one. Let x, y X n such that f(x) = f(y). If both x and y are nondegenerate elements, then x = y by the assumption. Suppose that one of x, y is a degenerate element. We may assume that y = s j z for some z X n 1 and some 0 j n 1. Then f(x) = f(y) = f(s j z) = s j f(z). By the first assumption, x must be a degenerate element. Let x = s i w.

43 1. SIMPLICIAL SETS 43 If i = j, then f(w) = d i s i f(w) = d i f(s i w) = d i f(x) = d i f(y) = d i f(s i z) = d i s i f(z) = f(z). By induction, w = z and so x = s i w = s i z = y. If i j, we may assume that i < j. Then f(w) = d i f(s i w) = d i f(x) = d i f(y) = d i f(s j z) = f(d i s j z). By induction, w = d i s j z and so It follows that x = s i w = s i d i s j z = s i s j 1 d i z = s j s i d i z. f(s i d i z) = d j f(s j s i d i z) = d j f(x) = d j f(y) = d j f(s j z) = d j s j f(z) = f(z) and so s i d i z = z by induction. Thus x = s j s i d i z = s j z = y and hence the result. Now let X be a simplicial set and let 2 X0 be the set of all subsets of X 0. Define φ: 2 X0 n 0 X D n by setting φ(x) = {f x (0), f x (1),..., f x (n)} for x X n. Theorem Let X be a simplicial set. Then X is polyhedral if and only if the following holds: (1). There exists an order of X 0 such that, for each x X n, f x (0) f x (1) f x (n). (2). The function φ: n 0 XD n 2 X0 is one-to-one. Proof. = follows from the construction. =. By the proof of Theorem 3.6 of Chapter 1, for each x Xn D, we have f x (0) < f x (1) < < f x (n) and the abstract simplicial complex K with vertex set X 0 is defined by requiring that: {a 0,..., a n } is a simplex of K if and only if the set {a 0,..., a n } Im φ. Moreover by the proof of Theorem 3.6 of Chapter 1, the map φ induces a -map g : K X given by g({f x (0), f x (1),..., f x (n)}) = x for each x X D n. The -map g extends to a simplicial map g : K S X. Since g maps onto nondegenerate elements of X, the simplicial map g is onto. Since g sends each nondegenerate element of K S (that are given by the elements in K ) to a nondegenerate element of X and g : Kn Xn D is one-to-one, the map g is one-to-one by the above lemma. Thus g is a simplicial isomorphism and hence the result.

44 44 2. SIMPLICIAL SETS AND HOMOTOPY 1.4. Relations Between -Sets and Simplicial Sets. Let X be a simplicial 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 X S by adding degeneracies. Let σ n = (0, 1,..., n) [n] be the nondegenerate element. Recall that we have simplicial maps d i = f diσ n : [n 1] [n] s i = f siσ n : [n + 1] [n] be the representing maps of d i σ n and s i σ n for 0 i n. Given a -set X, we define a sequence of sets by setting X S n = k 0 [k] n X k /, where is the equivalence relation generated by (a, d i x) (d i (a), x) for any a [k 1] n, x X k, k 1 and 0 i k. Define the faces d j : X S n X S n 1 and the degeneracies s j : Xn S Xn+1 S by setting d j (a, x) = (d j a, x) s j (a, x) = (s j a, x) for 0 j n, that is, taking the faces and degeneracies on the first coordinate. We check that d j and s j preserve the equivalence relation: d j (a, d i x) = (d j a, d i x) d j (d i (a), x) = (d j d i (a), x) = (d i (d j a), x) because d i is a simplicial map s j (a, d i x) = (s j a, d i x) s j (d i (a), x) = (s j d i (a), x) = (d i (s j a), x) because d i is a simplicial map. Thus d j and s j preserve the equivalence relation and so d j : Xn S Xn 1 S and s j : Xn S Xn+1 S are well-defined. The simplicial identities automatically hold as they are induced from the simplicial structure of [k] s. The construction looks complicated. But the output simplicial set is not so big as the equivalence relation identifies many elements. Let φ n : X n X S n be the function defined by φ n (x) to be the equivalence class [σ n, x] in X S n. Then Thus φ = {φ n }: X X S is a -map. d j (φ n (x)) = d j [σ n, x] = [d j σ n, x] = [d j (σ n 1, x] = [σ n 1, d j x] = φ(d j (x)).

45 1. SIMPLICIAL SETS 45 Lemma The -map φ: X X S is one-to-one. Proof. Observe that the equivalence relation is generated by connecting some elements in [k] n X k and [k 1] n X k 1 for each k. Thus [σ n, x] [σ n, y] for x y X n, and so φ is one-to-one. Theorem The construction X X S is a functor from the category of -sets and -maps to the category of simplicial sets and simplicial maps. Moreover suppose that X is a -set such that one of the faces d i : X n X n 1 is onto for each n 1. Then the simplicial set X S is generated by φ(x) with all nondegenerate elements given in φ(x). Proof. For showing the first assertion, let f : X Y be a -map. Then the maps id [k] f : [k] n X k [k] n Y k induces a function f S n : X S n Y S n because from id [k] f(d i a, x) = (d i a, f(x)) id [k 1] f(a, d i x) = (a, f(d i x)) = (a, d i (f(x)) the functions {id [k] f} preserves the equivalence relations. Moreover f S = {fn S } X S Y S is a simplicial map because the functions {id [k] f} commutes with faces and degeneracies. Clear (g f) S = g S f S for -maps f : X Y and g : Y Z. Thus X X S is a functor. For the second assertion, we first show that every equivalence class in Xn S can be represented by an element in [n] n X n. Let (a, x) [k] n X k. If k < n, since there exists a face d i : X k+1 X k which is onto, x = d i y for some y X k+1. Thus (a, x) = (a, d i y) (d i a, y) with (d i, y) [k + 1] n X k+1. By iterating this procedure, (a, x) (b, z) for some (b, z) [n] X n. If k > n, then all elements in [k] n are given by iterated faces σ k. Thus a = d i1 d i2 d ik n σ k for some 0 i 1 < i 2 < < i k n k. It follows that and so d i k n d i2 d i1 (σ n ) = d i1 d i2 d ik n σ k = a (a, x) = (d i k n d i2 d i1 (σ n ), x) (d i k n 1 d i2 d i1, d ik n x) (σ n, d i1 d i2 d ik n x). Thus every equivalence class in X S n can be represented by an element in [n] n X n. Now we let (a, x) [n] n X n. Note that σ n is only nondegenerate element in [n] n = {(i 0, i 1,..., i n ) 0 i 0 i 1 i n n}. If a = σ n, then (a, x) φ(x). Otherwise a = s l1 s l2 s lk d j1 d j2 d jk σ n for some 0 l k < < l 2 < l 1 n 1 and 0 j 1 < j 2 < < j k n with k 1. (Recall that d j is given by deleting the coordinates in the sequences (i 0,..., i n ) and

46 46 2. SIMPLICIAL SETS AND HOMOTOPY s j is given by doubling the coordinates. Since a = (i 0,..., i n ) (0, 1,..., n), we can apply d j operations and then apply degeneracies backwards to a.) Then [a, x] = [s l1 s l2 s lk d j1 d j2 d jk σ n, x] = s l1 s l2 s lk [d j1 d j2 d jk σ n, x] = s l1 s l2 s lk [d j k d j1 (σ n k ), x] = s l1 s l2 s lk [σ n k, d j1 d j2 d jk x] with [σ n k, d j1 d j2 d jk x] φ(x) and hence the result. Let f : X Y be a -map. Then there is a commutative diagram X φ X S because, for any x X n, f Y S f φ Y S f S φ(x) = f S [σ n, x] = [σ n, f(x)] = φ f(x). Thus φ: X X S is a natural transformation. More precisely φ is a natural transformation from the identity functor of the category of -sets to the functor ( ) S composing with the forgetful functor from the category of simplicial sets to the category of -sets. Proposition The -map φ: X X S 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 g : X S Z such that g = g φ. (2). Suppose that X satisfies the property that one of the faces d i : X n X n 1 is onto for each n 1. Then g is uniquely determined by g. Proof. The second assertion follows from the first assertion because X S is generated by X in this case. For proving assertion (1), let Z n = k 0 [k] n Z k / where is the equivalence relation generated by (a, d i z) (d i (a), z) (b, s i z) (s i b, z) for any a [k 1] n, b [k + 1] n, z Z k and 0 i k. Then Z = { Z n } n 0 is a simplicial set with the faces and the degeneracies defined by d j : Z n Z n 1 s j : Z n Z n+1 d j (a, z) = (d j a, z) s j (a, z) = (s j a, z) for 0 j n. (Similar to the construction X S, one can check that faces and degeneracies defined above preserve the equivalence relation.)

47 1. SIMPLICIAL SETS 47 Let ψ : Z n Z n be the function defined by ψ(z) = [σ n, z]. Then the sequence of functions ψ = {ψ n }: Z Z is a simplicial map. Let ḡ n : X S n Z n be induced by the functions id [k] g : [k] n X k [k]n Z k. From the lines in the proof of the above theorem, it follows that the function ḡ n is well-defined and ḡ = {ḡ n }: X S Z is a simplicial map. Now there is a commutative diagram X φ X S because for any x X n g ḡ ψ Z Z ḡ φ(x) = ḡ([σ n, x]) = [σ n, g(x)] = ψ g(x). The assertion will follow by proving that ψ : Z Z is an isomorphism because if so, the composite g = ψ 1 ḡ is the desired map. To show that ψ : Z Z is onto, from the proof of the above theorem, every equivalence class in Z n can be represented by an element in [n] n Z n. Let (a, x) [n] n Z n. If a σ n, then a = s i1 s i2 s ik d j1 d j2 d jk σ n for some 0 l k < < l 2 < l 1 n 1 and 0 j 1 < j 2 < < j k n with k 1. Then Thus d jk d j2 d j1 s ik s i2 s i1 (σ n ) = d jk d j2 d j1 s ik s i2 (s i1 σ n 1 ) = d jk d j2 d j1 s ik s i3 (s i1 s i2 (σ n 1 )) = d jk d j2 d j1 s ik s i3 (s i1 s i2 σ n 1 ) = d jk d j2 d j1 (s i1 s i2 s ik σ n k ) = d jk d j2 (s i1 s i2 s ik d i1 (σ n k )) = d jk d j2 (s i1 s i2 s ik d i1 σ n k+1 ) = d jk d j3 (s i1 s i2 s ik d i1 d j2 (σ n k+1 )) = s i1 s i2 s ik d j1 d j2 d jk σ n = a. (a, x) = (d jk d j2 d j1 s ik s i2 s i1 (σ n ), x) (σ n, s i1 s i2 s ik d j1 d j2 d jk x) and so ψ is onto. To see that ψ is one-to-one, similar to the above argument, every element in set k=0 [k] n is the image of the iterated composites of the functions d i and s j on σ n [n] n. The equivalence relation is to identify the elements (a, z) [k] n Z k with an element (σ n, z ) where x is obtained by the correspondent iterated composites of faces and degeneracies on z. Thus (σ n, z) gives the unique representative in the equivalence class [σ n, z] and so ψ is one-to-one. The proof is finished now. Note. We can also directly construct ψ 1 : Z Z as follows: Let (a, z) [k] n Z k. Since a [k] n, a = s i1 s i2 s iq d j1 d j2 d jt σ k,

