Chapter 2 Homology. 2.1 Homological Algebra. f n 1 f 2

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1 Chapter 2 Homology 2.1 Homological Algebra Let F be a abelia group for each 0, ad let f : F F 1 be homomorphisms such that f f +1 = 0 (the trivial homomorphism) for all 0, with f 0 = 0. The the sequece F give as f +1 F f F 1 f 1 f 2 f 1 f 0 F1 F0 0 (1) is called a chai complex. I this geeral algebraic settig othig compels the idex to cosist strictly of the whole umbers, but uless otherwise specified we will always assume that a chai complex eds at idex value (or dimesio ) 0, with f 0 = 0. Thus, to say somethig holds for all i this cotext is iteded to mea for all 0. There will be times, i certai topological settigs, whe it will be coveiet to exted a chai complex out to dimesio = 1. From f f +1 = 0 we obtai Im f +1 Ker f F, so Im f +1 is a ormal subgroup of Ker f ad we ca meaigfully costruct a quotiet group H (F ) = Ker f / Im f +1, called the th homology group of F, for each 0. The elemets of H (F ) are cosets of the form x + Im f +1, usually called homology classes ad deoted by [x] whe it does ot lead to ambiguity. If x, y Ker f are such that [x] = [y], the x ad y are said to be homologous ad it follows that x y Im f +1. Suppose that the diagram f F +1 f +1 F F 1 ϕ +1 ϕ ϕ 1 g G +1 g +1 G G 1 has sequeces F ad G that are chai complexes, ad suppose also that the diagram is commutative give the maps ϕ : F G. The the maps ϕ take together defie a chai map F G, ad it s coveiet to deote the chai map by either ϕ : F G or {ϕ }. Now, if ψ : F G is aother chai map, ad if there also exist maps λ : F G +1 as i the diagram f F +1 f +1 F F 1 ψ +1 ϕ +1 λ ψ ϕ λ 1 ψ 1 ϕ 1 g G +1 g +1 G G 1 such that ϕ ψ = g +1 λ + λ 1 f

2 for all, the we call the collectio of maps {λ } a chai homotopy betwee the chai maps {ϕ } ad {ψ }, ad say that {ϕ } ad {ψ } are chai homotopic. Propositio 2.1. If {ϕ } is a chai map F G, the each map ϕ iduces a well-defied homomorphism ϕ : H (F ) H (G) give by for each x Ker f. ϕ (x + Im f +1 ) = ϕ (x) + Im g +1 Propositio 2.2. If {ϕ } is a chai map F G such that each map ϕ is a isomorphism, the each map ϕ is a isomorphism. Proof. Suppose that each ϕ : F G of the chai map F G is a isomorphism. Fix 0. Suppose that ϕ (x + Im f +1 ) = Im g +1. The ϕ (x) Im g +1, so there exists some y G +1 such that g +1 (y) = ϕ (x). Sice ϕ +1 is oto, there exists some z F +1 such that ϕ +1 (z) = y. Now, ϕ f +1 = g +1 ϕ +1, so ϕ (f +1 (z)) = g +1 (ϕ +1 (z)) = ϕ (x) ad the oe-to-oeess of ϕ implies that f +1 (z) = x. Hece x Im f +1, from which it follows that x + Im f +1 = Im f +1 ad therefore ϕ is oe-to-oe. Next, let y + Im g +1 H (G), so y Ker g. Sice y G ad ϕ is oto, there exists some x F such that ϕ (x) = y. If = 0 the x Ker f 0 also, so suppose that > 0. We obtai ϕ 1 (f (x)) = g (ϕ (x)) = g (y) = 0, whece the oe-to-oeess of ϕ 1 gives f (x) = 0 so that x Ker f. Thus, x + Im f +1 H (F ), ad ϕ (x + Im f +1 ) = ϕ (x) + Im g +1 = y + Im g +1 shows that ϕ is oto. Therefore ϕ is a isomorphism. Oe useful result that derives easily from a chai homotopy is the followig propositio, which will be used i later developmets. 2 Propositio 2.3. If {ϕ } ad {ψ } are chai-homotopic chai maps F G, the ϕ = ψ for all. Proof. Suppose that {ϕ } ad {ψ } are chai-homotopic maps. Fix ad x Ker f. For simplicity deote ϕ, ψ : F G by ϕ ad ψ, so ϕ, ψ : H (F ) H (G). It must be demostrated that ϕ (x + Im f +1 ) = ϕ(x) + Im g +1 = ψ(x) + Im g +1 = ψ (x + Im f +1 ), or equivaletly (ϕ(x) ψ(x)) + Im g +1 = Im g +1.

3 3 But by hypothesis there exist maps λ 1 ad λ such that so ϕ ψ = g +1 λ + λ 1 f, (ϕ ψ)(x) = g +1 (λ (x)) + λ 1 (f (x)) = g +1 (λ (x)) + λ 1 (0) = g +1 (λ (x)). Hece (ϕ ψ)(x) Im g +1, which implies that (ϕ ψ)(x) + Im g +1 = Im g +1 ad therefore ϕ (x + Im f +1 ) = ψ (x + Im f +1 ) as desired. I what follows, for ay chai complex F (see the top row of the diagram above) let F : F F be idetity maps, which take together form the chai map F : F F. Also let H (F ) be the idetity map o H (F ) for all. Propositio 2.4. Let θ : E F ad ϕ : F G be chai maps. The, for all, (1) F = H(F ) (2) (ϕ θ ) = ϕ θ Referrig to the diagram above, if there exists a chai map θ : G F such that {θ ϕ } is chai homotopic to { F } ad {ϕ θ } is chai homotopic to { G }, the {ϕ } is called a chai-homotopy equivalece. Propositio 2.5. If {ϕ } is a chai-homotopy equivalece, the ϕ is a isomorphism for all. Proof. Suppose that the chai map ϕ : F G is a chai-homotopy equivalece. The there exists a chai map θ : G F such that {θ ϕ } is chai homotopic to { F } ad {ϕ θ } is chai homotopic to { G }. By Propositio 2.3, (θ ϕ ) = F ad (ϕ θ ) = G, ad so by Propositio 2.4 we obtai θ ϕ = H(F ) ad ϕ θ = H(G). Therefore, by Propositio 1.1, ϕ is a isomorphism. The chai complex (1) is said to be a exact sequece if Im f +1 = Ker f for all, i which case H (F ) = 0. A exact sequece of the form 0 A i B j C 0 is called a short exact sequece. Let E, F, ad G be chai complexes i the mold of (1), ad let i : E F ad j : F G be chai maps such that i 0 E j F G 0 is a short exact sequece for every. The the diagram i Figure 1 is commutative ad is called a short exact sequece of chai complexes. By Propositio 2.1 there are well-defied homomorphisms i : H (E) H (F ) ad j : H (F ) H (G), ad what we re iterested

4 e E +1 e +1 E E 1 (E) i +1 i i 1 f F +1 f +1 F F 1 (F ) j +1 j j 1 g G +1 g +1 G G 1 (G) Figure 1 i doig is costructig homomorphisms ρ : H (G) H 1 (E) such that we obtai a log exact sequece of homology groups H (E) i H (F ) j H (G) ρ H 1 (E) i 1 H 1 (F ) Let [z] H (G), so z Ker g. Sice j is oto, z = j (y) for some y F. Now, f (y) F 1, ad sice j 1 (f (y)) = g (j (y)) = g (z) = 0 it follows that f (y) Ker j 1 = Im i 1 ad so there exists some x E 1 such that i 1 (x) = f (y). Sice i 2 (e 1 (x)) = f 1 (i 1 (x)) = f 1 (f (y)) = 0 (F is a chai complex so f 1 f = 0) ad i 2 is oe-to-oe, we coclude that e 1 (x) = 0 ad hece x represets a homology class [x] H 1 (E). We defie ρ ([z]) = [x]. Theorem 2.6. The sequece is exact. H (E) i H (F ) j H (G) ρ H 1 (E) i 1 H 1 (F ) We roud out the sectio with two last results that are purely a matter of homological algebra but will have wide utility whe dealig with topological matters. Lemma 2.7 (The Five-Lemma). I a commutative diagram of abelia groups as give below, if the two rows are exact ad the maps α, β, δ, ɛ are isomorphisms, the γ is also a isomorphism. i j k l A B C D E α β γ δ ɛ A i B j C k D l E

5 5 Lemma 2.8 (The Splittig Lemma). For a short exact sequece 0 A i B j C 0 of abelia groups the followig statemets are equivalet: (1) There exists a homomorphism p : B A such that p i = A. (2) There exists a homomorphism s : C B such that j s = C. (3) If f : A A C is give by f(a) = (a, 0) ad g : A C C is give by g(a, c) = c, the there is a isomorphism Φ : B A C such that the followig diagram is commutative: i B 0 A Φ C 0 j f A C g

