Chain complexes
Homology is defined using algebraic objects called chain complexes.
A chain complex is a sequence of abelian groups, (C n )n∈ Z, together with homomorphisms dn: Cn →Cn−1 for n∈Z, such thatdn−1◦dn = 0.
LetR be an (associative) ring with unit 1 R A chain complex of R-modules can analogously be defined as a sequence of R-modules (C n )n∈ Z with R-linear maps d n : C n →Cn−1 such that d n−1 ◦d n = 0.
We fix the following terminology:
• The homomorphisms d n are called differentials or boundary operators.
• The elements x∈C n are called n-chains.
• Any x∈Cn such thatdnx= 0 is called an n-cycle We denote the group of n-cycles by
• Any x∈C n of the formx=d n+1 y for somey ∈C n+1 is called an n-boundary.
The cycles and boundaries form subgroups of the group of chains The identity dn◦dn+1 = 0 implies that the image of d n+1 is a subgroup of the kernel ofd n and thus
We often drop the subscript n from the boundary maps and just write C ∗ for the chain complex.
The abelian group H n (C) := Z n (C)/B n (C) is called the nth homology group of the complex
In homology theory, we represent the equivalence class of a cycle \( c \) in \( Z_n(C) \) as \([c] \in H_n(C)\) Two chains \( c \) and \( c_0 \) in \( C_n \) are considered homologous if their difference \( c - c_0 \) is a boundary This relationship establishes an equivalence relation among chains A complex is classified as acyclic if its homology groups are trivial, except in degree 0.
0 otherwise Here, the only non-zero differential is d 1 ; let it be the multiplication with N ∈N, then
2 Take C n =Z for all n ∈Zand consider differentials dn (id Z n odd
The homology of this chain complex vanishes in all degrees.
3 ConsiderC n =Zfor alln∈Z again, but let all boundary maps be trivial The homology of this chain complex equals Z in all degrees.
We need morphisms of chain complexes:
Let C∗ and D∗ be two chain complexes A chain map f: C∗ → D∗ is a sequence of homomor- phisms f n : C n →D n such thatd D n ◦f n =fn−1◦d C n for all n, i.e., the diagram
A chain mapf sends cycles to cycles, since d D n f n (c) =fn−1(d C n c) = 0 for a cycle c , and boundaries to boundaries, since f n (d C n+1 λ) =d D n+1 f n+1 (λ)
We therefore obtain an induced map of homology groups
1 There is a chain map from the chain complex mentioned in Example 1.1.4.1 to the chain complex D ∗ that is concentrated in degree zero and has D 0 =Z/NZ.
Note that (f 0 )∗ is an isomorphism on the zeroth homology group; all homology groups are isomorphic.
2 Are there chain maps between the complexes from Examples 1.1.4.2 and 1.1.4.3?
If f: C∗ →D∗ and g: D∗ →E∗ are two chain maps, then H n (g)◦H n (f) = H n (g◦f) for all n.
We next study situations in which two chain maps induce the same map on homology.
A chain homotopy H between two chain maps f, g: C∗ →D∗ is a sequence of homomorphisms (H n )n∈ Z with H n : C n →D n+1 such that for alln d D n+1 ◦Hn+Hn−1◦d C n =fn−gn. d
If such an H exists, then the chain maps f and g are said to be (chain) homotopic We write f 'g.
We will see in Section 1.4 geometrically defined examples of chain homotopies.
1 Being chain homotopic is an equivalence relation on chain maps.
2 If f and g are homotopic, then H n (f) =H n (g) for all n.
In the context of homotopy theory, if H represents a homotopy from the function f to g, then the negative homotopy −H serves as a homotopy from g back to f Additionally, every chain map f is homotopic to itself through a chain homotopy where H equals 0 Furthermore, if f is homotopic to g through H and g is homotopic to h through K, it follows that f is homotopic to h via the combined homotopy H + K.
2 We have for every cyclec∈Zn(C∗):
Here, the class of the first term vanishes; in the second term d C n c= 0, since cis a cycle.
Let f: C∗ → D∗ be a chain map We call f a chain homotopy equivalence, if there exists a chain map g: D ∗ →C ∗ such that g◦f 'id C ∗ and f◦g ' id D ∗ The chain complexes C ∗ and
D∗ are said to be chain homotopically equivalent.
Chain homotopically equivalent chain complexes exhibit isomorphic homology, but the converse is not necessarily true; chain complexes with isomorphic homology may not be chain homotopically equivalent For instance, as demonstrated in Example 1.1.6.1, there exists no non-zero morphism of abelian groups from ZN to Z.
If C∗ and C ∗ 0 are chain complexes, then their direct sum, C∗⊕C ∗ 0 , is the chain complex with
(C∗⊕C ∗ 0 ) n =C n ⊕C n 0 =C n ×C n 0 with differential d=d⊕d 0 given by d ⊕ (c, c 0 ) = (dc, d 0 c 0 ).
Similarly, if (C∗ (j) , d (j) )j∈J is a family of chain complexes, then we can define their direct sum as follows:
C n (j) as abelian groups and the differential d⊕ is defined via the property that its restriction to the jth summand is d (j)
Singular homology
The fundamental group of a topological space X is defined by examining the homotopy classes of maps from the circle S¹ to X In contrast, singular homology is defined using maps from higher-dimensional objects, specifically simplices.
Letv 0 , , v n be n+ 1 points in R n+1 Consider the convex hull
If the vectors v 1 −v 0 , , v n −v 0 are linearly independent, then K(v 0 , , v n ) is the simplex generated by v0, , vn We denote such a simplex by simp(v0, , vn).
Note that simplex really means “simplex with an ordering of its vertices”.
1 Denote by e i ∈R n+1 the vector that has an entry 1 in coordinate i+ 1 and is zero in all other coordinates The standard topological n-simplex is ∆ n := simp(e 0 , , e n ).
2 The first examples of standard topological simplices are:
• ∆ 1 is the line segment in R 2 between e 0 = (1,0)∈R 2 and e 1 = (0,1)∈R 2
• ∆ 2 is a triangle in R 3 with vertices e 0 , e 1 and e 2 and ∆ 3 is homeomorphic to a tetrahedron.
3 The coordinate description of the standard n-simplex in R n+1 is
We consider the standard simplex ∆ n as a subset ∆ n ⊂R n+1 ⊂R n+2 ⊂
The boundary of ∆ 1 consists of two copies of ∆ 0 , the boundary of ∆ 2 consists of three copies of ∆ 1 In general, the boundary of ∆ n consists of n+ 1 copies of ∆ n−1
We describe the boundary by the following (n+ 1) face maps for 06i6n di =d n−1 i : ∆ n−1 ,→∆ n ; (t0, , tn−1)7→(t0, , ti−1,0, ti, , tn−1).
The image ofd n−1 i in ∆ n is the face that is opposite to the vertexe i It is the (n−1)-simplex generated by the n−1-tuple e 0 , , ei−1, e i+1 , , e n of vectors in R n+1
Concerning the composition of face maps, the following rule holds: d n−1 i ◦d n−2 j =d n−1 j ◦d n−2 i−1 , for all 06j < i6n.
Example: face maps for ∆ 0 and composition into ∆ 2 : d 2 ◦d 0 =d 0 ◦d 1
Both expressions yield d n−1 i ◦d n−2 j (t 0 , , tn−2) = (t 0 , , tj−1,0, t j , ti−2,0, ti−1, , tn−2)
Let X be an arbitrary topological space, X 6= ∅ A singular n-simplex in X is a continuous map α: ∆ n →X.
It is important to note that the function α only needs to be continuous; there is no necessity for it to be smooth or injective Unlike the definition of the fundamental group, we do not identify simplices, nor do we designate a specific base point.
The boundary of a 1-simplex, represented as an interval, is defined by the difference between its endpoints To understand this concept, it is essential to explore the addition and subtraction of 0-simplices.
We recall some algebraic notions:
1 Any abelian group A can be seen as a Z-module with n.a :=a+ .+a
Abelian groups are closely related to Z-modules, as they can be represented in a bijective manner An abelian group A is considered free over a subset B, which serves as a Z-basis, if every element a in Z can be expressed uniquely as a Z-linear combination of the elements in B.
2 The group Z r is free abelian with basis {e 1 , , e r } with e i = (0, ,0,1,0, ,0) The group Z2 is not free, since it does not admit a basis: the vector 1 ∈ Z2 is not free since
A free abelian group F with basis B is uniquely characterized by its universal property, which states that any function f: B → A, where A is any abelian group, can be uniquely extended to a group homomorphism h: F → A This means that for every element b in the basis B, the homomorphism h satisfies h(b) = f(b).
4 Any subgroup of a free abelian group F is a free abelian group of smaller rank.
Let X be a topological space Let S n (X) be the free abelian group generated by all singular n-simplices inX We call S n (X)the n-th singular chain module of X.
1 Elements of the singular chain group S n (X) are thus sums P i∈Iλ i α i with λ i ∈ Z and λ i = 0 for almost all i∈I and α i : ∆ n →X a singular n-simplex All sums are effectively finite sums.
For every positive integer n, there exist non-trivial elements in the space S n (X), given that X is not empty By selecting a point x₀ from X, we can define a constant map κ x₀ : ∆ n → X, which represents a singular n-simplex α It is important to note that we define S n (∅) as 0 for all n greater than zero.
3 By the universal property 1.2.5.3, to define group homomorphisms from S n (X) to some abelian group, it suffices to define such a map on generators.
LetXbe any topological space As an example, we computeS0(X): a continuous mapα: ∆ 0 →
The value α(e 0 ) =: x α ∈ X determines the point in the set X, allowing a singular 0-simplex P i∈Iλ i α i to be expressed as a formal sum of points P i∈Iλ i x α i with λ i ∈ Z This concept is prevalent in complex analysis, particularly in counting the zeros and poles of a meromorphic function, where multiplicities contribute to an element in S 0 (X) Additionally, in algebraic geometry, a divisor is recognized as an element within S 0 (X).
Using the face mapsd n−1 i : ∆ n−1 →∆ n from Example 1.2.3.3, we define a group homomorphism
∂i:Sn(X)→Sn−1(X)on generators by precomposition with the face map
∂i(α) =α◦d n−1 i and call it the ith face of the singular simplex α.
On S n (X), we thus get by Z-linear extension ∂ i (P jλ j α j ) = P jλ j (α j ◦d n−1 i ).
The face maps on S n (X) satisfy the simplicial relations
The relation follows immediately from the relation d n−1 i ◦d n−2 j =d n−1 j ◦d n−2 i−1 , for all 06j < i6n. in Lemma 1.2.3
We define the boundary operator on singular chains as∂:S n (X)→Sn−1(X)as the alternating sum ∂ =Pn i=0(−1) i ∂ i
The map ∂ is a boundary operator, i.e ∂◦∂ = 0.