48 48 2. SIMPLICIAL SETS AND HOMOTOPY for unique sequences 0 l k < < l 2 < l 1 n 1 and 0 j 1 < j 2 < < j t k with k 0. Define ˆψ n (a, z) = s i1 s i2 s iq d j1 d j2 d jt z. Exercise 1.5. Let ˆψ n : [k] n Z k Z n be defined as above. Show that (1). ˆψn induces a function ψ : Z n Z n. (2). The sequence of functions ψ = { ψ n }: Z Z is a simplicial map. (3). ψ is inverse of ψ : Z Z More Examples of Simplicial Sets. Example 1.12 (Classifying spaces of categories). A category C is called small if the class of morphisms in C is a set. Let C be any small category. The simplicial set BC is defined by setting BC n is the set of the strings A 0 f 1 A 1 f 2 f n A n of compose-able arrows of length n of morphisms f j Mor(C) with d i is given by deleting A i in the sequence and s i is given by doubling A i is the sequence. That is d i on the above sequence is A 0 f 1 A 1 f 2 and s i on the above sequence is f i 1 A i 1 f i+1 f i f i+2 A i+1 f n A n A 0 f 1 A1 f 2 f i Ai id Ai f i+1 Ai+1 f i+2 f n An. The simplicial set BC is called the classifying space of the category C. Example 1.13 (Classifying spaces of groups). Let M be a monoid. The simplicial set W M is defined by setting W M n = {(g 0, g 1,..., g n ) g i M} = M n+1 with faces and degeneracies given by { (g0, g d i (g 0, g 1,..., g n ) = 1,..., g i 1, g i g i+1, g i+2..., g n ) if i < n (g 0,..., g n 1 ) if i = n s i (g 0, g 1,..., g n ) = (g 0, g 1,..., g i, e, g i+1..., g n ). For checking simplicial identities, one could describe the sequence (g 0, g 1,..., g n ) as a string g 0 g1 g2 g n. The face operation d i is given by removing ith heart for 0 i n. In this sense, the last face d n deletes g n because the heart of g n is removed. The degeneracy operation s i is given by doubling ith heart for 0 i n. Define the left action of M on M n+1 by setting g (g 0, g 1,..., g n ) = (gg 0, g 1,..., g n ). Then d i (g (g 0, g 1,..., g n )) = g d i (g 0, g 1,..., g n ) and s i (g (g 0, g 1,..., g n )) = g s i (g 0, g 1,..., g n ), that is, the action M W M W M is simplicial, where M is regarded as a discrete in the sense that M n = M and d i = s i = id M. By modulo the action of M, we obtain the simplicial set W M = W M/M.

49 1. SIMPLICIAL SETS 49 Define s 1 : W M n W M n+1 by setting s 1 (g 0, g 1,..., g n ) = (e, g 0, g 1,..., g n ). Then Identities (1.5) hold. This means that the geometric realization W M is contractible. If M is a group, then the geometric realization of W M is the classifying of M. Exercise 1.6. Let M be a monoid. Using the strings g0 g1 gn 1 gn show that the sequence of sets {M n+1 } n 0 with { (g1,..., g d i (g 0, g 1,..., g n ) = n ) if i = 0 (g 0, g 1,..., g i 2, g i 1 g i, g i+1..., g n ) if i < n s i (g 0, g 1,..., g n ) = (g 0, g 1,..., e, g i, g i+1..., g n ). forms a simplicial set with a right action of M. Example 1.14 (Another construction of classifying spaces). Let X be any set. Define EX by setting EX n = X n+1 with d i (x 0, x 1,..., x n ) = (x 0,..., x i 1, x i+1,..., x n ) s i (x 0, x 1,..., x n ) = (x 0,..., x i 1, x i, x i, x i+1,..., x n ). Then EX is a simplicial set. This defines a functor from the category of sets to the category of simplicial sets. Moreover the functions EX n EY n E(X Y )n ((x 0,..., x n )) ((y 0,..., y n )) ((x 0, y 0 ), (x 1, y 1 ),..., (x n, y n )) induces a simplicial isomorphism EX EY = E(X Y ). Thus, if X is a group (monoid), then EX is a simplicial group (monoid). Let X be a pointed set with basepoint. Define s 1 on EX by setting s 1 (x 0,..., x n ) = (, x 0,..., x n ). Then identities (1.5) hold. Thus the geometric realization EX is contractible. Suppose that X = G is a group. Observe that (EG) 0 = G and so G is a simplicial subgroup EG, where G is regarded as a discrete simplicial group. The resulting simplicial coset BG = EG/G is the quotient of a contractible simplicial group EG by its simplicial subgroup G. Thus the geometric realization of BG is the classifying space of G. Example 1.15 (Classifying spaces of simplicial monoids). Let M be a simplicial monoid. The simplicial set W M is defined by setting W M n = {(g n, g n 1,..., g 0 ) g i M i } = M n M n 1 M 0 with faces and degeneracies given by d i (g n, g n 1,..., g 0 ) = { (di g n, d i 1 g n 1,..., d 1 g n i+1, d 0 g n i g n i 1, g n i 2..., g 0 ) if i < n (d n g n, d n 1 g n 1..., d 1 g 1 ) if i = n s i (g n, g n 1,..., g 0 ) = (s i g n, s i 1 g n 1,..., s 0 g n i, e, g n i 1..., g 0 ). The left action of M on W M given by h (g n, g n 1,..., g 0 ) = (hg n, g n 1,..., g 0 )

50 50 2. SIMPLICIAL SETS AND HOMOTOPY for h M n and (g n, g n 1,..., g 0 ) W M n is simplicial. The classifying space W M = W M/M is the quotient of W M by the action of M. Example 1.16 (Simplicial structure on braids). Let X n = B n+1 be the braid group of (n + 1)-strands with faces and degeneracies given by: d i β is the braid obtained by removing the (i + 1)st strand of β and s i β is the braid obtained by doubling the (i+1)st strand of β (that is the (i+1)st strand is replaced by two untwisted strands in its small neighborhood). Then X = {X n } is a simplicial set. Since the faces and degeneracies are not group homomorphism, X is a not a simplicial group. (In fact, X is so-called crossed simplicial group as it satisfies d i (ββ ) = d i βd β(i) β and s i (ββ ) = s i βs β(i) β, where β(i) means that the ith strand goes from ith point to β(i)th point.) The sequence of pure braid groups {P n+1 } n 0 with the above faces and degeneracies is a simplicial group. 2. Geometric Realization of Simplicial Sets The standard geometric n-simplex n is defined by n n = {(t 0, t 1,..., t n ) t i 0 and t i = 1}. Define d i : n 1 n and s i : n+1 n by setting i=0 d i (t 0, t 1,..., t n 1 ) = (t 0,..., t i 1, 0, t i,..., t n 1 ), s i (t 0, t 1,..., t n+1 ) = (t 0,..., t i 1, t i + t i+1,..., t n+1 ) for 0 i n. Let X be a simplicial set. Then its geometric realization X is a CW -complex defined by X = ( n, x)/ = n X n /, x X n n 0 n=0 where ( n, x) is n labeled by x X n and is generated by (z, d i x) (d i z, x) for any x X n and z n 1 labeled by d i x, and (z, s i x) (s i z, x) for any x X n and z n+1 labeled by s i x. Note that the points in ( n+1, s i x) and ( n 1, d i x) are identified with the points in ( n, x). Exercise 2.1. Prove that [n] = n and S n = S n. Let f : X Y be a simplicial map. Then its geometric realization f is defined by f (z, x) = (z, f(x)) for any x X n and z n labeled by x. Clearly f is continuous. Proposition 2.1. Let X be a simplicial set. Then X is a CW -complex. Thus the geometric realization gives a functor from the category of simplicial sets to the category of CW -complexes.

51 2. GEOMETRIC REALIZATION OF SIMPLICIAL SETS 51 Proof. From the push-out diagram [n] x X n nondegenerate fx [n] skn 1 X there is a push-out diagram x X n nondegenerate x X n nondegenerate [n] [n] = n fx skn X, fx [n] skn 1 X x X n nondegenerate [n] = n fx skn X. Thus X is obtained by attaching cell-by-cell and so X is a CW -complex. Proposition 2.2. Let K be an abstract simplicial complex. Then there is a homeomorphism K S = K. Proof. Exercise. Exercise 2.2. Show that [m] [n] = m n. Theorem 2.3. Let X and Y be simplicial sets. Then there is a one-to-one and onto continuous map η : X Y X Y. Proof. The coordinate projections p 1 : X Y X and p 2 : X Y Y induce continuous maps p 1 : X Y X and p 2 : X Y Y and so a continuous map η = (p 1, p 2 ): X Y X Y. We show that η is bijective. From the above exercise, this holds for X = [m] and Y = [n]. By induction on the skeleton filtration of Y and using the push-out [m] [n] [m] skn 1 Y [m] y Y n nondegenerate y Y n nondegenerate [n] [m] skn Y, we have that η : [m] Y [m] Y

52 52 2. SIMPLICIAL SETS AND HOMOTOPY is bijective for any simplicial set Y. The proof is then finished by induction on the skeleton filtration of X and using the push-out [m] Y sk m 1 X Y x X m nondegenerate x X n nondegenerate [m] Y skm X Y. Definition 2.4. A topological space X is called compactly generated if X is Hausdorff and each subset A of X with the property that A C is closed for any every compact subspace C of X is itself closed. Exercise 2.3. Show that (1). every locally compact Hausdorff space is compactly generated. (2). every Hausdorff space satisfying the first axiom of countability is compactly generated. (3). every metric space is compactly generated. Given a Hausdorff space X, we can redefine a topology on X as follows: Let C = {C X C compact} be the collection of compact spaces. Define a subset A of X to be closed if and only if A C is closed for every C C. The new topological space is denoted by k(x) which has the same underline set as X with the same compact sets. Note that the identity map k(x) X is continuous. Namely possibly k(x) has more closed sets. If X is already compactly generated, then k(x) = X. Let X and Y be two Hausdorff spaces. Let X CG Y = k(x Y ). The category of compactly generated spaces means the subcategory of topological spaces, where the objects are compactly generated spaces and the morphisms are continuous maps between two compactly generated spaces. The Cartesian product in the category of compactly generated spaces is given by X CG Y. (Note. If X and Y are compactly generated, then X Y may not be compactly generated in general and so we need to add more closed sets in X Y to make X CG Y such that the Cartesian product X CG Y is still compactly generated.) Exercise 2.4. Let X and Y be simplicial sets. X CG Y if and only if (f x f y ) 1 (A) Show that A is closed in is closed in m n = [m] [n] for every x X m and y Y n with m, n 0. By this exercise, you can then prove the following: Theorem 2.5. Let X and Y be simplicial sets. Then is a homeomorphism. η : X Y X CG Y

53 3. HOMOTOPY AND FIBRANT SIMPLICIAL SETS 53 Corollary 2.6. Let G be a simplicial group. Then G is a topological group in the category of compactly generated spaces. Proof. The multiplication µ: G G G induces a continuous multiplication such that G is a topological group. with G G = G CG G G Exercise 2.5. Prove the following statement: (1). For any simplicial set X, CX is homeomorphic to the unreduced cone C X of X. (2). If X is a pointed simplicial set, CX is homeomorphic to the reduced cone C X of X. (3). If X is a pointed simplicial set, ΣX = Σ X. 3. Homotopy and Fibrant Simplicial Sets Let I = [1]. As a simplicial set, n+1 i I n = {t i = ( {}}{{}}{ 0,..., 0, 1,..., 1) 0 i n + 1} n+1 d j t i n+1 s j t i n+1 = { t i n t i 1 n if if = t i θ(i+j n 1) n = { t i n+2 t i+1 n+2 if if, i 0 j < n + 1 i n + 1 i j n 0 j < n + 1 i n + 1 i j n = t i+θ(i+j n 1), n+2 { 0 if x < 0 where θ(x) = Given a simplicial set X, the simplicial subsets 1 if x 0. X 0 and X 1 of X I is given by (X 0) n = {(x, t 0 ) x X n }, (X 1) n = {(x, t 1 ) x X n }. Definition 3.1. Let f, g : X Y be simplicial maps. We call f homotopic to g if there is a simplicial map F : X I Y such that F X 0 = f and F X 1 = g in which case write f g. If A is a simplicial subset of X and f, g : X Y are simplicial maps such that f A = g A, we callf homotopic to g relative to A, denoted by f g rel A, if there is a homotopy F : X I Y such that F X 0 = f, F X 1 = g and F A I = f. A pointed simplicial set means a simplicial set with a choice of a basepoint. Let X and Y be pointed simplicial sets. A pointed simplicial map f : X Y means a simplicial map which sends the basepoint of X to the basepoint of Y. We usually denote by the basepoint. Let f, g : X Y be pointed maps. A pointed homotopy means a simplicial map F : X I Y such that F X 0 = f, F X 1 = g and

