Book 7A

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Course 7Book 7A: Topology and the Fundamental GroupChapter 10

Singular Homology and Mayer–Vietoris

Singular chains, homotopy invariance, exact sequences, excision and Mayer–Vietoris, the homology of spheres and lens spaces, Jordan–Brouwer and invariance of domain, cellular homology and the Lefschetz fixed point theorem.

39 min read · Updated Oct 24, 2026

Read with Hatcher, Algebraic Topology, §2.1 (singular homology, homotopy invariance, exact sequences, excision, the equivalence of simplicial and singular homology), §2.2 (degree, cellular homology, Mayer–Vietoris, the Euler characteristic) and, for Jordan–Brouwer and invariance of domain, Proposition 2B.1 and Theorem 2B.3; or Lee, Introduction to Topological Manifolds, chapter 13.

In this chapter · 9 sections
  1. 10.1Inside or outside
  2. 10.2Singular homology
  3. 10.3Exact sequences
  4. 10.4The homology of spheres
  5. 10.5Jordan–Brouwer and invariance of domain
  6. 10.6Cellular homology
  7. 10.7The Lefschetz fixed point theorem
  8. 10.8History
  9. 10.9Exercises
  10. ↗Free resources

7A.9 Homology in Brief introduced homology for spaces with a triangulation, computed it by linear algebra, and stated the facts needed for the Poincaré conjecture. This chapter builds the theory properly, for all spaces, and proves those facts. The construction is singular homology: chains are formal sums of continuous maps of simplices, so no triangulation is needed and functoriality is automatic. Its power comes from three properties, proved here:

  • Homotopy invariance: homotopic maps induce the same maps on homology.
  • Exactness: a pair (X,A)(X, A) gives a long exact sequence linking the homology of AA, of XX, and of XX relative to AA.
  • Excision: relative homology does not see what happens deep inside AA. Equivalently, the Mayer–Vietoris sequence computes the homology of a union from the pieces and their intersection.

With these three, the homology of spheres follows by induction, and with it the classical theorems of topology in every dimension: Brouwer's fixed point theorem, the Jordan–Brouwer separation theorem, invariance of domain and of dimension. Cellular homology then computes everything from a CW structure (7A.6 CW Complexes and Cellular Approximation) with almost no work, and proves that the Euler characteristic is a topological invariant. The homology of lens spaces and of the Poincaré homology sphere, which 7A.12 The Poincaré Conjecture, Precisely uses to explain what the conjecture is about, comes out in a few lines.

By the end of this chapter you will be able to:

  • define singular homology, prove its functoriality and homotopy invariance (with the prism operator), and compute H0H_0;
  • prove the zig-zag lemma and use the long exact sequence of a pair; state excision and use it for good pairs, H∗(X,A)≅H~∗(X/A)H_*(X, A) \cong \tilde H_*(X/A);
  • derive and apply the Mayer–Vietoris sequence: spheres, wedges, connected sums;
  • define the degree of a map Sn→SnS^n \to S^n, compute it for reflections and the antipodal map, and prove the hairy ball and Brouwer theorems with it;
  • prove the Jordan–Brouwer separation theorem and invariance of domain;
  • compute cellular homology of RPn\mathbb{R}P^n, lens spaces and surfaces, prove that χ\chi is a topological invariant, and state the Lefschetz fixed point theorem.

Inside or outside

In the world In use The point-in-polygon test

Every map application, geographic information system and video game constantly asks whether a point lies inside a polygon: is this address inside the city limits, has the character walked into a building? The standard answer is the ray-casting test: draw a ray from the point in any direction and count how many times it crosses the polygon's edges. An odd count means inside, an even count outside. The test is correct because of the Jordan curve theorem: a simple closed curve in the plane separates it into exactly two components, a bounded inside and an unbounded outside, and crossing the curve transversally always switches between them. For polygons the theorem is elementary; for arbitrary continuous curves it is surprisingly hard, and its natural proof, in every dimension, is by homology.

Where the picture breaks

Implementations must handle the cases the theorem's idealisation ignores: rays through vertices, edges collinear with the ray, floating-point rounding of nearly degenerate polygons, and self-intersecting "polygons", for which the parity rule and the winding-number rule (7A.4 The Fundamental Group) give different answers. The topology says what the right answer is for a genuine simple closed curve; robust geometry code is needed to compute it.

Singular homology

Let Δn={(t0,…,tn)∈Rn+1:ti≥0,∑ti=1}\Delta^n = \{(t_0, \dots, t_n) \in \mathbb{R}^{n+1} : t_i \geq 0, \sum t_i = 1\} be the standard nn-simplex, with vertices e0,…,ene_0, \dots, e_n and faces Δn−1→Δn\Delta^{n-1} \to \Delta^n given by inserting a zero in position ii.

Definition 10.1 Singular homology

A singular nn-simplex in a space XX is a continuous map σ:Δn→X\sigma : \Delta^n \to X. The singular chain group Cn(X)C_n(X) is the free abelian group on all singular nn-simplices. The boundary of σ\sigma is the alternating sum of its faces,

∂σ=∑i=0n(−1)iσ∣[e0,…,e^i,…,en],\partial\sigma = \sum_{i=0}^n(-1)^i\sigma|_{[e_0, \dots, \hat e_i, \dots, e_n]},

extended linearly. Then ∂∘∂=0\partial\circ\partial = 0 (Exercise 10.14), and the singular homology is Hn(X)=ker⁡∂n/im⁡∂n+1H_n(X) = \ker\partial_n/\operatorname{im}\partial_{n+1}.