This is an immediate consequence of the simplicial relations in Lemma 1.2.10
We therefore obtain for a topological space X a complex of (free) abelian groups,
.→S n (X)−→ ∂ Sn−1(X)−→ ∂ −→ ∂ S 1 (X)−→ ∂ S 0 (X)→0, the singular chain complex, S∗(X) We abbreviate the group Zn(S∗(X)) of cycles by Zn(X), the group B n (S∗(X)) of boundaries by B n (X) and the n-th homology group H n (S∗(X)) by
For a space X, the abelian group H n (X)is called the nth singular homology group of X.
1 Note that all 0-chains are 0-cycles, Z 0 (X) =S 0 (X).
2 The boundary of a 1-chain α: ∆ 1 →X is
∂α=α◦d 0 −α◦d 1 =α(e 1 )−α(e 0 ) which justifies the name “boundary”.
3 To find an example of a 1-cycle, consider a 1-chain c=α+β+γ, where we take singular 1-simplices α, β, γ: ∆ 1 → X such that α(e 1 ) = β(e 0 ), β(e 1 ) = γ(e 0 ) and γ(e 1 ) = α(e 0 ). Calculate ∂α=∂ 0 α−∂ 1 α=α(e 1 )−α(e 0 ) and similarly for β and γ to find ∂c= 0 This motivates the word “cycle”.
We need to understand how continuous maps of topological spaces interact with singular chains and singular homology.
Let f: X → Y be a continuous map The map f n = S n (f) : S n (X) → S n (Y) is defined on generators α: ∆ n →X by postcomposition fn(α) =f ◦α: ∆ n −→ α X −→ f Y.
For any continuous map f: X →Y we have commuting diagrams
Sn−1(X) f n−1 // Sn−1(Y), i.e (f n )n∈ Z is a chain map and hence induces by the remarks following Definition 1.1.5, a map
By definition, we have for a singular n-simplex α: ∆ n →X by the associativity of the compo- sition of maps
1 The identity map on X induces the identity map on Hn(X) for all n>0 and if we have a composition of continuous maps
In categorical terms, S n (−) and H n (−) function as functors that map topological spaces and continuous maps to the category of abelian groups When combined, these functors create S∗(−), which operates as a functor transforming topological spaces and continuous maps into chain complexes of abelian groups, utilizing chain maps as morphisms.
3 One implication of Lemma 1.2.16 is that homeomorphic spaces have isomorphic homology groups:
In Theorem 1.4.7, we will see the stronger statement that homotopic maps induce the same morphism in homology.
Our first (not too exciting) calculation is the following:
The homology groups of a one-point space pt are trivial but in degree zero,
For every n > 0 there is precisely one continuous map α: ∆ n → pt, the constant map We denote this map by κ n Then the boundary of κ n is
For all n we have Sn(pt)∼= Z generated by κn and therefore the singular chain complex looks as follows:
The homology groups H 0 and H 1
We start with the following observation:
For any topological space X, there is a homomorphism ε: H 0 (X)→Z with ε6= 0 for X 6=∅.
• If X 6= ∅, we have a unique morphism X → pt of topological spaces which induces by Lemma 1.2.16 a morphism of chain complexes S∗(X) → S∗(pt) It maps any 0-simplex α: ∆ 0 →X to
∆ 0 → α X →pt, the generator of H0(pt), the constant map κ0 : ∆ 0 →pt, cf Proposition 1.2.18.
The map \( \tilde{\epsilon}: S_0(X) \to \mathbb{Z} \), defined by \( \tilde{\epsilon}(\alpha) = 1 \) for any generator \( \alpha: \Delta^0 \to X \), demonstrates that \( \tilde{\epsilon} \left( \sum_{i \in I} \lambda_i \alpha_i \right) = \sum_{i \in I} \lambda_i \) establishes a well-defined homology map on \( S_0(X) \) Since only finitely many \( \lambda_i \) are non-trivial, this results in a finite sum, ensuring clarity in the homological structure.
LetS0(X)3c=∂bbe a boundary and writeb=P i∈Iνiβi withβi: ∆ 1 →X and a finite set I Then we get
We said thatS 0 (∅) is zero, so H 0 (∅) = 0 In this case, we define ε to be the zero map.
If X 6= ∅, then any singular 0-simplex α: ∆ 0 → X can be identified with its image point, so the map ε on S 0 (X) counts points in X with multiplicities.
If X is a path-connected, non-empty space, thenε:H 0 (X)→ ∼ = Z.
1 As X is non-empty, there is a point x∈ X and the constant map κ x with value x is an element in S 0 (X) with ε(κ x ) = 1 Therefore, the group homomorphism ε is surjective.
2 For any other point y ∈ X there is a continuous path ω: [0,1]→ X with ω(0) =x and ω(1) =y We define a singular 1-simplexα ω : ∆ 1 →X as α ω (t 0 , t 1 ) =ω(1−t 0 ) for t 0 +t 1 = 1, 06t 0 , t 1 61 Then
∂(α ω ) =∂ 0 (α ω )−∂ 1 (α ω ) =α ω (e 1 )−α ω (e 0 ) =α ω (0,1)−α ω (1,0) =κ y −κ x , and the two singular 0-simplices κ x , κ y in the path connected space X are homologous. This shows that ε is injective.
Note that in the proof, we associated to a continuous pathω inX from xto y a 1-simplex α ω on X with ∂α ω =κ y −κ x In the sequel, we will identify them frequently.
IfX is a disjoint union,X =F i∈IX i , such that allX i are non-empty and path-connected, then
This gives an interpretation of the zeroth homology group ofX: it is the free abelian group generated by the path-components of X.
The singular chain complex of X splits as the direct sum of chain complexes of the X i :
S n (X i ) for all n Boundary summands ∂i stay in a component, in particular,
S 0 (X i )∼=S 0 (X) is the direct sum of the boundary operators ∂: S 1 (X i ) → S 0 (X i ) and the claim follows from
Next, we relate the homology group H 1 to the fundamental group π 1 To this end, we see continuous paths ω inX as 1-simplicesα ω , as in the proof of Proposition 1.3.2.
Let ω1, ω2, ω be paths in a topological space X.
1 Constant paths are null-homologous.
2 If ω 1 (1) = ω 2 (0), we can define the concatenation ω 1 ∗ω 2 of ω 1 followed by ω 2 Then α ω 1 ∗ω 2 −α ω 1 −α ω 2 is a boundary.
3 If ω1(0) = ω2(0), ω1(1) = ω2(1) and if ω1 is homotopic to ω2 relative to {0,1}, then αω 1 and α ω 2 are homologous as singular 1-chains.
4 Any 1-chain of the formαω∗ω ¯ is a boundary Here, ¯ω(t) :=ω(1−t).
1 Denote by c x the constant path on x ∈ X Consider the constant singular 2-simplex α(t 0 , t 1 , t 2 ) = x Then∂α =c x −c x +c x =c x
2 We define a singular 2-simplex β: ∆ 2 →X onX as follows.
We define β on the boundary components of ∆ 2 as indicated and prolong it constantly along the sloped inner lines Then
3 Let H: [0,1]×[0,1]→X a homotopy from ω 1 toω 2 As we have that H(0, t) =ω 1 (0) ω 2 (0), we can factor H over the quotient [0,1]×[0,1]/{0} ×[0,1]∼= ∆ 2 with induced map h: ∆ 2 →X Then
∂h=h◦d 0 −h◦d 1 +h◦d 2 The first summand is null-homologous by 1., because it is constant (with value ω 1 (1) ω 2 (1)), the second one is ω 2 and the last is ω 1 , thus ω 1 −ω 2 is null-homologous.
4 Consider a singular 2-simplexγ: ∆ 2 →X as indicated below.
Let X be path-connected and x ∈ X Let h: π 1 (X, x) → H 1 (X) be the map, that sends the homotopy class [ω] π 1 of a closed path ω to its homology class [ω] = [α ω ] H 1 This map is called the Hurewicz-homomorphism 1
Lemma 1.3.4.3 ensures that h is well-defined and by Lemma 1.3.4.2 h([ω1][ω2]) =h([ω1∗ω2]) 1.3.4.2 = [ω1] + [ω2] =h([ω1]) +h([ω2]) thushis a group homomorphism For a closed pathωwe have by Lemma 1.3.4.4 that [¯ω] =−[ω] in H 1 (X).
Recall that the commutator subgroup [G, G] of G is the smallest subgroup of a group G containing all commutators [g, h] :=ghg −1 h −1 for all g, h∈G.
A normal subgroup of G is defined such that if c belongs to the commutator subgroup [G, G], then for any element g in G, the expression gcg^(-1)c^(-1) is a commutator Additionally, due to the closure property of subgroups, the element gcg^(-1)c is also included in the commutator subgroup [G, G].
Let Gbe an arbitrary group, then its abelianization, Gab, is the quotient group G/[G, G].
The abelianization comes with a projection G→G ab It can be characterized by the universal property that any group homomorphism G→A with A abelian factorizes uniquely as
Let X be a path-connected non-empty space Since H1(X) is abelian, the Hurewicz homomor- phism factors over the abelianization of π 1 (X, x) It induces an isomorphism π 1 (X, x) ab ∼=H 1 (X) , i.e. π 1 (X, x) h // p
We develop an inverse function φ to the mapping hab by selecting a path uy from the base point x to any point y in the space X For the base point x, we define ux as a constant path that remains at x.
Let α represent an arbitrary singular 1-simplex, and define y i as α(e i ) We introduce the map φ˜: S 1 (X) → π 1 (X, x) ab, which acts on the generator α of S 1 (X) by associating it with the class of the closed path ˜φ(α) = [u y 0 ∗ω α ∗u¯ y 1 ] This map φ is then extended linearly across all of S 1 (X), with the understanding that the composition in the fundamental group π 1 is expressed multiplicatively.
• We have to show that ˜φ is trivial on boundaries, so let β: ∆ 2 →X a singular 2-simplex. Then φ(∂β) = ˜˜ φ(β◦d 0 −β◦d 1 +β◦d 2 ) = ˜φ(β◦d 0 ) ˜φ(β◦d 1 ) −1 φ(β˜ ◦d 2 ).
Abbreviating β◦d i with α i , we get as a result
In this article, we demonstrate that the image of the homomorphism φ is abelian By simplifying the paths ¯u y 1 ∗u y 1 and ¯u y 2 ∗u y 2, we arrive at the expression [u y 0 ∗α 2 ∗α 0 ∗α¯ 1 ∗u¯ y 0] Notably, the term α 2 ∗α 0 ∗α¯ 1 represents a closed path that delineates the boundary of the singular 2-simplex β, confirming its null-homotopic nature in the space X Consequently, we conclude that φ(∂β˜) = 1, indicating that the map φ can be extended to a homomorphism φ: H 1 (X) → π 1 (X, x) that is abelian.