6 6 2.2 Simplicial Homology Recall that a set A R m is covex if tv + (1 t)w A wheever v, w A ad 0 < t < 1, which is to say that the lie segmet coectig ay two poits i A must also lie i A. The followig result will prove useful later o. Propositio 2.9. Let A R m be a covex set. If v 1,..., v A ad t 1,..., t 0 such that k t k = 1, the k t kv k A. Proof. The statemet is clearly true i the case = 1. Suppose that it is true for some N. Let v 1,..., v +1 A ad t 1,..., t +1 0 such that t t +1 = 1. If t +1 = 1 the we must have t k = 0 for all 1 k, whece k t kv k = v +1 A ad we re doe. Assumig that t +1 1, observe that from k=1 t k = 1 t +1 we have t k = 1. 1 t +1 Now, +1 t k v k = k=1 k=1 t k v k + t +1 v +1 = (1 t +1 ) k=1 where by the iductive hypothesis k=1 t k 1 t +1 v k k=1 t k 1 t +1 v k + t +1 v +1, is a elemet of A. Thus, sice v +1 A ad A is covex, we coclude that is also i A. +1 t k v k k=1 The covex hull of a set A R m, deoted by C(A), is the itersectio of all covex sets that cotai A; that is, C(A) = {C : A C ad C is covex} Propositio If A = {v 0,..., v }, the { C(A) = t k v k : k (t k 0) ad k=0 } t k = 1. k=0

7 7 Figure 2. The covex hull of poits i space. Proof. Suppose that A = {v 0,..., v }, ad let { S = t kv k : k (t k 0) ad } t k = 1. k k It is clear that A S. Let p, q S. The p = k s kv k ad q = k t kv k for some oegative reals s k, t k such that k s k = k t k = 1. Now, for ay 0 < r < 1, rp + (1 r)q = r k s kv k + (1 r) k t kv k = k (rs k + (1 r)t k )v k, where rs k + (1 r)t k 0 for each k, ad k (rs k + (1 r)t k ) = r k s k + (1 r) k t k = r + (1 r) = 1. Hece rp + (1 r)q S ad S is a covex set cotaiig A. From this coclusio it follows that C(A) S. Next, let C be ay covex set such that A C. For ay q S there exist scalars t k 0 such that k t kv k = 1 ad q = k t kv k. Now, sice v k C for each k ad C is covex, it follows from Propositio 2.9 that q C. Hece S C, ad sice C is a arbitrary covex set that cotais A, we obtai S C(A). The stereoscopic figure pair i Figure 2 illustrates the covex hull for a set A of te poits i R 3. Of course, if the three poits that appear i the iterior of C(A) were removed from A, the same covex hull would result. We will be particularly iterested i fiite sets of poits for which o oe poit ca be removed without alterig the covex hull. If A = {v 0,..., v } R m for some m + 1 ad the vectors v 1 v 0,..., v v 0 are liearly idepedet, the C(A) is called a -simplex, the poits v k are called the vertices of the simplex, ad C(A) is deoted by [v 0,..., v ]. The orderig of the vertices i the symbol [v 0,..., v ] further specifies a orietatio o the -simplex that is cosidered a essetial part of its defiitio: amely, ay edge [v i, v j ] of [v 0,..., v ] is orieted i the directio of the vector v j v i if i < j, or v i v j if i > j. Propositio 2.10 alog with the defiitio of a -simplex make clear that ay poit i [v 0,..., v ] is uiquely expressible i the form k t kv k, with the barycetric coordiates t k of the poit beig oegative scalars that sum to 1.

8 8 u 2 u 2 u 1 u 1 u 0 u 0 Figure 3. The stadard 2-simplex 2. Defiig u i 1 = (0,..., 1,..., 0) R +1 with the 1 i the ith positio for 1 i + 1, the stadard -simplex is the -simplex [u 0,..., u ] R +1, ad so sice k t ku k = (t 0,..., t ), { } = (t 0,..., t ) R +1 : k (t k 0) ad t k = 1. See Figure 3 for a illustratio of the stadard 2-simplex. The caoical liear homeomorphism from to ay -simplex [v 0,..., v ] is the liear trasformatio k t ku k k t kv k that preserves orietatio. k=0

9 9 2.3 Sigular Homology Give a topological space X, a sigular -simplex is a cotiuous map σ : X. For each 0 defie C (X) to be the free abelia group with basis the set of all sigular -simplices associated with X. The elemets of C (X) are called -chais, ad are writte as fiite formal sums k i=1 iσ i, where k, i Z ad σ i : X. The sigular boudary maps : C (X) C 1 (X) are homomorphisms defied by (σ) = i=0 ( 1) i σ [u0,...,û i,...,u ] δ 1 i (2) for each basis elemet σ C (X), where δ 1 i : 1 [u 0,..., û i,..., u ] is the caoical liear homeomorphism as discussed i the previous sectio (it is ofte suppressed i the iterests of brevity). As with the simplicial boudary maps it ca be show that +1 = 0, so we obtai a chai complex C +1 (X) +1 C (X) C 1 (X), called the sigular chai complex of X ad deoted by C(X), which gives rise to homology groups H (X) = Ker / Im +1 called the sigular homology groups of X. Now, suppose Y is aother topological space, ad let ϕ : X Y be cotiuous. For each the map ϕ iduces a homomorphism 1 ϕ : C (X) C (Y ) defied by ϕ (σ) = ϕ σ for each basis elemet σ : X of C (X). Thus, for ay -chai i iσ i we have ( ) ϕ iσ i = iϕ (σ i ) = i(ϕ σ i ). i i i Deotig the sigular boudary maps C (X) C 1 (X) ad C (Y ) C 1 (Y ) by X Y, respectively, we fid that ( ) (ϕ 1 X )(σ) = ϕ 1 i ( 1)i σ [u0,...,û i,...,u ] δ 1 i ad therefore the diagram = i ( 1)i ((ϕ σ) [u0,...,û i,...,u ] δ 1 i ) = Y (ϕ σ) = Y (ϕ (σ)) = ( Y ϕ )(σ), +1 C +1 (X) X C (X) X C 1 (X) ad ϕ +1 ϕ ϕ 1 +1 C +1 (Y ) Y C (Y ) Y C 1 (Y ) 1 Other books would represet ϕ by ϕ, ad I oce was tempted to use ϕ ; but ϕ is uiformative for obvious reasos, ad ϕ coveys o more iformatio tha ϕ already does.

10 is commutative ad it follows that {ϕ } forms a chai map C(X) C(Y ). By Propositio 2.1 the maps ϕ iduce homomorphisms ϕ : H (X) H (Y ) betwee homology groups. The first sigificat geeral result we are i a positio to obtai i sigular homology is the followig. Propositio Let {X α } α A be the path-compoets of a topological space X. The H (X) = α A H (X α ). Proof. For [σ] := σ + +1 (C +1 (X)) H (X) we have σ = a i=1 iσ i for sigular -simplices σ i : X. Each image σ i ( ) i X is path-coected, so for each 1 i a there exists some α i A such that σ i : X αi. Let i 1 = 1, i 2 = mi{1 < i a : α i α 1 }, ad i geeral i k = mi{i : α i α i1,..., α ik 1 } for 1 k b, where b a. The X αi1,..., X αib are the distict path compoets of X that cotai the images of the σ i, ad for coveiece we ca desigate each X αik by X ik. Let B k = {i σ i : X ik } for each k, ad defie ϕ k = i B k i σ i so that ϕ k C (X ik ). Now σ = k ϕ k, ad σ Ker implies k ϕ k = 0 ad therefore ϕ k = 0 for each k sice C (X ir ) C (X is ) = wheever r s. It follows that each ϕ k is i the kerel of restricted to C (X ik ), so that [ϕ k ] ik := ϕ k + +1 (C +1 (X ik )) is i H (X ik ) ad we ca defie a map Ω : H (X) α A H (X α ) by b Ω([σ]) = [ϕ k ] ik. Suppose that [σ] = [τ], so σ τ (C +1 (X)) ad there s some ξ C +1 (X) such that ξ = σ τ. As before, we ca write σ = a k=1 ϕ α k such that ϕ αk C (X αk ) ad α i α j wheever i j. Similarly, τ = b k=1 ψ β k such that ψ βk C (X βk ) ad β i β j wheever i j. By defiitio Ω([σ]) = ([ϕ α ]) α A, where [ϕ α ] is a class i H (X α ) with ϕ α = 0 if α α k for all 1 k a. I similar fashio Ω([τ]) = ([ψ α ]) α with ψ α = 0 if α β k for all 1 k b. With this kid of arragemet we ca write σ = α A k=1 ϕ α ad τ = α A ψ α, so ξ = α (ϕ α ψ α ). However, ξ itself is expressible as α A ξ α with ξ α C +1 (X α ) for each α, ad clearly we must have ξ α = ϕ α ψ α C (X α ) sice the X α are disjoit. Sice ϕ α ψ α (C +1 (X α )), we fid that [ϕ α ] = [ψ α ] as classes i H (X α ), whece Ω([σ]) = Ω([τ]) ad Ω is well-defied. Now assume simply that [σ], [τ] H (X). Oce agai rearrage to write σ = a k=1 ϕ α k ad τ = b k=1 ψ β k, oly such that α k = β k for all 1 k c a (assumig a b for defiiteess) ad {α k : c + 1 k a} {β k : c + 1 k b} =. 10

