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cells of 1 skeleton then gives a covering space X projecting homeomorphically onto cells of X. Restricting p to the e e X 1→X 1 over the 1 skeleton of X. Show: e (a) Two such covering spaces X1→X and 2→X 1 are isomorphic. X 1 e 1→X 1 and X 1 X→X is a normal covering space iff e (b) e e e X 1→X 1 is normal. X2→X are iso... |
, so the closure eα of an edge eα is homeomorphic either to I or S 1. α Iα, a subset of X is open (or closed) iff it intersects the closure eα of each edge eα in an open (or closed) set in eα. One says that X has the weak topology with respect to the subspaces eα. In this topology a sequence of points in the interiors o... |
graph X is a subspace Y ⊂ X that is a union of vertices and edges of X, such that eα ⊂ Y implies eα ⊂ Y. The latter condition just says that Y is a closed subspace of X. A tree is a contractible graph. By a tree in a graph X we mean a subgraph that is a tree. We call a tree in X maximal if it contains all the vertices... |
Y →Y is continuous since it is ⊔⊓ continuous on the closure of each edge and Y has the weak topology. S Given a maximal tree T ⊂ X and a base vertex x0 ∈ T, then each edge eα of X − T determines a loop fα in X that goes first from x0 to one endpoint of eα by a path in T, then across eα, then back to x0 by a path in T. ... |
α. These X is the lifts define the edges of a graph structure on e X. The resulting topology on same as its original topology since both topologies have the same basic open sets, the ` e e covering projection X→X being a local homeomorphism. e e ⊔⊓ We can now apply what we have proved about graphs and their fundamental ... |
would not be monotone and hence would have local maxima, occurring when the edgepath backtracked, retracing some edge it had just crossed. Thus in a tree there is a unique nonbacktracking edgepath joining any two points. All the vertices and edges along this edgepath are distinct. A tree can contain no subgraph homeom... |
, show that π1(X, x0) has a basis in which one element is represented by the loop Y. 5. Construct a connected graph X and maps f, g : X→X such that f g = 11 but f and g do not induce isomorphisms on π1. [Note that f∗g∗ = 11 implies that f∗ is surjective and g∗ is injective.] 6. Let F be the free group on two generators... |
finite index such that x ∉ H. Hence x has nontrivial image in a finite quotient group of F. In this situation one says F is residually finite. 12. Let F be a finitely generated free group, H ⊂ F a finitely generated subgroup, and x ∈ F − H. Show there is a subgroup K of finite index in F such that K ⊃ H and x ∉ K. [Apply Ex... |
ite fundamental group must be R2 since it is noncompact. Nonclosed surfaces deformation retract onto graphs, so such surfaces are K(G, 1) ’s with G free. Example 1B.3. The infinite-dimensional projective space RP∞ is a K(Z2, 1) since its universal cover is S ∞, which is contractible. To show the latter fact, a homotopy ... |
�s for arbitrary finitely generated abelian groups G. For example the n dimensional torus T n, the product of n circles, is a K(Zn, 1). Example 1B.6. For a closed connected subspace K of S 3 that is nonempty, the complement S 3 −K is a K(G, 1). This is a theorem in 3 manifold theory, but in the special case that K is a ... |
[e]. (The notation ∆ b b b The group G acts on EG by left multiplication, an element g ∈ G taking the simplex [g0, ···, gn] linearly onto the simplex [gg0, ···, ggn]. Only the identity e takes any simplex to itself, so by an exercise at the end of this section, the action of G on EG is a covering space action. Hence t... |
the other hand, BG has the virtue of being functorial: A homomorphism f : G→H induces a map Bf : BG→BH sending a simplex [g1| ··· |gn] to the simplex [f (g1)| ··· |f (gn)]. A different construction of a K(G, 1) is given in §4.2. Here one starts with any 2 dimensional complex having fundamental group G, for example 90 C... |