• The composition φ◦h ab evaluated on the class of a closed path ω gives φ◦h ab [ω] π 1 =φ[ω] H 1 = [u x ∗ω∗u¯ x ] π 1
But we chose u x to be constant, thus φ◦h ab = id.
Ifc=P λ i α i is a 1-cycle, thenh ab ◦φ(c) is of the form [c+D ∂c ] where theD ∂c -part comes from the contributions of the uy i The fact that ∂(c) = 0 implies that the summands in
D ∂c cancel off and thus h ab ◦φ= id H 1 (X)
Note that abelianization of an abelian group is the group itself: G ∼= G ab Whenever the fundamental group is abelian, we thus have H1(X)∼=π1(X, x).
Standard results on the fundamental group π 1 yield explicit results for the following first ho- mology groups:
Homotopy invariance
The main goal of this section is to show that two continuous maps that are homotopic induce identical maps on homology groups.
• Let α: ∆ n → X a singular n-simplex; consider two homotopic maps f, g : X → Y The homotopy
We consider a map from the prism over ∆ n, represented as ∆ n ×[0,1] α×id → X×[0,1]→ H Y, which transitions from f◦α to g◦α Our goal is to derive a chain homotopy between the chain maps S(f) and S(g) within the context of singular chain complexes.
• To that end we have to cut the (n+ 1)-dimensional prism ∆ n ×[0,1] into (n+ 1)-simplices.
In low dimensions this is intuitive:
– The one-dimensional prism ∆ 0 ×[0,1] is homeomorphic to the standard 1-simplex
– The two-dimensional prism ∆ 1 ×[0,1] ∼= [0,1] 2 (which has the shape of a square) can be cut in two triangles, i.e into two copies of the standard 2-simplex ∆ 2
– ∆ 2 × [0,1] is a 3-dimensional prism and that can be glued together from three tetrahedrons, e.g like
In general, we compose the (n+ 1)-dimensional prism ∆ n ×[0,1] from n+ 1 copies of the standard simplex ∆ n+1 :
For a given n ∈N 0 , definen+ 1 injections pi: ∆ n+1 → ∆ n ×[0,1]
(t 0 , , t n+1 ) 7→ ((t 0 , , ti−1, t i +t i+1 , t i+2 , , t n+1 ), t i+1 + .+t n+1 ) with i= 0, , n These are(n+ 1)-simplices on the prism on the topological space∆ n ×[0,1].
• The image of the standard basis vectors e k with k = 0,1, n+ 1 is p i (e k ) ((e k ,0), for 06k 6i, (ek−1,1), for k > i.
For example, in the case n = 1, we have p 0 e 0 7→ e 0 p 1 e 0 7→ e 0 e 1 7→ e 0 +e 2 e 1 7→ e 1 e 2 7→ e 1 +e 2 e 2 7→ e 1 +e 2
• For all n>0, we obtain n+ 1 group homomorphisms
P i : S n (X)→S n+1 (X×[0,1]) fori= 0,1, nwhich is defined on a generatorα: ∆ n →XofS n (X) via precomposition:
• For k= 0,1 let j k :X →X×[0,1] be the inclusion x7→(x, k).
The group homomorphisms P i satisfy the following relations:
• For the first point, note that for α: ∆ n →X, ∂ 0 ◦P 0 (α) is the singular n-simplex
∆ n d → 0 ∆ n+1 → p 0 ∆ n ×[0,1] α×id → X×[0,1]. The composition of the first two maps on ∆ n evaluates to p 0 ◦d 0 (t 0 , , t n ) = p 0 (0, t 0 , , t n ) = ((t 0 , , t n ), n
X i=0 t i ) = ((t 0 , , t n ),1) and thus the whole map equals
• Similarly, we compute p n ◦d n+1 (t 0 , , t n ) = p n (t 0 , , t n ,0) = ((t 0 , , t n ),0) and deduce ∂ n+1 ◦P n =S n (j 0 ).
• For the third identity, one checks that pi ◦ di = pi−1 ◦ di on ∆ n : both give ((t 0 , , t n ),Pn j=it j ) on (t 0 , , t n ).
• For d) in the case i>j+ 1, consider the following diagram
Checking coordinates one sees that this diagram commutes: both give ((t 0 , , tj−1,0, ti−1+t i , t n ),Pn j=it j ) on (t 0 , , t n ).
The remaining case follows from a similar observation.
For each n >0, we define a group homomorphism
P: S n (X)→S n+1 (X×[0,1]) as the alternating sum P =Pn i=0(−1) i P i
The group homomorphismsP provide a chain homotopy between the chain maps S(j 0 ), S(j 1 ) :
We evaluate the left hand side on a singularnsimplexα: ∆ n →X and find from the definitions
If we single out the terms involving the pairs of indices (0,0) and (n, n+ 1) in the first sum, we are left with by Lemma 1.4.4.1 and 2.
According to Lemma 1.4.4, only the first two summands remain, as the terms in the initial sum where i equals j and i equals j minus one cancel out, as demonstrated in Lemma 1.4.4.3 The other remaining terms also cancel using the same method outlined in the proof of Lemma 1.2.12.
So, finally we can prove the main result of this section:
If f, g: X →Y are homotopic maps, then they induce the same map on homology.
Let H: X×[0,1]→Y be a homotopy from f tog, i.e H◦j 0 =f and H◦j 1 =g Set
We claim that (K n ) n is a chain homotopy between the two chain maps (S n (f)) n and (S n (g)) n Note that H :X×I →Y induces a chain map (S n (H)) n Therefore we get
Hence the two chain maps S(f) and S(g) are chain homotopic; by Proposition 1.1.9.2, we have
1 If two spaces X, Y are homotopy equivalent, thenH∗(X)∼=H∗(Y).
2 In particular, if X is contractible, then
1 Since R n is contractible for all n, the above corollary implies that its homology is trivial but in degree zero where it consists of the integers.
2 As the M¨obius strip is homotopy equivalent to S 1 , we know that their homology groups are isomorphic.
3 The zero section of a vector bundle induces a homotopy equivalence between the base and the total space, hence these two have isomorphic homology groups.
The long exact sequence in homology
In a typical scenario, we consider a subspace A within a topological space X, where we possess certain information about either A or X Our goal is to compute the homology of the other space by leveraging this partial knowledge.
To explore topological applications, it's essential to understand algebraic techniques related to chain complexes Specifically, a short exact sequence of chain complexes leads to the formation of a long exact sequence in homology.
1 Let A, B, C be abelian groups and
A f // B g // C a sequence of homomorphisms Then this sequence is exact, if the image of f equals the kernel of g.
f i+1 // A i f i // A i+1 f i−1 // of homomorphisms of abelian groups (indexed over the integers) is called (long) exact, if it is exact at every A i , i.e the image of f i+1 equals the kernel of f i for all i.
3 An exact sequence of the form
0 // A f // B g // C // 0 is called a short exact sequence.
0 // U ι // A is exact, iff ι: U →A is a monomorphism The sequence
B % // Q // 0 is exact, iff %: B →Q is an epimorphism Finally, Φ : A→A 0 is an isomorphism, iff the sequence 0 // A φ // A 0 // 0 is exact.
0 // A f // B g // C // 0 is exact, iff f is injective, the image of f equals the kernel of g and g is an epimorphism.
4 Another equivalent description is to view a long exact sequence as a chain complex with vanishing homology groups Homology measures the deviation from exactness.
If A∗, B∗, C∗ are chain complexes andf∗: A∗ →B∗, g∗: B∗ →C∗ are chain maps, then we call the sequence of chain complexes
A∗ f ∗ //B∗ g ∗ //C∗ exact, if the image of f n is the kernel ofg n for all n ∈Z.
Thus such an exact sequence of chain complexes is a commuting double ladder
in which every row is exact and where in the columns we have differentials, i.e d◦d= 0.
Let pbe a prime, then the diagram
0 0 0 has exact rows and columns It is an exact sequence of chain complexes Here, π denotes the appropriate canonical projection map.
If 0 // A∗ f //B∗ g //C∗ //0 is a short exact sequence of chain complexes, then there exists a homomorphism δ n : H n (C∗)→Hn−1(A∗) for alln ∈Z which is natural, i.e if
0 //C ∗ 0 // 0 is a commutative diagram of chain complexes in which the rows are exact, thenH n−1 (α)◦δ n δ n 0 ◦H n (γ),
The method of proof is an instance of a diagram chase The homomorphism δn is called connecting homomorphism.
We show the existence of a δ first and then prove that the constructed map satisfies the naturality condition. a) Definition of δ:
We work with the following maps:
For any element \( c \) in \( C_n \) with \( d(c) = 0 \), we can select a preimage \( b \) in \( B_n \) such that \( g_n(b) = c \), leveraging the surjectivity of \( g_n \) Given that \( dg_n(b) = d(c) = 0 \), it follows that \( g_{n-1}(db) \) is also zero, indicating that \( db \) is in the kernel of \( g_{n-1} \) and consequently in the image of \( f_{n-1} \) Therefore, there exists an \( a \) in \( A_{n-1} \) such that \( f_{n-1}(a) = db \) Since \( f_{n-2}(da) = 0 \) and \( f_{n-2} \) is injective, it can be concluded that \( a \) is a cycle.
To verify that δ is well-defined, we consider two distinct preimages, b and b0, such that g n b = g n b0 = c This leads to the equation g n (b - b0) = 0, implying the existence of an element ˜a in A n for which f n ˜a = b - b0 We define a0 as a - d˜a, which belongs to A n-1 Consequently, we find that f n-1 a0 = f n-1 a - f n-1 d˜a b + d b0, with the condition that f n-1 d˜a is small compared to n˜a db0 Given that f n-1 is injective, a0 is uniquely determined Since a is homologous to a0, we conclude that [a] = [a0] = δ[c], demonstrating that δ is independent of the chosen preimage b.
To ensure the value remains consistent when adding a boundary term to c, we define c₀ as c + d˜c for some ˜c in Cn+1 By selecting preimages of c and ˜c under the surjective maps gₙ and gₙ₊₁, we identify b and ˜b such that gₙ(b) = c and gₙ₊₁(˜b) = ˜c Consequently, the element b₀ = b + d˜b has a boundary of db₀, guaranteeing that both choices lead to the same outcome, a.
Therefore the connecting morphism δ n : H n (C∗)→Hn−1(A∗) is well-defined. b) We have to show that δ is natural with respect to maps of short exact sequences.
