Split (1)

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. and B2n−+1 In particular, when n = 2, B2n−1 j is a lens-shaped 3 ball and L is obtained from this ball by identifying its two curved disk faces via ρr, which may be described as the composition of the reﬂection across the plane con- taining the rim of the lens, taking one face of the lens to the other, followed by a ...
Lm(ℓ1, ···, ℓn) as the quotient of S 2n−3, and Lm(ℓ1, ···, ℓn) is obtained from this subspace by attaching two cells, of dimensions 2n − 2 and 2n − 1, coming from the interiors of B2n−1 and and B2n−2 its two identiﬁed faces B2n−2 j+1. Lm(ℓ1, ···, ℓn) with one cell ek in each dimension k ≤ 2n − 1. Inductively this give...
, ···, ℓn) for n = 1, 2, ···, each of which is a subcomplex of the next in the cell structure we have just constructed, so Lm(ℓ1, ℓ2, ···) is also a CW complex. Its cellular chain complex consists of a Z in each dimension with boundary maps alternately 0 and m, so its reduced homology consists of a Zm in each odd dimen...
of the choice of CW structure on X. Theorem 2.44. χ (X) = n(−1)n rank Hn(X). Here the rank of a ﬁnitely generated abelian group is the number of Z summands P when the group is expressed as a direct sum of cyclic groups. We shall need the following fact, whose proof we leave as an exercise: If 0→A→B→C→0 is a short exac...
this it follows that the boundary maps in the long exact sequence for (X, A) are zero, so the long exact sequence breaks up into short exact sequences 0 -→ Hn(A) i∗-----→ Hn(X) j∗-----→ Hn(X, A) -→ 0 The relation r∗i∗ = 11 actually gives more information than this, by the following piece of elementary algebra: Splitti...
short exact sequences satisfying (b) only determine B as a semidirect product of A and C. The diﬃculty is that s(C) might not be a normal subgroup of B. In the nonabelian case one deﬁnes ‘splitting’ to mean that (b) is satisﬁed. i-----→ B In both the abelian and nonabelian contexts, if C is free then every exact seque...
1→S 1, z ֏ z2, this says that the M¨obius band does not retract onto its boundary circle. Homology of Groups In §1.B we constructed for each group G a CW complex K(G, 1) having a con- tractible universal cover, and we showed that the homotopy type of such a space K(G, 1) is uniquely determined by G. The homology group...
n. Reﬂecting the richness of group theory, the homology of groups has been studied quite extensively. A good starting place for those wishing to learn more is the text- book [Brown 1982]. At a more advanced level the books [Adem & Milgram 1994] and [Benson 1992] treat the subject from a mostly topological viewpoint. M...
B) ψ------------→ Cn(A + B) -→ 0 150 Chapter 2 Homology where ϕ(x) = (x, −x) and ψ(x, y) = x + y. The exactness of this short exact sequence can be checked as follows. First, Ker ϕ = 0 since a chain in A ∩ B that is zero as a chain in A (or in B ) must be the zero chain. Next, Im ϕ ⊂ Ker ψ since ψϕ = 0. Also, Ker ψ ⊂ I...
of the van Kampen theorem, and H1 is the abelianization of π1 for pathconnected spaces, as we show in §2.A. H1(A)⊕ H1(B) / Im Φ There are also Mayer–Vietoris sequences for decompositions X = A ∪ B such that A and B are deformation retracts of neighborhoods U and V with U ∩V deformation retracting onto A ∩ B. Under the...
of a M¨obius band wraps is injective we obtain H2(K) = 0. Furthermore, twice around the core circle. Since we have H1(K) ≈ Z⊕ Z2 since we can choose (1, 0) and (1, −1) as a basis for Z⊕ Z. All the higher homology groups of K are zero from the earlier part of the Mayer–Vietoris Φ Φ sequence. Example 2.48. Let us descri...
× {0} and X × {1}. Surjectivity of the maps i∗ in the upper row implies that the next maps are 0, which in turn implies that the maps ∂ are injective. Thus the map ∂ in the upper row gives an isomorphism of Hn+1(X × I, X × ∂I) onto the kernel of i∗, which consists of the pairs (α, −α) for α ∈ Hn(X). This kernel is a c...
× I attached at one of its ends. By Proposition 0.18 the sliding operation preserves homotopy type, so we obtain a homotopy equivalence Z ≃ A ∪ B. The exact sequence (∗) in this case is the Mayer–Vietoris sequence. A relative form of the Mayer–Vietoris sequence is sometimes useful. If one has a pair of spaces (X, Y ) =...
+D)→Cn(X, Y ) also induce isomorphisms on homology. The relative Mayer–Vietoris sequence is then the long exact sequence of homology groups associated to the short exact sequence of chain complexes given by the third row of the diagram. Homology with Coeﬃcients There is an easy generalization of the homology theory we...
cients is often the most natural tool in the absence of orientability. All the theory we developed in §2.1 for Z coeﬃcients carries over directly to general coeﬃcient groups G with no change in the proofs. The same is true for Mayer– Vietoris sequences. Diﬀerences between Hn(X; G) and Hn(X) begin to appear only when on...
pair (X, A) with G1 coeﬃcients to the corresponding sequence with G2 coeﬃcients. Also, the maps ϕ∗ commute with homomorphisms f∗ induced by maps f : (X, A)→(Y, B). Now let f : S k→S k have degree m and let ϕ : Z→G take 1 to a given element g ∈ G. Then we have a commutative diagram as at the right, where commu- tativit...
ﬃcients in §2.B. As another illustration, we will now give an example of a map f : X→Y with the property that the induced maps f∗ are trivial for homology with Z coeﬃcients but not for homology with Zm coeﬃcients for suitably chosen m. Thus homology with Zm coeﬃcients tells us that f is not homotopic to a constant map,...
ﬁxed point. Construct maps RP2n−1→RP2n−1 without ﬁxed points from linear transformations R2n→R2n without eigenvectors. 3. Let f : S n→S n be a map of degree zero. Show that there exist points x, y ∈ S n with f (x) = x and f (y) = −y. Use this to show that if F is a continuous vector ﬁeld deﬁned on the unit ball Dn in ...
identifying north and south poles to a point. (b) S 1 × (S 1 ∨ S 1). (c) The space obtained from D2 by ﬁrst deleting the interiors of two disjoint subdisks in the interior of D2 and then identifying all three resulting boundary circles together via homeomorphisms preserving clockwise orientations of these circles. (d)...
n is odd, show there exist even maps of any given even [Hints: If f is even, it factors as a composition S n→RPn→S n. Using the degree. calculation of Hn(RPn) in the text, show that the induced map Hn(S n)→Hn(RPn) sends a generator to twice a generator when n is odd. It may be helpful to show that the quotient map RPn...
RPm as its m skeleton. n ≈ Hn. 20. For ﬁnite CW complexes X and Y, show that χ (X × Y ) = χ (X) χ (Y ). If a ﬁnite CW complex X is the union of subcomplexes A and B, show that 21. χ (X) = χ (A) + χ (B) − χ (A ∩ B). 22. For X a ﬁnite CW complex and p : χ ( X) = n χ (X). X→X an n sheeted covering space, show that e e 23...
Show that if A is contractible in X then Hn(X, A) ≈ fact that (X ∪ CA)/X is the suspension SA of A. Hn(X)⊕ Hn−1(A), using the 27. The short exact sequences 0→Cn(A)→Cn(X)→Cn(X, A)→0 always split, but why does this not always yield splittings Hn(X) ≈ Hn(A)⊕ Hn(X, A)? 28. (a) Use the Mayer–Vietoris sequence to compute th...
. (e) The map S 1 × S 1→S 1 × S 1 that interchanges the two factors and then reﬂects one of the factors. Hn(X ∨ Y ) ≈ Hn(Y ) if the basepoints of X and Y that are identiﬁed in X ∨ Y are defor- 31. Use the Mayer–Vietoris sequence to show there are isomorphisms Hn(X)⊕ mation retracts of neighborhoods U ⊂ X and V ⊂ Y. e 3...
Mayer–Vietoris sequence in simplicial ho- mology for a complex X decomposed as the union of subcomplexes A and B. ∆ Computations and Applications Section 2.2 159 38. Show that a commutative diagram with the two sequences across the top and bottom exact, gives rise to an exact sequence ··· -→ En+1 -→ Bn -→ Cn ⊕ Dn -→ E...
. Show that if G is a ﬁnite group of homeomorphisms of X, then the homomorphism G→GLn(Z) assigning to g : X→X the induced homomorphism g∗ : H1(X; Z)→H1(X; Z) is injective. Show the same result holds if the coeﬃcient group Z is replaced by Zm with m > 2. What goes wrong when m = 2? 43. (a) Show that a chain complex of f...
∗ : hn(X)→ hn(Y ) such that (f g)∗ = f∗g∗ and 11∗ = 11, and so that the following three axioms are satisﬁed. e e e hn(X)→ (1) If f ≃ g : X→Y, then f∗ = g∗ : (2) There are boundary homomorphisms ∂ : hn(Y ). hn(X/A)→ e pair (X, A), ﬁtting into an exact sequence e q∗------------→ hn(X/A) hn(X) hn(A) i∗------------→ ∂-----...
point Section 2.3 161 obvious relative form, and axiom (2) is broken into two parts, the ﬁrst hypothesizing a long exact sequence involving these relative groups, with natural boundary maps, the second stating some version of excision, for example hn(X, A) ≈ hn(X/A, A/A) if one is dealing with CW pairs. In axiom (3) th...
� hn(x0) for any point x0 ∈ X, since the long exact sequence of the hn(X) ≈ hn(x0) = 0 for all n, pair (X, x0) splits via the retraction of X onto x0. Note that as can be seen by looking at the long exact sequence of reduced homology groups of the pair (x0, x0). e e The groups hn(x0) ≈ hn(S 0) are called the coeﬃcients...
the form of a Mayer-Vietoris sequence ··· -→ hn(A ∩ B) ϕ-----→ hn(A) ⊕ hn(B) ψ-----→ hn(X) ∂-----→ hn−1(A ∩ B) -→ ··· Categories and Functors Formally, singular homology can be regarded as a sequence of functions Hn that assign to each space X an abelian group Hn(X) and to each map f : X→Y a homomorphism Hn(f ) = f∗ :...
such as CW complexes, keeping continuous maps as the morphisms. We could also restrict the morphisms, for example to homeomorphisms. The category of groups, with homomorphisms as morphisms. Or the subcategory of abelian groups, again with homomorphisms as the morphisms. Generalizing The Formal Viewpoint Section 2.3 16...
diagrams. Going a step further, there is a category whose objects are short exact sequences of chain complexes and whose morphisms are commutative diagrams of maps between such short exact sequences. F (X), F (Y ) A functor F from a category C to a category D assigns to each object X in C an object F (X) in D and to e...
category whose objects are sequences of abelian groups and whose morphisms are sequences of homomorphisms. The composition of the two preceding functors is the functor assigning to a space its singular homology groups. The ﬁrst example above, the singular chain complex functor, can itself be re- garded as the composit...
omorphism G1→G2, as in the proof of Lemma 2.49. In general, if one has two functors F, G : C→D then a natural transformation T from F to G assigns a morphism TX : F (X)→G(X) to each object X ∈ C, in such a way that for each morphism f : X→Y in C the square at the right commutes. The case that F and G are contravariant ...
the torsion subgroup of Hn(X, A; Z), show that the functors (X, A) ֏ Tn(X, A), with the obvious induced homomorphisms Tn(X, A)→Tn(Y, B) and boundary maps Tn(X, A)→Tn−1(A), do not deﬁne a homology theory. Do the same for the ‘mod torsion’ functor MTn(X, A) = Hn(X, A; Z)/Tn(X, A). 2. Deﬁne a candidate for a reduced homo...
). Proof: Recall the notation f ≃ g for the relation of homotopy, ﬁxing endpoints, between paths f and g. Regarding f and g as chains, the notation f ∼ g will mean that f is homologous to g, that is, f − g is the boundary of some 2 chain. Here are some facts about this relation. (i) If f is a constant path, then f ∼ 0....
class of a loop f to the homology class of the 1 cycle f. Homology and Fundamental Group Section 2.A 167 P To show h is surjective when X is path-connected, let i niσi be a 1 cycle representing a given element of H1(X). After relabeling the σi ’s we may assume each ni is ±1. By (iv) we may in fact take each ni to be +...
i0 − τi1 + τi2 for singular 1 simplices τij, then the formula ∆ f = ∂ i niσi = i ni∂σi = i,j(−1)j niτij P P P implies that we can group all but one of the τij ’s into pairs for which the two coeﬃcients (−1)jni in each pair are +1 and −1. The one remaining τij is equal to 2 j ’s corresponding to the paired τij ’s, prese...
with boundary consisting of a single circle formed by the edge corresponding to f. This is because any pattern of identiﬁcations of pairs of edges of a ﬁnite collection of disjoint 2 simplices produces a compact sur- formula f = ∂ face with boundary. We leave it as an exercise for the reader to check that the algebrai...
the other hand, are trivial in homology since the portion of M on one side of γi is a compact surface bounded by γi, so γi is homotopic to a loop that is a product of commutators, as we saw a couple paragraphs earlier. The loop α′ i represents the same homology class as αi since the region between γi and αi ∪ α′ i pro...
S n is equivalent to the standard S n−1 ⊂ S n, equivalent in the sense that there is a homeomorphism of S n taking the given embedded S n−1 onto the standard S n−1. In particular, both complementary regions are homeomorphic to open balls. See [Brown 1960] for a precise statement and proof. When n = 2 it is a classical...
boundary in S n − h(Ik), then α is also not a boundary in at least one of A and B. (When i = 0 the word ‘cycle’ e e S n − h(Ik) e here is to be interpreted in the sense of augmented chain complexes since we are dealing with reduced homology.) By iteration we can then produce a nested sequence of closed intervals I1 ⊃ ...
ends with the terms = 0 H0(A) and which appears to contradict the fact that S n − h(S n−1) has two path-components. e The only way out of this dilemma is for h to be surjective, so that A ∩ B is empty and H−1(∅) which is Z rather than 0. In particular, this shows that S n cannot be embedded in Rn since this would yiel...
and Bn is obtained from Bn−1 by attaching 2n horns. There are homeomorphisms hn : Bn−1→Bn that are the identity outside a small neighborhood of Bn − Bn−1. As n goes to inﬁnity, the composition hn ··· h1 approaches a map f : B0→R3 which is continuous since the convergence is uniform. The set of points in B0 where f is ...
commutator of meridian and longitude circles in the torus, which correspond to α1 and α2. Van Kampen’s theorem now implies that the inclusion Y0 ֓ Y1 induces an injection of π1(Y0) into π1(Y1) as the inﬁnite cyclic subgroup generated by [α1, α2]. In a similar way we can regard Yn+1 as being obtained from Yn by adjoini...
ant. This result is known classically as Invariance of Domain, the word ‘domain’ being an older designation for an open set in Rn. Theorem 2B.3. If U is an open set in Rn and h : U→Rn is an embedding, or more generally just a continuous injection, then the image h(U) is an open set in Rn. Proof: Viewing S n as the one-...
n dimensional generalization of the Jordan curve theorem were ﬁrst proved by Brouwer around 1910, at a very early stage in the development of algebraic topology. Division Algebras Here is an algebraic application of homology theory due to H. Hopf: Theorem 2B.5. R and C are the only ﬁnite-dimensional division algebras ...
that a ﬁnite-dimensional division algebra over R must have dimension a power of 2. The fact that the dimension can be at most 8 is a famous theorem of [Bott & Milnor 1958] and [Kervaire 1958]. See §4.B for a few more comments on this. Proof: Suppose ﬁrst that Rn has a commutative division algebra structure. Deﬁne a ma...
multiple of the identity element 1 ∈ A and we write j2 = a + bj for a, b ∈ R, then (j − b/2)2 = a + b2/4 so by rechoosing j we may assume that j2 = a ∈ R. If a ≥ 0, say a = c2, then j2 = c2 implies (j + c)(j − c) = 0, so j = ±c, but this contradicts the choice of j. So j2 = −c2 and by rescaling j we may assume j2 = −1...
0 n→X always lift to The map p♯ is surjective since singular simplices σ : is simply-connected. Each σ has in fact precisely two lifts σ1 and σ2. Because we e are using Z2 coeﬃcients, the kernel of p♯ is generated by the sums σ1 + σ2. So if we e n, then the image of deﬁne τ to send each σ : e τ is the kernel of p♯. Ob...
isms or zero as indicated. An odd map f : S n→S n induces a quotient map f : RPn→RPn. These two maps induce a map from the transfer sequence to itself, and we will need to know that the squares in the resulting diagram commute. This follows from the naturality of the long exact sequence of homology associated to a shor...
S n−1. ⊔⊓ Exercises 1. Compute Hi(S n − X) when X is a subspace of S n homeomorphic to S k ∨ S ℓ or to S k ∐ S ℓ. 2. Show that Hi(S n − X) ≈ Hn−i−1(X) when X is homeomorphic to a ﬁnite connected graph. [First do the case that the graph is a tree.] e e 3. Let (D, S) ⊂ (Dn, S n−1) be a pair of subspaces homeomorphic to ...
produce an embedding S 2 ֓ R3 for which neither component of R3 − S 2 is simply-connected. Simplicial Approximation Section 2.C 177 7. Analyze what happens when the number of handles in the basic building block for the Alexander horned sphere is doubled, as in the ﬁgure at the right. 8. Show that R2n+1 is not a divisi...
simplex to a simplex is uniquely determined by its values on vertices, this means that a simplicial map is uniquely P determined by its values on vertices. It is easy to see that a map from the vertices i tivi ֏ P of K to the vertices of L extends to a simplicial map iﬀ it sends the vertices of each simplex of K to th...
σ of X, in which case st v1 ∩ ··· ∩ st vn = st σ. Proof: The intersection st v1 ∩ ··· ∩ st vn consists of the interiors of all simplices τ whose vertex set contains {v1, ···, vn}. If st v1 ∩ ··· ∩ st vn is nonempty, such a τ exists and contains the simplex σ = [v1, ···, vn] ⊂ X. The simplices τ containing ⊔⊓ {v1, ···,...
the lemma, and we can extend g linearly over [v1, ···, vn]. Both f (x) and g(x) lie in a single simplex of L since g(x) lies in [g(v1), ···, g(vn)] and f (x) lies in the star of this simplex. So taking the linear path (1−t)f (x)+tg(x), 0 ≤ t ≤ 1, in the simplex containing f (x) and g(x) deﬁnes a homotopy from f to g. ...
deﬁne tr ϕ to be the trace of the induced homomorphism ϕ : A/Torsion→A/Torsion. For a map f : X→X of a ﬁnite CW complex X, or more generally any space whose homology groups are ﬁnitely generated and vanish in high dimensions, the Lefschetz number τ(f ) is deﬁned to be n(−1)n tr is the identity, or is homotopic to the ...
for example for RPn if n is even. The case of projective spaces is interesting because of its connection with linear algebra. An invertible linear transformation f : Rn→Rn takes lines through 0 to lines through 0, hence induces a map f : RPn−1→RPn−1. Fixed points of f are equivalent to eigenvectors of f. The character...
.2, using cup products in cohomology. One could go further and consider the quaternionic case. The antipodal map of S 4 = HP1 has no ﬁxed points, but every map HPn→HPn with n > 1 does have a ﬁxed point. This is shown in Example 4L.4 using considerably heavier machinery. Proof of 2C.3: The general case easily reduces to...
� we have d x, f (x) > ε, while g(σ ) lies within distance ε/2 of f (x) and σ lies within distance ε/2 of x, as a consequence of the fact that σ is contained in a simplex of L, K being a subdivision of L. The Lefschetz numbers τ(f ) and τ(g) are equal since f and g are homotopic. Since g is simplicial, it takes the n s...
ﬁgure below, with f : X→X the 180 degree rotation about a vertical axis passing through the central hole of X. Since f has no ﬁxed points, we should have τ(f ) = 0. The induced map f∗ : H0(X)→H0(X) is the identity, as always for a path-connected space, so this contributes 1 to τ(f ). For H1(X) we saw in Example 2A.2 t...
. Examples for even genus are described in one of the exercises. Fixed point theory is a well-developed side branch of algebraic topology, but we touch upon it only occasionally in this book. For a nice introduction see [Brown 1971]. Simplicial Approximations to CW Complexes The simplicial approximation theorem allows ...
and cylinders that have all the essential features of actual mapping cones and cylinders. ∆ 2→ Let us ﬁrst construct the simplicial analog of a mapping cylinder. For a simplicial map f : K→L this will be a simplicial complex M(f ) containing both L and the barycentric subdivision K′ of K as subcomplexes, and such that...
onto L, the inﬁnite concatenation of the previous deformation retractions, with the deformation retraction of M(f \|\| Kn) onto M(f \|\| Kn−1) performed in the t interval [1/2n+1, 1/2n]. The map r1 \|\| K may not equal f, but it is homotopic to f via the linear homotopy tf +(1−t)r1, which is deﬁned since r1(σ ) ⊂ f (σ ) for...
have a simplicial complex Yn+1 = Yn α C(fα), where C(fα) is obtained from M(fα) α. Since Dn+1 × I deformation retracts onto by attaching a cone on the subcomplex S n ∂Dn+1 × I ∪ Dn+1 × {1}, we see that Zn+1 deformation retracts onto Zn ∪ Yn+1, which in turn deformation retracts onto Yn ∪ Yn+1 = Yn+1 by induction. Like...
��xed points if the ﬁxed points are isolated. [Hint: Barycentrically subdivide X to make the ﬁxed point set a subcomplex.] 5. Let M be a closed orientable surface embedded in R3 in such a way that reﬂection across a plane P deﬁnes a homeomorphism r : M→M ﬁxing M ∩ P, a collection of circles. Is it possible to homotope ...
a natural product, called cup product, which makes the cohomology groups of a space into a ring. This is an extremely useful piece of additional structure, and much of this chapter is devoted to studying cup products, which are considerably more subtle than the additive structure of cohomology. How does contravariance...
, x). This and there is an obvious candidate for this map, the diagonal map turns out to work very nicely, giving a well-behaved product in cohomology, the cup ∆ product. Another sort of extra structure in cohomology whose existence is traceable to contravariance is provided by cohomology operations. These make the coh...
whose value on an oriented 1(X; G). We will be interested in the homomorphism δ : 0(X; G) to the function δϕ ∈ ∆ 0(X; G)→ ∆ group sending ϕ ∈ ∆ ∆ ∆ ∆ ∆ The Idea of Cohomology 187 edge [v0, v1] is the diﬀerence ϕ(v1) − ϕ(v0). For example, X might be the graph formed by a system of trails on a mountain, with vertices at...
= 1(X; G)/ Im δ will be trivial iﬀ the equation 1(X; G). Solving this equation 0(X; G) for each ψ ∈ ∆ means deciding whether specifying the change in ϕ across each edge of X determines ∆ 0(X; G). This is rather like the calculus problem of ﬁnding a an actual function ϕ ∈ function having a speciﬁed derivative, with the...
as the relation between H 0(X; G) and H0(X; G), with H 0(X; G) being a direct product of copies of G and H0(X; G) a direct sum, with one copy for each component of X in either case. ∆ ∆ ∆ ∆ Deﬁne 0(X; G) and 1(X; G)→ ∆ to the abelian group G, and deﬁne Now let us move up a dimension, taking X to be a 2 dimensional com...
]. Thus δψ measures the deviation of ψ from being additive. 1(X; G) ϕ(v1)−ϕ(v0) ∆ ∆ ϕ(v2)−ϕ(v0) ϕ(v2)−ϕ(v1) 1(X; G)→ δ-----→ − ∆ ∆ ∆ ∆ From another point of view, δψ can be regarded as an obstruction to ﬁnding 0(X; G) with ψ = δϕ, for if ψ = δϕ then δψ = 0 since δδϕ = 0 as we ϕ ∈ saw above. We can think of δψ as a loca...
the value of ϕ on all vertices in the region. The Idea of Cohomology 189 When G = Z we can reﬁne this construction by building Cψ from a number of arcs in each 2 simplex, each arc having a transverse orientation, the orientation which agrees or disagrees with the orientation of each edge according to the sign of the v...
called the dual group of of duality between the homomorphism δ : ∆ homomorphism ∂ : i(X; G) with the group Hom( i+1(X)→ ∆ ∆ δϕ([v0, ···, vi+1]) = ∆ ∆ Xj ∆ ∆ (−1)jϕ([v0, ···, vj, ···, vi+1]) b and the latter sum is just ϕ(∂[v0, ···, vi+1]). Thus we have δϕ = ϕ∂. In other words, ϕ-----→ G, which δ sends each ϕ ∈ Hom( i(...
�cient Theorem Let us begin with a simple example. Consider the chain complex where Z 2-----→ Z is the map x ֏ 2x. If we dualize by taking Hom(−, G) with G = Z, we obtain the cochain complex In the original chain complex the homology groups are Z ’s in dimensions 0 and 3, together with a Z2 in dimension 1. The homology...
� = ∂∗ : C ∗ n. The reason why δ goes in the opposite direction from ∂, increasing rather than decreasing dimension, is purely formal: For a homomorphism α : A→B, the dual homomorphism ϕ-----→ G to the α∗ : Hom(B, G)→Hom(A, G) is deﬁned by α∗(ϕ) = ϕα, so α∗ sends B ϕ-----→ G. Dual homomorphisms obviously satisfy (αβ)∗ ...
n(C), G). If ϕ is in Im δ, say ϕ = δψ = ψ∂, then ϕ is zero on Zn, so ϕ0 = 0 and hence also ϕ0 = 0. Thus there is a well-deﬁned quotient map h : H n(C; G)→Hom(Hn(C), G) sending the cohomology class of ϕ to ϕ0. Obviously h is a homomorphism. It is not hard to see that h is surjective. The short exact sequence 0 -→ Zn -→ ...
(i) where the vertical boundary maps on Zn+1 and Bn are the restrictions of the boundary map in the complex C, hence are zero. Dualizing (i) gives a commutative diagram (ii) The rows here are exact since, as we have already remarked, the rows of (i) split, and the dual of a split short exact sequence is a split short ...
→G that vanish on the subgroup Bn, and such homoKer i∗ morphisms are the same as homomorphisms Zn/Bn→G. Under this identiﬁcation of Cohomology Groups Section 3.1 193 n with Hom(Hn(C), G), the map H n(C; G)→ Ker i∗ Ker i∗ considered earlier. Thus we can rewrite (iv) as a split short exact sequence n in (iv) becomes the ...
ness at the left end of a short exact sequence after dualization is in fact all that goes wrong, in view of the following: Exercise. If A→B→C→0 is exact, then dualizing by applying Hom(−, G) yields an exact sequence A∗← B∗← C ∗← 0. However, we will not need this fact in what follows. The exact sequence (vi) has the spe...
�s will be constructed inductively. Since the Fi ’s are free, it suﬃces to deﬁne each αi on a basis for Fi. To deﬁne α0, observe that surjectivity of f ′ 0 implies that for each basis element x of F0 there exists x′ ∈ F ′ 0(x′) = αf0(x), so we deﬁne α0(x) = x′. We would like to deﬁne α1 in the same way, sending a basis...
that f ′ lies in Im f ′ i βi = βi−1fi and the relation βi−1 = f ′ f ′ have i+1 i+1(x′) = βi(x) − λi−1fi(x). This will be possible if βi(x) − λi−1fi(x) i+1 = Ker f ′ i (βi − λi−1fi) = 0. Using the relation i λi−1 + λi−2fi−1 which holds by induction, we i, which will hold if f ′ 1 such that f ′ 1 = Ker f ′ 0 and f ′ i (...
with a free resolution F ′′ of H ′′ also given, since for a composition H Cohomology Groups Section 3.1 195 one can choose the compositions βnαn of extensions αn of α and βn of β as an extension of βα. In particular, if we take α to be an isomorphism and β to be its inverse, with F ′′ = F, then α∗β∗ = (βα)∗ = 11, the ...
in books on homological algebra, for example [Brown 1982], [Hilton & Stammbach 1970], or [MacLane 1963]. However, this interpretation of Ext(H, G) is rarely needed in algebraic topology. Summarizing, we have established the following algebraic result: Theorem 3.2. If a chain complex C of free abelian groups has homolo...
itely generated, with torsion subgroups Tn ⊂ Hn and Tn−1 ⊂ Hn−1, then H n(C; Z) ≈ (Hn/Tn)⊕ Tn−1. ⊔⊓ It is useful in many situations to know that the short exact sequences in the universal coeﬃcient theorem are natural, meaning that a chain map α between chain complexes C and C ′ of free abelian groups induces a commuta...
and the coeﬃcient group G is also an R module. If R is a ﬁeld, for example, then R modules are always free and so the ExtR term is always zero since we may choose free resolutions of the form 0→F0→H→0. It is interesting to note that the proof of Lemma 3.1 on the uniqueness of free res- olutions is valid for modules ov...
With this slightly modiﬁed R(H, G) = H 0(F ; G) = H ∗ = HomR(H, G) by the exactness of deﬁnition we have Ext0 0 ← H ∗← 0. The real reason why unreduced Ext groups are better than re1 ← F ∗ F ∗ duced groups is perhaps to be found in certain exact sequences involving Ext and 1 ← F ∗ 1 ← F ∗ Hom derived in §3.F, which wo...
G) in the cochain complex ···←------ C n+1(X; G) δ←------------- C n(X; G) δ←------------- C n−1(X; G)←------ ···←------ C 0(X; G)←------ 0 Elements of Ker δ are cocycles, and elements of Im δ are coboundaries. For a cochain ϕ to be a cocycle means that δϕ = ϕ∂ = 0, or in other words, ϕ vanishes on boundaries. Since t...
theorem gives an isomorphism H 1(X; G) ≈ Hom(H1(X), G) since Ext(H0(X), G) = 0, the group H0(X) being free. If X is path-connected, H1(X) is the abelianization of π1(X) and we can identify Hom(H1(X), G) with Hom(π1(X), G) since G is abelian. The universal coeﬃcient theorem has a simpler form if we take coeﬃcients in I...
ology, this gives H n(X; G) can be deﬁned by dualizing ε-----→ Z→0, then taking Ker / Im. As with H n(X; G) = H n(X; G) for n > 0, and the universal coeﬃcient H0(X), G). We can describe the diﬀerence beH 0(X; G) and H 0(X; G) more explicitly by using the interpretation of H 0(X; G) tween as functions X→G that are const...
Every function from singular n simplices in A to G can be extended to be deﬁned on all singular n simplices in X, for example by assigning the value 0 to all singular n simplices not in A, so i∗ is surjective. The kernel of i∗ consists of cochains taking the value 0 on singular n simplices in A. Such cochains are the ...
A to be a point x0, this exact sequence gives an identiﬁcation of H n(X; G) with H n(X, x0; G). e More generally there is a long exact sequence for a triple (X, A, B) coming from e the short exact sequences 0←------ C n(A, B; G) i∗←------ C n(X, B; G) j∗←------ C n(X, A; G)←------ 0 The long exact sequence of reduced ...
+ 1) chains in X whose boundary lies in A. On such chains we have ϕ∂ = ϕ∂ since the extension of ϕ to ϕ is irrelevant. The net result of all this is that hδ(α) Cohomology Groups Section 3.1 201 is represented by ϕ∂. Let us compare this with ∂∗h(α). Applying h to ϕ restricts its domain to cycles in A. Then applying ∂∗ ...
commutative diagram This follows from the naturality of the algebraic universal coeﬃcient sequences since the vertical maps are induced by the chain maps f♯ : Cn(X, A)→Cn(Y, B). When the subspaces A and B are empty we obtain the absolute forms of these results. Homotopy Invariance. The statement is that if f ≃ g : (X,...
A + B) such that ρι = 11 and 11 − ιρ = ∂D + D∂ for a chain homotopy D. Dualizing by taking Hom(−, G), we have maps 202 Chapter 3 Cohomology ρ∗ and ι∗ between C n(A + B; G) and C n(X; G), and these induce isomorphisms on cohomology since ι∗ρ∗ = 11 and 11 − ρ∗ι∗ = D∗δ + δD∗. By the ﬁve-lemma, the maps C n(X, A; G)→C n(A ...
α with inclusions iα : Xα ֓ X, the product map W hn(Xα) is an isomorphism for each n. α i∗ α : We have already seen that the ﬁrst axiom holds for singular cohomology. The sec- ond axiom follows from excision in the same way as for homology, via isomorphisms H n(X/A; G) ≈ H n(X, A; G). Note that the third axiom involves...
, A; G). ∆ ∆ ∆ ∆ ∆ ∆ Cellular Cohomology. For a CW complex X this is deﬁned via the cellular cochain ∆ ∆ complex formed by the horizontal sequence in the following diagram, where coeﬃcients in a given group G are understood, and the cellular coboundary maps dn are Cohomology Groups Section 3.1 203 the compositions δnjn...
ality of h, and commu- tativity of the second square was shown in the discussion of the long exact sequence of cohomology groups of a pair (X, A). ⊔⊓ Mayer–Vietoris Sequences. In the absolute case these take the form ··· -→ H n(X; G) -----→ H n(A; G) ⊕ H n(B; G) -----→ H n(A ∩ B; G) -→ H n+1(X; G) -→ ··· where X is the...
iors of C and D. To derive this, consider ﬁrst the map of short exact sequences of cochain complexes Here C n(A + B, C + D; G) is deﬁned as the kernel of C n(A + B; G) -→ C n(C + D; G), the restriction map, so the second sequence is exact. The vertical maps are restrictions. The second and third of these induce isomorp...
? More specifG, Cn(X) G, Cn−1(X) ically, what are the groups hn(X; G) when G = Z, Zm, and Q? 5. Regarding a cochain ϕ ∈ C 1(X; G) as a function from paths in X to G, show that if ϕ is a cocycle, then →Hom (a) ϕ(f g) = ϕ(f ) + ϕ(g), (b) ϕ takes the value 0 on constant paths, (c) ϕ(f ) = ϕ(g) if f ≃ g, (d) ϕ is a cobound...
). e 9. Show that if f : S n→S n has degree d then f ∗ : H n(S n; G)→H n(S n; G) is multiplication by d. 10. For the lens space Lm(ℓ1, ···, ℓn) deﬁned in Example 2.43, compute the cohomology groups using the cellular cochain complex and taking coeﬃcients in Z, Q, Zm, and Zp for p prime. Verify that the answers agree wi...

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