54 54 2. SIMPLICIAL SETS AND HOMOTOPY F I = (that is F I is the constant map to the basepoint of Y ) in which we write f g rel. Exercise 3.1. Let f, g : X Y be simplicial maps. Show that f g if and only if there is a family of functions F i/(n+1) : X n Y n for n 0 and 0 i n + 1 satisfying the following conditions: 1) d j F i/(n+1) = F (i θ(i+j n 1))/n d j, 2) s j F i/(n+1) = F (i+θ(i+j n 1))/(n+2) s j, 3) F 0 = f and 4) F 1 = g. We need fibrant assumption on Y such that homotopy relation is an equivalence relation on the set of simplicial maps from X to Y. Definition 3.2. Let X be a simplicial set. The elements x 0,..., x i 1, x i+1,..., x n X n 1 are called matching faces with respect to i if d j x k = d k x j+1 for j k and k, j + 1 i. A simplicial set X is called fibrant (or Kan complex) if it satisfies the following homotopy extension condition for each i: Let x 0,..., x i 1, x i+1,..., x n X n 1 be any elements that are matching faces with respect to i. Then there exists an element w X n such that d j w = x j for j i. Let Λ i [n] be the simplicial subset of [n] generated by all d j σ n for j i, where σ n = (0, 1,..., n) [n] n is the nondegenerate element. Proposition 3.3. Let X be a simplicial set. Then X is fibrant if and only if every simplicial map f : Λ i [n] X has an extension for each i. Proof. Suppose that X is fibrant. Let f : Λ i [n] X. The elements f(d 0 σ n ), f(d 1 σ n ),..., f(d i 1 σ n ), f(d i+1 σ n ),..., f(d n σ n ) have matching faces and so there exists an element w X n such that d j w = f(d j σ n ) for j i. Then the representing map g = f w : [n] X is an extension of f. Conversely let x 0,..., x i 1, x i+1,..., x n X n 1 be any elements that are matching faces with respect to i. Then the representing maps f xj : [n 1] X for j i defines a simplicial map f : Λ i [n] X such that the diagram [n 1] dj = f djσ n Λ i [n] fx j X commutes for each j. (Note. The matching faces condition is used here for having f well-defined.) By the assumption, there exists an extension g : [n] X such that g Λi [n] = f. Let w = g(σ n ). Then d j w = d j g(σ n ) = g(d j σ n ) = f(d j σ n ) = x j f

55 3. HOMOTOPY AND FIBRANT SIMPLICIAL SETS 55 for j i. Thus X is fibrant. Lemma 3.4. Let A B denote any of the following pairs: [n] Λ i [m] [n] [m] ( [n] Λ i [m] [n] [m]) [n] [m]. Let X be a fibrant simplicial set. Then any map f : A X can be extended to a simplicial map g : B X. Proof. For each such a pair, B can be obtained from A by successively adjoining a simplex and one of its faces, all other faces already lying in a simplicial subset. Iterated application of the extension condition gives the lemma. Proposition 3.5. Let Y be a fibrant simplicial set and let X be any simplicial set. Let A be a simplicial subset of X. Then any map f : (A [m]) (X Λ i [m]) Y can be extended to a simplicial map g : X [m] Y. Proof. Let sk n (X) = A sk n (X). First f can be extended to a simplicial map sk 0 X Y. For n > 0, there is a push-out diagram ( ( [n] [m]) ( [n] Λ i [m]) ) skn 1 X [m] x X D n x X D n ( [n] [m]) push skn X [m], where X n D is the set of nondegenerate elements in X n A n. The assertion follows by induction, where the step from sk n 1 X to sk n X is shown in Lemma 3.4. Let X and Y are pointed simplicial sets. Let Map (X, Y ) be the simplicial subset of Map(X, Y ) with for each n 0. Map (X, Y ) n = {f : X [n] Y f(s n 0 [n]) = } Proposition 3.6. Let Y be a fibrant simplicial set (with a basepoint) and let X be any simplicial set (with a basepoint). Then the mapping space Map(X, Y ) and Map (X, Y ) are fibrant. Proof. Let f 0,..., f i 1, f i+1,..., f n Map (X, Y ) n 1 have matching faces, that is, each f j : X [n 1] Y, and if we consider each such [n 1] as the jth face of [n], the f j agree on sk n 2 [n] to give a simplicial map f : (X Λ i [n]) ( [n]) Y, where f [n] is the constant map to the basepoint of Y. By Proposition 3.5, f extends to a simplicial map g : X [n] Y, where A =. Then g Map (X, Y ) n with d j g = f j for j i. Thus Map (X, Y ) is fibrant. Similarly Map(X, Y ) is fibrant.

56 56 2. SIMPLICIAL SETS AND HOMOTOPY Theorem 3.7. Let Y be a fibrant simplicial set (with a basepoint) and let X be any simplicial set (with a basepoint). Then the homotopy relation ( rel ) is an equivalence relation. In such case, denote by [X, Y ] the set of the pointed homotopy classes of all pointed simplicial maps from X to Y. Proof. f g rel = g f rel. Let f g rel : X Y under a homotopy F : X I Y with F I =. Consider the mapping space Map (X, Y ). Then f, g Map (X, Y ) 0 and F Map (X, Y ) 1 with d 1 F = f and d 0 F = g. The elements x 1 = s 0 d 1 F and x 2 = F have matching faces with respect to 0 because d 1 x 1 = d 1 s 0 d 1 F = d 1 F = d 1 x 2. Since Map (X, Y ) is fibrant, there exists w Map (X, Y ) 2 such that d 1 w = s 0 d 1 F and d 2 w = F. Let F = d 0 w. Then d 0 F = d 0 d 0 w = d 0 d 1 w = d 0 s 0 d 1 F = d 1 F, d 1 F = d 1 d 0 w = d 0 d 2 w = d 0 F and so F is a homotopy from g to f. Since F Map (X, Y ) 1, F I is the constant simplicial map. Thus F is a pointed homotopy from g to f. f F g rel, g G h rel = f h rel. We have F, G Map (X, Y ) 1 such that d 1 F = f, d 0 F = d 1 G = g and d 0 G = g. The elements x 0 = G and x 2 = F have matching faces with respect to 1 because d 1 x 0 = d 1 G = d 0 F = d 0 x 2. Thus there exists w Map (X, Y ) 2 such that d 0 w = G and d 2 w = F. Let F = d 1 w. Then d 1 F = d 1 d 1 w = d 1 d 2 w = d 1 F = f, d 0 F = d 0 d 1 w = d 0 d 0 w = d 0 G = h and so f h rel. Note. The ideas in the above proof can be shown as in the picture 0 d1(0, 1, 2) d 2 (0, 1, 2) 1 d0 (0, 1, 2) 2 Exercise 3.2. Show that [n], [n] and S n are NOT fibrant. Exercise 3.3. Let Y be a fibrant simplicial set and let X be any simplicial set. Let A be a simplicial subset of X. Show that the homotopy relation ( rel A) is an equivalence relation. Example 3.1. Let X be any topological space and let S n (X) = Map( n, X) be the set of all continuous maps from n X. Then S (X) = {S n (X)} n 0 is a simplicial set, called singular simplicial set, with faces and degeneracies d i = d i : Map( n, X) Map( n 1, X), s i = s i : Map( n, X) Map( n+1, X) induced by the maps d i : n 1 n and s i : n+1 n. Let x 0,..., x i 1, x i+1,..., x n S n 1 (X) have matching faces with respect to i. Namely each x j is a continuous map x j : [n 1] X.

57 3. HOMOTOPY AND FIBRANT SIMPLICIAL SETS 57 Let Λ i [n] be the subcomplex of n obtained by removing i-th face of the n- simplex. The matching faces condition allows us to define a continuous map f : Λ i [n] X such that f restricted to jth face of the n-simplex is x j. Observe that Λ i [n] is a retract of n. Let r : n Λ i [n] be a retraction map. Then g = f r : n X is an extension of f. Consider g as an element in S n (X) = Map( n, X). Then d j g = x j for j i. Thus S (X) is fibrant. Example 3.2. Let X be a fibrant simplicial set and let A be a simplicial retract of X. We show that A is also fibrant. Let r : X A be a simplicial retract. For any simplicial map f : Λ i [n] A, there exists a simplicial map g : [n] X such that g Λi [n] = f because X is fibrant. Let g = r g : [n] A. Then g Λ i [n] = r g Λ i [n] = r f = f because f(λ i [n]) A. Hence A is fibrant. An important example of fibrant simplicial sets is as follows. Theorem 3.8 (Moore). Any simplicial group G is a fibrant simplicial set. Proof. Let G be a simplicial group and let y 0,..., ŷ k,..., y n G n 1 have matching faces. Construct the elements in G n as follows: w 0 = s 0 y 0, w i = w i 1 (s i d i w i 1 ) 1 s i y i for 0 < i < k, w n = w k 1 (s n 1 d n w k 1 ) 1 s n 1 y n, w i = w i+1 (s i 1 d i w i+1 ) 1 s i y i for k < i < n. We check that d j w k+1 = y j for j k. First we show by induction that d j w i = y j for j i < k. Note that d 0 w 0 = d 0 s 0 y 0 = y 0. Assume that d j w i 1 = y i 1 for j i 1 < k with i < k. Then, for j < i and d j w i = d j (w i 1 (s i d i w i 1 ) 1 s i y i ) = y j (d j s i d i w i 1 ) 1 d j s i y i = y j (s i 1 d j d i w i 1 ) 1 s i 1 d j y i = y j (s i 1 d i 1 d j w i 1 ) 1 s i 1 d i 1 y j by matching faces condition = y j (s i 1 d i 1 y j ) 1 s i 1 d i 1 y j by induction = y j d i w i = d i (w i 1 (s i d i w i 1 ) 1 s i y i ) = d i w i 1 (d i s i d i w i 1 ) 1 d i s i y i = d i w i 1 (d i w i 1 ) 1 y i = y i Hence d j w i = y j for j i < k. If k = n, then w n 1 satisfies the property that d j w i = y i for i < n. Thus we may assume that k < n now. For 0 t n k 1, we show by induction that