Let c ∈ Z n (C ∗ ), then δ[c] = [a] for some b ∈ B n with g n b = c and a ∈ A n−1 with fn−1a Therefore, Hn−1(α)(δ[c]) = [αn−1(a)].
In the context of the second long exact sequence, we find that the connecting homomorphism δ 0 relates the elements δ 0 [γ n (c)] and [αn−1(a)], demonstrating that δ 0 ◦ H n (γ) is equal to Hn−1(α) ◦ δ This relationship highlights the coherence between the homomorphisms involved, affirming the structure of the sequence.
With this auxiliary result at hand we can now prove the main result in this section:
Proposition 1.5.6 (Long exact sequence in homology).
For any short exact sequence
0 // A∗ f //B∗ g //C∗ //0 of chain complexes we obtain a long exact sequence of homology groups
We haveH n (g)◦H n (f)[a] = [g n (f n (a))] = 0, because the composition ofg n andf n is zero. This proves that the image of H n (f) is contained in the kernel of H n (g).
For the converse, let [b] ∈ H n (B∗) with [g n b] = 0 Since g n b is a boundary, there exists c∈C n+1 with dc=g n b As g n+1 is surjective, we find ab 0 ∈B n+1 with g n+1 b 0 =c Hence g n (b−db 0 ) = g n b−dg n+1 b 0 dc= 0.
Exactness at B n indicates that we can find an element a in A n such that f n a equals b minus db 0 Since b is a cycle, we have that fn−1(da) is small and equals d(b−db 0), resulting in db equaling zero The injectivity of the map f n−1 implies that da must also be zero Consequently, f n a is homologous to b, leading to the conclusion that H n (f)[a] equals [b] Thus, we can assert that the kernel of H n (g) is contained within the image of H n (f).
Let [b] ∈ H n (B∗), then δ[g n b] = 0 because b is a cycle, so 0 is the only preimage under the injective mapfn−1 of db= 0 Therefore the image of H n (g) is contained in the kernel of the connecting morphism δ.
Now assume that δ[c] = 0, thus in the construction of δ, the a is a boundary, a = da 0 Then for a preimage b of cunder g n , we have by the definition of a d(b−f n a 0 ) df n a 0 fn−1a= 0.
Thusb−f n a 0 is a cycle and g n (b−f n a 0 ) = g n b−g n f n a 0 =g n b−0 =g n b=c, so we found a preimage for [c] under H n (g) and the kernel ofδ is contained in the image ofH n (g). c) Exactness at H n−1 (A ∗ ):
Let c be a cycle in Z n (C ∗ ) Again, we choose a preimage b of c under g n and an a with fn−1(a) Then Hn−1(f)δ[c] = [fn−1(a)] = [db] = 0 Thus the image ofδ is contained in the kernel of Hn−1(f).
If a ∈ Zn−1(A∗) with Hn−1(f)[a] = 0 Then fn−1a = db for some b ∈ Bn Take c = gnb. Then by definition δ[c] = [a].
The long exact sequence of a pair of spaces
Let X be a topological space andA ⊂X a subspace of X.
The inclusion map \( i: A \to X \), defined by \( i(a) = a \), induces a map of chain complexes \( S_n(i): S_n(A) \to S_n(X) \) It is important to note that the inclusion of spaces does not necessarily result in a monomorphism on homology groups For example, when we include \( A = S^1 \) into \( X = D^2 \), this illustrates the point.
By Corollary 1.4.8.2, since D is contractible, we know that Hn(D) = 0 for n >1 and by Corollary 1.3.9 that H 1 (S 1 ) = Z.
2 Consider the quotient groupsS n (X, A) :=S n (X)/S n (A) Sinced n (S n (A))⊂Sn−1(A), the differential induces a well-defined map dn Sn(X)/Sn(A) → Sn−1(X)/Sn−1(A) c n +S n (A) 7→ d n (c n ) +Sn−1(A) that squares to zero.
3 Alternatively,S n (X, A) is isomorphic to the free abelian group generated by alln-simplices β: ∆ n →X whose image is not completely contained inA, i.e β(∆ n )∩(X\A)6=∅.
We consider pairs of spaces (X, A).
The relative chain complex of the pair (X, A) is
S∗(X, A) :=S∗(X)/S∗(A) with the differentials described in Remark 1.6.1.2.
• Elements in S n (X, A)are called relative chains in (X, A).
• Cycles inS n (X, A)are chainscwith ∂ X (c)a linear combination of generators with image in A These are called relative cycles.
• Boundaries in S n (X, A)are chains c inX such thatc=∂ X b+awhere a is a chain inA. These are called relative boundaries.
The following facts are immediate from the definition:
The relative homology groups of the pair of spaces (X, A) are the homology groups of the relative chain complex S ∗ (X, A)from Definition 1.6.2:
Theorem 1.6.5 (Long exact sequence for relative homology).
1 For any pair of topological spaces A⊂X we obtain a long exact sequence
2 For a map of spaces f: X → Y with f(A) ⊂ B ⊂ Y, we get an induced map of long exact sequences
A map f: X →Y with f(A)⊂B is denoted by f: (X, A)→(Y, B).
1 By definition of the relative chain complex S∗(X, A) the sequence
0 // S∗(A) S ∗ (i) // S∗(X) π // S∗(X, A) // 0 is an exact sequence of chain complexes and by Proposition 1.5.6 we obtain the long exact sequence in the first claim.
0 // S n (B) S n (i) // S n (Y) π // S n (Y, B) // 0 commutes We now use Proposition 1.5.5.
We obtain a long exact sequence
.→Hj(S n−1 )→Hj(D n )→Hj(D n ,S n−1 )→ δ Hj−1(S n−1 )→Hj−1(D n )→ .
The disc D n is contractible and by Corollary 1.4.8, we have H j (D n ) = 0 for j > 0 From the long exact sequence we get that δ: H j (D n ,S n−1 )∼=H j−1 (S n−1 ) for j >1 and n >1.
1 A subspace ι: A ,→X is a weak retract, if there is a map r: X →A with r◦ι'id A
2 A subspaceι:A ,→X is a deformation retract, if there is a homotopyR:X×[0,1]→X such that
Any deformation retract is a weak retract: take r :=R(−,1) :X →A Condition (c) then amounts to r◦ι= id A
If i: A ,→X is a weak retract, then
From the defining identity of a weak retract r◦ι'id A , we get by Theorem 1.4.7 thatH n (r)◦
Hn(i) = Hn(idA) = id H n (A) for all n Hence Hn(i) is injective for all n This implies that
0→H n (A) H −→ n (i) H n (X) is exact Injectivity ofHn−1(i) yields that the image ofδ: H n (X, A)→
Hn−1(A) is trivial Therefore, the long exact sequence of Theorem 1.6.4 decomposes into short exact sequences
0→H n (A) H −→ n (i) H n (X)−→ π ∗ H n (X, A)→0 for all n As H n (r) is a left-inverse for H n (i) we obtain a splitting
Hn(X) → Hn(A)⊕Hn(X, A) [c] 7→ ([rc], π∗[c]) with inverse
In the context of homology, the equation H n (i)[a] + [a 0 ]−H n (i◦r)[a 0 ] holds for any [a 0 ] in H n (X) such that π∗[a 0 ] = [b] The second mapping is well-defined; if [a 00 ] is another element satisfying π∗[a 00 ] = [b], then the difference [a 0 −a 00 ] can be expressed as Hn(i)[˜a] Since this element lies in the kernel of π∗, it follows that [a 0 −a 00 ]−H n (i◦r)[a 0 −a 00 ] = H n (i)[˜a]−H n (i◦r◦i)[˜a] is trivial.
For any ∅6=A ⊂X such that A⊂X is a deformation retract, then
The map r := R(−,1) : X → A defines a homotopy from the identity map on X to the composition of i and r This establishes that r serves as a retraction, as indicated by the condition r◦i = id A Consequently, this leads to the conclusion that spaces A and X are homotopically equivalent Furthermore, by Corollary 1.4.8, it follows that the homology groups H n (i) are isomorphic for all n > 0.
If X has two subspacesA, B ⊂X, then(X, A, B) is called a triple, if B ⊂A⊂X.
Any triple gives rise to three pairs of spaces (X, A), (X, B) and (A, B) and accordingly we have three long exact sequences in homology But there is another long exact sequence:
For any triple (X, A, B), there is a natural long exact sequence
This sequence is part of the following braided commutative diagram displaying four long exact sequences
In particular, the connecting homomorphism δ: H n (X, A) → Hn−1(A, B) is the composite δ =π∗ (A,B) ◦δ (X,A)
Note that S n (B)⊂S n (A)⊂S n (X); by the homomorphism theorem, the sequence
0 // S n (A)/S n (B) // S n (X)/S n (B) // S n (X)/S n (A) // 0. is exact Now apply Proposition 1.5.6 to obtain the long exact sequence.
Excision
The goal is to clarify the concept of relative homology groups For a subspace A within a space X, it is evident that the relative homology group H∗(X, A) differs from the homology group H∗(X\A) A clear example of this distinction is seen in the figure eight, where A represents the point that connects the two loops of S¹.
• X\A has two connected components By Corollary 1.3.3 we haveH0(X\A)∼=Z⊕Z.
• Anyx∈X\Ais a generator for the group of 0-cycles Since the spaceXis path connected, it is homologous to the point a ∈ A and thus vanishes in relative homology The group
To simplify the relative homology group H∗(X, A) through excision, careful consideration is essential The initial step involves refining singular simplices, a process known as barycentric subdivision, which is a commonly utilized technique in this context.
First, we construct cones Let v ∈∆ p and let α: ∆ n →∆ p be a singular n-simplex on ∆ p
The cone ofα: ∆ n →∆ p with respect tov ∈∆ p is the singular(n+1)-simplexK v (α) : ∆ n+1 →
This map is well-defined and continuous On the standard basis vectorsKv givesKv(ei) =α(ei) for 06i6n but K v (e n+1 ) = v Extending K v linearly gives a map on chain groups
1 For c∈S 0 (∆ p ), the boundary of the cone K v (c) is the 0-chain
∂Kv(c) = ε(c).κv−c with κ v (e 0 ) =v and ε the augmentation as introduced in Proposition 1.3.1.
2 For n >0 we have that ∂◦K v −K v ◦∂ = (−1) n+1 id.
1 For a singular 0-simplexα: ∆ 0 →∆ p we know thatε(α) = 1 and we calculate
∂Kv(α)(e0) = Kv(α)◦d0(e0)−Kv(α)◦d1(e0) = Kv(α)(e1)−Kv(α)(e0) = v−α(e0). Extending linearly shows the claim.