K(G, 1). Then every homomorphism π1(X, x0)→π1(Y, y0) is induced by a map (X, x0)→(Y, y0) that is unique up to homotopy fixing x0. To deduce the theorem from this, let X and Y be CW complex K(G, 1) ’s with isomorphic fundamental groups. The proposition gives maps f : (X, x0)→(Y, y0) and g : (Y, y0)→(X, x0) inducing inve... |
�β]) being zero in π1(Y, y0). We have i∗([ψβ]) = 0 since the cell e2 β provides a nullhomotopy of ψβ in X. Hence f∗([ψβ]) = ϕi∗([ψβ]) = 0, and so f can be extended over e2 β. γ with n > 2 is possible since the attaching maps ψγ : S n−1→X n−1 have nullhomotopic compositions f ψγ : S n−1→Y. This is because f ψγ lifts to ... |
and restricting to a deformation retraction of T onto x0. (It is easy to extend such a deformation retraction to a homotopy defined on all of X 1.) We can construct a homotopy from f0|X 1 to f1|X 1 by first deforming f0|X 1 and f1|X 1 to take T to y0 by composing with ht, then applying the earlier argument to obtain a h... |
. In fact, the cell structure on B at the ends of e. The resulting space B at each vertex v of each edge e of Γ Γ Γ Γ canonically subdivided into a complex structure using the prism construction from the proof of Theorem 2.10, but we will not need to do this here. Γ ∆ More generally, instead of BGv one could take any C... |
for this group is A ∗C B, the free product of A and B amalgamated along the subgroup C. According to the theorem, A ∗C B contains both A and B as subgroups. Γ Γ For example, a free product with amalgamation Z ∗Z Z can be realized by mapping cylinders of the maps S 1← S 1→S 1 that are m sheeted and n sheeted covering s... |
for the Exercises. can be obtained from Γ Γ Γ Γ As a very special case, taking ϕ = ψ = 11 gives A∗A = A× Z since we can take = K(A, 1)× S 1 in this case. More generally, taking ϕ = 11 with ψ an arbitrary K automorphism of A, we realize any semidirect product of A and Z as A∗A. For example, the Klein bottle occurs this... |
so we have injective of A. To see this, note that p : X) = 0, hence maps π1( A) = 0. For example, if X is the torus S 1 × S 1 and A is the circle S 1 × {x0}, then π1( p−1(A) consists of infinitely many parallel lines in R2, each a universal cover of A. Mf →Mf be the Mf is itself the mapping cylinder universal cover of ... |
) ’s at the ends of Mf. We build universal covers of K(Gv, 1) ’s a copy of the universal cover of each mapping cylinder meeting Mf at the end of Mf in question. Now repeat the process to Mg of K construct K3 by attaching universal covers of mapping cylinders at all the universal Γ covers of K(Gv, 1) ’s created in the p... |
structure of, has one vertex for each e Γ Γ a mapping cylinder of K construction of T Corresponding to. The inductive construction of K is reflected in an inductive as a union of an increasing sequence of subtrees T1 ⊂ T2 ⊂ ···. consisting of a central vertex with a number e Γ K1 is a subtree T1 ⊂ T Γ e K1 to of edges ... |
nullhomotopic. 3. Show that every graph product of trivial groups is free. 4. Use van Kampen’s theorem to compute A∗C as a quotient of A ∗ Z, as stated in the text. having one vertex, Z, and one edge, the map Z→Z 5. Consider the graph of groups that is multiplication by 2, realized by the 2 sheeted covering space S 1→... |
verse is also true: A finitely generated group having a free subgroup of finite index is isomorphic to such a graph product. For a proof of this Γ see [Scott & Wall 1979], Theorem 7.3.] Γ Γ Γ The fundamental group π1(X) is especially useful when studying spaces of low dimension, as one would expect from its definition whi... |
have the advantage that Hi(S n) = 0 for i > n. The computability of homology groups does not come for free, unfortunately. The definition of homology groups is decidedly less transparent than the definition of homotopy groups, and once one gets beyond the definition there is a certain amount of technical machinery to be ... |
. One could take the viewpoint that these rather algebraic axioms are all that really matters about homology groups, that the geometry involved in the definition of homology is secondary, needed only to show that the axiomatic theory is not vacuous. The extent to which one adopts this viewpoint is a matter of taste, and... |