58 58 2. SIMPLICIAL SETS AND HOMOTOPY d j w n t = y j for 0 j < k 1 and n t j n. When t = 0, for j < k, d j w n = d j (w k 1 (s n 1 d n w k 1 ) 1 s n 1 y n ) = d j w k 1 (d j s n 1 d n w k 1 ) 1 d j s n 1 y n = d j w k 1 (s n 2 d j d n w k 1 ) 1 s n 2 d j y n as j < k n 1 = d j w k 1 (s n 2 d n 1 d j w k 1 ) 1 s n 2 d n 1 y j by matching faces condition = y j (s n 2 d n 1 y j ) 1 s n 2 d n 1 y j = y j and d n w n = d n (w k 1 (s n 1 d n w k 1 ) 1 s n 1 y n ) = d n w k 1 (d n s n 1 d n w k 1 ) 1 d n s n 1 y n = d n w k 1 (d n w k 1 ) 1 y n = y n Assume that d j w n t+1 = y j for 0 j < k 1 and n t + 1 j n with 0 t n k 1. Then, for 0 j k 1, d j w n t = d j (w n t+1 (s n t 1 d n t w n t+1 ) 1 s n t 1 y n t ) = d j w n t+1 (d j s n t 1 d n t w n t+1 ) 1 d j s n t 1 y n t = y j (s n t 2 d j d n t w n t+1 ) 1 s n t 2 d j y n t since j k 1 < k n t 1 = y j (s n t 2 d n t 1 d j w n t+1 ) 1 s n t 2 d n t 1 y j by matching faces condition = y j (s n t 2 d n t 1 y j ) 1 s n t 2 d n t 1 y j = y j and, for n t + 1 j n, Now d j w n t = d j (w n t+1 (s n t 1 d n t w n t+1 ) 1 s n t 1 y n t ) = d j w n t+1 (d j s n t 1 d n t w n t+1 ) 1 d j s n t 1 y n t = y j (s n t 1 d j 1 d n t w n t+1 ) 1 s n t 1 d j 1 y n t since n t 1 < n t j 1 = y j (s n t 2 d n t d j w n t+1 ) 1 s n t 2 d n t y j by matching faces condition = y j (s n t 2 d n t y j ) 1 s n t 2 d n t y j. = y j d n t w n t = d n t (w n t+1 (s n t 1 d n t w n t+1 ) 1 s n t 1 y n t ) = d n t w n t+1 (d n t s n t 1 d n t w n t+1 ) 1 d n t s n t 1 y n t = d n t w n t+1 (d n t w n t+1 ) 1 y n t = y n t. The induction is finished and so d j w k+1 = y j for j k. 4. The Relations between Spaces and Simplicial Sets Theorem 4.1 (Simplicial Extension Theorem). Let X be a simplicial set and let A be a simplicial subset of X. Let Y be a fibrant simplicial set and let f : A Y be a simplicial map. Suppose that there is a continuous map φ: X Y such that φ A = f : A Y. Then there exists a simplicial map φ : X Y such that φ A = f : A Y and φ φ: X Y rel A. We state this theorem without proof. One can find a proof in [30].

59 4. THE RELATIONS BETWEEN SPACES AND SIMPLICIAL SETS 59 Corollary 4.2. Let X be a simplicial set with a simplicial subset A and let Y be a fibrant simplicial set. Suppose that A is a retract of X, that is there is a continuous map r : X A such that r A = id A. Then any simplicial map f : A Y can be extended to a simplicial map g : X Y. Proof. Let φ = f r. Then g = φ is an extension of f. Let f, g : X Y be pointed simplicial maps between pointed simplicial sets. Suppose that f g rel under a pointed homotopy F : X I Y. By taking geometric realization, we have the commutative diagram X I = X I ( I) ( X 0) ( X 1) F Y f g = ( I) (X 0) (X 1) and so F is a pointed homotopy from f to g, that is f g rel. Thus the geometric realization f f induces a function : [X, Y ] [ X, Y ]. Theorem 4.3. Let X be any pointed simplicial set and let Y be any pointed fibrant simplicial set. Then the geometric realization induces a one-to-one correspondence : [X, Y ] [ X, Y ]. Proof. First we show that is onto. Let φ: X Y be any pointed map. Let A = be the simplicial subset of X generated by the basepoint. Let f : A Y be the constant map to the basepoint of Y. By Theorem 4.1, there exists a simplicial map φ such that φ A = f and φ φ rel A. The first condition means that φ is a pointed map and the second means that φ φ rel. This proves that is onto. Now we show that is one-to-one. Let f, g : X Y be pointed maps such that f g rel. Let A = ( I) (X 0) (X 1) and let h = f g : A = ( I) (X 0) (X 1) Y. Since f g rel, there exists a continuous map φ: X I = X I Y such that φ A = h. By Theorem 4.1, there exists a simplicial map F : X I Y such that F A = h and so f g rel. The proof is finished. Now consider the singular simplicial functor S : X S (X) for any topological space X, where S n (X) = Map( n, X) as sets with faces d i and degeneracies s i induced from the continuous maps and d i : n 1 n (t 0,..., t n 1 ) (t 0,..., t i 1, 0, t i,..., t n 1 ) s i : n+1 n (t 0,..., t n+1 ) (t 0,..., t i 1, t i + t i+1, t i+1,..., t n+1 ), respectively. Let X be any topological space. There is a canonical continuous map e X : S (X) X

60 60 2. SIMPLICIAL SETS AND HOMOTOPY defined as follows: Since n is compact and Hausdorff, the evaluation map e n : n Map( n, X) X is continuous, where Map( n, X) is given by compact-open topology. The set Map( n, X) can be regarded as a topological space with discrete topology and so e n is a continuous map by regarding Map( n, X) as a set. Lemma 4.4. Let X be any topological space. The maps e n induces a continuous map e X : S (X) X. Moreover Φ X is functorial with respect to X, that is, there is a commutative diagram S (X) e X X S (φ) φ ey S (Y ) Y for any continuous map φ: X Y. Proof. Recall that S (X) = n S n (X)/, n 0 where is generated by (t, d i f) (d i t, f) for t n 1 (t, s i f) (s i t, f) for t n+1 and f S n (X). Note that and f S n (X) and e n 1 (t, d i f) = e n 1 (t, f d i ) = f d i (t) = f(d i t) = e n (d i t, f) for t n and f S n (X), and e n+1 (t, s i f) = e n+1 (t, f s i ) = f s i (t) = f(s i (t)) = e n (s i t, f) for t n+1 and f S n (X). Thus the maps e n induces a continuous map e X : S (X) X. Clearly e X is functorial with respect to X. Theorem 4.5. Let X be a topological space with any choice of basepoint. The map e X : S (X) X induces an isomorphism on homotopy groups. Proof. First we show that e X : π n ( S (X) ) π n (X) is onto. Let g : S n X be a continuous map. Let g be the composite g : n pinch p S n = n / D n g X. Then g is an element in S n (X) = Map( n, X) with d i g = d i = for 0 i n. Thus the representing map f g : [n] S (X)

61 4. THE RELATIONS BETWEEN SPACES AND SIMPLICIAL SETS 61 factors through the quotient simplicial set S n = [n]/ [n], that is there exists a simplicial map f g : S n S (X) such that the diagram [n] S n commutes. We check the composite n = [n] f g S (X) f g f g S (X) e X X. Let σ n be the nondegenerate element in [n]. Note any t n = [n] = q is represented by (t, σ n ) n [n]. Since the composite which gives the commutative diagram Since p is onto and q [n] q / id n f g (t, σ n ) = (t, g), e X f g (t) = g(t) [n] pinch p S n = S n g f g X e X f g S (X). e X f g p = g = g p, we have e X barf g = g and so π n ( S (X) ) π n (X) is onto. Next we show that π n ( S (X) ) π n (X) is one-to-one. By Theorem 4.3, it suffices to show that the composite [S n, S (X)] π n ( S (X) ) e X π n (X) is one-to-one. Let f, g : S n S (X) be pointed maps such that e X f e X g rel under a pointed homotopy F : S n I X. Thus F Sn 0 = e x f, F Sn 1 = e X g and F I =. Let F be the simplicial map given by the composite F : S n I i1 i2 S ( S n ) S ( I ) = S ( S n I ) S (F ) S (X), where i 1 is the representing map for the pinch map n S n as an element in S n ( S n ), and i 2 is the representing for the identity map of 1 as an element in S 1 ( I ). Then F I =, F Sn 0 is the representing map for the composite (4.1) n pinch S n F S n 0=e X f X,

62 62 2. SIMPLICIAL SETS AND HOMOTOPY and F S n 1 is the representing map for the composite (4.2) n pinch S n F S n 1=e X g X. Let σ n S n n be the nondegenerate element. Then f(σ n ) S n (X) is a continuous map from n to X with f(σ n ) d i = for 0 i n. Consider the composite n = q q [n] q / pinch p S n f S (X) Let σ n be the nondegenerate element in [n]. For any t n, e X f p(t) = e X f p(t, σ n ) = e X (t, f(σ n )) = f(σ n )(t). e X X. It follows that the composite in Equation 4.1 is the representing map for the element f(σ n ) S n (X) and so F S n 0 = f. Similarly F S n 1 = g. Thus the map F give a pointed homotopy f g rel. The proof is finished. Now given a simplicial set X. There is a canonical simplicial map defined as follows: i X : X S ( X ) For any x X n, let f x : [n] X be the representing map of x. Then f x : n = [n] X gives an element in S n ( X ). The functions i X : X S (X) is defined by setting i(x) = f x. Lemma 4.6. The sequence of functions i X : X S ( X ) is a simplicial map. Moreover i X is functorial with respect to simplicial sets X and simplicial maps. Proof. The assertion follows from the following commutative diagram n+1 ==== [n + 1] n s i s i ======= [n] d i d i f six fx X f dix n 1 ==== [n 1]. Proposition 4.7. The composite is the identity map. Proof. Recall that i X e X X S ( X ) X X = n X n /, n 0

63 5. HOMOTOPY GROUPS 63 where (t, d i x) (d i t, x) and (t, s i x) (s i t, x). Thus any element α X is represented by an element (t, x) with t n and x X n. Let σ n be the nondegenerate element in [n]. Then f x (σ n ) = x. Thus e X i X (α) is given by the point e X f X (t, σ n ) = f x (t) = {(t, x)} and hence the result. Corollary 4.8. Let X be a pointed simplicial set. Then the map i X induce an isomorphism i X : π n ( X ) πn ( S ( X ) ). for each n 0. By the Whitehead Theorem which states that a pointed map between pathconnected CW -complexes which induces isomorphisms on all homotopy groups is a homotopy equivalence. Thus if π 0 (X) is trivial, then i X : X S ( X ) is a homotopy equivalence because both X and S ( X ) are CW -complexes. The relations between simplicial sets and spaces can be given as in the picture simplicial sets CW-complexes i ( ) id ( ) S simplicial sets S e ( ) S id ( ) topological spaces. 5. Homotopy Groups Let X be a pointed fibrant simplicial set. The homotopy group π n (X), as a set, is defined by π n (X) = [S n, X] and so π n (X) = π n ( X ) as sets. We are going to give the combinatorial description for the group structure on π n (X) for n 1 such that combinatorially defined group structure coincides with the group structure geometrically defined. An element x X n is called spherical if d i x = for all 0 i n. Given a spherical element x X n, then its representing map f x : [n] X factors through the simplicial quotient set S n = [n]/ [n]. Conversely any simplicial map f : S n X gives a spherical element f(σ n ) X n, where σ n is the nondegenerate element in S n n. This gives a one-to-one correspondence spherical elements in X n simplicial maps S n X Since we have the notion of homotopy on simplicial maps, we also want to see how to give homotopy on the spherical elements in X n.