2 Forn >0 we have to calculate∂ i K v (α) and it is straightforward to see that∂ n+1 K v (α) = α and ∂ i (K v (α)) =K v (∂ i α) for all 06i < n+ 1.
In the context of an n-simple α: ∆ n → ∆ p on ∆ p, the barycenter v(α) is defined as v := (1/(n+1)) ∑_{i=0}^{n} α(e_i), where the additional vertex is the barycenter of the vertices The barycentric subdivision B: S n (∆ p) → S n (∆ p) is defined inductively, with B(α) = α for α ∈ S0(∆ p) and B(α) = (-1)^n Kv(B(∂α)) for n > 0 For n > 1, this can be expressed as B(α) = ∑_{i=0}^{n} (-1)^{n+i} Kv(B(∂_i α)).
If we taken=pand α= id∆ n , then for small n this looks as follows: You cannot subdivide a point any further For n = 1 we get
And for n= 2 we get (up to tilting)
The barycentric subdivision is a chain map
We have to show that ∂B=B∂.
• If α is a singular zero chain, then the fact Bα = α from the definition implies ∂Bα ∂α = 0 andB∂α=B(0) = 0.
But the boundary terms are zero chains on whichB is the identity, so we get with Lemma 1.7.2.1
In the last step, we used that B is the identity on the 0-chain ∂α Note, that the v is v(α), not a v(∂ i α).
• We prove the claim inductively, so let α ∈Sn(∆ p ) Then
Here, the first equality is by definition, the second one follows by Lemma 1.7.2.2 and then we use the induction hypothesis and the fact that ∂∂ = 0.
Our goal is to demonstrate that barycentric subdivision B has no impact on the homology groups To achieve this, we establish that the chain map B: S∗(∆ p) → S∗(∆ p) is chain homotopic to the identity.
To this end, we construct ψ n :S n (∆ p )→S n+1 (∆ p ) again inductively on generators as ψ 0 (α) := 0, ψ n (α) := (−1) n+1 K v (Bα−α−ψn−1∂α) with v := n+1 1 Pn i=0α(e i ) the barycenter.
The sequence (ψ n ) n is a chain homotopy from B to the identity on S∗(∆ p ).
• For n= 0 we have ∂ψ 0 = 0 and this agrees with B−id in that degree.
With Lemma 1.7.2.2 we can transform the latter to B +Kv∂B −∂Kv and as B is a chain map, this equals B+K v B∂−∂K v In chain degree oneB∂ agrees with∂, thus this reduces to
B+Kv∂ −∂Kv =B−(∂Kv−Kv∂) 1.7.2.2 = B −id.
• So, finally we can do the inductive step:
=B−id−ψn−1∂+ remaining terms The equation
K v ∂ψn−1∂+K v ψn−2∂ 2 =K v (∂ψn−1+ψn−2∂)∂ ind ass = K v B∂−K v ∂ from the inductive assumption ensures that these terms give zero.
A singular n-simplex α: ∆ n →∆ p is called affine, if α( n
We abbreviate v i :=α(e i ), so α(Pn i=0t i e i ) = Pn i=0t i v i and we call the elements v i ∈ ∆ p the vertices of α.
Let A be a subset of a metric space(X, d) The diameter of A is sup{d(x, y)|x, y ∈A} and we denote it by diam(A).
Accordingly, the diameter of an affinen-simplex α in∆ p is the diameter of its image, and we abbreviate that with diam(α).
For any affine singularn-simplexαevery simplex in the chainBα has diameter6 n+1 n diam(α).
Thus barycentric subdivision of affine simplices decreases the diameter Either you believe this lemma, or you prove it, or you check Bredon, Proof of Lemma 13.7 (p 226).
Each simplex in the chain Bα is affine, enabling us to repeatedly apply B and reduce the diameter of individual simplices Consequently, the k-fold iteration, B^k(α), has a diameter that does not exceed (n+1)/n^k * diam(α).
In the following we use the easy, but powerful trick to express the singular n-simplex α :
The mapping of an n-simplex on ∆ n is represented as ∆ n → X, where α = α◦id∆ n = Sn(α)(id∆ n) This framework allows for the application of barycentric subdivision to general spaces, as the identity map id∆ n can be interpreted as an n-simplex on ∆ n By applying barycentric subdivision to this simplex, we obtain a chain B(id ∆ n) that belongs to S n (∆ n) Consequently, a singular n-simplex on X, defined by a map α: ∆ n → X, leads to the creation of a morphism of abelian groups.
2 Similarly,ψ X n : S n (X)→S n+1 (X) is defined as ψ X n (α) := S n+1 (α)◦ψ n (id ∆ n ).
1 The maps B X are natural in X , i.e for any map X → f Y of topological spaces the diagram
B X :S∗(X)→S∗(X) are homotopic to the identity on S n (X).
• Let f: X →Y be a continuous map We have
In the first step, we used the definition of B n X ; in the second step the functorality of
S n (−) In the last step, we used the definition of S n (f) Thus the maps B X are natural in X.
• The calculation for ∂ψ X n +ψ n−1 X ∂ = B n X −id S n (X ) uses that α induces a chain map and thus we get
∂ψ X n (α) defn = ∂◦S n+1 (α)◦ψ n (id ∆ n ) S chain map = S n (α)◦∂◦ψ n (id ∆ n ).
Now we consider singular n-chains that are spanned by ’small’ singular n-simplices Here,
“smallness’ is defined in terms of an open covering.
Let \( U = \{ U_i, i \in I \} \) represent an open covering of the space \( X \) The free abelian group \( S_n U(X) \) is generated by all singular n-simplices \( \alpha: \Delta^n \to X \), where the image of \( \Delta^n \) under \( \alpha \) is entirely contained within one of the open sets \( U_i \) in the covering \( U \).
Note thatS n U (X) is an abelian subgroup of the singular chain group S n (X) The restriction of the differential of S n (X) gives a chain complex
We denote its homology byH n U (X) As we will see now, these chains suffice to detect everything in singular homology.
1 For any subspace A⊂X, the barycentric subdivision ofc∈S n (A) is again in S n (A), i.e.
2 If c ∈ S n (X) is a cycle relative A ⊂ X, then B(c) is a cycle relative A as well that is homologous to c relativeA.
3 Let U be an open covering of X Then every cycle in S n (X) is homologous to a cycle in
1 This follows at once from the definition of barycentric subdivision.
2 We note that the map ψn : Sn(X) → Sn+1(X) maps α ∈ Sn(A) to ψn(α) ∈ Sn+1(A). Now consider for a relative cycle cthe equation cf 1.7.10.2
Since c is a relative cycle, ∂c ∈ Sn−1(A) and by part 1, ψn−1∂c ∈ Sn(A) Thus Bc is homologous to c relativeA Its boundary is
∂Bc=∂c+∂ψn−1∂c Thus Bcis a relative cycle as well.
3 Consider a singular n-chain α =Pm j=1λ j α j ∈ S n (X) onX and let L j for 16 j 6m be the Lebesgue numbers for the m coverings{α j −1 (U i ), i∈I} of the simplex ∆ n Choose k, such that n+1 n k
6L 1 , , L m ThenB k α 1 up toB k α m are all chains inS n U (X) Therefore
From part 2, we know that B k α is a cycle as well that is homologous toα.
For any open covering U, the injective chain map
S ∗ U (X),→S∗(X) induces an isomorphism in homology, H n U (X)∼=H n (X).
The map on homology is surjective, since for any cycle c∈S n (X), we find by Lemma 1.7.12.3 a homologous cycle c 0 ∈S n U (X).
The map is injective as well: suppose c∈ S n U (X) is a boundary in S n (X), i.e c =∂e with e∈S n+1 (X) Find k ∈Nsuch that B k (e)∈S n+1 U (X) and
∂B k (e)−∂e=∂ψ˜n−1(c) is a boundary in S n U (X) Thus, c=∂e=∂(B k (e)−ψ˜n−1(c)) is a boundary in S n U (X) as well
We remark that this isomorphism actually comes from a homotopy equivalence of chain complexes.
With this we get the main result of this section:
Let W ⊂A ⊂ X such that ¯W ⊂A Then the inclusion˚ i: (X\W, A\W),→ (X, A) induces an isomorphism of relative homology groups
Hn(i) : Hn(X\W, A\W)∼=Hn(X, A) for all n>0.
• We first prove that H n (i) is surjective.
Let c ∈ Sn(X, A) be a relative cycle, i.e ∂c ∈ Sn−1(A) Consider the open covering
Let U be defined as the set {A, X˚ \W¯} and denote it as {U, V} within X We then subdivide to identify k such that c₀ := Bk c forms a chain in Sₙ U(X) According to Lemma 1.7.12.2, it can be concluded that c₀ is homologous to c relative to A Furthermore, we can express c₀ as the sum of two chains, c₁ and c₂, where c₁ and c₂ correspond to the respective open sets U and V It is important to note that this decomposition is not unique.
The boundary of c 0 is ∂c 0 = ∂B k c = B k ∂c; by assumption this is a chain in Sn−1(A). Moreover, we find from the decomposition c 0 =c U +c V
∂c 0 =∂c U +∂c V with ∂c U ∈ Sn−1(U) ⊂ Sn−1(A) Thus, ∂c V = ∂c 0 −∂c U ∈ Sn−1(A) Since ∂c V ∈ V is supported in X \W, we have ∂c V ∈ Sn−1(A\W) Therefore, c V is a relative cycle in
In H n (X, A), we find [c] = [c 0 ] = [c U +c V ] = [c V ], where we used in the first step Lemma 1.7.12.2 This shows that Hn(i)[c V ] = [c] ∈ Hn(X, A) Thus [c V ] is a preimage of [c] in
• The injectivity of Hn(i) is shown as follows.
Assume there exists a chain \( c \in S_n(X \setminus W) \) with \( \partial c \in S_{n-1}(A \setminus W) \) such that \( H_n(i)[c] = 0 \) This implies that \( c \) can be expressed as \( c = \partial b + a_0 \) for some \( b \in S_{n+1}(X) \) and \( a_0 \in S_n(A) \), with the ability to select all summands to avoid \( W \).
We write b as b U +b V with b U ∈Sn+1(U)⊂Sn+1(A) and b V ∈ Sn+1(V)⊂Sn+1(X\W). Then c=∂b U +∂b V +a 0
Note thata 0 and∂b U are chains inS n (A\W) So we have writtencas a boundary of a chain b V inSn+1(X\W) plus a chaina 0 +∂b U inSn(A\W) Thus [c] = 0∈Hn(X\W, A\W).