11 Redesigate idices as follows: β c+1 = α a+1, β c+2 = α a+2,..., β b = α a+r, where b = c + r. Now, sice Ω is well-defied, ([ a ]) b Ω([σ] + [τ]) = Ω ϕ αk + ψ βk k=1 k=1 ([ c = Ω (ϕ αk + ψ αk ) + k=1 a k=c+1 ϕ αk + a+r k=a+1 ψ αk ]) We ca let ξ k = ϕ αk + ψ αk for 1 k c, ξ k = ϕ αk for c + 1 k a, ad ξ k = ψ αk for a + 1 k a + r, so that Sice Ω([σ] + [τ]) = Ω = ([ a+r ]) a+r ξ k = [ξ k ] αk k=1 k=1 c [ϕ αk + ψ αk ] αk + k=1 a [ϕ αk ] αk + k=c+1 [ϕ αk + ψ αk ] αk = [ϕ αk ] αk + [ψ αk ] αk, α k = β k for 1 k c, ad ψ αk = ψ βk a+c for a + 1 k a + r, Ω([σ] + [τ]) = = a [ϕ αk ] αk + k=1 a [ϕ αk ] αk + k=1 Recallig that b = c + r, we obtai Ω([σ] + [τ]) = c [ψ αk ] αk + k=1 c [ψ βk ] βk + k=1 a [ϕ αk ] αk + k=1 a+r k=a+1 c+r k=c+1 a+r k=a+1 [ψ αk ] αk. [ψ βk a+c ] βk a+c [ψ βk ] βk. b [ψ βk ] βk = Ω([σ]) + Ω([τ]) ad see that Ω is a homomorphism. Next, suppose that Ω([σ]) = 0. Proceedig as with the well-defiedess argumet, we obtai ([ϕ α ]) α A = 0 ad hece ϕ α (C +1 (X α )) for all α A. But the ϕ α (C +1 (X)) for all α ad [ ] [ a ] a a [σ] = ϕ α = ϕ αk = [ϕ αk ] = (ϕ αk + (C +1 (X))) = 0, α A k=1 k=1 where the third equality holds sice each ϕ αk is, by costructio, i the kerel of ad so represets a class i H (X). Therefore Ω is ijective. Now suppose that ([ϕ α ]) α A α A H (X α ), so the set S = {α A : [ϕ α ] 0} must be fiite. For each α S we have k=1 k=1 ϕ α Ker[ : C (X α ) C 1 (X α )], 11

12 12 ad thus ϕ := α S ϕ α Ker[ : C (X) C 1 (X)] so that [ϕ] := ϕ + (C +1 (X)) H (X). Now, Ω([ϕ]) = α S [ϕ α ] = α A[ϕ α ] = ([ϕ α ]) α A sice [ϕ α ] = 0 for α A S. Therefore Ω is surjective. I what follows recall the otatioal covetio whereby [u 0 ] ad [u 0, u 1 ] deote the stadard simplices 0 ad 1, respectively. Propositio If X is path-coected, the H 0 (X) = Z. Proof. Defie the homomorphism ɛ : C 0 (X) Z by ɛ( i iσ i ) = i i, so that ɛ(σ) = 1 for each sigular 1-simplex σ. If σ : 0 X is give by σ(u 0 ) = x 0 for some x 0 X, the for ay k Z we have ɛ(kσ) = k for kσ C 0 (X), showig that ɛ is surjective ad thus C 0 (X)/ Ker ɛ = Z. Fix τ : 1 X i C 1 (X). The ɛ( 1 τ) = ɛ(τ [u1 ] τ [u0 ]) = ɛ(τ [u1 ]) ɛ(τ [u0 ]) = 1 1 = 0. Thus, if σ Im 1, the there exists some i iτ i C 1 (X) such that 1 ( i iτ i ) = σ, whece ) ɛ(σ) = ɛ ( i i 1 τ i = i ɛ( 1 τ i ) = 0 i ad we coclude that Im 1 Ker ɛ. Now suppose that ϕ = i iσ i Ker ɛ, so i i = 0. Fix x 0 X. Defie σ 0 C 0 (X) by σ 0 (u 0 ) = x 0. For each i, sice σ i (u 0 ), x 0 X ad X is path-coected, there exists a path τ i : 1 X such that τ i (u 0 ) = x 0 ad τ i (u 1 ) = σ i (u 0 ). The ( ) 1 iτ i = i 1 τ i = i(τ i [u1 ] τ i [u0 ]) = iτ i [u1 ] iτ i [u0 ] i i i i i = i iσ i i iσ 0 = i iσ i σ 0 i i = i iσ i = ϕ, which shows that ϕ Im 1 ad hece Ker ɛ Im 1. Therefore H 0 (X) = C 0 (X)/ Im 1 = C 0 (X)/ Ker ɛ = Z. Combiig the two propositios above it follows that for ay space X, if {X α } α A are the path-compoets of X, the H 0 (X) = α A Z. Propositio If X = {p}, the H (X) = 0 for all 1. Proof. For X a sigle poit p we fid that C (X) = Z with geerator p, where p : X is the sigular -simplex give by p (x) = p for all x. Fix σ Ker C (X), so σ = mp for some m Z such that (mp ) = 0

13 13 Suppose that is odd. Now, mp +1 C +1 (X), ad (mp +1 ) = m ( 1) i p +1 [u0,...,û i,...,u +1 ] = m ( 1) i p = mp = σ i=0 shows that σ Im +1. Therefore [σ] = 0. Suppose that is eve. σ = 0 implies that m p = 0; that is, m ( 1) i p [u0,...,û i,...,u ] = m ( 1) i p 1 = mp 1 = 0, i=0 which implies that m = 0. Thus σ = 0 ad oce agai we obtai [σ] = 0. It s cocluded, the, that H (X) = 0 for ay 1. i=0 i=0 So for X = {p} Propositio 2.12 implies that H 0 (X) = Z, ad thus there s a exceptio to the patter i Propositio To fix this (if ideed a fix is desired), we ca modify the defiitio for the 0th homology class to make it also trivial i the case whe X is a sigle poit. The way to do this is to replace the map 0 : C 0 (X) 0 with ɛ : C 0 (X) Z to form the augmeted chai complex C 2 (X) 2 C1 (X) 1 C0 (X) ɛ Z 0, the defie H 0 (X) = Ker ɛ/ Im 1. Sice it was foud i the proof of Propositio 2.12 that Ker ɛ = Im 1 whe X is oempty ad path-coected, we fid i particular H 0 ({p}) = 0. Settig H (X) = H (X) for 1, we obtai what are kow as reduced homology groups which have the virtue of beig trivial i all oegative dimesios for oe-poit spaces. If X is oempty but ot path-coected we still have Im 1 Ker ɛ. Defie ɛ : H 0 (X) Z by ɛ([σ]) = ɛ(σ), ad ote that ɛ is surjective ad Ker ɛ = H 0 (X). Hece H 0 (X)/ H 0 (X) = Z, or equivaletly H 0 (X) = H 0 (X) Z. Theorem If the maps f, g : X Y are homotopic, the f = g : H (X) H (Y ) for all. Proof. Suppose that f, g : X Y are homotopic. I light of Propositio 2.3 it will suffice to show that the chai maps f, g : C (X) C (Y ) are chai-homotopic. That is, it must be show that there exists, for all 0, maps λ : C (X) C +1 (Y ) such that g f = Y +1 λ + λ 1 X. Let F : X I Y be a homotopy from f to g, so F (, 0) = f( ) ad F (, 1) = g( ). As usual will be desigated by [u 0,..., u ], ad for the product space I defie {0} = [v 0,..., v ] ad {1} = [w 0,..., w ]. Fially, defie the homomorphism λ : C (X) C +1 (Y ) by λ (σ) = ( 1) i F (σ ) [v0,...,v i,w i,...,w ] i=0