64 64 2. SIMPLICIAL SETS AND HOMOTOPY 5.1. Path Product and fundamental groupoids. Let σ 1 = (0, 1) [1] 1. A simplicial map λ: [1] X is called a path. Since Hom( [1], X) = X 1, the path are one-to-one correspondence to the elements in X 1 under the function given by λ x λ = λ(σ 1 ). The initial point of λ is λ(0) = λ(d 1 σ 1 ) = d 1 x λ X 0, and the ending point of λ is λ(1) = λ(d 0 σ 1 ) = d 0 x λ X 0. Given two paths λ and µ such that λ(1) = µ(0). Then d 0 x λ = d 1 x µ and so the elements x 0 = x µ and x 2 = x λ have matching faces. By the fibrant assumption, there is an element w X 2 such that d 0 w = x µ and d 2 w = x λ, see the picture 0 d 2 1 (5.1) d1 2. d0 Define λ µ = f d1w. Since w may not be unique, the product is only a relation. Let λ: [1] X be a path. Let [λ] = {µ: [1] X λ µ rel [1]}, called path homotopy class represented by λ. We are going to show that the above product induces a well-defined product structure on path-homotopy classes. Lemma 5.1. Let w, w X 2 such that d 0 w = d 0 w and d 2 w = d 2 w. Then Proof. Define the map f d1w f d1w rel [1]. F : (Λ 1 [2] I) ( [2] I) X by setting F Λ 1 [2] I(x, t) = f w Λ 1 [2](x) = f w Λ 1 [2](x), F [2] 0 = f w, F [2] 1 = f w. Then F can be extended to a simplicial map F : [2] I X because the geometric realization (Λ 1 [2] I) ( [2] I) is a retract of [2] I = 2 I. Now the composite [1] I d1 id I [2] I F X is a homotopy between f d1w and f d1w relative to [1]. Lemma 5.2. Let λ 1, λ 1, λ 2, λ 2 : [1] X be paths such that λ i λ i for i = 1, 2 and λ 1 = λ 2. Then λ 1 λ 2 λ 1 λ 2 rel [1]. rel [1] Proof. Let w, w X 2 such that d 0 w = λ 2, d 2 w = λ 1, d 0 w = λ 2 and d 2 w = λ 1. Let F i : [1] I X be a relative homotopy from λ i to λ i for i = 1, 2. Then the map F : (Λ 1 [2] I) ( [2] I) X

65 5. HOMOTOPY GROUPS 65 by setting F d 0 [1] I = F 2, F d 2 [1] I = F 1, F [2] 0 = f w, F [2] 1 = f w. is a well-defined simplicial map. So F can be extended to a simplicial map F : [2] I X. The composite F (d 1 id I ) is a homotopy from d 1 w to d 1 w relative to [1]. Lemma 5.3 (Associativity). Let X be a fibrant simplicial set and let λ 1, λ 2 and λ 3 be paths in X such that λ 1 (1) = λ 2 (0) and λ 2 (1) = λ 3 (0). Then (λ 1 λ 2 ) λ 3 λ 1 (λ 2 λ 3 ) rel [1]. Proof. Since d 0 x λ2 = λ 2 (1) = λ 3 (0) = d 1 x λ3, there exists w 0 X 2 such that d 0 w 0 = x λ3 and d 2 w 0 = x λ2. Since d 2 (d 1 w 0 ) = d 1 d 2 w = d 1 x λ2 = λ 2 (0) = λ 1 (1) = d 0 x λ1, there exists w 2 X 2 such that d 0 w 2 = d 1 w 0 and d 2 w = x λ1. Since d 0 x λ1 = λ 1 (1) = λ 2 (0) = d 1 x λ2, there exists w 3 X 2 such that d 0 w 3 = x λ2 and d 2 w 3 = x λ1. These elements can be shown in the picture w 0 w 2 x λ3 d0 x λ2 d 2 d 0 d1 x λ2 λ 3 d0 d 2 x λ1 d 2 d1 x λ1 (λ 2 λ 3) d 1 w 3 x λ1 λ 2. Since d 1 w 0 = d 0 w 2, d 2 w 0 = d 0 w 3 and d 2 w 2 = d 2 w 3, the elements w 0, w 2, w 3 have matching faces with respect to 1 and so there exists u X 3 such that d i u = w i for i 1. Now x λ1 (λ 2 λ 3) = d 1 w 2 = d 1 d 2 u = d 1 (d 1 u) with d 2 (d 1 u) = d 1 d 3 u = d 1 w 3 = x λ1 λ 2 and d 0 (d 1 u) = d 0 d 0 u = d 0 w 0 = x λ3. Thus d 1 (d 1 u) is a representative for (λ 1 λ 2 ) λ 3. The assertion follows. Given a point y Y 0, denote by ɛ y : X Y for the constant simplicial map ɛ(x) = s n 0 (y) for x X n. Lemma 5.4 (Path Inverse). Let X be a fibrant simplicial set and let λ be a path in X. Then there exists a path λ 1 such that λ λ 1 ɛ λ(0) rel [1].

66 66 2. SIMPLICIAL SETS AND HOMOTOPY Proof. Let x 2 = x λ and x 1 = s 0 λ(0). Then d 1 x 2 = d 1 x λ = λ(0) = d 1 x 1. Thus x 1, x 2 have matching faces and so there exists w X 2 such that d 2 w = x λ and d 1 w = s 0 λ(0). Note that f s0λ(0) is the constant path. Let λ 1 = f d0w. By definition, λ λ 1 ɛ λ(0) rel [1]. For a fibrant simplicial set X, denote by P(X) the set of path homotopy classes. For a topological space X, write P(X) for the fundamental groupoid of X. Theorem 5.5. Let X be a fibrant simplicial set. Then P(X) is the quotient set of X 1 subject to the relation generated by x x if d 0 x = d 0 x, d 1 x = d 1 x and there exists w X 2 such that d 2 w = x, d 1 w = x and d 0 w = ɛ d0x. Moreover product structure on the pair of elements in P(X) which have matching faces is given by: If x, x X 1 such that d 0 x = d 1 x, then [x] [x ] = [d 1 w] for some w X 2 with the property that d 0 w = x and d 2 w = x. Moreover the map : P(X) P( X ) is an isomorphism of groupoids. Proof. We only need to show that f x f x rel [1] if and only if x x. In geometry, one can easily see that as in the picture: ( 1 1 )/(1 1 ) = 2 X' 0 2 X' x 1 x 1 But in simplicial situation, ( [1] [1])/(1 [1]) = [2] and so this statement is less obvious in simplicial situation. We will use simplicial extension theorem to give a proof. Suppose that f x f x rel [1]. Define a simplicial map by setting f : [2] X f d 2 [1] = f x, f d 1 [1] = f x, f d0 [1] = ɛ d0x. Since f x f x rel [1], there is a continuous map φ: [2] X such that φ [2] = f from the above picture. By simplicial extension theorem, f can be extended to a simplicial map g : [2] X. Let w = g(σ 2 ). Then d 2 w = x, d 1 w = x and d 0 w = s 0 d 0 x and so x x.

67 5. HOMOTOPY GROUPS 67 Conversely suppose that x x. Define a simplicial map f : (I I) = ( I I) (I I) X by setting f I 0 = f x, f I 1 = f x, f 0 I = ɛ d1x, f 1 I = ɛ d0x. From the above picture, f : (I I) = ( I I) X can be extended to a continuous map from I I to X. Thus, by simplicial extension theorem, f can be extended to a simplicial map F : I I X which is a homotopy from f x to f x relative to I. Exercise 5.1. Let X be a fibrant simplicial set and let x, x X 1. Show that x x if and only if d 0 x = d 0 x, d 1 x = d 1 x and there exists w X 2 such that d 0 w = x, d 1 w = x and d 2 w = ɛ d1x Fundamental Groups. Whence the path product is decided, the fundamental group will follow. So we have the following: Theorem 5.6. Let X be a pointed fibrant simplicial set. Then the fundamental group π 1 (X) is the quotient set of the spherical elements in X 1 subject to the relation generated by x x if there exists w X 2 such that d 2 w = x, d 1 w = x and d 0 w =. The product structure in π 1 (X) is given by: [x] [x ] = [d 1 w], where w is any element in X 2 with the property that d 0 w = x and d 2 w = x. Moreover the map : π 1 (X) π 1 ( X ) preserves the product structure Higher Homotopy Groups. The ideas of higher homotopy groups follow the same lines and so we have the following: Theorem 5.7. Let X be a pointed fibrant simplicial set. Then the homotopy group π n (X) is the quotient set of the spherical elements in X n subject to the relation generated by x x if there exists w X n+1 such that d 0 w = x, d 1 w = x and d j w = for j > 1. The product structure in π n (X) is given by: [x] + [x ] = [d 1 w], where w is any element in X n+1 with the property that d 0 w = x, d 2 w = x and d j w = for j > 2. Moreover the map : π n (X) π n ( X ) preserves the product structure. Proposition 5.8. Let X be a pointed fibrant simplicial set. Then π n (X) is commutative for n 2. By using the fact that π n ( X ) is commutative, the proof is straightforward as π n (X) = π n ( X ). A combinatorial proof is given as follows. It will take several steps: For being careful in computation, we write [x][x ] for the product instead of [x] + [x ]. The definition of product is equivalent to say that: Let w X n+1 such that all d i w are spherical elements and d j w = for j > 2. Then [d 1 w] = [d 2 w][d 0 w].

68 68 2. SIMPLICIAL SETS AND HOMOTOPY Recall that the simplicial identity for faces is d j d i = d i d j+1 for j i. We can consider this relation as the symmetric matrix below the first row given by: d 0 d 1 d 2 d 3 d 4 d 0 d 1 d 0 d 2 d 0 d 3 d 0 d 4 d 0 d 0 d 1 d 2 d 1 d 3 d 1 d 4 d 1 d 0 d 1 d 1 d 2 d 3 d 2 d 4 d 2 d 0 d 2 d 1 d 2 d 2 d 3 d 4 d 3 d 0 d 3 d 1 d 3 d 2 d 3 d 3 For checking whether the elements x 0, x 1,... are matching faces, we can then look at whether the following matrix below first row is symmetric x 0 x 1 x 2 x 3 x 4 d 0 x 1 d 0 x 2 d 0 x 3 d 0 x 4 d 0 x 0 d 1 x 2 d 1 x 3 d 1 x 4 d 1 x 0 d 1 x 1 d 2 x 3 d 2 x 4. d 2 x 0 d 2 x 1 d 2 x 2 d 3 x 4 d 3 x 0 d 3 x 1 d 3 x 2 d 3 x 3 Proof. Step 1. Suppose that w X n+1 such that all faces d i w are spherical elements and d i w = for i 1, 2, 3. Then [d 2 w] = [d 1 w][d 3 w]. Let x 0 = s 2 d 3 w, x 1 = w, x 2 = s 1 d 1 w, x 4 = s 0 d 3 w and x j = for j > 4. We check that the elements x i are matching faces with respect to 3. This is done by the matrix: x 0 x 1 x 2 x 3 x 4 x 5 d 3 w d 1 w d 3 w d 1 w d 3 w d 2 w d 1 w. d 3 w d 3 w Thus there exists an element u X n+2 such that d j u = x j for j 3. Now and so [d 2 w] = [d 1 w][d 3 w]. d 0 (d 3 u) = d 2 d 0 u = d 2 x 0 = d 3 w d 1 (d 3 u) = d 2 d 1 u = d 2 x 1 = d 2 w d 2 (d 3 u) = d 2 d 2 u = d 2 x 2 = d 1 w Step 2. Suppose that w X n+1 such that all faces d i w are spherical elements and d i w = for i 0, 2, 3. Then [d 0 w][d 2 w] = [d 3 w]. Since a 0 = d 0 w, a 1 =, a 3 =, and a j = for j 4 are matching faces with respect to 2, there exists x 1 X n+1 such that d 0 x 1 = d 0 w, d 1 x 1 =, and d j x 1 = for j 3. Let t = d 2 x 1. 1 = [d 2 x 1 ][d 0 w].