Mayer-Vietoris sequence
We consider the following situation: there are subspaces X 1 , X 2 ⊂X such that X 1 and X 2 are open inX and such thatX =X 1 ∪X 2 We consider the open covering U={X 1 , X 2 } We need the following maps:
Note that by definition, the sequence of complexes
0 // S∗(X1∩X2) (i 1 ,i 2 ) // S∗(X1)⊕S∗(X2) // S ∗ U (X) // 0 (1) is exact Here, the second map is
Note that here the open sets are ordered to define the difference.
There is a long exact sequence
The proof follows from the exact sequence (1) of chain complexes by Lemma 1.7.12, because
1 As an application, we calculate the homology groups of spheres Let X = S m and let
X ± :=S m \{∓e m+1 } The subspacesX + andX − are contractible and thereforeH ∗ (X ± ) 0 for all positive ∗.
The Mayer-Vietoris sequence is as follows
δ // H n (X + ∩X − ) // H n (X + )⊕H n (X − ) // H n (S m ) δ // Hn−1(X + ∩X − ) // For n >1 we can deduce fromH n (X ± ) = 0
The first map is the connecting homomorphism δ and the second map is
Hn−1(i) : Hn−1(S m−1 )→Hn−1(X + ∩X − ) wherei is the inclusion of S m−1 into X + ∩X − and this inclusion is a homotopy equivalence Thus define
This D is an isomorphism for all n >2.
We have to control what is going on in small degrees and dimensions.
2 We know from the Hurewicz isomorphism thatH 1 (S m ) is trivial form >1, cf Corollary 1.3.9 Here, we show this directly via the Mayer-Vietoris sequence:
We have to understand the map in the second line Let 1 be a base point of X + ∩X − Then the map on H 0 is
This map is injective and therefore the connecting homomorphismδ: H 1 (S m )→H 0 (X + ∩
We compute H1(S1) using a Mayer-Vietoris argument, focusing on the scenario where n1 equals m In this case, the intersection X+ ∩ X− divides into two distinct components We select base points P+ in X+ and P− in X−, and analyze the corresponding exact sequence.
0 // H1(S 1 ) δ // Z⊕Z // Z⊕Z // Z. The kernel of the last map, the difference of H0(κ1) and H0(κ2),
H 0 (X + )⊕H 0 (X − )→H 0 (S 1 ) is spanned by ([P + ],[P − ]) and thus isomorphic toZ This is the image of (H 0 (i 1 ), H 0 (i 2 )). Therefore, the sequence
0 // H1(S 1 ) δ // Z⊕Z // Z // 0 is short exact; thus H 1 (S 1 ) ∼= Z is a free abelian group of rank 1 We already knew this from the Hurewicz isomorphism.
4 We now combine the arguments.
• For 0< n < m we get by applying D repeatedly,
H n (S m ) ∼ = // Hn−1(S m−1 ) ∼ = // ∼ = // H 1 (S m−n+1 )∼=π 1 (S m−n+1 ). and the latter is trivial by 2.
• Similarly, for 0< m < n we have similarly
The last claim follows directly by another simple Mayer-Vietoris argument.
• The remaining case 0< m=n gives a non-vanishing result:
We can summarize the result as follows.
The homology groups of spheres are:
Let à 0 := [P + ]−[P−]∈H 0 (X + ∩X − )∼=H 0 (S 0 ) and letà 1 ∈H 1 (S 1 )∼=π 1 (S 1 ) be given by the degree one map (i.e the class of the identity on S 1 , i.e the class of the loop t 7→e 2πit ).
Define the higher à n via the map D from 1.8.2.1 as Dà n = àn−1 Then à n is called the fundamental class in H n (S n ).
In order to obtain a relative version of the Mayer-Vietoris sequence, we need a tool from homological algebra.
B1 β 1 //B2 β 2 //B3 β 3 //B4 β 4 //B5 be a commutative diagram of exact sequences If the four maps f 1 , f 2 , f 4 , f 5 are isomorphisms, then so is f 3
Again, we are chasing diagrams.
• We show that f 3 is injective.
Assume that there is an a ∈A 3 with f 3 a = 0 Then β 3 f 3 a =f 4 α 3 a = 0, as well Butf 4 is injective, thus α3a = 0 Exactness of the top row gives, that there is an a 0 ∈ A2 with α 2 a 0 =a This implies f 3 α 2 a 0 =f 3 a= 0 =β 2 f 2 a 0
Exactness of the bottom row gives us ab ∈B 1 withβ 1 b =f 2 a 0 , butf 1 is an isomorphism so we can lift b toa 1 ∈A 1 with f 1 a 1 =b.
Thusf 2 α 1 a 1 =β 1 b =f 2 a 0 and asf 2 is injective, this implies thatα 1 a 1 =a 0 So finally we get that a =α 2 a 0 =α 2 α 1 a 1 , but the latter is zero, thus a= 0.
To establish the surjectivity of the function f3, consider an element b in the set B3 Transfer b to the set B4 using the mapping β3, and define a as the inverse image of b under f4, denoted as f4⁻¹(β3(b)) Notably, if β3(b) equals zero, a more straightforward path emerges: there exists an element b2 in B2 such that β2(b2) equals b, and consequently, an element a2 in A2 exists with f2(a2) equating to b2 Thus, it follows that f3(α2(a2)) equals β2(b2), which ultimately equals b.
Consider f 5 α 4 a This is equal to β 4 β 3 b and hence trivial Therefore α 4 a = 0 and thus there is an a 0 ∈A 3 with α 3 a 0 =a Then b−f 3 a 0 is in the kernel of β 3 , because β 3 (b−f 3 a 0 ) =β 3 b−f 4 α 3 a 0 =β 3 b−f 4 a= 0.
Hence we get a b 2 ∈ B 2 with β 2 b 2 =b−f 3 a 0 Define a 2 as f 2 −1 (b 2 ), so a 0 +α 2 a 2 is in A 3 and f 3 (a 0 +α 2 a 2 ) = f 3 a 0 +β 2 f 2 a 2 =f 3 a 0 +β 2 b 2 =f 3 a 0 +b−f 3 a 0 =b.
In a topological space X, let A and B be open sets within X, forming an open covering of their union A∪B The set U, defined as {A, B}, represents this covering The relationship between absolute and relative chains is illustrated through the following diagram of exact sequences.
In this context, ψ is generated by the inclusion ϕ: S n U (A∪B) → Sn(A∪B), while ∆ represents the diagonal map and diff signifies the difference map The exactness of the first two rows is evident, and the exactness of the third row can be established through a variant of the nine-lemma or by performing a direct diagram chase.
Consider the two right-most non-trivial columns in this diagram Each gives a long exact sequence in homology and we focus on five terms:
According to the five-lemma 1.8.5, since Hn(ϕ) and Hn−1(ϕ) are isomorphic as established by Corollary 1.7.13, it follows that Hn(ψ) is also isomorphic This finding, combined with the non-trivial exact row at the bottom of the first diagram, leads to a significant conclusion.
Theorem 1.8.6 (Relative Mayer-Vietoris sequence).
If A, B ⊂X are open in A∪B, then the following sequence is exact:
Reduced homology and suspension
For any path-connected space, the zeroth homology is isomorphic to the integers, so this copy of
Zis superfluous information and we want to get rid of it Let pt denote the one-point topological space Then for any space X there is a unique continuous map ε: X →pt.
We defineHe n (X) := ker(H n (ε) : H n (X)→H n (pt))and call it the reduced nth homology group of the space X.
1 Note that He n (X)∼=H n (X) for all positive n.
2 If X is path-connected, thenHe0(X) = 0, cf Proposition 1.3.1.
3 Choose a base point x∈X Then the composition
{x},→X → {x} is the identity Because of H n (pt) ∼=H n ({x}), we get from proposition 1.6.8 about weak retracts
The retraction r: X → {x}splits the long exact sequence of relative homology for{x} → X
H n ({x})→H n (X)→H n (X,{x})→ . and thus we identify reduced homology as relative homology, He n (X)∼=H n (X,{x}).
4 We can prolong the singular chain complexS∗(X) and consider the chain complex of free abelian groups Se∗(X):
.→S 1 (X)→S 0 (X)−→ ε Z→0. where ε(α) = 1 for every singular 0-simplex α This is precisely the augmentation we considered in Proposition 1.3.1 Then for all n>0,
For every continuous mapf: X →Y induces a chain map S∗(f) : S∗(X)→S∗(Y); for the evaluation, we have ε Y ◦S 0 (f) = ε X We thus obtain the following result:
The functorial assignment X 7→ H∗(Se∗(X)) establishes a relationship between spaces and their homology, where for any continuous map f: X → Y, there is an induced map H∗(Se∗(f)): H∗(Se∗(X)) → H∗(Se∗(Y)) This ensures that the identity map on X corresponds to the identity in homology, and the composition of maps is preserved, highlighting the coherent structure of homological relationships in topological spaces.
As a consequence,He∗(−) is a functor.
The identification of reduced homology groups with relative homology groups H n (X,{x}) in Remark 1.9.2.3 allows us to derive a reduced form of the Mayer-Vietoris sequence This observation also extends to the long exact sequence for a pair of spaces.
For each pair (X, A) of spaces, there is a long exact sequence
// Hen(A) // Hen(X) // Hen(X, A) // Hen−1(A) // and a reduced Mayer-Vietoris sequence, if X 1 ∩X 2 6=∅, which is identical in positive degrees and ends as
1 Recall that we can express the real projective plane RP 2 as the quotient space of S 2 modulo antipodal points or as a quotient of D 2 :
We use the latter definition and setX =RP 2 ,A=X\ {[0,0]} (which is an open M¨obius strip and hence homotopically equivalent to S 1 ) and B = ˚D 2 Then
Thus we know that H1(A) ∼= Z, H1(B) ∼= 0 and H2(A) = H2(B) = 0 We choose generators for H 1 (A) andH 1 (A∩B) as follows:
Leta represents the path along the outer circle in a positive mathematical direction, beginning at the point (1,0), serving as the generator for H1(A) Conversely, γ denotes the loop traversing the inner circle in a positive direction, acting as the generator for H1(A∩B), where A∩B is defined as D\{0} Consequently, the inclusion map from A∩B to A induces a significant relationship between these two spaces.
This suffices to compute H∗(RP 2 ) up to degree two because the long exact sequence is
H˜ 2 (A)⊕H˜ 2 (B) = 0 →He 2 (X)→He 1 (A∩B)∼=Z→ ã2 He 1 (A)∼=Z→He 1 (X)→He 0 (A∩B) = 0.
On the two copies of the integers, the map is given by multiplication by two and thus we obtain:
The higher homology groups are trivial, because there H n (RP 2 ) is located in a long exact sequence between trivial groups.