14 for each σ : X, where F (σ ) : I X I Y ad each term i the sum we take to be precomposed by the caoical liear homeomorphism +1 [v 0,..., v i, w i,..., w ] I. It s istructive to examie the = 0 case ad show that g 0 f 0 = 1 Y λ 0 + λ 1 0 X. Desigatig λ 1 0, this etails showig 1 Y λ 0 = g 0 f 0. Lettig δ deote appropriate caoical liear homeomorphisms [z 0,..., z ] for ay -simplex [z 0,..., z ], we have for ay σ : 0 X, ( Y 1 λ 0 )(σ) = Y 1 (F (σ ) [v0,w 0 ] δ 1 ) = Y 1 (F (σ ) δ 1 ), where the secod equality holds sice [v 0, w 0 ] = 0 I. Pressig o, Now, ( Y 1 λ 0 )(σ) = F (σ ) δ 1 [u1 ] δ 0 F (σ ) δ 1 [u0 ] δ 0. (F (σ ) δ 1 [u1 ] δ 0 )(u 0 ) = (F (σ ) δ 1 )(u 1 ) = (F (σ ))(u 0, 1) = F (σ(u 0 ), 1), where F (σ(u 0 ), 1) = g(σ(u 0 )) = (g 0 (σ))(u 0 ) ad thus F (σ ) δ 1 [u1 ] δ 0 = g 0 (σ). I similar fashio we fid F (σ ) δ 1 [u0 ] δ 0 = f 0 (σ), ad so ( 1 Y λ 0 )(σ) = g 0 (σ) f 0 (σ) ad we re doe. The = 2 case is the highest dimesioed case that ca be readily visualized, so let s cosider it ext. Let σ : 2 X. For brevity let H = F (σ ) ad [j] = [u 0,..., û j,..., u 3 ]. Also it will be coveiet to defie [z 0,..., z ] j = [z 0,..., ẑ j,..., z ]. Now, Y (λ 2 (σ)) = ( 1) j H [v0,w 0,w 1,w 2 ] δ 3 [j] δ 2 ( 1) j H [v0,v 1,w 1,w 2 ] δ 3 [j] δ 2 = j=0 + 3 ( 1) j H [v0,v 1,v 2,w 2 ] δ 3 [j] δ 2 j=0 3 ( 1) [ j H [v0,w 0,w 1,w 2 ] j δ 2 H [v0,v 1,w 1,w 2 ] j δ 2 + H [v0,v 1,v 2,w 2 ] j δ 2], j=0 ad from λ 1 ( 2 X (σ)) = 2 j=0 ( 1)j λ 1 (σ [u0,...,û j,...,u 2 ] δ 1 ) we obtai 2 λ 1 ( 2 X (σ)) = ( 1) j[ F (σ [u0,...,û j,...,u 2 ] δ 1 ) [v0,w 0,w 1 ] δ 2 Cosider the workigs of j=0 j=0 F (σ [u0,...,û j,...,u 2 ] δ 1 ) [v0,v 1,w 1 ] δ 2] (3) (σ [u1,u 2 ] δ 1 ) [v0,w 0,w 1 ] δ 2 : 2 1 I X I. 14

15 The 2-simplex [v 0, w 0, w 1 ] is regarded as a subspace of 1 I, so for a poit p 2 we have δ 2 (p) = (q, t) for q 1 ad 0 t 1. Now, (σ [u1,u 2 ] δ 1 )(q, t) = (σ [u1,u 2 ] )(δ 1 (q), t) = (σ )(δ 1 (q), t), where δ 1 (q) [u 1, u 2 ] with [u 1, u 2 ] regarded as a subspace of 2 so that (δ 1 (q), t) [v 1, w 1, w 2 ] 2 I. Thus (σ [u1,u 2 ] δ 1 ) [v0,w 0,w 1 ] δ 2 is equal to (σ ) [v1,w 1,w 2 ] δ 2 sice ((σ ) [v1,w 1,w 2 ] δ 2 )(p) = (σ ) [v1,w 1,w 2 ](δ 1 (q), t) = (σ )(δ 1 (q), t) (ote that δ 2 : 2 [v 1, w 1, w 2 ] 2 I will caoically carry p directly to (δ 1 (q), t)). Expadig (3) (ad suppressig the caoical homeomorphisms), we obtai λ 1 ( 2 X (σ)) = [ F (σ [u1,u 2 ] ) [v0,w 0,w 1 ] F (σ [u1,u 2 ] ] ) [v0,v 1,w 1 ] [ F (σ [u0,u 2 ] ) [v0,w 0,w 1 ] F (σ [u0,u 2 ] ] ) [v0,v 1,w 1 ] + [ F (σ [u0,u 1 ] ) [v0,w 0,w 1 ] F (σ [u0,u 1 ] ] ) [v0,v 1,w 1 ] ad thus λ 1 ( X 2 (σ)) = H [v1,w 1,w 2 ] H [v1,v 2,w 2 ] H [v0,w 0,w 2 ] + H [v0,v 2,w 2 ] + H [v0,w 0,w 1 ] H [v0,v 1,w 1 ]. Now at last we add: Y 3 (λ 2 (σ)) + λ 1 ( X 2 (σ)) = F (σ ) [w0,w 1,w 2 ] δ 2 F (σ ) [v0,v 1,v 2 ] δ 2. For p 2, (F (σ ) [w0,w 1,w 2 ] δ 2 )(p) = F (σ ) [w0,w 1,w 2 ](p, 1) = F (σ(p), 1) = g(σ(p)) = (g σ)(p) = (g 2 (σ))(p), so F (σ ) [w0,w 1,w 2 ] δ 2 = g 2 (σ), ad similarly F (σ ) [v0,v 1,v 2 ] δ 2 = f 2 (σ). Hece ( Y 3 λ 2 + λ 1 X 2 )(σ) = (g 2 f 2 )(σ) for the basis elemet σ of C 2 (X). The geeral case for arbitrary requires careful bookkeepig ad will be addressed at a later time. Lemma Let X, Y, Z be topological spaces, ad let f : X Y ad g : Y Z be cotiuous maps. The, for all, (1) ( X ) = H(X) (2) (g f) = g f 15

16 Proof. Fix. For σ : X we have ( X ) (σ) = X σ = σ so that ( X ) = C(X). Now, by Propositio 2.4, ( X ) = C(X) = H(X), which proves part (1). Next, from g f : X Z we obtai the map (g f) : C (X) C (Z), where (g f) (σ) = (g f) σ = g (f σ) = g (f σ) = g (f (σ)) = (g f )(σ), 16 ad so we obtai (g f) = (g f ) = g f by oce agai appealig to Propositio 2.4. This proves part (2). Corollary If f : X Y is a homotopy equivalece, the f : H (X) H (Y ) is a isomorphism for all. Proof. Suppose that f : X Y is a homotopy equivalece. The there is a map g : Y X such that g f X ad f g Y. By Theorem 2.14, (g f) = ( X ) ad (f g) = ( Y ), ad thus by Lemma 2.15 we obtai g f = H(X) ad f g = H(Y ). Therefore f is a isomorphism by Propositio 1.1. If X is a space ad A is a oempty closed subspace that is a deformatio retract of some eighborhood of X, the the pair (X, A) is kow as a good pair. The followig theorem will be prove over course of the ext two sectios, with the maps ˆ to be determied alog the way. Theorem If (X, A) is a good pair, the there is a exact sequece H (A) i H (X) q H (X/A) ˆ H 1 (A) H 0 (X/A) 0, where i is iduced by the iclusio map i : A X ad q is iduced by the quotiet map q : X X/A.

17 Relative Homology Give a topological space X, let A X be a subspace. From the free abelia groups C (X) ad C (A) we form the quotiet group C (X, A) = C (X)/C (A), ad defie to be the homomorphism iduced by : (C (X), C (A)) (C 1 (X), C 1 (A)) so that (σ + C (A)) = (σ) + C 1 (A) for each σ C (X). It s easy to check that +1 = 0 holds for all, ad so a chai complex C +1 (X, A) +1 C (X, A) C 1 (X, A) results. The homology groups associated with this chai complex, give as H (X, A) = Ker[ : C (X, A) C 1 (X, A)] Im[ +1 : C +1 (X, A) C (X, A)], are called relative homology groups. If A X ad B Y, a cotiuous map f : (X, A) (Y, B) iduces homomorphisms f : (C (X), C (A)) (C (Y ), C (B)) i the usual fashio give i sectio 2.1, ad these i tur iduce homomorphisms f : C (X, A) C (Y, B) give by f (σ + C (A)) = f (σ) + C (B) = f σ + C (B) for each basis elemet σ C (X). The maps f costitute a chai map C(X, A) C(Y, B), ad so by Propositio 2.1 they iduce well-defied homomorphisms f : H (X, A) H (Y, B) give by f ( (σ + C (A)) + +1 (C +1 (X, A)) ) = f (σ + C (A)) + +1 (C +1 (Y, B)). Slightly more compactly we ca write f ( (σ + C (A)) + Im X +1) = (f σ + C (B)) + Im Y +1. Let i : A X be the iclusio map. This map iduces homomorphisms i : C (A) C (X) give by i (σ) = i σ for each map σ : A. Also we itroduce the homomorphism j : C (X) C (X, A) give by j (σ) = σ + C (A) for each σ : X. Clearly each i is ijective ad each j is surjective. If σ Im i, the σ C (A) ad so j (σ) = σ + C (A) = C (A) (the zero elemet of C (X, A)), which shows that σ Ker j. If σ Ker j, the σ C (A) ad it follows that i (σ) = i σ = σ, which shows that σ Im i. Thus, Im i = Ker j, ad we coclude that 0 C (A) i j C (X) C (X, A) 0