69 5. HOMOTOPY GROUPS 69 In addition to x 1, let x 0 = s 0 d 0 w, x 2 = w, x 4 = s 2 d 3 w and x j = for j 5. From the table x 0 x 1 x 2 x 3 x 4 x 5 d 0 w d 0 w d 0 w d 0 w d 3 w d 2 x 1 d 2 w d 3 w, d 3 w the elements x j are matching faces and there exists u X n+2 such that d j u = x j for j 3. From d 0 (d 3 u) = d 2 d 0 u = d 2 x 0 = d 1 (d 3 u) = d 2 d 1 u = d 2 x 1 d 2 (d 3 u) = d 2 d 2 u = d 2 x 2 = d 2 w d 3 (d 3 u) = d 3 d 4 u = d 3 x 4 = d 3 w, we have [d 2 w] = [d 2 x 1 ][d 3 w] = [d 0 w] 1 [d 3 w] and hence the statement. Step 3. Suppose that w X n+1 such that all faces d i w are spherical elements and d i w = for i 0, 1, 2, 3. Then [d 1 w][d 3 w] = [d 0 w][d 2 w]. Let x 4 = w. Since a 1 =, a 2 =, a 3 = d 0 w and a j = for j > 3 are matching faces with respect to 0, there exists x 0 such that d j x 0 = a j for j > 0. According to Step 2, [d 0 x 0 ] = [d 3 x 0 ] = [d 0 w]. Since b 0 = d 0 x 0, b 1 =, b 3 = d 1 w and b j = for j > 3 are matching faces with respect to 1, there exits x 1 X n+1 such that d j x 1 = b j for j 2. From Step 2, [d 0 x 1 ][d 2 x 1 ] = [d 3 x 1 ], that is [d 0 x 0 ][d 2 x 1 ] = [d 1 w] or [d 0 w][d 2 x 1 ] = [d 1 w]. Let x 2 = s 2 d 2 w and x j = for j > 4. The elements x 0, x 1, x 2, x 4,... are matching faces by the table x 0 x 1 x 2 x 3 x 4 x 5 d 0 x 0 d 0 w d 0 x 0 d 1 w d 2 w d 2 x 1 d 2 w d 3 w d 0 w d 1 w d 2 w It follows that there exists an element u X n+2 such that d j u = x j for j 3. From d 0 (d 3 u) = d 2 d 0 u = d 2 x 0 = d 1 (d 3 u) = d 2 d 1 u = d 2 x 1 d 2 (d 3 u) = d 2 d 2 u = d 2 x 2 = d 2 w d 3 (d 3 u) = d 3 d 4 u = d 3 x 4 = d 3 w, we have [d 2 w] = [d 2 x 1 ][d 3 w] = [d 0 w] 1 [d 1 w][d 3 w] and hence the statement. Final Step. Let w X n+1 such that all faces d i w are spherical elements and d i w = for i 0, 1, 2,. From Step 3, since [d 3 w] = 1, [d 1 w] = [d 1 w][d 3 w] = [d 0 w][d 2 w].

70 70 2. SIMPLICIAL SETS AND HOMOTOPY and by definition [d 1 w] = [d 2 w][d 0 w]. It follows that [d 0 w][d 2 w] = [d 2 w][d 0 w]. The proof is finished now Homotopy Addition Theorem. Lemma 5.9. Let X be a pointed fibrant simplicial set and let x, x, x be spherical elements in X n with n 2. Suppose that there exists y X n+1 such that d q y = x, d q+1 y = x, d q+2 y = x, and d j y = for j q, q + 1, q + 2. Then in π n (X). [x] [x ] + [x ] = 0 Proof. The proof is given by induction on q. When q = 0, this is the definition of the product in π n (X). For q > 0, we use the extension condition on X to move the relation one step down. Consider the elements x i = for i < q 1, x q 1 = s q+1 d q+2 y, x q = y, x q+1 = s q d q y, x q+3 = s q 1 d q+2 y and x i = for i > q + 3. These elements have matching faces by the table x q 2 x q 1 x q x q+1 x q+2 x q+3 x q+4 d q 1 x q = d q+2 y d q 1 x q 1 = d q y d q+2 y d q y = x d q+2 y d q+1 y d q y d q+2 y d q+2 y So there exists w X n+2 such that d i w = x i for i q + 2. Let z = d q+2 w. Then d q+1 d j w = d q+1 x j = if j < q 1 d q+1 d q 1 w = d q+1 x q 1 = d q+2 y = x if j = q 1 d d j z = d j d q+2 w = q+1 d q w = d q+1 x q = d q+1 y = x if j = q d q+1 d q+1 w = d q+1 x q+1 = d q y = x if j = q + 1 d q+2 d q+3 w = d q+2 x q+3 = if j = q + 2 d q+2 d j+1 w = d q+2 x j+1 = if j > q + 2 By induction, we have and hence the result. [x ] [x ] + [x] = 0 Lemma Let X be a pointed fibrant simplicial set and let x X n+1 with d i x = x i for 0 i n + 1 such that all x i are spherical. Let w K n+1 with d i w = x i for q i, i + 1, d q w = and d q+1 w = z for some 0 q n 1. Then in π n (X) [x q ] [x q+1 ] + [z] = 0..

71 5. HOMOTOPY GROUPS 71 Proof. We check that the elements v i = s q+1 x i for i q 1, v q = s q x q, v q+2 = x, v q+3 = w and v i = s q+2 x i 1 for i q + 4 are matching faces with respect to q + 1. Note that each x i is spherical. We have d j v i = if 0 i q 1 and 0 j q 0 i q 1 and j > q + 2 i = q and 0 j q 1 i = q and j > q + 1 i q + 4 and j q + 1 i q + 4 and j > q + 3 x i if 0 i q 1 and j = q + 1, q + 2 x q if i = q and j = q, q + 1 x j if i = q + 2 d j w if i = q + 3 x i 1 if i q + 4 and j = q + 2, q + 3. Note that d i w = x i for q i, i + 1 and d q w =. This gives the table v 0 v q 1 v q v q+1 v q+2 = x v q+3 = w v q+4 v n+2 x 0 x 0 x q 1 x q 1 x q x q x q+1 d q+1 w x 0 x q 1 x q x q+2 x q+3 x. n+1 x 0 x q 1 x q+2 x q+3 x n+1 x q+3 x q+3 x q+4 x q+4 x n+1 x n+1 Thus there exists u K n+2 such that d i u = v i for i q + 1. Now d q d i u = d q v i = if i < q d q d q u = d q v q = x q if i = q d i (d q+1 u) = d q+1 d q+2 u = d q+1 v q+2 = x q+1 if i = q + 1 d q+1 d q+3 u = d q+1 v q+3 = d q+1 w = z if i = q + 2 d q+1 d i+1 u = d q+1 v i+1 = if i > q + 2. Thus [x q ] [x q+1 ] + [z] = 0 which is the assertion. Theorem 5.11 (Homotopy Addition Theorem). Let X be a pointed fibrant simplicial set. Let y i X n be spherical elements for 0 i n + 1 with n 2. Then in π n (X) the equation [y 0 ] [y 1 ] + [y 2 ] + ( 1) n+1 [y n+1 ] = 0 holds if and only if there exists y K n+1 such that d i y = y i for 0 i n + 1. Note. If n = 1, Homotopy Addition Theorem holds by definition of the product in the fundamental group. Proof. = Let y X n+1 such that d i y = y i is spherical for each 0 i n+1. For each 0 q n 1, since the spherical elements are always matching faces as

72 72 2. SIMPLICIAL SETS AND HOMOTOPY their faces are all trivial, there exists w q such that { if 0 i q d i w q = y i if i q + 2 Let z q = d q+1 w q. By applying Lemma 5.10 to y and w 0 for q = 0, we have [y 0 ] [y 1 ] + [z 0 ] = 0. For q > 0, note that d j w q = d j w q 1 for j q, q + 1, d q w q = and d q+1 w q = z q, d q w q 1 = z q 1 and d q+1 w q = y q+1, by Lemma 5.10, [z q 1 ] [y q+1 ] + z q = 0 for 1 q n 1. This gives that equations and so Since [y 0 ] [y 1 ] + [z 0 ] = 0 [z 0 ] + [y 2 ] [z 1 ] = 0 [z 1 ] [y 3 ] + [z 2 ] = 0 ( 1) n 1 [z n 2 ] + ( 1) n [y n ] + ( 1) n+1 [z n 1 ] = 0 ( 1) n+1 [z n 1 ] + n ( 1) i [y i ] = 0. i=0 if j < n 1 if j = n 1 d j w n 1 = d n w n 1 = z n 1 if j = n y n+1 if j = n + 1, we have [z n 1 ] + [y n+1 ] = 0. Thus n+1 ( 1) i [y i ] = 0. i=0 = Let y i X n be spherical elements for 0 i n + 1 such that [y 0 ] [y 1 ] + [y 2 ] + ( 1) n+1 [y n+1 ] = 0. Since y 1,..., y n+1 are matching faces with respect to 0, there exists w K n+1 such that d i w = y i for i 0. Note that d 0 w is spherical as d i d 0 w = d 0 d i+1 w =. From the previous step, we have n+1 ( 1) i [d i w] = 0 i=0 and so [d 0 w] = [y 0 ]. Let f : [n + 1] X be the simplicial map such that f d0 [n] = f d0w, f d1 [n] = f y0 and f di [n] = ɛ for i > 1. We check that the geometric realization f : [n + 1] = n+1 X can be extended to a continuous map g : [n + 1] = n+1 X. Let F : I [n] pinch I S n = I ( [n]/ [n]) F X, where F is a pointed homotopy between f d0w and f y0, that is, F 0 S n = f d0 w, F 1 S n = f y0 and F I = ɛ. Consider { } n+2 n+1 = (t 0, t 1,..., t n+2 ) R n+2 t i 0 t i = 1 i=0

73 5. HOMOTOPY GROUPS 73 and I n = { (s, t 1, t 2,..., t n+2 ) R n+2 t i 0 } n+1 t i = 1 0 s 1. i=0 Observe that the first two faces of n+1 are the subspaces d 0 [n] = {(0, t 1, t 2,..., t n+2 ) R n+2 } n+1, d 1 [n] = {(t 0, 0, t 2,..., t n+2 ) R n+2 } n+1. Define the continuous map θ : I n n+1 by setting θ(s, t 1, t 2,..., t n+2 ) = (st 1, (1 s)t 1, t 2,..., t n+2 ). Then θ is onto and θ 1 (d 0 n ) (0 n ) (I d 0 n 1 ) θ 1 (d 1 n ) = (1 n ) (I d 0 n 1 ) θ 1 (d i n ) = I d i 1 n 1 for i 2 Define a function g : n+1 X by setting g(x) = F (θ 1 (x)) for x n+1. We check that g is well-defined with g n+1 = f. Case I. x n+1 n+1. Then the pre-image of θ has exactly one point and g(x) is well-defined in this case. Case II. x d i n for i 2. Then θ 1 (x) I n. Since F I n =, g(x) = = f (x). Case III. x d 0 n. Consider the equation θ(s, t 1, t 2,..., t n+2) = (st 1, (1 s)t 1, t 2,..., t n+2) = x = (0, t 1, t 2,..., t n+2 ). If t i > 0 for all 1 i n + 2, then s = 0 and t i = t i. In this case, g(x) = F (0, t 1,..., t n+2 ) = f d0w (x) = f (x). If one of t i is zero, then θ 1 (x) I n. In this case, F (θ 1 (x)) = = f (x). Case IV. x d 1 n. Consider the equation θ(s, t 0, t 1,..., t n+1) = (st 0, (1 s)t 0, t 1,..., t n+1) = x = (t 0, 0, t 2,..., t n+2 ). If t i > 0 for all i 1, then s = 1, t i = t i for 0 i n + 1. In this case, g(x) = F (1, t 0, t 2..., t n+1 ) = f y0 (x) = f (x). If one of t i is zero, then θ 1 (x) I n. In this case, F (θ 1 (x)) = = f (x). Hence g is well-defined. To show that g is continuous, let A be a closed set of X. Then F 1 (A) is closed in I n. Since I n is Hausdorff and compact, F 1 (A) is compact and so g 1 (A) = θ( F 1 (A)) is compact. It follows that g 1 (A) is closed and hence g is continuous.