We can compute the homology groups of bouquets of topological spaces by relating them to the homology groups of individual spaces, particularly in favorable scenarios Given a family of topological spaces (Xi)i∈I, each with a designated basepoint x i ∈X i, we can analyze the bouquet formed by these spaces to derive meaningful homological information.
If the inclusion ofx i intoX i is pathological, we cannot apply the Mayer-Vietoris sequenceHowever, we get the following:
If there are neighbourhoods U i of x i ∈ X i together with a deformation of U i to{x i }, then we have for any finite E ⊂I
In the situation above, we say that the space X i is well-pointed with respect to the point x i ∈X i
In the case of two bouquet summands, we analyze the open covering of \(X_1 \vee U_2 \cup U_1 \vee X_2\) for \(X_1 \vee X_2\) The intersection \((X_1 \vee U_2) \cap (U_1 \vee X_2) = U_1 \cap U_2\) is contractible, which allows us to apply the Mayer-Vietoris sequence This leads to the conclusion that \(H_n(X) \cong H_n(X_1 \vee U_2) \oplus H_n(U_1 \vee X_2)\) for \(n > 0\) Additionally, for \(H_0\), we derive an exact sequence.
By induction we obtain the case of finitely many bouquet summands
He n (X i ) but for this one needs a colimit argument We postpone that for a while.
In the context of relative topology, if A is a closed subspace of X and serves as a strong deformation retract of an open neighborhood U containing A, we can extend our findings to the full relative case The canonical projection π: X → X/A maps A to its image b in the quotient space X/A, which is well-pointed at b due to the neighborhood π(U).
The canonical projection π :X → X/A induces a homeomorphism of pairs (X\A, U \A) ∼ (X/A\ {b}, π(U)\ {b}) Consider the following diagram:
The upper and lower left arrows are isomorphisms due to A being a deformation retract of U The upper right isomorphism arises from excision, as A equals ¯A and is a subset of U, in accordance with Theorem 1.7.14 Similarly, the lower right arrow is also a result of excision Additionally, the right vertical arrow is an isomorphism because there exists a homeomorphism of pairs.
1 The cone of a topological space X is the topological space
2 The suspension of a topological space X is the topological space ΣX :=X×[0,1]/(x 1 ,0)∼(x 2 ,0)and (x 1 ,1)∼(x 2 ,1) for all x 1 , x 2 ∈X
1 The cone over a pointpis an interval The cone over an interval is a triangle, a 2-simplex. The cone over ann-simplex is an (n+ 1)-simplex The cone overS n is a closed (n+ 1)-ball.
2 Note that for any topological space X, the cone CX is contractible to its apex Thus
H˜n(CX) = 0 for alln>0 Similarly, forA⊂X, we haveCA⊂CX and ˜Hn(CX, CA) = 0 for all n >0.
4 We have natural embeddings X →CX with x 7→[x,1] and CX →ΣX with x7→ [x, 1 2 ].
We can see the suspension as two cones, glued together at their bases.
Let A ⊂ X be a closed subspace and assume that A is a deformation retract of an open neighbourhood A⊂U Then
1 We first note two equivalences:
X∪CA/CA'X/A, where the coneCA is attached to X by identifyingA⊂X and the base A⊂CA:
A A A A A A rr rr rr rr rr r rr rr rr rr r rr rr rr rr rr rr rr rr rr r rr r rr
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2 Consider the triple (CX, X∪CA, CA) We obtain from Proposition 1.6.11 the long exact sequence on homology groups
// H n (CX, CA) // H n (CX, CA∪X) δ // H˜n−1(X∪CA, CA) // Since cones are contractible, the connecting morphism δ gives us isomorphisms
H n (CX, CA∪X)∼= ˜Hn−1(X∪CA, CA)
3 Using Proposition 1.9.8 and the equivalences from part 1 of the proof, we compute the right hand side:
H˜n−1(X∪CA, CA) 1.9.8 ∼= ˜Hn−1(X∪CA/CA)∼= ˜ 1 Hn−1(X/A) 1.9.8 ∼= ˜Hn−1(X, A) Similarly, we get for the left hand side
Homotopy groups do not adhere to the straightforward suspension isomorphism seen in homology groups For instance, while ΣS² is homotopically equivalent to S³, the third homotopy group π₃(S²) is isomorphic to Z, and π₄(S³) is isomorphic to Z/2Z This discrepancy highlights the complexity of homotopy groups, which is addressed by the Freudenthal suspension theorem This theorem indicates that π₁₊₃(S³) is isomorphic to π₁₊₄(S⁴), leading to the concept of stable homotopy groups, denoted as π₁ₛ, which represents the first stable homotopy group of the sphere.
Mapping degree
Recall that we defined in Definition 1.8.4 fundamental classes à n ∈H˜ n (S n )∼=Z for all n>0.
H˜ n (f)à n = deg(f)à n with deg(f)∈Z We call this integer the degree of f.
In the case of n = 1, we can connect the concept of mapping degree to the fundamental group of the 1-sphere Specifically, the generator of π1(S1, 1) can be represented by the loop ω: [0,1] → S1, defined as t ↦ e^(2πit) The abelianized Hurewicz map, denoted as hab: π1(S1, 1) → H1(S1), maps the class of ω to the element 1 in H1(S1) This demonstrates the naturality of the Hurewicz map in relation to the fundamental group, establishing a clear relationship between these mathematical constructs.
H 1 (S 1 ) H 1 (f ) // H 1 (S 1 ) shows that deg(f)à 1 def = H 1 (f)à 1 nat = h ab (π 1 (f)[w]) def = h ab (k[w]) =kà 1 where k is the degree of f defined via the fundamental group Thus both notions coincide for n = 1.
The connecting homomorphism in the long exact sequence of relative homology establishes an isomorphism between Hn(D n, S n−1) and ˜Hn−1(S n−1) This relationship allows us to define degrees of maps f: (D n, S n−1) → (D n, S n−1) by introducing a fundamental class ¯à n, which is represented as δ −1 à n.
H n (D n ,S n ) ThenH n (f)(¯à n ) := deg(f)¯à n gives a well-defined integer deg(f)∈Z.
The degree of self-maps ofS n satisfies the following properties:
1 If f is homotopic to g, then deg(f) = deg(g).
2 The degree of the identity onS n is one.
3 The degree is multiplicative,i.e., deg(g◦f) = deg(g)deg(f).
4 If f is not surjective, then deg(f) = 0.
The first three properties follow directly from the definition of the degree Iff is not surjective, then it is homotopic to a constant map and this has degree zero
It is true that the group of (pointed) homotopy classes of self-maps ofS n is isomorphic to
Z and thus the first statement in Proposition 1.10.2 can be upgraded to an ’if and only if’, but we will not prove that here.
Recall that ΣS n ∼=S n+1 Iff: S n →S n is continuous, then the suspension Σ(f) : ΣS n →ΣS n is given as ΣS n 3[x, t]7→[f(x), t].
Suspensions leave the degree invariant, i.e for f: S n →S n we have deg(Σ(f)) = deg(f).
In particular, for every integer k ∈Z there is a continuous map f: S n →S n with deg(f) =k.
The suspension isomorphism of Theorem 1.9.11 is induced by a connecting homomorphism which is functorial by Proposition 1.5.5 Using the isomorphism H n+1 (S n+1 ) ∼= H n+1 (ΣS n ), the connecting homomorphism sends à n+1 ∈ H n+1 (S n+1 ) to ±à n ∈ H˜ n (S n ) But then the commutativity of
H˜ n (S n ) H n (f) // H˜ n (S n ) ensures that ±deg(f)àn =±deg(Σf)àn with the same sign
The degree of a self-map on S^1 exhibits an additivity relation, expressed as deg(ω₀₀ ? ω₀) = deg(ω₀₀) + deg(ω₀), which applies to the concatenation of paths This concept can be extended to higher dimensions through the consideration of the pinch map T: S^n → S^n/S^(n−1) ∨ S^n and the fold map F: S^n ∨ S^n → S^n.
F is induced by the identity ofS n
For f, g: S n →S n based, we have deg(F ◦(f∨g)◦T) = deg(f) + deg(g).
The map H n (T) sends à n to (à n , à n ) ∈ H˜ n S n ⊕H˜ n S n ∼= ˜H n (S n ∨S n ) Under this isomor- phism, the map H n (f ∨ g) corresponds to (à n , à n ) 7→ ( ˜H n (f)à n ,H˜ n (g)à n ) and this yields (deg(f)à n ,deg(g)à n ) which under the fold map is sent to the sum
We use the mapping degree to show some geometric properties of self-maps of spheres.
We prove the claim by induction.à 0 was by definition 1.8.4 the difference class [+1]−[−1], and f (0) ([+1]−[−1]) = [−1]−[+1] =−à 0
We defined àn in such a way thatDàn =àn−1 Therefore, as D is obtained from a connecting homomorphism and thus by proposition 1.5.5 natural,
Letf i (n) : S n →S n be the map (x 0 , , x n )7→(x 0 , , xi−1,−x i , x i+1 , , x n ) As in Proposition 1.10.5, one shows that the degree of f i (n) is −1 AsA=fn (n) ◦ .◦f 0 (n) , the claim follows from
In particular, for even n, the antipodal map cannot be homotopic to the identity.
Let f, g: S n → S n with f(x) 6=g(x) for all x ∈ S n , then f is homotopic to A◦g, with A the antipodal map In particular, deg(f) = deg(A◦g) = deg(A)ãdeg(g) = (−1) n+1 deg(g).
By assumption, for all x∈ S n the segment t 7→(1−t)f(x)−tg(x) does not pass through the origin for 0 6t61 Thus the homotopy
||(1−t)f(x)−tg(x)|| with values in S n connects f to −g =A◦g
For any f: S n →S n with deg(f) = 0 there exists a point x + ∈S n withf(x + ) = x + and a point x− with f(x−) =−x−.
If f(x)6= x= id(x) for all x, then by Proposition 1.10.7, f is homotopic to A◦id = A Thus deg(f) = deg(A) 6= 0 If f(x) 6= −x for all x, then f is homotopic to A ◦(−id) and thus deg(f) = (−1) n+1 deg(A)6= 0
If n is even, then for any continuous map f : S n → S n , there is an x ∈ S n with f(x) = x or f(x) =−x.
In the case of an even n, the degree of the mapping A is −1 If the function f(x) is not equal to x for all x in the sphere S^n, it follows from the proof of Corollary 1.10.8 that deg(f) equals deg(A), which is also −1 Conversely, if f(x) is not equal to −x for all x in S^n, then the degree of f becomes deg(A) multiplied by the degree of the negative identity on S^n, resulting in a degree of 1 However, it is impossible for both conditions to hold simultaneously.