18 C +1 (A) A C (A) A C 1 (A) i +1 i i 1 +1 C +1 (X) X C (X) X C 1 (X) j +1 j j 1 C +1 (X, A) +1 C (X, A) C 1 (X, A) Figure 4 is a short exact sequece. The claim is that i : C(A) C(X) ad j : C(X) C(X, A) are chai maps, which is to say the diagram i Figure 4 is commutative ad therefore a short exact sequece of chai complexes. We have, for ay σ : A, ) (i 1 A )(σ) = i 1 ( i ( 1)i σ [u0,...,û i,...,u ] = ( ) i ( 1)i i 1 σ [u0,...,û i,...,u ] ad for ay σ : X, = i ( 1)i σ [u0,...,û i,...,u ] = X (σ) = X (i (σ)) = ( X i )(σ); (j 1 X )(σ) = j 1 ( i ( 1)i σ [u0,...,û i,...,u ] = ( i ( 1)i σ [u0,...,û i,...,u ] + C 1 (A) ) ( ) = i ( 1)i σ [u0,...,û i,...,u ] + C 1 (A) = (σ) + C 1 (A) = (σ + C (A)) = (j (σ)) = ( j )(σ). We ow defie homomorphisms ρ : H (X, A) H 1 (A) i the same maer as the maps ρ i sectio 2.1, where H 1 (A) = Ker A 1/ Im A. Let (σ + C (A)) + Im +1 H (X, A), so σ + C (A) Ker C (X, A). Sice σ C (X) we have j (σ) = σ + C (A), ad from j 1 X = j comes X σ Ker j 1 = Im i 1. So there is some τ C 1 (A) such that i 1 (τ) = X σ, which immediately implies τ = X σ. Defie ρ ( (σ + C (A)) + Im +1 ) = X σ + Im A. By Theorem 2.6 we obtai a log exact sequece H (A) i H (X) j H (X, A) ) ρ H 1 (A) i 1 H 1 (X)

19 Propositio For A X ad B Y, let f : (X, A) (Y, B) be a map such that f : X Y ad f A : A B are each homotopy equivaleces. The for all the maps f : H (X, A) H (Y, B) are isomorphisms. 19 Proof. By Corollary 2.16 the maps f : H (X) H (Y ) ad f A := (f A ) : H (A) H (B) are isomorphisms for all. Fix 0. The diagram i H (A) X j H (X) X ρ H (X, A) X i H 1 (A) X 1 H 1 (X) f A f f f A 1 f 1 i H (B) Y j H (Y ) Y ρ H (Y, B) Y i H 1 (B) Y 1 H 1 (Y ) has exact rows, ad the claim is it s also commutative. It s easily verified that the diagram i C (A) X j C (X) X C (X, A) is commutative. For istace f A f f i C (B) Y j C (Y ) Y C (Y, B) ( f j X )(σ) = f (σ + C (A)) = f (σ) + C (B) = j Y (f (σ)) = (j Y f )(σ) cliches commutativity of the secod square, ad a similar routie will show commutativity of the first square. Now, {f }, { f }, {j X } ad {j Y } are chai maps that iduce the well-defied homomorphisms foud i the first diagram, ad sice the fuctor preserves commutativity, we coclude that the secod ad first squares i the first diagram are likewise commutative. Next, for (σ + C (A)) + Im +1 H (X, A) we obtai (f A 1 ρ X )((σ + C (A)) + Im X +1) = f A 1 ( X σ + Im A ) = f A 1( X (σ)) + Im B = f 1 ( X (σ)) + Im B = Y (f (σ)) + Im B = ρ Y ((f (σ) + C (B)) + Im Y ) = (ρ Y f )((σ + C (A)) + Im X +1), where the third equality holds sice X σ C 1 (A) C 1 (X) ad f 1 A is the restrictio of f 1 to C 1 (A); ad the fourth equality holds sice {f } is a chai map C(X) C(Y ). This shows commutativity of the third square. Therefore, by the Five-Lemma, the maps f are isomorphisms.

20 Example Show a map f : (D, S 1 ) (D, D {0}) caot be a homotopy equivalece of pairs; that is, there s o map g : (D, D {0}) (D, S 1 ) such that f g ad g f are homotopic to D through maps (D, D {0}) (D, D {0}) ad (D, S 1 ) (D, S 1 ), respectively. Solutio. Suppose there is such a map g. Let {ϕ t : (D, S 1 ) (D, S 1 )} t I be a homotopy such that ϕ 0 = D ad ϕ 1 = g f. Sice f(x) = x for all x D, we have ϕ 1 (x) = g(f(x)) = g(x) so that ϕ 1 = g ad hece g D. Let i : S 1 D be the iclusio map. Sice g : D {0} S 1 is cotiuous ad 0 is i the closure of D {0}, we must have g(0) S 1 so that g : D S 1 ad hece g i = g S 1 : S 1 S 1. Now, for each 1 we have H 1 (S 1 ) i H 1 (D g ) H 1 (S 1 ), where H 1 (S 1 ) = Z ad H 1 (D ) = 0 so that g i = 0. O the other had the maps {ϕ t S 1 : S 1 S 1 } t I costitute a homotopy g i S 1, so by Theorem 2.14 ad Lemma 2.15 we obtai g i = (g i) = S 1 = H 1 (S 1 ), 20 ad hece for 1 H 1 (S 1 ) we obtai (g i )(1) = 1; that is, g i cotradictio. 0, which is a

21 The Excisio Theorem Let A ad B be subspaces of X. The iclusio map j : (B, A B) (X, A) iduces homomorphisms j : C (B) C (X) give by j (β) = j β for a basis elemet β : B of C (B). Each j, i tur, iduces a homomorphism of quotiet groups give by j : C (B, A B) C (X, A) j (β + C (A B)) = j (β) + C (A) = j β + C (A) = β + C (A). (4) The Excisio Theorem states circumstaces whe the maps j will iduce isomorphisms of relative homology groups. Theorem 2.20 (Excisio Theorem). If A, B X such that X = A B, the the iclusio j : (B, A B) (X, A) iduces isomorphisms j : H (B, A B) H (X, A) for all. The reaso for the use of the term excisio is perhaps made clearer by the followig result. Corollary Give subspaces Z A X such that Z A, the the iclusio j : (X Z, A Z) (X, A) iduces isomorphisms j : H (X Z, A Z) H (X, A) for all. Proof. Let B = X Z, ad ote that A B = A Z. It remais to show that A B = X. Let x X, ad suppose that x / A. Sice Z A, we have x / Z ad thus x X Z. Now, X Z is ope, so there exists some ope set O such that x O X Z X Z = B. Hece x B ad we coclude that X = A B. Therefore, by the Excisio Theorem, j : (X Z, A Z) (X, A) iduces isomorphisms H (X Z, A Z) = H (X, A). To prove the Excisio Theorem there is a techical result that first eeds to be established. Let U = {U k } be a collectio of subspaces of X such that X = k U k, ad let C U (X) be the subgroup of C (X) cosistig of -chais i iσ i such that for each i there is some k for which σ i ( ) U k. It s easy to see that if α C U (X), the α C U 1(X); thus, if we let U deote the restrictio of to C U (X), ad defie ι : C U (X) C (X) to be the iclusio map, we obtai a commutative diagram of chai complexes: C+1(X) U +1 U C U (X) U C 1(X) U ι +1 ι ι 1 C +1 (X) +1 C (X) C 1 (X)