The groups Cn(X)C_n(X) are enormous (one generator for each continuous map), but homology is computable, and for simplicial complexes agrees with the simplicial homology of 7A.9 Homology in Brief (Hatcher, Theorem 2.27). The advantage is functoriality: a continuous map f:X→Yf : X \to Y sends σ\sigma to f∘σf\circ\sigma, commuting with ∂\partial, so it induces f∗:Hn(X)→Hn(Y)f_* : H_n(X) \to H_n(Y), with (g∘f)∗=g∗∘f∗(g\circ f)_* = g_*\circ f_* and id∗=id\mathrm{id}_* = \mathrm{id}. Homeomorphic spaces have isomorphic homology.

H0H_0 counts path components. A 0-simplex is a point, a 1-simplex a path, and ∂σ=σ(e1)−σ(e0)\partial\sigma = \sigma(e_1) - \sigma(e_0). So two points are homologous iff they are joined by a path, and H0(X)H_0(X) is free abelian on the path components. The reduced homology H~∗\tilde H_* replaces H0H_0 by the kernel of the augmentation ∑nixi↦∑ni\sum n_ix_i \mapsto \sum n_i; for nonempty XX, H0=H~0⊕ZH_0 = \tilde H_0\oplus\mathbb{Z} and H~n=Hn\tilde H_n = H_n for n>0n > 0, and a point has H~∗=0\tilde H_* = 0.

Theorem 10.2 Homotopy invariance

If f≃g:X→Yf \simeq g : X \to Y, then f∗=g∗:Hn(X)→Hn(Y)f_* = g_* : H_n(X) \to H_n(Y). Consequently homotopy equivalent spaces have isomorphic homology, and contractible spaces have H~∗=0\tilde H_* = 0.

Proof. Let F:X×I→YF : X\times I \to Y be a homotopy. Subdivide the prism Δn×I\Delta^n\times I into n+1n + 1 simplices [v0,…,vi,wi,…,wn][v_0, \dots, v_i, w_i, \dots, w_n], where vj=(ej,0)v_j = (e_j, 0) and wj=(ej,1)w_j = (e_j, 1), and define the prism operator P:Cn(X)→Cn+1(Y)P : C_n(X) \to C_{n+1}(Y) by

P(σ)=∑i=0n(−1)iF∘(σ×id)∣[v0,…,vi,wi,…,wn].P(\sigma) = \sum_{i=0}^n(-1)^iF\circ(\sigma\times\mathrm{id})|_{[v_0, \dots, v_i, w_i, \dots, w_n]}.

A direct computation of faces (Exercise 10.15) gives the chain homotopy identity

∂P+P∂=g#−f#,\partial P + P\partial = g_\# - f_\#,

the algebraic version of "the boundary of the prism is the top, minus the bottom, minus the prism over the boundary". For a cycle zz, g#z−f#z=∂Pzg_\#z - f_\#z = \partial Pz is a boundary, so [g#z]=[f#z][g_\#z] = [f_\#z].

Exact sequences

For A⊆XA \subseteq X, the relative chains Cn(X,A)=Cn(X)/Cn(A)C_n(X, A) = C_n(X)/C_n(A) inherit ∂\partial, and their homology Hn(X,A)H_n(X, A) consists of chains whose boundary lies in AA, modulo chains in AA and boundaries. Intuitively, it is the homology of XX with AA "collapsed".

Lemma 10.3 The zig-zag lemma

A short exact sequence of chain complexes 0→A∙→iB∙→jC∙→00 \to A_\bullet \xrightarrow{i} B_\bullet \xrightarrow{j} C_\bullet \to 0 gives a long exact sequence

⋯→Hn(A)→i∗Hn(B)→j∗Hn(C)→δHn−1(A)→i∗Hn−1(B)→⋯ .\cdots \to H_n(A) \xrightarrow{i_*} H_n(B) \xrightarrow{j_*} H_n(C) \xrightarrow{\delta} H_{n-1}(A) \xrightarrow{i_*} H_{n-1}(B) \to \cdots.

Proof. The connecting map. Let c∈Cnc \in C_n be a cycle. Lift it to b∈Bnb \in B_n with j(b)=cj(b) = c. Then j(∂b)=∂c=0j(\partial b) = \partial c = 0, so ∂b=i(a)\partial b = i(a) for a unique a∈An−1a \in A_{n-1}, and i(∂a)=∂∂b=0i(\partial a) = \partial\partial b = 0, so aa is a cycle. Set δ[c]=[a]\delta[c] = [a]. Different choices of lift change bb by i(a′)i(a') and aa by ∂a′\partial a', and changing cc by a boundary ∂c′\partial c' changes aa by a boundary too, so δ\delta is well defined.

Exactness is checked at each of the three spots by diagram chasing; for instance at Hn(C)H_n(C): δj∗[b]=0\delta j_*[b] = 0 since a lift of j(b)j(b) is bb itself, a cycle; and if δ[c]=0\delta[c] = 0, then a=∂a′a = \partial a', so b−i(a′)b - i(a') is a cycle mapping to cc (Exercise 10.16 for the rest).

Applied to 0→C(A)→C(X)→C(X,A)→00 \to C(A) \to C(X) \to C(X, A) \to 0, the lemma gives the long exact sequence of the pair

⋯→Hn(A)→Hn(X)→Hn(X,A)→∂Hn−1(A)→⋯ ,\cdots \to H_n(A) \to H_n(X) \to H_n(X, A) \xrightarrow{\partial} H_{n-1}(A) \to \cdots,

where ∂\partial takes a relative cycle to its boundary, a cycle in AA.

Excision. The deep property of singular homology is that relative homology is local.

Theorem 10.4 Excision (statement)

If Z⊆A⊆XZ \subseteq A \subseteq X with the closure of ZZ contained in the interior of AA, then the inclusion (X∖Z,A∖Z)→(X,A)(X\setminus Z, A\setminus Z) \to (X, A) induces isomorphisms Hn(X∖Z,A∖Z)≅Hn(X,A)H_n(X\setminus Z, A\setminus Z) \cong H_n(X, A). Equivalently, if X=int⁡A∪int⁡BX = \operatorname{int}A\cup\operatorname{int}B, then Hn(B,A∩B)≅Hn(X,A)H_n(B, A\cap B) \cong H_n(X, A).

The proof (Hatcher, §2.1) shows that every chain is homologous to one made of small simplices, each lying in AA or in BB, by repeated barycentric subdivision: subdividing a simplex into smaller ones is chain homotopic to the identity, and after enough subdivisions every simplex is small (the Lebesgue number lemma, 2B.3 Compactness). The idea is the same as in van Kampen's theorem (7A.5 Computing π₁): break loops, or chains, into small pieces that each lie in one open set.

For a good pair (a closed AA that is a deformation retract of a neighbourhood, for instance a CW pair), excision gives

Hn(X,A)≅H~n(X/A).H_n(X, A) \cong \tilde H_n(X/A).

For instance (Dn,Sn−1)(D^n, S^{n-1}) is good with Dn/Sn−1≅SnD^n/S^{n-1} \cong S^n.

Mayer–Vietoris. If X=int⁡A∪int⁡BX = \operatorname{int}A\cup\operatorname{int}B, the chains in XX made of simplices lying in AA or in BB compute H∗(X)H_*(X) (the subdivision argument again), and 0→C(A∩B)→C(A)⊕C(B)→C(A)+C(B)→00 \to C(A\cap B) \to C(A)\oplus C(B) \to C(A) + C(B) \to 0, with maps x↦(x,−x)x \mapsto (x, -x) and (y,z)↦y+z(y, z) \mapsto y + z, is exact. The zig-zag lemma gives the Mayer–Vietoris sequence

⋯→Hn(A∩B)→Hn(A)⊕Hn(B)→Hn(X)→∂Hn−1(A∩B)→⋯ ,\cdots \to H_n(A\cap B) \to H_n(A)\oplus H_n(B) \to H_n(X) \xrightarrow{\partial} H_{n-1}(A\cap B) \to \cdots,

the homological van Kampen theorem. The connecting map takes a cycle z=a+bz = a + b (aa in AA, bb in BB) to ∂a=−∂b\partial a = -\partial b, a cycle in A∩BA\cap B: cut the cycle along the intersection and take the cut. The same holds with reduced homology when A∩B≠∅A\cap B \neq \emptyset.

The homology of spheres

Theorem 10.5 Homology of spheres

H~k(Sn)=Z\tilde H_k(S^n) = \mathbb{Z} for k=nk = n and 00 otherwise.

Proof. Induct on nn. For S0S^0, two points, H~0=Z\tilde H_0 = \mathbb{Z}. Cover SnS^n by AA and BB, the complements of the north and south poles, each contractible, with A∩B≃Sn−1A\cap B \simeq S^{n-1} (a band around the equator). Reduced Mayer–Vietoris gives 0=H~k(A)⊕H~k(B)→H~k(Sn)→∂H~k−1(Sn−1)→00 = \tilde H_k(A)\oplus\tilde H_k(B) \to \tilde H_k(S^n) \xrightarrow{\partial} \tilde H_{k-1}(S^{n-1}) \to 0, so H~k(Sn)≅H~k−1(Sn−1)\tilde H_k(S^n) \cong \tilde H_{k-1}(S^{n-1}).

Immediate consequences:

  • SnS^n is not contractible, and Sn≄SmS^n \not\simeq S^m for n≠mn \neq m.
  • Invariance of dimension: Rn≇Rm\mathbb{R}^n \not\cong \mathbb{R}^m for n≠mn \neq m, because Rn∖{0}≃Sn−1\mathbb{R}^n\setminus\{0\} \simeq S^{n-1}.
  • No retraction Dn→Sn−1D^n \to S^{n-1}: it would give H~n−1(Sn−1)=Z\tilde H_{n-1}(S^{n-1}) = \mathbb{Z} factoring through H~n−1(Dn)=0\tilde H_{n-1}(D^n) = 0. Hence Brouwer's fixed point theorem in every dimension, by the argument of 7A.4 The Fundamental Group (a fixed-point-free map gives a retraction), without the smooth approximation used in 7A.8 Smooth Topology.

Degree. A map f:Sn→Snf : S^n \to S^n induces multiplication by an integer on Hn(Sn)=ZH_n(S^n) = \mathbb{Z}, its degree deg⁡f\deg f. It is a homotopy invariant, multiplicative, and deg⁡id=1\deg\mathrm{id} = 1. It agrees with the smooth degree of 7A.8 Smooth Topology (counting preimages of a regular value with signs; Hatcher, Proposition 2.30).

Proposition 10.6 Reflections and the antipodal map

A reflection of SnS^n in a hyperplane through the origin has degree −1-1. The antipodal map x↦−xx \mapsto -x has degree (−1)n+1(-1)^{n+1}.

Proof. Give SnS^n the CW structure with two nn-cells, the hemispheres Δ1\Delta_1, Δ2\Delta_2 on either side of the hyperplane, glued along Sn−1S^{n-1}. Then Δ1−Δ2\Delta_1 - \Delta_2 (suitably oriented) is a cycle generating HnH_n (Exercise 10.17), and the reflection swaps the two hemispheres, sending the generator to its negative. The antipodal map is the composite of n+1n + 1 reflections, one in each coordinate.

The hairy ball theorem. If vv is a nowhere vanishing tangent vector field on SnS^n, then ft(x)=cos⁡(πt)x+sin⁡(πt)v(x)∣v(x)∣f_t(x) = \cos(\pi t)x + \sin(\pi t)\frac{v(x)}{|v(x)|} is a homotopy from the identity to the antipodal map. So 1=(−1)n+11 = (-1)^{n+1}, and nn is odd. On even-dimensional spheres every vector field vanishes somewhere, as 7A.8 Smooth Topology proved with Poincaré–Hopf.

Jordan–Brouwer and invariance of domain

Proposition 10.7 Complements of discs and spheres

(a) If h:Dk→Snh : D^k \to S^n is an embedding, then H~∗(Sn∖h(Dk))=0\tilde H_*(S^n\setminus h(D^k)) = 0. (b) If h:Sk→Snh : S^k \to S^n is an embedding, k<nk < n, then H~i(Sn∖h(Sk))=Z\tilde H_i(S^n\setminus h(S^k)) = \mathbb{Z} for i=n−k−1i = n - k - 1 and 00 otherwise.

Proof. (a) Induct on kk; for k=0k = 0, SnS^n minus a point is Rn\mathbb{R}^n. Write Dk=Ik−1×ID^k = I^{k-1}\times I and split II into halves I1I_1, I2I_2, giving A=Sn∖h(Ik−1×I1)A = S^n\setminus h(I^{k-1}\times I_1) and B=Sn∖h(Ik−1×I2)B = S^n\setminus h(I^{k-1}\times I_2), open, with A∪B=Sn∖h(Ik−1×{12})A\cup B = S^n\setminus h(I^{k-1}\times\{\frac12\}) (acyclic by induction) and A∩B=Sn∖h(Dk)A\cap B = S^n\setminus h(D^k). Mayer–Vietoris gives H~i(A∩B)≅H~i(A)⊕H~i(B)\tilde H_i(A\cap B) \cong \tilde H_i(A)\oplus\tilde H_i(B). If a class α∈H~i(Sn∖h(Dk))\alpha \in \tilde H_i(S^n\setminus h(D^k)) were nonzero, it would be nonzero in AA or in BB; repeating the halving produces nested intervals Im→{t}I_m \to \{t\} with α\alpha nonzero in Sn∖h(Ik−1×Im)S^n\setminus h(I^{k-1}\times I_m) for all mm. But α\alpha is carried by a compact chain, and Sn∖h(Ik−1×{t})S^n\setminus h(I^{k-1}\times\{t\}), the increasing union of these open sets, is acyclic by induction; a cycle bounding there bounds in some Sn∖h(Ik−1×Im)S^n\setminus h(I^{k-1}\times I_m), by compactness of the bounding chain. Contradiction.

(b) Induct on kk; S0S^0 is two points, and SnS^n minus two points ≃Sn−1\simeq S^{n-1}. Write Sk=D1∪D2S^k = D_1\cup D_2, hemispheres meeting in Sk−1S^{k-1}. With A=Sn∖h(D1)A = S^n\setminus h(D_1) and B=Sn∖h(D2)B = S^n\setminus h(D_2), acyclic by (a), A∪B=Sn∖h(Sk−1)A\cup B = S^n\setminus h(S^{k-1}) and A∩B=Sn∖h(Sk)A\cap B = S^n\setminus h(S^k). Mayer–Vietoris gives H~i(Sn∖h(Sk))≅H~i+1(Sn∖h(Sk−1))\tilde H_i(S^n\setminus h(S^k)) \cong \tilde H_{i+1}(S^n\setminus h(S^{k-1})), and induction finishes.

Theorem 10.8 Jordan–Brouwer separation theorem

An embedded (n−1)(n - 1)-sphere Σ\Sigma in SnS^n separates it into exactly two components, each of which has Σ\Sigma as its boundary. (The case n=2n = 2 is the Jordan curve theorem.)

Proof. By (b) with k=n−1k = n - 1, H~0(Sn∖Σ)=Z\tilde H_0(S^n\setminus\Sigma) = \mathbb{Z}: two path components (open subsets of SnS^n are locally path-connected, so these are the components). For the boundary statement: each component UU is open with ∂U⊆Σ\partial U \subseteq \Sigma. If ∂U\partial U missed a point xx, it would miss a small open disc D∋xD \ni x in Σ\Sigma, chosen so that A=Σ∖DA = \Sigma\setminus D is an embedded closed disc. By (a), Sn∖AS^n\setminus A is connected, but it is the disjoint union of UU, the other component, and DD, and UU would be both open and closed in it (Exercise 10.18).

In dimension 22 more is true, the Schoenflies theorem: each component is homeomorphic to an open disc. In dimension 33 this fails: Alexander's horned sphere (1924) is an embedded S2S^2 in S3S^3 whose outside is not simply connected. Wild embeddings like this are one reason three-dimensional topology needs care, and why the smooth or piecewise-linear categories, where Alexander's theorem says every embedded 2-sphere bounds a ball on each side, are used (10A.5 The Prime Decomposition).

Theorem 10.9 Invariance of domain (Brouwer 1912)

If U⊆RnU \subseteq \mathbb{R}^n is open and h:U→Rnh : U \to \mathbb{R}^n is continuous and injective, then h(U)h(U) is open and hh is a homeomorphism onto it.

Proof. Regard Rn⊆Sn\mathbb{R}^n \subseteq S^n. For x∈Ux \in U take a small closed ball DD around xx in UU, with boundary sphere SS. By (b), Sn∖h(S)S^n\setminus h(S) has two components; by (a), Sn∖h(D)S^n\setminus h(D) is connected. Write Sn∖h(S)=(Sn∖h(D))∪h(int⁡D)S^n\setminus h(S) = (S^n\setminus h(D))\cup h(\operatorname{int}D), a disjoint union of two connected sets, so these are its two components, and components of an open set are open. So h(int⁡D)h(\operatorname{int}D) is an open neighbourhood of h(x)h(x). Hence hh is an open map, so a homeomorphism onto its image.

Consequences: a manifold has a well-defined dimension; a manifold's boundary points (those with half-space neighbourhoods) are distinct from its interior points, so the boundary of a manifold is well defined (7A.3 Manifolds and Surfaces); and a continuous injection of a closed nn-manifold into another closed connected nn-manifold is onto (Exercise 10.19).

Cellular homology

For a CW complex XX, the long exact sequences of the pairs (Xn,Xn−1)(X^n, X^{n-1}) and the fact that Xn/Xn−1X^n/X^{n-1} is a wedge of nn-spheres (one for each nn-cell) give:

Theorem 10.10 Cellular homology (statement)

Let CnCW(X)=Hn(Xn,Xn−1)C_n^{CW}(X) = H_n(X^n, X^{n-1}), free abelian on the nn-cells, with boundary dnd_n the composite Hn(Xn,Xn−1)→Hn−1(Xn−1)→Hn−1(Xn−1,Xn−2)H_n(X^n, X^{n-1}) \to H_{n-1}(X^{n-1}) \to H_{n-1}(X^{n-1}, X^{n-2}). Then d∘d=0d\circ d = 0 and the homology of (CCW,d)(C^{CW}, d) is H∗(X)H_*(X). Explicitly, d(eαn)=∑βdαβeβn−1d(e^n_\alpha) = \sum_\beta d_{\alpha\beta}e^{n-1}_\beta, where dαβd_{\alpha\beta} is the degree of the map Sn−1→Xn−1→Xn−1/(Xn−1∖eβn−1)≅Sn−1S^{n-1} \to X^{n-1} \to X^{n-1}/(X^{n-1}\setminus e^{n-1}_\beta) \cong S^{n-1}: attach, then collapse everything but the cell eβe_\beta.

The proof is a diagram chase with the long exact sequences, using Hk(Xn,Xn−1)=0H_k(X^n, X^{n-1}) = 0 for k≠nk \neq n (Hatcher, Theorem 2.35). In practice the degrees are easy to read off.

Example 10.11 Cellular computations
  1. Spheres: Sn=e0∪enS^n = e^0\cup e^n, n≥2n \geq 2: all boundary maps are 00, recovering H~n=Z\tilde H_n = \mathbb{Z}.
  2. Surfaces: for Σg\Sigma_g with one 0-cell, 2g2g 1-cells and one 2-cell attached along ∏[ai,bi]\prod[a_i, b_i], each 1-cell appears twice with opposite signs in the attaching word, so d2=0d_2 = 0; d1=0d_1 = 0 since each 1-cell is a loop. So H0=ZH_0 = \mathbb{Z}, H1=Z2gH_1 = \mathbb{Z}^{2g}, H2=ZH_2 = \mathbb{Z}. For the nonorientable surface NkN_k (word a1a1⋯akaka_1a_1\cdots a_ka_k), d2=(2,…,2)d_2 = (2, \dots, 2), so H1=Zk−1⊕Z/2H_1 = \mathbb{Z}^{k-1}\oplus\mathbb{Z}/2 and H2=0H_2 = 0.
  3. Real projective space: one cell in each dimension, with dk:Ck→Ck−1d_k : C_k \to C_{k-1} multiplication by 1+(−1)k1 + (-1)^k (the attaching map is the double cover Sk−1→RPk−1S^{k-1} \to \mathbb{R}P^{k-1}, which on the top cell is the sum of the identity and the antipodal map). So dk=0d_k = 0 for kk odd and 22 for kk even, and
Hk(RPn)=Z (k=0),Z/2 (k odd,0<k<n),0 (k even,0<k≤n),Z (k=n odd).H_k(\mathbb{R}P^n) = \mathbb{Z}\ (k = 0),\quad \mathbb{Z}/2\ (k \text{ odd}, 0 < k < n),\quad 0\ (k \text{ even}, 0 < k \leq n),\quad \mathbb{Z}\ (k = n \text{ odd}).

In particular H∗(RP3)=Z,Z/2,0,ZH_*(\mathbb{R}P^3) = \mathbb{Z}, \mathbb{Z}/2, 0, \mathbb{Z}. 4. Lens spaces: L(p,q)L(p, q) has one cell in each dimension with d1=0d_1 = 0, d2=pd_2 = p, d3=0d_3 = 0 (7A.6 CW Complexes and Cellular Approximation), so H∗(L(p,q))=Z,Z/p,0,ZH_*(L(p, q)) = \mathbb{Z}, \mathbb{Z}/p, 0, \mathbb{Z}, independent of qq. Homology cannot tell L(5,1)L(5, 1) from L(5,2)L(5, 2): 7A.11 Cohomology and Poincaré Duality distinguishes them with the linking form. Subtler still, L(7,1)L(7, 1) and L(7,2)L(7, 2) are homotopy equivalent but not homeomorphic. 5. Complex projective space: cells only in even dimensions, so all d=0d = 0 and H2k(CPn)=ZH_{2k}(\mathbb{C}P^n) = \mathbb{Z} for 0≤k≤n0 \leq k \leq n, zero otherwise.

Corollary 10.12 The Euler characteristic is a topological invariant

For a finite CW complex, χ(X)=∑(−1)ncn=∑(−1)nrank⁡Hn(X)\chi(X) = \sum(-1)^nc_n = \sum(-1)^n\operatorname{rank}H_n(X). In particular χ\chi does not depend on the cell structure, and homotopy equivalent complexes have the same χ\chi.

Proof. For a chain complex of finitely generated free groups, rank–nullity in each degree gives cn=rank⁡Zn+rank⁡Bn−1c_n = \operatorname{rank}Z_n + \operatorname{rank}B_{n-1} and rank⁡Hn=rank⁡Zn−rank⁡Bn\operatorname{rank}H_n = \operatorname{rank}Z_n - \operatorname{rank}B_n; the alternating sums agree, with the BB terms telescoping.

This is the invariance promised in 7A.3 Manifolds and Surfaces and 7A.6 CW Complexes and Cellular Approximation: Euler's V−E+F=2V - E + F = 2 for every polyhedral sphere, and χ=2−2g\chi = 2 - 2g for every cell structure on Σg\Sigma_g.

Connected sums. For closed connected nn-manifolds MM, NN, n≥3n \geq 3: removing a ball does not change HkH_k for k<n−1k < n - 1, and Mayer–Vietoris over the gluing sphere gives Hk(M#N)≅Hk(M)⊕Hk(N)H_k(M\#N) \cong H_k(M)\oplus H_k(N) for 0<k<n0 < k < n (for orientable MM, NN; Exercise 10.20). In dimension three, H1(M#N)=H1(M)⊕H1(N)H_1(M\#N) = H_1(M)\oplus H_1(N), the abelianisation of the free product π1(M)∗π1(N)\pi_1(M)*\pi_1(N) (7A.5 Computing π₁, 10A.5 The Prime Decomposition).

The Lefschetz fixed point theorem

For a map f:X→Xf : X \to X of a finite CW complex, the Lefschetz number is

L(f)=∑n(−1)ntr⁡(f∗:Hn(X;Q)→Hn(X;Q)).L(f) = \sum_n(-1)^n\operatorname{tr}\big(f_* : H_n(X; \mathbb{Q}) \to H_n(X; \mathbb{Q})\big).
Theorem 10.13 Lefschetz fixed point theorem (statement)

If XX is a finite CW complex (or a retract of one) and L(f)≠0L(f) \neq 0, then ff has a fixed point.

For f=idf = \mathrm{id}, L=χ(X)L = \chi(X); for a contractible XX, L(f)=1L(f) = 1 for every ff, recovering Brouwer. Every map RP2k→RP2k\mathbb{R}P^{2k} \to \mathbb{R}P^{2k} has a fixed point (its rational homology is that of a point), and a map Sn→SnS^n \to S^n without fixed points has degree (−1)n+1(-1)^{n+1}, the antipodal map's. The proof approximates ff by a cellular map that moves every cell off itself if there are no fixed points, so that the trace on cellular chains, which equals L(f)L(f) by the same telescoping as for χ\chi, is zero. For smooth maps on closed manifolds, L(f)L(f) is the signed count of fixed points, the analogue of Poincaré–Hopf (7A.8 Smooth Topology), which is the case of the flow of a vector field for a short time.

In the world In use Counting holes in materials and in data

Homology groups computed from discretised data are routine in materials science and imaging. Betti numbers of segmented 3D images, the numbers of connected components, tunnels and cavities, measure the connectivity of porous rocks (which controls permeability), bone microstructure (trabecular connectivity is an indicator in osteoporosis research), and foams. They are computed from voxel data by exactly the cellular homology of this chapter, applied to the cubical complex of occupied voxels, together with the Euler characteristic, which is cheap to compute locally by counting cells.

Where this goes Duality and the Poincaré conjecture

7A.11 Cohomology and Poincaré Duality dualises: cohomology, cup products and Poincaré duality, which turn the homology computations of this chapter into the statement that a closed simply connected 3-manifold has the homology of S3S^3. With 7A.6 CW Complexes and Cellular Approximation (Whitehead) and 10A.4 Homotopy Groups, Degree and the Hurewicz Theorem (Hurewicz), that makes it a homotopy sphere. 7A.12 The Poincaré Conjecture, Precisely then states the conjecture precisely.

History

Poincaré introduced homology in Analysis Situs (1895) and its supplements, with Betti numbers and torsion; Emmy Noether, around 1925, recast it in terms of groups. Singular homology was developed by Lefschetz and by Eilenberg (1944), who gave its modern definition; Eilenberg and Steenrod axiomatised homology in 1945 by homotopy invariance, exactness and excision. Walther Mayer (1929) and Leopold Vietoris (1930) found their sequence. Camille Jordan stated his curve theorem in 1887; Oswald Veblen gave the first rigorous proof in 1905, and Brouwer proved the higher-dimensional separation theorem and invariance of domain in 1911–12. Lefschetz proved his fixed point theorem in 1926–37, and Alexander's horned sphere appeared in 1924.

Recall Where we stand

Singular homology uses formal sums of maps of simplices; it is functorial, H0H_0 counts path components, and the prism operator makes it a homotopy invariant. Short exact sequences of chain complexes give long exact sequences (zig-zag lemma), in particular for pairs; excision, proved by barycentric subdivision, makes relative homology local, so H∗(X,A)=H~∗(X/A)H_*(X, A) = \tilde H_*(X/A) for good pairs, and Mayer–Vietoris computes the homology of unions. Hence H~∗(Sn)\tilde H_*(S^n) is Z\mathbb{Z} in degree nn, which gives Brouwer in all dimensions, invariance of dimension, degree theory (reflections have degree −1-1, the antipodal map (−1)n+1(-1)^{n+1}, so only odd spheres have nonvanishing vector fields), the Jordan–Brouwer separation theorem and invariance of domain. Cellular homology computes H∗H_* from cells and degrees: RPn\mathbb{R}P^n, surfaces, CPn\mathbb{C}P^n, and lens spaces H∗=Z,Z/p,0,ZH_* = \mathbb{Z}, \mathbb{Z}/p, 0, \mathbb{Z} independent of qq. The Euler characteristic is a topological invariant, and a nonzero Lefschetz number forces a fixed point.

Exercises

Exercise 10.14 ∂∘∂=0\partial\circ\partial = 0

Show that ∂∂σ=0\partial\partial\sigma = 0 for a singular nn-simplex: each face of codimension two, σ∣[…e^i…e^j… ]\sigma|_{[\dots\hat e_i\dots\hat e_j\dots]} with i<ji < j, appears twice in ∂∂σ\partial\partial\sigma, with opposite signs.

Solution

Removing eje_j then eie_i (i<ji < j, where eie_i is still in position ii) gives sign (−1)j(−1)i(-1)^j(-1)^i. Removing eie_i first, eje_j moves to position j−1j - 1, so removing it gives (−1)i(−1)j−1(-1)^i(-1)^{j-1}. The two terms cancel.

Exercise 10.15 The prism identity in low dimensions

For a singular 0-simplex (a point xx), P(x)P(x) is the path t↦F(x,t)t \mapsto F(x, t). Verify ∂P(x)=g(x)−f(x)\partial P(x) = g(x) - f(x), which is the identity ∂P+P∂=g#−f#\partial P + P\partial = g_\# - f_\# in degree 00. For a 1-simplex σ\sigma, describe P(σ)P(\sigma) as two triangles filling the square σ×I\sigma\times I, and verify the identity.

Solution

∂(path)=end−start=F(x,1)−F(x,0)=g(x)−f(x)\partial(\text{path}) = \text{end} - \text{start} = F(x, 1) - F(x, 0) = g(x) - f(x), and P∂x=0P\partial x = 0. For σ\sigma: Pσ=F(σ×id)∣[v0,w0,w1]−F(σ×id)∣[v0,v1,w1]P\sigma = F(\sigma\times\mathrm{id})|_{[v_0, w_0, w_1]} - F(\sigma\times\mathrm{id})|_{[v_0, v_1, w_1]}. Its boundary consists of the top edge [w0,w1][w_0, w_1] (=gσ= g\sigma), the bottom edge [v0,v1][v_0, v_1] with a minus sign (−fσ-f\sigma), the vertical edges [v0,w0][v_0, w_0] and [v1,w1][v_1, w_1], which are PP of the endpoints with signs giving −P∂σ-P\partial\sigma, and the diagonal [v0,w1][v_0, w_1] twice with opposite signs. So ∂Pσ=gσ−fσ−P∂σ\partial P\sigma = g\sigma - f\sigma - P\partial\sigma.

Exercise 10.16 Exactness at Hn(B)H_n(B) and Hn−1(A)H_{n-1}(A)

Complete the proof of the zig-zag lemma: show im⁡i∗=ker⁡j∗\operatorname{im}i_* = \ker j_* at Hn(B)H_n(B) and im⁡δ=ker⁡i∗\operatorname{im}\delta = \ker i_* at Hn−1(A)H_{n-1}(A).

Solution

At Hn(B)H_n(B): j∗i∗=(ji)∗=0j_*i_* = (ji)_* = 0. If j∗[b]=0j_*[b] = 0, then j(b)=∂c′j(b) = \partial c'; lift c′=j(b′)c' = j(b'), then j(b−∂b′)=0j(b - \partial b') = 0, so b−∂b′=i(a)b - \partial b' = i(a), and aa is a cycle since ii is injective and i(∂a)=∂b=0i(\partial a) = \partial b = 0; [b]=i∗[a][b] = i_*[a]. At Hn−1(A)H_{n-1}(A): i∗δ[c]=[i(a)]=[∂b]=0i_*\delta[c] = [i(a)] = [\partial b] = 0. If i∗[a]=0i_*[a] = 0, i(a)=∂bi(a) = \partial b, and c=j(b)c = j(b) is a cycle with δ[c]=[a]\delta[c] = [a].

Exercise 10.17 A generator of Hn(Sn)H_n(S^n)

Using the CW structure of SnS^n with two nn-cells (hemispheres) glued along Sn−1S^{n-1} (with its own structure with two cells in each dimension), show that the cellular chain D1−D2D_1 - D_2 is a cycle generating Hn(Sn)H_n(S^n), and that the reflection swapping the hemispheres sends it to −(D1−D2)-(D_1 - D_2).

Solution

Orient both hemispheres by projecting them to the same disc in Rn\mathbb{R}^n; then both have boundary the equator with the same orientation, d(D1)=d(D2)=[Sn−1]d(D_1) = d(D_2) = [S^{n-1}], and d(D1−D2)=0d(D_1 - D_2) = 0. The nn-cycles are multiples of D1−D2D_1 - D_2 (the kernel of Z2→Z\mathbb{Z}^2 \to \mathbb{Z}, (a,b)↦a+b(a, b) \mapsto a + b), and there are no (n+1)(n + 1)-cells, so Hn=Z(D1−D2)H_n = \mathbb{Z}(D_1 - D_2). The reflection maps D1D_1 to D2D_2 and vice versa, preserving the orientations induced from the equator, so D1−D2↦D2−D1D_1 - D_2 \mapsto D_2 - D_1.

Exercise 10.18 Every point of the curve is on the boundary

Complete the proof that each component UU of S2∖ΣS^2\setminus\Sigma has boundary all of Σ\Sigma. Suppose x∈Σx \in \Sigma is not in Uˉ\bar U. Choose an arc DD of Σ\Sigma around xx with DD disjoint from Uˉ\bar U, and let A=Σ∖int⁡DA = \Sigma\setminus\operatorname{int}D, an embedded arc. Show that UU and the other component VV both lie in the connected set S2∖AS^2\setminus A, and derive a contradiction.

Solution

S2∖AS^2\setminus A is connected by part (a) of Proposition 10.7. It is the disjoint union of UU, VV and int⁡D\operatorname{int}D. UU is open in S2∖AS^2\setminus A, and closed in it: its closure in S2S^2 is U∪∂UU\cup\partial U, and ∂U⊆Σ\partial U \subseteq \Sigma with ∂U∩D=∅\partial U\cap D = \emptyset by the choice of DD, so ∂U⊆A\partial U \subseteq A; and the closure of UU in S2∖AS^2\setminus A is UU. A nonempty proper clopen subset of a connected space is impossible, since V≠∅V \neq \emptyset lies outside UU. So every point of Σ\Sigma is in ∂U\partial U.

Exercise 10.19 Injections of closed manifolds

Let MM, NN be closed connected nn-manifolds and f:M→Nf : M \to N continuous and injective. Show that ff is a homeomorphism.

Solution

By invariance of domain in charts, ff is open, so f(M)f(M) is open; it is compact, hence closed. NN is connected, so f(M)=Nf(M) = N, and an open continuous bijection is a homeomorphism.

Exercise 10.20 Homology of a connected sum

Let MM, NN be closed connected orientable nn-manifolds, n≥3n \geq 3. (a) Show that Hk(M∖ball)≅Hk(M)H_k(M\setminus\text{ball}) \cong H_k(M) for k<n−1k < n - 1 and that Hn−1(M∖ball)→Hn−1(M)H_{n-1}(M\setminus\text{ball}) \to H_{n-1}(M) is an isomorphism, using the pair sequence and excision, H∗(M,M∖ball)≅H∗(Dn,Sn−1)H_*(M, M\setminus\text{ball}) \cong H_*(D^n, S^{n-1}), and the fact (proved in 7A.11 Cohomology and Poincaré Duality) that Hn(M)=ZH_n(M) = \mathbb{Z} maps onto Hn(M,M∖ball)H_n(M, M\setminus\text{ball}). (b) Apply Mayer–Vietoris to M#NM\#N to get Hk(M#N)≅Hk(M)⊕Hk(N)H_k(M\#N) \cong H_k(M)\oplus H_k(N) for 0<k<n0 < k < n. (c) Compute H∗(RP3#RP3)H_*(\mathbb{R}P^3\#\mathbb{R}P^3) and H∗(S2×S1#S2×S1)H_*(S^2\times S^1\#S^2\times S^1).

Solution

(a) Hk(M,M∖B)=Hk(Dn,Sn−1)H_k(M, M\setminus B) = H_k(D^n, S^{n-1}) is Z\mathbb{Z} for k=nk = n and 00 otherwise. The sequence Hn(M)→Hn(M,M∖B)→Hn−1(M∖B)→Hn−1(M)→0H_n(M) \to H_n(M, M\setminus B) \to H_{n-1}(M\setminus B) \to H_{n-1}(M) \to 0 with the first map onto gives the isomorphism in degree n−1n - 1; lower degrees are isomorphisms since the relative groups vanish. (b) M#N=M′∪N′M\#N = M'\cup N' with intersection a collar ≃Sn−1\simeq S^{n-1}, contributing only in degrees 00 and n−1n - 1; the map Hn−1(Sn−1)→Hn−1(M′)⊕Hn−1(N′)H_{n-1}(S^{n-1}) \to H_{n-1}(M')\oplus H_{n-1}(N') is zero, because the gluing sphere is the boundary of M′M' (it is the boundary of a chain representing the fundamental class of M′M' relative to its boundary) and likewise of N′N'; this gives the splitting. (c) Z,(Z/2)2,0,Z\mathbb{Z}, (\mathbb{Z}/2)^2, 0, \mathbb{Z}; and Z,Z2,Z2,Z\mathbb{Z}, \mathbb{Z}^2, \mathbb{Z}^2, \mathbb{Z}.

Exercise 10.21 Rehearsal: fixed points on projective space

(a) Compute H∗(RP2k;Q)H_*(\mathbb{R}P^{2k}; \mathbb{Q}) from the cellular computation, and deduce that every map RP2k→RP2k\mathbb{R}P^{2k} \to \mathbb{R}P^{2k} has a fixed point. (b) Give a fixed-point-free map of RP3\mathbb{R}P^3 (consider [x0:x1:x2:x3]↦[−x1:x0:−x3:x2][x_0 : x_1 : x_2 : x_3] \mapsto [-x_1 : x_0 : -x_3 : x_2]). Why is there no contradiction?

Solution

(a) With rational coefficients the torsion disappears: H0=QH_0 = \mathbb{Q} and all others vanish (the top group H2k=0H_{2k} = 0 since 2k2k is even). So L(f)=1≠0L(f) = 1 \neq 0. (b) A fixed point would need (−x1,x0,−x3,x2)=λ(x0,x1,x2,x3)(-x_1, x_0, -x_3, x_2) = \lambda(x_0, x_1, x_2, x_3), so x1=−λx0x_1 = -\lambda x_0 and x0=λx1=−λ2x0x_0 = \lambda x_1 = -\lambda^2x_0, and similarly for x2x_2, x3x_3; this forces λ2=−1\lambda^2 = -1, impossible over R\mathbb{R} unless all coordinates vanish. H3(RP3;Q)=QH_3(\mathbb{R}P^3; \mathbb{Q}) = \mathbb{Q}, and the map acts by 11 there (it comes from a rotation of S3S^3), so L=1−1=0L = 1 - 1 = 0, consistent.

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