Finally, we can say the following about hairstyles of hedgehogs of arbitrary even dimension:
Any tangential vector field on an even-dimensional sphere S 2k vanishes in at least one point.
Recall that we can describe the tangent space at a point x∈S 2k ⊂R 2k+1 as
Assume that V is a tangential vector field which does nowhere vanish, i.e V(x) 6= 0 for all x∈S 2k and V(x)∈T x (S 2k ) for all x Consider the continuous map f : S 2k → S 2k x 7→ ||V V (x) (x)||
Assuming f(x) = x implies that V(x) is aligned with the direction of x, which contradicts the requirement for V(x) to be tangential Consequently, f(x) cannot equal x for any x in S 2k Similarly, the assumption that f(x) = -x leads to the same contradiction, reinforcing that the existence of such a V is not feasible.
CW complexes
A topological space X is called ann-cell, if X is homeomorphic toR n The number n is called the dimension of the cell.
1 Every point is a zero cell The spaces ˚D n ∼=R n ∼=S n \N are n-cells.
2 Note that an n-cell cannot be an m-cell for n 6=m, because R n R m for n 6= m This follows, since R n ∼=R m would imply
S n−1 'R n \ {0} ∼=R m \ {0} 'S m−1 , but ˜Hn−1(S n−1 ) ∼=Z for all n and ˜Hn−1(S m−1 ) = 0 for n 6=m Hence the dimension of a cell is well-defined.
A cell decomposition of a space X is a decomposition of X into subspaces, each of which is a cell of some dimension, i.e.,
X i , X i ∼=R n i Here, this decomposition is meant as a set, not as a topological space.
1 The boundary of a 3-dimensional cube has a cell decomposition into 8 points, 12 open edges, and 6 open faces.
2 The standard 3-simplex can be decomposed into 4 zero-cells, six 1-cells, four 2-cells, and a 3-cell.
3 The n-dimensional sphere (for n > 0) has a cell decomposition into the north pole and its complement, thus into a single zero-cell and n-cell.
A topological Hausdorff space X together with a cell decomposition is called a CW complex, if it satisfies the following conditions:
(a) [Characteristic maps] For every n-cell σ ⊂ X, there is a continuous map Φσ: D n → X such that the restriction of Φ σ to the interior ˚D n is a homeomorphism Φσ|˚ D n : ˚D n
−→σ and such that Φ σ maps S n−1 ∼=∂D n to the union of cells of dimension at most n−1.
(b) [Closure finiteness] For every n-cell σ, the closure σ¯ ⊂ X has a non-trivial intersection with only finitely many cells of X.
(c) [Weak topology] A subsetA ⊂X is closed, if and only if A∩σ¯ ⊂σ is closed for all cells σ inX.
1 The map Φ σ as in (a) is called a characteristic map of the cell σ Its restriction Φ σ | S n−1 to the boundary ∂D n ∼=S n−1 is called an attaching map.
2 Property (b) is the closure finite condition: the closure of every cell is contained in finitely many cells This is the ’C’ in CW.
Since ¯σ is closed in X, the requirement that A∩σ¯ is closed in X is equivalent to condition (c) This condition can be rephrased as the axiom stating that a subset A ⊂ X is open if and only if A∩σ¯ is open in ¯σ for every cell σ in X.
4 IfX is a CW complex with only finitely many cells, then we callX finite Conditions (b) and (c) are then automatically fulfilled.
Every non-empty CW complex must include at least one zero cell If a non-empty CW complex were to have a lowest dimension cell greater than zero, its boundary, which is an (n-1)-dimensional sphere, could not be represented using cells of dimension n-1 or lower.
6 It follows from axiom (a) that for every n-cell σ, we have σ = Φ σ (D n ).
Proof: From the general inclusion f(B)⊂f(B) for continuous maps, we conclude σ= Φ σ (˚D n )⊃Φ σ (D n )⊃σ
In a Hausdorff space, the compact subspace Φ σ (D n ) is closed, leading to the conclusion that Φ σ (D n ) equals σ Consequently, the closure σ is compact in X, as it is the continuous image of the compact set D n.
7 It follows that ¯σ\σ for ann-cell σ is contained in the union of cells of dimension at most n−1.
8 Every finite CW complex is compact, since it is the union of finitely many compact subspaces Φ σ (D n ).
1 The CW structures on a fixed topological space are not unique For example,S 2 with the
The CW structure derived from the cell-decomposition of S² \ {N} t {N} features a unique 0-cell located at the north pole and a single 2-cell Additionally, projecting shapes such as tetrahedrons, cubes, octahedrons, or even more irregular forms onto the sphere generates alternative CW structures.
2 Consider the following spaces with cell decomposition:
In the analysis of CW complexes, Figure 1 demonstrates a violation of axiom (a) due to one of its 1-cells having a boundary not contained within the 0-cells Conversely, Figure 2 adheres to the axiom with its four 1-cells and four 0-cells Figure 3 fails to qualify as a CW complex because one of its 1-cells lacks compact closure, while Figure 4 successfully meets the criteria to be classified as a CW complex.
3 Consider the topological space X =X 1 ∪X 2 ⊂R 2 with
X 1 :={(x,sin1 x)|0< x 0 and σ k 1 = ( k+1 1 , k 1 ) does not give a
CW structure on [0,1] Consider the following countable subsetA⊂[0,1]
Then A ∩σ¯ k 1 is precisely the point 1 2 ( 1 k + k+1 1 ) This is closed, but the subset A is not closed in [0,1], since it does not contain the limit point 0 ofA.
A simplicial complex, also known as a polyhedron, is a fundamental concept in topology that consists of a set K of simplices in R^n This structure is defined by the essential condition that if K includes a simplex, it must also contain all of its faces.
(b) The intersection of two simplices of K is either empty or a common face.
(c) K is locally finite, i.e every point of R n has a neighborhood that intersects only finitely many simplices of K.
A simplicial map is a map that takes anyk-simplex affinely into ak 0 -simplex withk 0 6k.
• The subspace |K|:=∪s∈Ks⊂R n is called the topological space underlying the complex
K Simplicial homology can be defined for a simplicial complex; it depends only on |K|.
Simplicial homology has notable limitations, as exemplified by the sphere S², which can be represented as a simplicial complex with 14 simplices derived from projecting a tetrahedron onto the sphere, whereas it can be simplified to a CW complex with just one 0-cell and one 2-cell Similarly, while a 2-torus S¹ × S¹ can be expressed as a CW complex consisting of only four cells, its smallest simplicial complex requires 42 cells, highlighting the inefficiency of simplicial representations compared to CW complexes.
1 The unionX n :=S σ⊂X,dim(σ)6nσ of cells of dimension at most n is called the n-skeleton of X.
2 If we haveX=X n , butX n−1 (X, then we say thatXisn-dimensional,i.e.,dim(X) = n.
A subset Y of a CW complex X is referred to as a subcomplex or sub-CW complex when it possesses a cell decomposition using the cells of X Additionally, for every cell σ within Y, the closure of σ in X must also be included in Y, ensuring that σ¯ is a subset of Y.
4 For any subcomplex Y ⊂X, (X, Y) is called a CW pair.
Let X be a CW complex and Y ⊂ X be a subspace, together with a cell decomposition by a subset of cells of X Then the following conditions are equivalent:
1 Y is a subcomplex, i.e for any cellσ ⊂Y, the closure σ in X is contained inY.
3 The cell decomposition (with the cells of X) endows Y with the structure of a CW complex.
2⇒ 1 is trivial: σ ⊂Y =Y (Here the bar denotes closure in X, of course.)
In the topology of X, according to axiom "W," a subset Y is considered closed if its intersection with any cell σ in X is also closed Given that X is closure finite, each cell σ intersects with only a finite number of cells within X Since Y is formed by the union of these cells, only a limited number of them will be present in the intersection σ∩Y, which consists of the intersection of σ with the union of the cells σ₁ through σᵣ that belong to Y Consequently, it follows that σ∩Y equals σ∩(σ₁ ∪ ∪ σᵣ).
The intersection of finite unions of closed subsets of X is closed in X, thus this is closed in X.
2 “Computing homology with simplicial chains is like computing integrals R b a f (x)dx with approximatingRiemann sums.” (Dold, Lectures in algebraic topology, 1972)
3⇒ 1 For any cellσ ⊂Y, a characteristic map Φ 0 σ forY exists by 3 It is also characteristic for
X Remark 1.11.6.6 that σ = Φ 0 σ (D) now implies that the closure of σ in Y agrees with the closure of σ inX.
A characteristic map for a cell σ ⊂ X in relation to the complex X is also characteristic for Y, indicating that characteristic maps exist for Y as well Consequently, the closure finiteness for Y is directly derived from that of X, confirming that axioms (a) and (b) are satisfied.
To demonstrate that for any subset A of Y, the condition that A intersected with the closure of any cell σ in Y is closed in Y implies that A itself is closed in Y Notably, the closure of σ in Y is denoted as σ Y Additionally, it is important to note that a set is considered closed in Y if and only if it is also closed in X.
The closure of each cell σ in Y aligns with its closure in X, demonstrating that it suffices to prove A∩σ is closed in X for all cells σ in X Closure finiteness leads to the conclusion that σ∩A equals σ∩(σ1∪ ∪σr)∩A, where σi are all cells in Y Since A∩σi is closed in X for every i, it follows that A∩σ is also closed in X.
1 Arbitrary intersections and arbitrary unions of subcomplexes are again subcomplexes.
3 Every union ofn-cells inX with X n−1 forms a subcomplex.
4 Every cell lies in a finite subcomplex.
1 The subcomplexes are closed in X by Lemma 1.11.10, hence their intersection is closed and by Lemma 1.11.10 a subcomplex The statement about the union follows directly from the definition of a subcomplex.
2 and 3 follow from the observation that for an n-cell σ we have thatσ = (σ\σ)∪σ is contained in X n−1 ∪σ.
4 Induction on the dimension of the cell; then use closure finiteness and σ = Φ σ (D n ).
We want to understand the topology of CW complexes.
1 Cells do not have to be open in X For example, in the CW structure on [0,1] with two zero cells 0 and 1, the 0-cells are not open in [0,1].
In a CW complex X, an n-cell σ is open within the n-skeleton X^n For any point x in σ, one can select a neighborhood U that is open in σ The intersection of U with any other cell σ^0 is empty, except when σ^0 is the same as σ, due to the absence of higher-dimensional cells; only the boundary of σ intersects with other cells Consequently, by the weak topology, U is also open in the entire space X.
The n-skeleton X n is by corollary 1.11.11.2 a subcomplex and thus by lemma 1.11.10.2 closed in X.