22 As usual H (X) = Ker / Im +1, ad ow we also defie H U (X) = Ker U / Im U +1. I accordace with the developmets i sectio 2.1, the iduced map ι : H U (X) H (X) is give by for each α Ker U. ι (α + Im U +1) = ι (α) + Im +1 Lemma Let U = {U k } be a collectio of subspaces of X such that X = k U k. If ι : C U (X) C (X) is the iclusio map, the the map ι : H U (X) H (X) is a isomorphism. To prove the lemma it will be show that the chai map ι : C U (X) C(X) is a chai homotopy equivalece, which meas there is a chai map ϕ : C(X) C U (X) such that {ι ϕ } is chai homotopic to { } (the idetity maps o C (X)) ad {ϕ ι } is chai homotopic to { U } (the idetity maps o C U (X)). The result follows from Propositio 2.5. Proof. We start with barycetric subdivisio of simplices. 2 By defiitio of a -simplex, { } [v 0,..., v ] = t i v i : i (t i 0) ad t i = 1 R m i=0 for some m + 1, where the set of vectors {v 1 v 0,..., v v 0 } is liearly idepedet. The baryceter of [v 0,..., v ] is the poit b = 1 i v +1 i, ad the barycetric subdivisio of [v 0,..., v ] is the subdivisio of [v 0,..., v ] ito smaller -simplices of the form [b, w 0,..., w 1 ] which we ow specify iductively. Whe = 0, the barycetric subdivisio of [v 0 ] is defied to be [v 0 ] itself. For = 1 we decompose [v 0, v 1 ] with baryceter b = 1v v 2 1 ito [b, v 0 ] ad [b, v 1 ]. For = 2, let b ij be the baryceter of the face [v i, v j ] of [v 0, v 1, v 2 ], ad decompose [v 0, v 1, v 2 ] ito [b, b 01, v 0 ], [b, b 01, v 1 ], [b, b 02, v 0 ], [b, b 02, v 2 ], [b, b 12, v 1 ], ad [b, b 12, v 2 ]. For = 3, if b ijk deotes the baryceter of the face [v i, v j, v k ] of [v 0, v 1, v 2, v 3 ], the a couple of the 24 members of the decompositio are [b, b 012, b 01, v 0 ] ad [b, b 012, b 01, v 1 ], which we could write as [b 0123, b 012, b 01, b 0 ] ad [b 0123, b 012, b 01, b 1 ] if we wished to employ our otatio to its fullest extet (the baryceter b i of [v i ] beig, of course, v i itself). See Figure 5. I geeral the barycetric subdivisio of [v 0,..., v ] is the collectio B[v 0,..., v ] of -simplices { [b, bl 1 0 l 1, b 1 l 2 0 l 2,..., b l 1, v ] } 2 0 l1 1 l 0 0 : l k i 1 < l k i & {l k 1 0,..., l k 1 k 1 } {lk 0,..., l k k}, where of course l k i {0,..., }, ad b l k 0 l k k = k i=0 1 k + 1 v l k i. is the baryceter of the k-dimesioal face [v l k 0,..., v l k k ] of [v 0,..., v ] for 1 k. Simply put, a member of B[v 0,..., v ] has as its vertices the baryceter of [v 0,..., v ], the baryceter of a ( 1)-dimesioal face F of [v 0,..., v ], the baryceter of a ( 2)-dimesioal face of F, ad so o dow the dimesios to coclude with a poit that is a vertex of F. 2 As with may results hereabouts, this proof is modeled alog the lies of the oe foud i Alle Hatcher s Algebraic Topology. i=0 22

23 23 v 3 v 3 b 03 b 03 b 23 b 23 b b v 0 v 0 b 01 b 12 v 2 b 01 b 12 v 2 v 1 v 1 Figure 5. Part of the barycetric subdivisio of a 3-simplex [v 0, v 1, v 2, v 3 ]. Lettig p q deote the Euclidea distace betwee poits p ad q i R m, we defie the diameter of [v 0,..., v ] to be diam[v 0,..., v ] = max{ p q : p, q [v 0,..., v ]}, which is a real value that is attaied for some ˆp, ˆq [v 0,..., v ] sice the set is compact. If p ad i t iv i are two poits i [v 0,..., v ], the p t iv i = t ip t iv i = t i(p v i ) i i i i i t i p v i = i t i max 0 i p v i = max 0 i p v i = p v i0 for some i 0 {0,..., }. Lettig p = i s iv i ad repeatig the process, we fid that p v i0 = v i0 s iv i vi0 v i1 i for some i 1. Hece for ay p, q [v 0,..., v ] we fid that p q v i v j for some i, j {0,..., }, ad therefore diam[v 0,..., v ] = max 0 i,j v i v j. What we will wat to show is that max{diam(w ) : W B[v 0,..., v ]} + 1 diam[v 0,..., v ]. (5) This is trivially true whe = 0 sice the oly member of B[v 0 ] is [v 0 ] itself, ad diam[v 0 ] = 0. For = 1 we have B[v 0, v 1 ] = {[b, v 0 ], [b, v 1 ]}, with diam[b, v 0 ] = b v 0 = 1 2 v v 1 v 0 = 1 2 v 1 v 0 = 1 2 diam[v 0, v 1 ] ad similarly diam[b, v 1 ] = 1 2 diam[v 0, v 1 ].

24 Employig iductio, let be arbitrary ad suppose (5) holds for every -simplex [v 0,..., v ]. Let V = [v 0,..., v +1 ] be ay ( + 1)-simplex, ad let BV be its barycetric subdivisio. Let W = [b, w 0,..., w ] BV, b beig the baryceter of V. First suppose that w i ad w j are vertices of W such that w i, w j b. The the poits w i ad w j must lie o the -dimesioal face F of V that has w 0 as its baryceter, ad sice [w 0,..., w ] is a member of BF ad F is a -simplex, by (5) we obtai w i w j diam([w 0,..., w ]) max{diam(u) : U BF } diam(f ). + 1 Of course, F V implies that diam(f ) diam(v ), so fially w i w j + 1 diam(v ) + 1 diam(v ). + 2 Next, take vertices w j ad b of W. Sice w j, b V we have b w j max 0 i +1 b v i = b v k for some k. Let b k be the baryceter of [v 0,..., ˆv k,..., v +1 ], so b k = +1 1 i=0,i k + 1 v i. Now, sice v k b k = v k i=0,i k + 2 v i = b, we obtai ( v k b = 1 v k + 2 v k ) + 2 b k = v k b k = v k b k diam(v ) ad therefore b w j + 1 diam(v ). + 2 Combiig the two cases aalyzed above leads to the geeral result + 1 diam(w ) = max p q diam(v ), p,q {b,w 0,...,w } + 2 which fially implies max{diam(w ) : W BV } + 1 diam(v ). + 2 We move o ow to the ext stage of the proof. Let Y R m be a covex set, ad let C(Y ) be the sigular chai complex C +1 (Y ) +1 C (Y ) C 1 (Y ) 1 A liear trasformatio l : Y ca be uiquely determied by defiig, for each 0 i, some w i Y for which l(u i ) = w i ; ideed, sice for each q there exist oegative scalars t i such that i t i = 1 ad q = i t iu i, we obtai l(q) = l( i t iu i ) = 24

25 i t il(u i ) = i t iw i Y by Propositio 2.9, ad therefore Im(l) Y as required. Now, if L(, Y ) is the collectio of all liear trasformatios l : Y, the we ca defie L (Y ) to be the subgroup of C (Y ) geerated by L(, Y ). The boudary maps : C (Y ) C 1 (Y ) the give rise to a chai complex L(Y ) L +1 (Y ) +1 L (Y ) L 1 (Y ) which is called a subcomplex of C(Y ). I what follows it will be coveiet to deote a map l L(, Y ) give by l(u i ) = w i by the symbol w 0,..., w. 3 It will also be coveiet to exted L(Y ) to dimesio 1 by lettig L 1 (Y ) be the free group geerated by the uique map [ ] Y, where [ ] is take to be the empty simplex that has o vertices; we ca deote this map by, so L 1 (Y ) = = Z. For ay poit y Y defie a homomorphism y : L (Y ) L +1 (Y ) by y( w 0,..., w ) = y, w 0,..., w, where y, w 0,..., w : +1 Y is give by y, w 0,..., w (u i ) = w i 1 for 1 i + 1, ad y, w 0,..., w (u 0 ) = y. Now, (y( w 0,..., w )) = ( 1) i y, w 0,..., w [u0,...,û i,...,u +1 ] δ i=0 +1 = y, w 0,..., w [u1,...,u +1 ] δ + ( 1) i y, w 0,..., w [u0,...,û i,...,u +1 ] δ i=1 +1 = w 0,..., w + ( 1) i y, w 0,..., ŵ i 1,..., w = w 0,..., w + = w 0,..., w = w 0,..., w i=1 ( 1) i+1 y, w 0,..., ŵ i,..., w i=0 ( 1) i y 1( w 0,..., ŵ i,..., w ) i=0 ( 1) i y 1( w 0,..., w [u0,...,û i,...,u ] δ 1 ) i=0 = w 0,..., w y 1( ( w 0,..., w )), where, for istace, the sixth equality is justified as follows: w 0,..., ŵ i,..., w takes u k 1 ad returs w k for k < i, ad w k+1 for k i, while w 0,..., w [u0,...,û i,...,u ] δ 1 maps as u k 1 u k w k for k < i, ad u k 1 u k+1 w k+1 for k i. 3 It seems to me highly iadvisable to deote l by [w 0,..., w ], sice this symbol is already take ad might lead oe to wrogly believe that Im(l) must ecessarily be a -simplex). 25

26 26 Sice w 0,..., w is a basis elemet for L (Y ), the precedig shows that ( +1 y)(σ) = σ (y 1 )(σ) = ( y 1 )(σ). (6) for ay σ L (Y ), with : L (Y ) L (Y ) beig the idetity map. Therefore we have +1 y + y 1 =. We ow defie a family of homomorphisms S : L (Y ) L (Y ) iductively. To start, we have S 1 : L 1 (Y ) L 1 (Y ) give by S 1 ( ) =. For ay k, give l = w 0,..., w k L( k, Y ), let b k be the baryceter of k, set l(b k ) = p, ad defie the homomorphism pk : L k (Y ) L k+1 (Y ) by The we defie S for 0 to be give by I particular pk(l) = p, w 0,..., w k. S (l) = (p 1 S 1 )(l). S 0 ( p ) = p 1(S 1 ( 0 ( p ))) = p 1(S 1 ( )) = p 1( ) = p shows that S 0 is the idetity o L 0 (Y ). If l = y 0, y 1 : 1 Y is i L 1 (Y ) ad has image equallig the simplicial 1-simplex [y 0, y 1 ], so that y 0 y 1, the l(b 1 ) = p with p y 0, y 1, ad S 1 (l) = (p0 S 0 1 )(l) = p0(s 0 (l [u1 ] l [u0 ])) = p0(l [u1 ] l [u0 ]) = p0( y 1 y 0 ) = p, y 1 p, y 0 shows that S 1 (l) equals a liear combiatio of sigular 1-simplices with images [p, y 0 ] ad [p, y 1 ], which are elemets of B[y 0, y 1 ]. Proceedig with a iductio argumet, let 1 ad suppose that for ay l = y 0,..., y L (Y ) with Im(l) = [y 0,..., y ], S (l) is a liear combiatio of sigular -simplices with images that are elemets of B[y 0,..., y ]. For ay l = y 0,..., y +1 L +1 (Y ) with ad l(b +1 ) = p, Im(l) = [y 0,..., y +1 ] := V S +1 (l) = (p S +1 )( y 0,..., y +1 ) = p(s ( +1 y 0,..., y +1 )) = +1 i=0 ( 1)i p(s ( y 0,..., ŷ i,..., y +1 ))

27 27 For each i, Im( y 0,..., ŷ i,..., y +1 := l i ) = [y 0,..., ŷ i,..., y +1 ] := V i, ad so by hypothesis S (l i ) is a liear combiatio of sigular -simplices with images that are elemets of BV i. Let W BV i be ay oe of these images. The W = [w 0,..., w ], where w 0 is the baryceter of V i, w 1 is the baryceter of a ( 1)-dimesioal face F 1 of V i, w 2 is the baryceter of a ( 2)-dimesioal face F 2 of F 1, ad so o util we arrive at w, which will be a vertex of F 1. Now p( w 0,..., w ) = p, w 0,..., w, which is a basis elemet of L +1 (Y ) with image [p, w 0,..., w ], ad sice p is the baryceter of V, w 0 is the baryceter of the -dimesioal face F = V i of V, ad all lower-dimesioed faces F 1,..., F 0 of V i are faces of V, it is clear that [p, w 0,..., w ] BV. Therefore S +1 (l) is a liear combiatio of sigular ( + 1)-simplices with images that are elemets of BV. To show that the maps S defie a chai map L(Y ) L(Y ), the commutativity property S 1 = S must be verified. The base case (whe = 1) is clear: (S 1 0 )( p ) = S 1 ( ) = = 0 ( p ) = ( 0 S 0 )( p ). For arbitary suppose that S = S 1 is true. Let l L( +1, Y ) with l(b) = p. Notig that +1 p = p 1, we obtai ( +1 S +1 )(l) = ( +1 p S +1 )(l) = (( p 1 ) (S +1 ))(l) = (S +1 )(l) (p 1 S +1 )(l) = (S +1 )(l) (p 1 S 1 +1 )(l) = (S +1 )(l), sice Now, defie homomorphisms T : L (Y ) L +1 (Y ) iductively as follows. Let T 1 ( ) = 0, ad for 0 set T (l) = p(l T 1 ( (l))) for each l L(, Y ), with p defied as above. Referrig to the diagram L 2 (Y ) 2 L 1 (Y ) 1 L 0 (Y ) 0 L 1 (Y ) 0 T 1 T 0 T 1 S S S S L 2 (Y ) 2 L 1 (Y ) 1 L 0 (Y ) 0 L 1 (Y ) 0

28 it will be show by iductio that the collectio of maps {T } = 1 is a chai homotopy betwee the chai maps {S } = 1 ad { } = 1, which etails demostratig that S = +1 T + T 1 (7) for 1. The base case is easy to secure sice T 1 0 ad T 2 0 (by defiitio), ad S 1 = 1. For arbitrary suppose that holds. The, for l L( +1, Y ), S = +1 T + T 1 ( +2 T +1 )(l) = +2 (p+1(l T ( +1 (l)))) = ( +2 p+1)(l T ( +1 (l))) = ( +1 p +1 )(l T ( +1 (l))) = ( +1 (l T ( +1 (l))) (p +1 )(l T ( +1 (l))) = ( +1 T +1 p +1 + p ( +1 T ) +1 )(l) = ( +1 T +1 p +1 + p ( S T 1 ) +1 )(l) = ( +1 T +1 p +1 + p +1 p S +1 )(l) = ( +1 T +1 S +1 )(l), where agai we make use of +1 0, ad therefore +1 S +1 = +2 T +1 + T +1. Sice T 1 0 o L 1 (Y ) = Z, we ca replace L 1 (Y ) with 0 ad obtai a trucated diagram i which {T } =0 is a chai homotopy betwee {S } =0 ad { } =0. Now begis the third part of the proof. Fix 0. For each geerator σ : X of C (X) there are iduced homomorphisms σ k : C k( ) C k (X) give by σ k (f) = σ f for each iteger k ad map f : k. Also there are the maps Sk : L k( ) L k ( ) that operate i the maer discussed above ad defie a chai map L( ) L( ). Fially there are the idetity maps = u 0,..., u k k : k k. Usig all these maps, we defie a ew homomorphism S : C (X) C (X) by S (σ) = σ (S ( )). To show the maps S defie a chai map C(X) C(X), we verify the commutativity property S 1 = S for each. I doig so, we use the symbol δ 1 i to deote the caoical liear homeomorphism 1 [u 0,..., û i,..., u ]. Thus, ( S )(σ) = ( σ (S ( ))) = ( σ )(S ( )) = ( σ 1 )(S ( )) = ( σ 1 ( S ))( ) = ( σ 1 (S 1 ))( ) = ( σ 1 S 1)( ) 28

29 29 = = = i=0 i=0 ( 1) i ( σ 1 S 1)( [u0,...,û i,...,u ] δ 1 i ) ( ) ( 1) i (σ [u0,...,û i,...,u ] δ 1 ) 1 1 S 1 1 i=0 = S 1 ( i=0 ( 1) i S 1 (σ [u0,...,û i,...,u ] δ 1 i ) ( 1) i σ [u0,...,û i,...,u ] δ 1 i = S 1 ( σ) = ( S 1 )(σ), i ) ( 1) where it must be admitted that there is a leap i goig from the fourth to the fifth row that would require some work to justify. Next, defie T : C (X) C +1 (X) by T (σ) = σ +1(T ( )), where for each k the map Tk : L k( ) L k+1 ( ) operates as described above. To be show is that { T } =0 is a chai homotopy betwee the chai maps { S } =0 ad { : C (X) C (X)} =0 (the idetity maps o C(X)). Thus it must be show that S = +1 T + T 1 (8) for all 0, which will be doe iductively. Let = 0. Sice S 0 0 : L 0 ( 0 ) L 0 ( 0 ) is the idetity map, we have for ay geerator σ : 0 X of C 0 (X), S 0 (σ) = σ 0 0(S 0 0( 0)) = σ 0 0( 0) = σ 0 = σ, so that S 0 = 0 : C 0 (X) C 0 (X) ad we obtai 0 S 0 = 0. Sice T k 1 0 for ay k, we have T 1 0 ad so ( 1 T 0 + T 1 0 )(σ) = 1 ( T 0 (σ)) = 1 ( σ 0 1(T 0 0 ( 0))) = ( 1 σ 0 1)(p0( 0 T 0 1( 0 ( 0)))) = ( 1 σ 0 1)(p0( 0)) = ( 1 σ 0 1)(p0( u 0 )) = ( 1 σ 0 1)( u 0, u 0 ) = 1 (σ u 0, u 0 ) = 0, where p( u 0 ) = u 0, u 0 sice the baryceter of 0 is u 0 ad 0(u 0 ) = u 0, ad the last equality follows from the observatio that σ u 0, u 0 : 1 0 X is a costat fuctio. It has ow bee established that 0 S 0 = 1 T 0 + T 1 0.