74 74 2. SIMPLICIAL SETS AND HOMOTOPY Now by the simplicial extension theorem, the simplicial map f : [n+1] X can be extended to a simplicial map f : [n + 1] X. Let z = f (σ n+1 ), where σ n+1 is the nondegenerate element in [n + 1]. Then d 0 z = d 0 w, d 1 z = y 0 and d j z = for j 2. Let v 0 = z, v 1 = w and v i = s 1 d i 1 w = s 1 y i 1 for i 3. These elements are matching faces by the table v 0 = z v 1 = w v 2 v 3 = s 1 d 2 w s 1 d 3 w s 1 d 4 w s 1 d n+1 w d 0 w d 0 w y 2 y 3 y 4 y n+1 y 0 y 1 y 2 y 3 y 4 y n+1 y 2 y 3 y 4 y n+1 Thus there exists u such that d i u = v i for i 2. Now d 1 v 0 = y 0 if i = 0 d i (d 2 u) = d 1 v 1 = y 2 if i = 1 d 2 v i+1 = d 2 s 1 y i = y i if i 2 The proof is finished now A Geometric Proof of Homotopy Addition Theorem. In this independent subsection, you are assumed to know some basic knowledge on algebraic topology, such as Allen Hatcher s book [17]. Observe that the boundary [n + 1] = n+1 d i [n]. Thus i=0 n+1 [n + 1]/ sk n 1 [n + 1] = (d i [n]/d i [n]) = i=0 Let X be a pointed simplicial set and let y 0, y 1,..., y n+1 be spherical elements in X n. Let f yi : S n X be the representing map for y i. Define the simplicial map g to be the composite [n + 1] This defines a continuous map We are going to show that q [n + 1]/ skn 1 [n + 1] = g : [n + 1] = S n n+1 [ g ] = ( 1) i [ f yi ]. i=0 n+1 i=0 X. S n n+1 i=0 n+1 i=0 S n. f yi X. From this, we will give a new proof for the Homotopy Addition Theorem. First we need to handle the homotopy class of the pinch map n+1 q : [n + 1] = n+1 n+1 S n = S n. i=0 i=0.

75 5. HOMOTOPY GROUPS 75 Recall that n = {(t 0, t 1,..., t n ) R n+1 t i 0 n t i = 1}. The symmetric group S n+1 acts on n is given by permuting coordinates. Let τ i,j : n n be the map by switching coordinates i and j for 0 i < j n, that is τ i,j (t 0, t 1,..., t n ) = (t 0, t 1,..., t i 1, t j, t i+1,..., t j 1, t i, t j+1,..., t n ). Then the boundary n is invariant under the map τ i,j. Note that n = S n 1 and so any self-map of n has a degree, see Hatcher s book [17, Section 2.2]. Lemma The map τ i,j : n n is of degree 1 for 0 i < j n. Proof. Let i=0 V i,j = {(t 0, t 1,..., t n ) R n+1 t i = t j } be the n-dimensional hyperspace R n+1 with the coordinates t i = t j. consisting of the points in which the i-coordinate is the same as the j-th coordinate. Then the action of τ i,j is the reflection on n with respect to the (n 1)-dimensional sphere given by n V i,j. By [17, p.134, Section 2.2, Part (e)], the map τ i,j is of degree 1. Exercise 5.2. Let S n+1 act on n by permuting coordinates. Show that the map σ : n n is of degree sign(σ) for each σ S n+1, where sign(σ) is the sign of the permutation σ, that is, sign(σ) = +1 if σ is an even permutation and sign(σ) = 1 if σ is an odd permutation. For each σ S n+1, since σ( n ) = n, the action of σ on n induces an action σ : n / n n / n. Note that n / n = S n. So any self-map of n / n has a degree. Lemma For each 0 i < j n, the map τ i,j : n / n n / n is of degree 1. Proof. Again let V i,j = {(t 0, t 1,..., t n ) R n+1 t i = t j }. Then τ i,j on n / n is the reflection with respect to the (n 1)-sphere given by (V i,j n )/(V i,j n ). Thus the degree of τ i,j is 1. Lemma Let n 2. Then the homotopy group ( q ) q π n S n = Z. i=1 Proof. This is a direct consequence of the Hurewicz Theorem [17, p.366, Theorem 4.32]: Since π 1 ( q i=1 Sn ) = 0, H j ( q i=1 Sn ) = 0 for 0 < j < n, ( q ) ( q ) q π n S n = H n S n = Z by the Hurewicz Theorem. i=1 i=1 i=1 i=1

76 76 2. SIMPLICIAL SETS AND HOMOTOPY By using this fact, we can introduce the concept of multi-degree for the maps from S n to q i=1 Sn. Let f : S n q i=1 Sn be a pointed map. Let l j be the degree of the composite q S n f S n π j S n, i=1 where π j is the projection to the j-th copy of S n for 1 j q. Then we call that f has the multi-degree (l 1,..., l q ). From the above lemma, any two (pointed) maps are homotopic if they have the same multi-degree. Lemma Let n 2. Then the pinch map n+1 q : [n + 1] = n+1 S n = i=0 n+1 is of multi-degree ɛ(1, 1, 1, 1,..., ( 1) n+1 ) for some ɛ = ±1. Proof. Let q has multi-degree (l 0, l 1,..., l n+1 ). Observe that the composite is the pinch map from n+1 q n+1 i=0 S n π j S n n+1 n+1 / Λ j [n + 1]. Since Λ j [n + 1] is contractible, the composite π j q is a homotopy equivalence. It follows that the degree n j = ±1. Now we consider the symmetric group action. Observe that there is a commutative diagram (5.2) n+1 n+1 τ j,j+1 for 0 j n. Recall that q n+1 S n = i=0 q n+1 S n = i=0 n+1 i=0 n+1 i=0 i=0 d i n /d i ( n ) τ j,j+1 d i n /d i ( n ) d i n = {(t 0, t 1,..., t i 1, 0, t i+1,..., t n+1 ) R n+2 } n+1. If i j, j + 1, then τ j,j+1 maps d i n onto itself by switching the coordinates t j and t j+1. By Lemma 5.12, the map τ j,j+1 di n : di n d i n is of degree 1. If i = j or j + 1, then τ j,j+1 switches the faces d j n and d j+1 n by keeping the same order of coordinates. It follows that the degree (5.3) deg(τ j,j+1 q ) = ( l 0, l 1,..., l j 1, l j+1, l j, l j+1, l j+2,..., l n+1 ). On the other hand, since τ j,j+1 : n+1 n+1 is of degree 1, (5.4) deg( q τ i,j ) = ( l 0, l 1,..., l n+1 ). S n

77 5. HOMOTOPY GROUPS 77 From Equations (5.2), (5.3) and (5.4), we obtain ( l 0, l 1,..., l j 1, l j+1, l j, l j+1, l j+2,..., l n+1 ) = ( l 0, l 1,..., l n+1 ) and so l j+1 = l j for each 0 j n. It follows that the map q has the multidegree ɛ(1, 1,..., ( 1) n+1 ) for some ɛ = ±1. Now we need to review how to define the product on the geometric homotopy groups π n ( X ). Consider S n as the subspace of unit vectors in R n+1. Let µ : S n S n S n be the map by pinching equator S n 1 = S n R n to a point, where the basepoint is chosen on the equator. Let f, g : S n X be pointed continuous maps. Then the composite S n µ S n S n f g X is defined as the product of f and g. The map µ has the property that the composite S n µ S n S n π j S n is of degree 1. In other words, µ has multi-degree (1, 1). Now given pointed continuous maps f 1,..., f q : S n X, the product f f q is given by the composite S n µ q i=1 where µ is of multi-degree (1, 1,..., 1). S n q i=1 fi X, Proposition Let f i : S n X be pointed map for 0 i n + 1 with n 2 and let g be the composite n+1 Then there exists ɛ = ±1 such that in π n ( X ). q n+1 ɛ n+1 i=0 S n n+1 i=0 fi X. ( 1) i [f i ] = [g] i=0 Proof. Since q has multi-degree ɛ(1, 1, 1, 1,..., ( 1) n+1 ), there is a commutative diagram up to homotopy n+1 n+1 q µ S n n+1 i=0 f i X i=0 n+1 θ i i=0 n+1 i=0 S n, where θ i : S n S n is of degree ɛ( 1) i. The assertion follows.

78 78 2. SIMPLICIAL SETS AND HOMOTOPY Another Proof of the Homotopy Addition Theorem. Let y 0, y 1,..., y n+1 be spherical elements in X n with n 2. Suppose that n+1 ( 1) i [y i ] = 0 in π n (X). Then i=0 n+1 ( 1) i [ f yi ] = 0 in π n ( X ). Define the simplicial map g to be the composite (5.5) [n + 1] By Proposition 5.16, in π n ( X ), i=0 q [n + 1]/ skn 1 [n + 1] = n+1 [ g ] = ɛ ( 1) i [ f yi ] = 0 i=0 n+1 i=0 S n n+1 i=0 f yi X. and so g : [n+1] X can be extended to a continuous map [n+1] X. By simplicial extension theorem, g can be extended to a simplicial map g : [n + 1] X. Let w = g (σ n+1 ), where σ n+1 is the nondegenerate element of [n + 1]. Since g d i [n] = f yi, we have d i w = d i g (σ n+1 ) = g (d i σ n+1 ) = f yi (d i σ n+1 ) = y i for each 0 i n + 1. Conversely there exists w X n+1 such that d i w = y i for each 0 i n + 1. Let g be defined as in Equation (5.5). Then f w [n+1] = g and so the map g : [n + 1] X is null-homotopic, that is [g] = 0 in π n (X). By Proposition 5.16, we have and hence the result. n+1 n+1 ( 1) i [y i ] = ( 1) i [ f yi ] = ɛ[ g ] = 0 i=0 i= Minimal Simplicial Sets. Given a space X, we can have fibrant simplicial set S (X). But S (X) seems too big as there are uncountable many elements in each S n (X). On the other hand, for having simplicial homotopy groups, we need fibrant assumption. This means that simplicial model for a given space can not be too small. We hope to have a kind of smallest fibrant simplicial sets. This will be the concept of minimal simplicial sets discussed below. First we define the homotopy for the elements in X n. Definition Let X be a fibrant simplicial set. Then for x, y X n call x y if the representing maps f x and f y are homotopic relative to [n]. Proposition Let X be a fibrant simplicial set and let x, y X n. Then x y if and only if d i x = d i y for all 0 i n and there exists w X n+1 such that d k w = x, d k+1 w = y, and d i w = d i s k x = d i s k y for i k, k + 1.

79 5. HOMOTOPY GROUPS 79 Proof. Let x, y X n be the elements such that d i x = d i y for all 0 i n. Consider the simplicial map : [n + 1] s k [n] f x X. Let σ n+1 be the nondegenerate element in [n + 1] n+1. Since f x s k (σ n+1 ) = f x (s k σ n ) = s k (f x (σ n )) = s k x, we have f x s k (d i σ n+1 ) = d i f x s k (σ n+1 ) = d i s k x for all 0 i n. By restricting to Λ k+1 [n + 1], f x s k Λ k+1 [n+1] : Λ k+1 [n + 1] X is the simplicial map such that f x s k d i [n] = f dis k x for i k + 1. Since [n + 1] = Λ k+1 [n + 1] d k+1 [n], we fill in the map on the missing face d k+1 [n] by f y, more precisely, let (5.6) f = (f x s k ) f y : [n + 1] = Λ k+1 [n + 1] d k+1 [n] X. To see f is well-defined, note that Λ k+1 [n + 1] d k+1 [n] = d k+1 [n]. The map f x s k on d k+1 [n] is given by f dk+1 s k x [n] = f x [n] Since d j x = d j y for all 0 j n, we have f x [n] = f y [n] and so the simplicial map f is well-defined. Observe that: There exists w X n+1 such that d k w = d k s k x = x, d k+1 w = y, and d i w = d i s k x = d i s k y for i k, k + 1 if and only if f can be extended to a simplicial map f w : [n + 1] X, if and only if the geometric realization f : [n + 1] = n+1 X can be extended to a continuous map n+1 X [ f ] = 0 in π n ( X ). In the above statement, the first equivalence follows from the construction of the map f and the fact that simplicial maps [n+1] X can be identified with the representing maps, and the second equivalence follows from the simplicial extension theorem. Now construct the second simplicial map (5.7) f : (I [n]) X by requiring that f 0 [n] = f x, f 1 [n] = f y and f I [n] (s, z) = f x (z). Observe that: f x f y rel [n] if and only if f can be extended to a simplicial map I [n] X, if and only if the geometric realization f : (I [n]) = (I n ) X can be extended to a continuous map I n X [ f ] = 0 in π n ( X ).