b−1a are to be regarded as equal if we make a commute with b−1. These two loops ab−1 and b−1a are really the same circle, just with a different choice of starting and ending point: x for ab−1 and y for b−1a. The same thing happens for all loops: Rechoosing the basepoint in a loop just permutes its letters cyclically, s... |
and let C0 be the free abelian group with basis the vertices x, y. Elements of C1 are chains of edges, or 1 dimensional chains, and elements of C0 are linear combinations of vertices, or 0 dimensional chains. Define a homomorphism ∂ : C1→C0 by sending each basis element a, b, c, d to y − x, the vertex at the head of th... |
1 dimensional cycles modulo those that are boundaries, the multiples of a − b. This quotient group is the homology group H1(X2). The previous example can be fit into this scheme too by taking C2 to be zero since there are no 2 cells in X1, so in this case H1(X1) = Ker ∂1/ Im ∂2 = Ker ∂1, which as we saw was free abelia... |
since the cycle A − B should be viewed as the boundary of C in the same way that the 1 dimensional cycle a − b is the boundary of A. Now we ∂1------------→ C0 and have a sequence of three boundary homomorphisms C3 the quotient H2(X4) = Ker ∂2/ Im ∂3 has become trivial. Also H3(X4) = Ker ∂3 = 0. The group H1(X4) is the... |
ices. Still, this is a rather restricted class of spaces, and the theory itself has a certain rigidity that makes it awkward to work with. The way around these obstacles is to step back from the geometry of spaces decomposed into simplices and to consider instead something which at first glance seems wildly more complic... |
us, the projective plane, and the Klein bottle can each be obtained from a square by identifying opposite edges in the way indicated by the arrows in the follow- ing figures: Cutting a square along a diagonal produces two triangles, so each of these surfaces can also be built from two triangles by identifying their edge... |
vertices also determines a canonical linear homeomorphism from the standard n simplex n onto any other n simplex [v0, ···, vn], preserving the order of vertices, namely, i tivi. The coefficients ti are the barycentric coordinates of the point (t0, ···, tn)֏ ∆ i tivi in [v0, ···, vn]. P If we delete one of the n + 1 vert... |
bottle and seven for the projective plane. The orientations on the edges in the pictures are compatible ∆ with a unique ordering of the vertices of each simplex, and these orderings determine the maps σα. of disjoint simplices A consequence of (iii) is that X can be built as a quotient space of a collection n→X, the q... |
plices, it is not hard to see that X must be a Hausdorff space. Condition (iii) then n is a homeomorphism onto its image, which is implies that each restriction σα It follows from Proposition A.2 in the Appendix that thus an open simplex in X. α of a CW complex structure on X with these open simplices σα( the σα ’s as ... |
orientations into account, so that all the faces of a simplex are coherently oriented, as indicated in P the following figure: i(−1)i[v0, ···, b b b b ∂[v0, v1] = [v1] − [v0] ∂[v0, v1, v2] = [v1, v2] − [v0, v2] + [v0, v1] ∂[v0, v1, v2, v3] = [v1, v2, v3] − [v0, v2, v3] + [v0, v1, v3] − [v0, v1, v2] In the last case, th... |
Chapter 2 Homology The latter two summations cancel since after switching i and j in the second sum, it becomes the negative of the first. ⊔⊓ The algebraic situation we have now is a sequence of homomorphisms of abelian groups ··· -→ Cn+1 ∂n+1 -----------------→ Cn ∂n------------→ Cn−1 -→ ··· -→ C1 ∂1------------→ C0 ∂... |
, the torus with the one vertex, three edges a, b, and c, and two 2 simplices U and L. As in the previous 0 (T ) ≈ Z. Since ∂2U = a + b − c = ∂2L and {a, b, a + b − c} is example, ∂1 = 0 so H 1 (T ) ≈ Z⊕ Z with basis the homology classes [a] ∆ a basis for ∆ and [b]. Since there are no 3 simplices, H 2 (T ) is equal to ... |
complex structure on S n by taking two copies of n(S n) ≈ Z for this ∆ n groups would be more difficult. ∆ ∆ Many similar examples could be worked out without much trouble, such as the other closed orientable and nonorientable surfaces. However, the calculations do tend to increase in complexity before long, particularl... |
advantage of simpler ∆ computations since fewer simplices are required. For example, to put a simplicial ∆ complex structure on the torus one needs at least 14 triangles, 21 edges, and 7 vertices, and for RP2 one needs at least 10 triangles, 15 edges, and 6 vertices. This would slow down calculations considerably! 108... |
�nition that homeomorphic spaces have isomorphic singular homology groups Hn, in contrast with the situation for H n. On the other hand, ∆ since the groups Cn(X) are so large, the number of singular n simplices in X usually being uncountable, it is not at all clear that for a complex X with finitely many simplices, Hn(X... |
such P ∆ Simplicial and Singular Homology Section 2.1 109 ∆ ∆ canceling pairs, construct an n dimensional complex Kξ from a disjoint union of n i, one for each σi, by identifying the pairs of (n−1) dimensional faces n simplices corresponding to the chosen canceling pairs. The σi ’s then induce a map Kξ→X. If n ξ is a ... |
Xα). components Xα there is an isomorphism of Hn(X) with the direct sum Proof: Since a singular simplex always has path-connected image, Cn(X) splits as the direct sum of its subgroups Cn(Xα). The boundary maps ∂n preserve this direct sum decomposition, taking Cn(Xα) to Cn−1(Xα), so Ker ∂n and Im ∂n+1 split similarly a... |
i is a Hence ∂ ⊔⊓ boundary, which shows that Ker ε ⊂ Im ∂1. P P i ni = 0. Thus i niσi since i niσ0 = i niσi − i niτi P P P P = Proposition 2.8. If X is a point, then Hn(X) = 0 for n > 0 and H0(X) ≈ Z. Proof: In this case there is a unique singular n simplex σn for each n, and ∂(σn) = i(−1)iσn−1, a sum of n + 1 terms, w... |
�]. The augmentation map ε is then the usual boundary map since ∂[v0] = [ e e Readers who know about the fundamental group π1(X) may wish to make a detour here to look at §2.A where it is shown that H1(X) is the abelianization of π1(X) whenever X is path-connected. This result will not be needed elsewhere in the chapte... |
utative diagrams can contain commutative triangles, pentagons, etc., as well as commutative squares. The fact that the maps f♯ : Cn(X)→Cn(Y ) satisfy f♯∂ = ∂f♯ is also expressed by saying that the f♯ ’s define a chain map from the singular chain complex of X to that of Y. The relation f♯∂ = ∂f♯ implies that f♯ takes cyc... |
I→ cases n = 1, 2. Proof of 2.10: The essential ingredient is a procedure for n × I into simplices. The figure shows the subdividing n × {0} = [v0, ···, vn] n × {1} = [w0, ···, wn], where vi and wi have the n. We can pass same image under the projection from [v0, ···, vn] to [w0, ···, wn] by interpolating a sequence of... |
0, ···, vi, wi, ···, wn] Xi We will show that these prism operators satisfy the basic relation ∂P = g♯ − f♯ − P ∂ Geometrically, the left side of this equation represents the boundary of the prism, and n × {0}, and the three terms on the right side represent the top n × {1}, the bottom the sides ∂ n × I of the prism. T... |
�(α) − f♯(α) is a boundary, so g♯(α) and f♯(α) determine the same homology class, which means ⊔⊓ that g∗ equals f∗ on the homology class of α. The relationship ∂P + P ∂ = g♯ − f♯ is expressed by saying P is a chain homotopy between the chain maps f♯ and g♯. We have just shown: Proposition 2.12. Chain-homotopic chain ma... |
to be computed in terms of lower-dimensional groups which may already be known, for example by induction. In order to formulate the relationship we are looking for, we need an algebraic definition which is central to algebraic topology. A sequence of homomorphisms ··· ---------→ An+1 αn+1 -------------------------→ An ... |
e e e i∗------------→ ··· -----→ e Hn−1(X) -----→ ··· H0(X/A) -----→ 0 where i is the inclusion A ֓ X and j is the quotient map X→X/A. e The map ∂ will be constructed in the course of the proof. The idea is that an Hn(X/A) can be represented by a chain α in X with ∂α a cycle in A element x ∈ whose homology class is ∂x... |
e Simplicial and Singular Homology Section 2.1 115 on Hn−1(∂Dn) ≈ Z. But i∗ and r∗ are both 0 since contradiction. The statement about fixed points follows as in Theorem 1.9. e Hn−1(Dn) = 0, and we have a ⊔⊓ e The derivation of the exact sequence of homology groups for a good pair (X, A) will be rather a long story. We... |
�nition of the relative boundary map we see: Elements of Hn(X, A) are represented by relative cycles: n chains α ∈ Cn(X) such that ∂α ∈ Cn−1(A). A relative cycle α is trivial in Hn(X, A) iff it is a relative boundary: α = ∂β + γ for some β ∈ Cn+1(X) and γ ∈ Cn(A). These properties make precise the intuitive idea that Hn... |
i∗------------→ Hn(B) j∗------------→ Hn(C) ∂-----→ Hn−1(A) i∗------------→ Hn−1(B) -→ ··· where Hn(A) denotes the homology group Ker ∂/ Im ∂ at An in the chain complex A, and Hn(B) and Hn(C) are defined similarly. The commutativity of the squares in the short exact sequence of chain complexes means that i and j are ch... |
within its homology class would have the form c + ∂c′. Since c′ = j(b′) for some b′, we then have c + ∂c′ = c + ∂j(b′) = c + j(∂b′) = j(b + ∂b′), so b is replaced by b + ∂b′, which leaves ∂b and therefore also a unchanged. Simplicial and Singular Homology Section 2.1 117 The map ∂ : Hn(C)→Hn−1(A) is a homomorphism sin... |
c′ = j(b′) for some b′ ∈ Bn+1. We have j(b − ∂b′) = j(b) − j(∂b′) = j(b) − ∂j(b′) = 0 since ∂j(b′) = ∂c′ = j(b). So b − ∂b′ = i(a) for some a ∈ An. This a is a cycle since i(∂a) = ∂i(a) = ∂(b − ∂b′) = ∂b = 0 and i is injective. Thus i∗[a] = [b − ∂b′] = [b], showing that i∗ maps onto Ker j∗. Ker ∂ ⊂ Im j∗. In the notat... |
�------------→ Hn−1(X) -→ ··· ··· -→ H0(X, A) -→ 0 The boundary map ∂ : Hn(X, A)→Hn−1(A) has a very simple description: If a class [α] ∈ Hn(X, A) is represented by a relative cycle α, then ∂[α] is the class of the cycle ∂α in Hn−1(A). This is immediate from the algebraic definition of the boundary homomorphism in the lo... |
= n 0 otherwise Example 2.18. Applying the long exact sequence of reduced homology groups to a pair (X, x0) with x0 ∈ X yields isomorphisms Hn(X, x0) ≈ Hn(X) for all n since Hn(x0) = 0 for all n. e There are induced homomorphisms for relative homology just as there are in the e nonrelative, or ‘absolute’, case. A map ... |
B ⊂ A ⊂ X : ··· -→ Hn(A, B) -→ Hn(X, B) -→ Hn(X, A) -→ Hn−1(A, B) -→ ··· This is the long exact sequence of homology groups associated to the short exact sequence of chain complexes formed by the short exact sequences 0 -→ Cn(A, B) -→ Cn(X, B) -→ Cn(X, A) -→ 0 Simplicial and Singular Homology Section 2.1 119 For examp... |
let C For a space X, let U = {Uj} be a collection of subspaces of X whose interiors U n (X) be the subgroup of Cn(X) consisting of i niσi such that each σi has image contained in some set in the cover U. The chains U boundary map ∂ : Cn(X)→Cn−1(X) takes C n (X) form a chain complex. We denote the homology groups of th... |
b The next two cases n = 1, 2 and part of the case n = 3 are shown in the figure. It follows from the inductive definition that the ver- tices of simplices in the barycentric subdivision of [v0, ···, vn] are exactly the barycenters of all ] of [v0, ···, vn] for 0 ≤ k ≤ n. When k = 0 this the k dimensional faces [vi0, ··... |
, ···, vn]. If neither wj nor wk is the barycenter b of [v0, ···, vn], then these two points lie in a proper face of [v0, ···, vn] and we are done by induction on n. So we may suppose wj, say, is the barycenter b, and then by the previous displayed inequalvi, ···, vn], ity we may take wk to be a vertex vi. Let bi be th... |
. We can uniquely designate a linear map n→Y by [w0, ···, wn] where wi is the image under λ of the ith vertex of n. To avoid having to make exceptions for 0 simplices it will be convenient to augment the complex LC(Y ) by setting LC−1(Y ) = Z generated by the empty simplex [∅], with ∂[w0] = [∅] for all 0 simplices [w0]... |
([∅]) = [∅], so S is the identity on LC−1(Y ). It is also the identity on LC0(Y ), since when n = 0 the formula for S becomes S([w0]) = w0(S∂[w0]) = w0(S([∅])) = w0([∅]) = [w0]. When λ is an embedding, with image a genuine n simplex [w0, ···, wn], then S(λ) is the sum of the n simplices in the barycentric subdivision o... |
� ∆ n × {1}, as indicated in the figure in the case n = 2. What T ∆ actually does is take the image of this subn. division under the projection n × I→ ∆ ∆ The chain homotopy formula ∂T + T ∂ = 11 − S is trivial on LC−1(Y ) where T = 0 and S = 11. Verifying the formula on LCn(Y ) with n ≥ 0 is done by the calculation ∆ ∆... |
Section 2.1 123 In similar fashion we define T : Cn(X)→Cn+1(X) by T σ = σ♯T n, and this gives a chain homotopy between S and the identity, since the formula ∂T + T ∂ = 11 − S holds by the calculation ∂T σ = ∂σ♯T n = σ♯∂T n = σ♯( ) = σ − Sσ − σ♯T ∂ = σ − Sσ − T (∂σ ) ∆ n ∆ ∆ where the last equality follows just as in th... |
σ )σ for each singular n→X. For this D we would like to find a chain map ρ : Cn(X)→Cn(X) U n (X) satisfying the chain homotopy equation with image in C ∆ (∗) ∂D + D∂ = 11 − ρ A quick way to do this is simply to regard this equation as defining ρ, so we let ρ = 11 − ∂D − D∂. It follows easily that ρ is a chain map since ∂... |
+ D∂ = U n (X)֓Cn(X) the inclusion. Furthermore, ρι = 11 since D is identically U n (X), hence the summation defining Dσ is ⊔⊓ empty. Thus we have shown that ρ is a chain homotopy inverse for ι. U n (X), as m(σ ) = 0 if σ is in C 11−ιρ for ι : C zero on C Proof of the Excision Theorem: We prove the second version, invo... |
for all n. e The upper left horizontal map is an isomorphism since in the long exact sequence of the triple (X, V, A) the groups Hn(V, A) are zero for all n, because a deformation retraction of V onto A gives a homotopy equivalence of pairs (V, A) ≃ (A, A), and Hn(A, A) = 0. The deformation retraction of V onto A indu... |
Let us find explicit cycles representing generators of the infinite Hn(S n). Replacing (Dn, ∂Dn) by the equivalent pair cyclic groups Hn(Dn, ∂Dn) and n→ n, viewed ( n). That it is a cycle is clear as a singular n simplex, is a cycle generating Hn( ∆ since we are considering relative homology. When n = 0 it certainly rep... |
generating n 1 − n/ ∆ ∆ ∆ ∆ ∆ ∆ Hn(S n) ∆ ≈------------→ Hn(S n, e n 2 ) ≈←------------- Hn( n 1, ∂ n 1 ) e where the first isomorphism comes from the long exact sequence of the pair (S n, n 2 ) ∆ and the second isomorphism is justified in the nontrivial cases n > 0 by passing to n 2 in the first group n 1 in the third g... |
’, which says in particular that Rm is not homeomorphic to Rn if m ≠ n. Theorem 2.26. If nonempty open sets U ⊂ Rm and V ⊂ Rn are homeomorphic, then m = n. Proof: For x ∈ U we have Hk(U, U − {x}) ≈ Hk(Rm, Rm − {x}) by excision. From the long exact sequence for the pair (Rm, Rm − {x}) we get Hk(Rm, Rm − {x}) ≈ Hk−1(Rm −... |
ity, not very interesting. We shall discuss the property now rather than interrupting later arguments to check it when it is needed, but the reader may prefer to postpone a careful reading of this discussion. The property is called naturality. For example, to say that the long exact sequence of a pair is natural means ... |
fact also implies naturality of the long exact sequence of a triple and the long exact sequence of reduced homology of a pair. Finally, there is the naturality of the long exact sequence in Theorem 2.13, that is, commutativity of the diagram where i and q denote inclusions and quotient maps, and f : X/A→Y /B is induce... |
2.1 129 ∆ istic maps The simplicial chain group Let us first show that the first and fourth vertical maps are isomorphisms for all n. n(X k, X k−1) is zero for n ≠ k, and is free abelian with n(X k, X k−1) has exactly the same basis the k simplices of X when n = k. Hence H description. The corresponding singular homolog... |
if β and δ are injective and α is surjective. The proofs of these two statements are straightforward diagram chasing. There is really no choice about how the argument can proceed, and it would be a good exercise for the reader to close the book now and reconstruct the proofs without looking. To prove (a), start with a... |
(b) = i′α(a) − β(b) = i′(a′)−β(b) = 0 implies i(a)−b = 0. Thus b = i(a), and hence c = j(b) = ji(a) = 0 ⊔⊓ since ji = 0. This shows γ has trivial kernel. 130 Chapter 2 Homology Returning to the proof of the theorem, we next consider the case that X is infinite- dimensional, where we will use the following fact: A compac... |
do the case of arbitrary X with A ≠ ∅, but this follows from the absolute case by applying the five-lemma to the canonical map from the long exact sequence of simplicial homology groups for the pair (X, A) to the corresponding long exact sequence of singular homology groups. ⊔⊓ is a We can deduce from this theorem that... |
] and [v1, v2], preserving the ordering of vertices? complex of a 2 simplex [v0, v1, v2] obtained 3 by performing the order-preserving 2. Show that the edge identifications [v0, v1] ∼ [v1, v3] and [v0, v2] ∼ [v2, v3] deformation retracts onto a Klein bottle. Also, find other pairs of identifications of edges that produce ... |
1, 2, 3 are Z, Zn, 0, Z, respectively. [The space X is an example of a lens space; see Example 2.43 for the general case.] n by identi9. Compute the homology groups of the fying all faces of the same dimension. Thus X has a single k simplex for each k ≤ n. ∆ 10. (a) Show the quotient space of a finite collection of dis... |
Hn(X, A) when X is S 2 or S 1 × S 1 and A is a finite set of points in X. (b) Compute the groups Hn(X, A) and Hn(X, B) for X a closed orientable surface of genus two with A and B the circles shown. [What are X/A and X/B?] 18. Show that for the subspace Q ⊂ R, the relative homology group H1(R, Q) is free abelian and find... |
, barycentric subdivision produces a simplicial complex. 24. Show that each n simplex in the barycentric subdivision of ∆ inequalities ti0 permutation of (0, ···, n). n is defined by n in its barycentric coordinates, where (i0, ···, in) is a ≤ ··· ≤ tin ≤ ti1 ∆ H1(X/A) if X = [0, 1] and A is the 25. Find an explicit, no... |
�� X have the property that f (A) ⊂ A for all homeomorphisms f : X→X. 29. Show that S 1 × S 1 and S 1 ∨ S 1 ∨ S 2 have isomorphic homology groups in all dimensions, but their universal covering spaces do not. ∆ ∆ ∆ 30. In each of the following commutative diagrams assume that all maps but one are isomorphisms. Show tha... |
as a composition S n→S n − {x0} ֓ S n and Hn(S n − {x0}) = 0 since S n − {x0} is contractible. Hence f∗ = 0. (c) If f ≃ g then deg f = deg g since f∗ = g∗. The converse statement, that f ≃ g if deg f = deg g, is a fundamental theorem of Hopf from around 1925 which we prove in Corollary 4.25. (d) deg f g = deg f deg g,... |
opy from f to Computations and Applications Section 2.2 135 the antipodal map. Note that the antipodal map has no fixed points, so the fact that maps without fixed points are homotopic to the antipodal map is a sort of converse statement. Here is an interesting application of degree: Theorem 2.28. S n has a continuous fie... |
KT] or [Husemoller 1966]. Another nice application of degree, giving a partial answer to a question raised in Example 1.43, is the following result: Proposition 2.29. Z2 is the only nontrivial group that can act freely on S n if n is even. Recall that an action of a group G on a space X is a homomorphism from G to the ... |
if f is a homeomorphism, then y can be any point and there is only one corresponding xi, so all the maps in the diagram are isomorphisms and deg f || xi = deg f = ±1. More generally, if f maps each Ui homeomorphically onto V, then deg f || xi = ±1 for each i. This situation occurs quite often in applications, and it i... |
flections k S n if necessary, we can make each local degree either +1 or of the summands of −1, whichever we wish. Thus we can produce a map S n→S n of degree ±k. W W Computations and Applications Section 2.2 137 Example 2.32. In the case of S 1, the map f (z) = zk, where we view S 1 as the unit circle in C, has degree ... |
��⊓ Note that for f : S n→S n, the suspension Sf maps only one point to each of the two ‘poles’ of S n+1. This implies that the local degree of Sf at each pole must equal the global degree of Sf. Thus the local degree of a map S n→S n can be any integer if n ≥ 2, just as the degree itself can be any integer when n ≥ 1.... |
→ Hk(X 1) -→ ··· -→ Hk(X k−1) -→ Hk(X k) -→ Hk(X k+1) -→ ··· By what we have just shown these are all isomorphisms except that the map to Hk(X k) may not be surjective and the map from Hk(X k) may not be injective. The first part of the sequence then gives statement (b) since Hk(X 0) = 0 when k > 0. Also, the last part ... |
to the finite-dimensional case. This reduction will be achieved by stretching X out to a complex that is at least locally finite-dimensional, using a special case of the ‘mapping telescope’ construction described in greater generality in §3.F. Consider X × [0, ∞) with its product cell structure, where we give [0, ∞) the... |
to i X i, a wedge sum of finite-dimensional complexes with n skeleton a point, so the finite-dimensional case of (∗) together with Corollary 2.25 describing the homology W Hk(Z/R) = 0 for k ≤ n. The same is therefore true for Z, of wedge sums implies that from the long exact sequence of the pair (Z, R), since R is contr... |
n−1) as linear combinations of n cells of X. The homology groups of this cellular chain complex are called the cellular homology groups of X. Temporarily we denote them H CW n (X). Theorem 2.35. H CW n (X) ≈ Hn(X). 140 Chapter 2 Homology Proof: From the diagram above, Hn(X) can be identified with Hn(X n)/ Im ∂n+1. Sinc... |
by induction on the dimension, without using cellular homology but only the basic results from the previous section. However, the viewpoint of cellular homology makes (i)–(iii) quite transparent. Next we describe how the cellular boundary maps dn can be computed. When n = 1 this is easy since the boundary map d1 : H1(... |
complement of Φ α] ∈ Hn(Dn α∗ takes a chosen generator [Dn α, ∂Dn The map summand of Hn(X n, X n−1) corresponding to en α. Letting en commutativity of the left half of the diagram then gives dn(en terms of the basis for Hn−1(X n−1, X n−2) corresponding to the cells en−1 is the projection of Commutativity of the diagra... |
, 2), we each have d2 injective and hence H2(Ng) = 0. If we change the basis for Zg by replacing the last standard basis element (0, ···, 0, 1) by (1, ···, 1), we see that H1(Ng) ≈ Zg−1 ⊕ Z2. 2 ··· a2 1a2 ∆ 142 Chapter 2 Homology These two examples illustrate the general fact that the orientability of a closed connecte... |
sending a to ρa and b to ρb. It is not hard to see that G is generated by ρa and ρb, so ρ is surjective. With more work one can compute that the kernel of ρ is Z2, generated by the element a5 = b3 = (ab)2, and this Z2 is in fact the center of π1(X). In particular, π1(X) has order 120 since G has order 60. a = ρ3 After... |
midway between them, and a reflection ∆ has degree −1. Since the cellular boundary maps are all zero, we deduce that Hi(T 3) is Z for i = 0, 3, Z3 for i = 1, 2, and 0 for i > 3. For K × S 1, when we compute local degrees for the front and back faces we find αβ on these two faces differs not by a reflection but by a rotati... |
finite cyclic summands of G. In the general nonfinitely generated case let F→G be a homomorphism of a free abelian group F onto G, sending a basis for F onto some set of generators of G. The kernel K of this homomorphism is a subgroup of a free abelian group, hence is itself free abelian. Choose bases {xα} for F and {yβ}... |
RPk−1 − RPk−2, and these two homeomorphisms are obtained from each other by precomposing with the antipodal map of S k−1, which has degree (−1)k. Hence deg qϕ = deg 11 + deg(−11) = 1 + (−1)k, and so dk is either 0 or multiplication by 2 according to whether k is odd or even. Thus the cellular chain complex for RPn is ... |
is relatively prime to m. We shall construct a CW structure on L with one cell ek for each k ≤ 2n − 1 and show that the resulting cellular chain complex is 0 -→ Z 0-----→ Z m------------→ Z 0-----→ ··· 0-----→ Z m------------→ Z 0-----→ Z -→ 0 with boundary maps alternately 0 and multiplication by m. Hence Hk Lm(ℓ1, ·... |
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