30 For the iductive step, let 0 be arbitrary ad suppose that equatio (8) holds. Take +1 to be the idetity map o C +1 ( +1 ) so that +1 ( +1) = +1. Referrig to the diagram C +2 ( ) C +1 ( ) C ( +1 ) σ σ σ C +2 (X) C +1 (X) C (X) we obtai, for ay basis elemet σ : +1 X of C +1 (X), ( +2 T +1 )(σ) = +2 ( σ +1 +2(T ( +1))) = ( +2 σ +1 +2)(T ( +1)) = ( σ )(T ( +1)) = ( σ ( +2 T ))( +1) = ( σ ( +1 T S +1 +1))( +1), by (7) = ( σ σ T σ S +1 +1)( +1) = σ +1 ( σ T +1 +1)( +1) S +1 (σ) = +1 (σ) ( T +1 )(σ) S +1 (σ) = ( +1 T +1 S +1 )(σ), where σ +1 = σ = +1 (σ), ad the remaiig justificatios for the eighth equality are left to the reader. Now the fourth ad last stage of the lemma s proof commeces. For m 0 let S m : C (X) C (X) be the mth iterate of S 0, with the uderstadig that S =. (The m otatio S is used here istead of S m simply because superscripts are already beig used liberally for idexig purposes i the proof.) Defie a homomorphism D m : C (X) C +1 (X) by m 1 D m (σ) = ( T i S )(σ) for σ : X ad m 0 (with D 0 (σ) = 0), resultig i the diagram i=0 C +1 (X) +1 C (X) C 1 (X) 30 D S m m S m 1 D m 1 S m C +1 (X) +1 C (X) C 1 (X) For fixed m it will be show that the maps D m provide a chai homotopy betwee { } =0 m ad { S } =0; that is, m S = +1 D m + D 1 m (9)

31 for all 0. Sice the maps it easily follows that = S i 1 S i S = S have bee verified to defie a chai map C(X) C(X), 1 holds for all i. Now, for ay σ : X ad m 1, ( +1 D m + D 1 m )(σ) = +1 (D m (σ)) + D 1( m (σ)) ( m 1 ) = +1 ( T m 1 i S )(σ) + ( T i 1 S 1)( (σ)) = m 1 i=0 i=0 i=0 i=0 [ ( +1 T i )( S (σ)) + ( T 1 S i 1 )(σ) ] m 1 [ = ( +1 T i )( S (σ)) + ( T i 1 )( S (σ)) ] m 1 = ( +1 T + T m 1 i 1 )( S (σ)) = i=0 m 1 i = ( S i=0 (i+1) 0 S )(σ) = ( S i=0 ( S )( m S )(σ) = ( S i (σ)) m S )(σ) which verifies (9). Recall that for ay y 0,..., y : Y i L (Y ) with image the -simplex [y 0,..., y ], S ( y 0,..., y ) is a chai of sigular -simplices with images that are elemets of B[y 0,..., y ], ad so S ( ) is a chai of maps with images that are elemets of B. Thus, S (σ) is a chai of maps of the form For ay of the maps σ α i1 I geeral σ α i1 : α i1 W i1 B σ σ(w i1 ) X. oto oto we fid that S (σ α i1 ) i tur yields a chai of maps of the form σ α i1 α i2 : α i2 W i2 B α i1 W i1 B σ σ(w i1 ) X. oto oto oto m S (σ) is a liear combiatio of maps of the form σ α i1 α im : α im Wim α im 1 Wim 1 α im 2 α i2 Wi2 α i1 Wi1 31 σ σ(w i1 ) X, (10) where W ij B ad α ij : W ij for each j. For ay -simplex V, defie B 1 V = BV, B 2 V = {BW 1 : W 1 B 1 V }, ad i geeral B V = { BW 1 : W 1 B 1 V } for 1. Recallig (5), it s see that if W 2 B 2 V, the W 2 BW 1 for some W 1 BV, ad so diam(w 2 ) ( ) diam(w 1) diam(v ). + 1 More geerally for W m B m V, ( ) m diam(w m ) diam(v ). (11) + 1

32 Now, for ay l = y 0,..., y : [y 0,..., y ], if W B the l(w ) B[y 0,..., y ]. This is easily see by oticig that if b is a baryceter for some k-dimesioal face of, where 0 k, the l(b) will be the baryceter of the correspodig face of [y 0,..., y ]. Referrig to (10), it follows that α im 1 maps W im B to some α im 2 maps W i m 1 B 2 to some W i m 1 BW im 1 B 2, W i m 2 B 2 W im 2 B 3, ad so o util we arrive at α i1 : W i1, which maps W i 2 B m 1 to some W i 1 B m 1 W i1 B m. m It is see, the, that S (σ) is a liear combiatio of maps ˆσ of the form σ α, where each α maps from oto some W m B m. Thus each ˆσ is effectively a restrictio of σ to some W m B m, where ( ) m diam(w m ) diam( ) + 1 by (11). Sice X = k U k ad σ : X is cotiuous, the collectio {σ 1 (Uk )} forms a ope cover for. Sice is compact there exists some ɛ σ > 0 (a Lebesgue umber for the cover) such that, for ay set W with diam(w ) < ɛ σ, there exists some k for which W σ 1 (Uk ). Let m be sufficietly large so that ( ) m diam( ) < ɛ σ. + 1 S m The (σ) is a chai of maps ˆσ, each havig image ˆσ(W ) i X for some set W with diam(w ) < ɛ σ, so that W σ 1 (Uk ) for some k. Hece, each sigular -simplex ˆσ i the chai S m m (σ) maps ito some U k X, ad therefore S (σ) C U (X). For each sigular -simplex σ let ad defie D : C (X) C +1 (X) by From (9) we have ( +1 D mσ m σ = mi{m Z : S m (σ) C U (X)}, σ S mσ D (σ) = D mσ (σ). (σ) = ( +1 D mσ )(σ) + (D mσ 1 )(σ) )(σ) + D 1 ( σ) = σ [ S m σ (σ) + (D mσ 1 )(σ) D 1 ( σ) ] ( +1 D )(σ) + ( D 1 )(σ) = (σ) ϕ (σ), 32

33 33 where ϕ (σ) is defied to be the expressio i the brackets, ad so +1 D + D 1 = ϕ. (12) S mσ It will be show that ϕ : C (X) C U (X). Clearly (D mσ 1 )(σ) D 1 ( σ). Let σ i = σ [u0,...,û i,...,u ] δ 1 i m σi = mi{m Z : S m 1 (σ i ) C U (X)}, ad observe that m σi m σ for each 0 i. Now, (D mσ 1 )(σ) D 1 ( σ) = ( 1) i D 1(σ mσ i ) Sice for j m σi ad = = i=0 ( mσ 1 ( 1) i ( T 1 i=0 i=0 it s readily see from (13) that m σ 1 j=0 j=m σi ( 1) i ( T 1 S j 1(σ i ) C U 1(X) T 1 : C U 1(X) C U (X), (σ) C U (X), so attetio turs to ad i=0 ( 1) i D m σi 1(σ i ) m σi 1 j S 1)(σ i ) ( T 1 j=0 (D mσ 1 )(σ) D 1 ( σ) C U (X), S j 1)(σ i ) S j 1)(σ i ). (13) ad therefore ϕ (σ) C U (X). To show that the maps ϕ costitute a chai map C(X) C U (X) as illustrated i the diagram C +1 (X) +1 C (X) C 1 (X) ) ϕ +1 ϕ ϕ 1 C+1(X) U +1 C U (X) C 1(X) U we show that ϕ 1 = ϕ. Usig (12), we obtai ad ϕ = +1 D D 1 = D 1 ϕ 1 = 1 D 1 D 2 1 = D 1, which verifies commutativity. Now, the iclusio maps ι costitute a chai map C U (X) C(X), ad sice (12) implies +1 D + D 1 = ι ϕ (14)

34 for all, it follows that {ι ϕ } is chai homotopic to { }. Next, we show that ϕ ι = U, where U : C U (X) C U (X) is the idetity map. Let σ : U k for some k, so that σ is a basis elemet for C U (X). Sice σ i : 1 U k for each i, we have m σi = 0 as well as m σ = 0. Hece, (ϕ ι )(σ) = ϕ (σ) = ( +1 D D 1 )(σ) = σ +1 (D 0 (σ)) = σ +1 (0) ( 1) i D 1(σ 0 i ) i=0 ( 1) i (0) = σ, which completes the argumet ad so U ϕ ι = 0. Defiig 0 U : C U (X) C U +1(X) to be the trivial homomorphism, what we have show is i= U + 0 U 1 = 0 = U ϕ ι, ad therefore {ϕ ι } is chai homotopic to { U }. At last we see that {ι } is a chai-homotopy equivalece, ad so by Propositio 2.5 each ι : H U (X) H (X) is a isomorphism. Proof of the Excisio Theorem. Let U = {A, B}, where A ad B are subspaces of X such that A B = X. Defie C U (X, A) = C U (X)/C U (A). The maps ι : C U (X) C (X) iduce homomorphisms o quotiet groups ῑ : C U (X, A) C (X, A) give by ῑ (α + C U (A)) = ι (α) + C (A) = α + C (A) for each α C U (X). C(X, A), It s easy to verify that the maps ῑ form a chai map C U (X, A) C+1(X, U +1 A) U C U (X, A) U C 1(X, U A) 34 ῑ +1 ῑ ῑ 1 C +1 (X, A) +1 C (X, A) C 1 (X, A) with the maps i the diagram beig defied as i sectio 2.4, ad the maps U beig the obvious restrictios. Thus each ῑ i tur iduces a homomorphism o homology groups ῑ : Ker U / Im U +1 := H U (X, A) Ker / Im +1 := H (X, A) defied accordig to the geeral algebraic formula give i sectio 2.1. Next, the maps ϕ : C (X, A) C U (X, A) defied by ϕ (α + C (X)) = ϕ (α) + C U (A) for each α C (X) give rise to a chai map C(X, A) C U (X, A); ad so, defiig maps D ad o C (X, A) i the caoical fashio from the maps D ad i the proof of Lemma 2.22, the diagram