80 80 2. SIMPLICIAL SETS AND HOMOTOPY In the above statement, the first equivalence follows from the definition of simplicial homotopy, and the second equivalence follows from simplicial extension theorem. For having the connections between the above two statements, we need construct the third simplicial map given by defining (5.8) f = f x f y : [n] [n] [n]: X, where [n] [n] [n] is the union of two copies of [n] by identifying the elements in the boundary [n]. Note that the geometric realization Note that in geometry [n] [n] [n] = n n n = S n. n n n = (I n )/, where is generated by (0, z) (t, z) for z n and 0 t 1. Let q 1 : (I n ) n n n be the quotient map. Clearly q 1 is a homotopy equivalence. Since f (t, z) = f x (z) for (t, z) I [n], we have f = f q 1. Let q 2 = s k id: S n = n+1 = Λ k+1 [n + 1] d k+1 n n n n = S n. From the commutative diagram n+1 = Λ k+1 [n + 1] d k+1 n p 1 n+1 / Λ k+1 [n + 1] = d k+1 n /d k+1 n q 2 = s k id n n n p 2 n n n / n = n / n, where p 1 is given by pinching Λ k+1 [n+1] to be the point and p 2 pinches the left n to the point. Since Λ k+1 [n + 1] and n are contractible, p 1 and p 2 are homotopy equivalences. Thus q 2 is a homotopy equivalence. Since f = (f x s k ) f y, we have f = f q 2. Now from the commutative diagram n+1 = Λ k+1 [n + 1] d k+1 n q 2 f n n n f X q 1 f (I n ), one of f, f, f is null homotopic implies that the other two are also null homotopic because q 1 and q 2 induce isomorphisms on π n ( X ). The assertion follows.

81 5. HOMOTOPY GROUPS 81 Definition A fibrant simplicial set is called minimal if it has the property that x y = x = y. Proposition Let X be a fibrant simplicial set. Then X is minimal if and only if, for any 0 k n + 1, whenever v, w X n+1 such that d i v = d i w for all i k, then d k v = d k w. In other words, a fibrant simplicial set is minimal if and only if for any two elements with all faces but one the same, then the missed face must be the same. Proof. = Let v, w X n+1 such that d i v = d i w for all i k. Note that Λ k [n + 1] = d k [n]. Since f v Λk [n+1] f dk v : Λ k [n + 1] d k [n] = [n + 1] X can be extended to the continuous map f v : [n + 1] X, we have f v Λk [n+1] f dk v rel n. Similarly f w Λ k [n+1] f dk w rel n. Since d i v = d i w for i k, f v Λ k [n+1] = f w Λ k [n+1] and so f v Λ k [n+1] = f w Λ k [n+1]. It follows that f dk v f dk w rel n and so f dk v f dk w rel [n], that is d k v d k w. Hence d k v = d k w by the minimal assumption. =. Let x, y X n such that x y. By Proposition 5.18, there exists w X n+1 such that d i w = d i s 0 x for i 1 and d 1 w = y. Let v = s 0 x. Then d i v = d i w for i 1. By the assumption, we have d 1 v = d 1 w and so x = y. Lemma Let X be a fibrant simplicial set. Then (1). Suppose that x X n and v X n 1 such that d k x v. Then there exists z X n such that d i z = d i x for i k and d k z = v. (2). Let F : [n] I X be a simplicial map and let x = F (σ n, 1), where σ n is the nondegenerate element in [n] n. Suppose that x y. Then there exists a simplicial map G: [n] I X such that G(σ n, 1) = y and Proof. (1) Since the map G ( [0] 0) ( ( [n]) I) = F ( [0] 0) ( ( [n]) I). f x Λ k [n] f dk x f v rel n, f x Λk [n] f v : n = Λ k [n] d k [n] X can be extended to a continuous map [n] X and so f x Λk [n] f v : [n] = Λ k [n] d k [n] X can be extended to a simplicial map g : [n] X. Let z = g(σ n ). Then d i z = d i x for i k and d k z = v. (2) Since F ( [0] 0) ( ( [n]) I) f x : ( [0] 0) ( ( [n]) I) [n] 1 X can be extended to the continuous map F : [n] I X, F ( [0] 0) ( ( [n]) I) f x f y rel n.

82 82 2. SIMPLICIAL SETS AND HOMOTOPY Thus F ( [0] 0) ( ( [n]) I) f y : ( [0] 0) ( ( [n]) I) [n] 1 X can be extended to a continuous map [n] I X and so F ( [0] 0) ( ( [n]) I) f y : ( [0] 0) ( ( [n]) I) ( [n] 1) X can be extended to a simplicial map G: [n] I X which has the desired property. Definition Let X a simplicial set. A simplicial subset A is called a strong deformation retract of X if there exits a homotopy F : X I X such that (1). F (a, t) = a for a A and t I, that is F A I is the composite A I proj. A X; (2). F X 0 = id X ; (3). F (X 1) A. A minimal simplicial set A is called a minimal subcomplex of X if A is a strong deformation retract of X. Theorem 5.23 (Existence of Minimal Subcomplex). Any fibrant simplicial set X contains a minimal subcomplex A. Proof. Let A 0 consist of a choice of one vertex in each homotopy class. Suppose that A j is defined for all 0 j n 1. Let A n consist of a choice of one simplex for each homotopy class [x] in X n of simplices all of whose faces are in A n 1, with requiring that if s i a [x] for some a A n 1 with 0 i n 1, then choose s i a in A n. 1. A n is well-defined: That is s i a s k a = s i a = s k a for a, a A n 1. If i = k, then a = d i s i a = d i s i a = a, because if x y then d j x = d j y for all j and so s i a = s k a when i = k. If i < k, then a = d i s i a = d i s k a = s k 1 d i a. Then It follows that a = d k s k a = d k s i a = d k s i s k 1 d i a = s i d k 1 s k 1 d i a = s i d i a. s i a = s i s k 1 d i a = s k s i d i a = s k a and so A n is well-defined. 2. A = {A n } n 0 is a simplicial subset of X. By the construction, d j (A n ) A n 1 and s j (A n+1 ) for 0 j n. Thus the sequence of subsets A = {A n } n 0 is a simplicial subset of X. 3. A is a strong deformation retract of X. Let sk A n (X) = A sk n (X). We are going to define inductively a homotopy F n : sk A n (X) I X satisfying the conditions: (1). F n A I (a, t) = a for a A and t I. (2). F n sk A n (X) 0 = id X sk A n (X). (3). F n sk A n 1 (X) I = F n 1. (4). F n (sk A n 1) A.

83 5. HOMOTOPY GROUPS 83 Note that every vertex x X 0 is homotopic to a vertex w A 0, that is, there is a path λ x : I X such that λ x (0) = x and λ x (1) A 0. Let p be the composite A I proj. A A. Let F 0 = p λ x : sk A 0 (X) I = (A I) I X. x X 0 A 0 x X 0 A 0 Then F 0 satisfies the required conditions. Suppose that F n 1 : sk A n 1(X) I X is defined satisfying the required conditions. Since F n 1 sk A n 1 (X) 0 = id X sk A n 1 (X), F n 1 = F n 1 (id X sk A n (X)): (sk A n 1(X) I) (sk A n (X) 0) X is a well-defined simplicial map. Let X D n Then there is a push-out diagram (5.9) ( [n] I) ( [n] 0) y X D n y X D n [n] I = {y X n A n y is nondegenerate}. y X D n f y (sk A n 1(X) I) (sk A n (X) 0) push j y X f n D y id I sk A n (X) I, where f y = (f y id I ) ( [n] I) ( [n] 0). Let y X n D. Since X is fibrant, there is an extension (sk A n 1(X) I) (sk A n 1 F n (X) 0) X f y F y ( [n] I) ( [n] 0) [n] I. Let x = F y (σ n, 1). Then all d j x A n 1 because F y restricted to [n] I is given by the composite F n 1 f y and F n 1 (sk A n 1(X) 1) A. Thus there exists an element z A n such that x z. According to Lemma 5.21, there exists a simplicial map G y : [n] I X such that G y (σ n, 1) = z A n and G y ( [n] I) ( [n] 0) = F y ( [n] I) ( [n] 0) and so we have the commutative diagram (sk A n 1(X) I) (sk A n (X) 0) f y F n 1 X G y ( [n] I) ( [n] 0) [n] I.

84 84 2. SIMPLICIAL SETS AND HOMOTOPY By the push-out diagram in Equation 5.9, the simplicial maps F n 1 and induces the simplicial map y X D n G y F n : sk A n (X) I X such that F n j = F n 1 and F n f y id I = G y. y X D n y X D n Thus F n satisfies Conditions (1)-(3). For each y X D n, since G y (σ n, 1) A n, we have G y ( [n] 1) A and so F n (sk A n (X) 1) A, that is F n satisfies Condition (4). The induction is finished. Now the simplicial map F = n F n : X I = n sk A n (X) I X makes A to be a strong deformation retract of X. 4. A is a minimal simplicial set. By Step 3, A is a retract of X and so A is also a fibrant simplicial set. Suppose that x, y A n with x y. Then x y in X n with all faces in A n 1. By the construction A n, x = y. Thus A is a minimal simplicial set. We finish the proof now. Let y 1, y 2 X n such that d j y 1 = d j y 2 for all 0 j n. Define θ y1,y 2 to be the composite θ y1,y 2 : S n θ [n] [n] [n] fy 1 fy 2 X, = where θ is given by identifying the upper hemisphere S n + with the left copy [n] = n and the lower hemisphere S n with the right copy of [n]. Lemma Let X be a fibrant simplicial set and let y 1, y 2, y 3 X n such that d j y 1 = d j y 2 = d j y 3 for all 0 j n. Then (1). If y 2 y 3, then θ y1,y 2 θ y1,y 3. (2). [θ y1,y 2 ] = [θ y2,y 1 ] in π n ( X ). (3). [θ y1,y 3 ] = [θ y1,y 2 ] + [θ y2,y 3 ] in π n ( X ). Proof. (1). Let F : [n] I X be a homotopy between f y2 and f y3 relative to [n]. Let G be the composite Then [n] I proj. f y1 [n] X. G F : ( [n] [n] [n]) I X is a homotopy between f y1 f y2 and f y1 f y3 and so θ y1,y 2 θ y1,y 3.

85 5. HOMOTOPY GROUPS 85 (2) Let τ be the self simplicial map of [n] [n] [n] by switching the left copy of [n] to the right copy of [n]. Then there is a commutative diagram S n θ [n] [n] [n] f y 1 f y2 X τ τ θ S n f y2 f y1 [n] [n] [n] X, where τ : S n S n is the reflection with respect to the equator S n 1 in S n. Since τ : S n S n is of degree 1, we have [θ y2,y 1 ] = [θ y1,y 2 ]. (3) Let n i be a copy of n for i = 1, 2, 3. Let Z be the push-out of the diagram n n 1 n 3 that is Z is obtained by gluing three copies of n together by identifying their boundaries with n 2 in the middle. There is a diagram n 2, n 1 n n φ 1 3 Z φ 2 D n S n 1 D n p ( n 1 n n 2 ) ( n 2 n n 3 ), where p is given by pinching the common boundary S n 1 to the point, φ 1 is induced by the inclusions from n 1 and n 2 into Z, and φ 2 is the map which folds the middle two copies of n 2 into one copy of n 2. Clearly this diagram is commutative up to homotopy, that is φ 1 φ 2 p, see the following picture: