Solutions to Homework # 1. Hatcher, Chap. 0, Problem 5. Suppose F : X I X is a deformation retraction of X onto a point x 0.

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1 Solutions to Homework # Htcher, Chp., Problem 4. Denote by i A the inclusion mp A X. Consider homotopy F : X I X such tht F := X, F (X) A, F t (A) A. We clim tht g := F is homotopy inverse of i A, i.e. g i A A, i A g X. To prove the first prt consider the homotopy g t = F t A. Observe tht g = g i A, g = F i A = A. To prove the second prt we consider the homotopy H t = F t : X X. Observe tht F = i A F since F (X) A. On the other hnd, F = X. Htcher, Chp., Problem 5. Suppose F : X I X is deformtion retrction of X onto point x. This mens F t (x ) = x, t, F = X, F (X) = {x }. We wnt to prove slightly stronger sttement, nmely, tht for ny neighborhood U of x there exists smller neighborhood V U of x such tht F t (V ) U, t I. - F (U) X x U t J t C t Figure : Constructing contrctible neighborhoods of x. Consider the pre-imge of U vi F, F (U) = { (x, t) X I; F t (x) U }. Note tht C := {x } I F (U) (see Figure ). For every t I we cn find neighborhood U t of x X, nd neighborhood J t of t I such tht (see Figure ) U t J t F (U).

2 The set C is covered by the fmily of open sets { U t J t, nd since C is compct, we }t I cn find t,..., t n I such tht C U tk J tk. k In prticulr, the set V := k U tk is n open neighborhood of x, nd V I F (U). This mens F t (V ) U, t, i.e. we cn regrd F t s mp from V to U, for ny t. If we denote by i V the inclusion V U we deduce tht the composition F t i V defines homotopy F : V I U between F = i V nd F = the constnt mp. In other words i V is null-homotopic. Htcher, Chp., Problem 9. Suppose X is contrctible nd A X is retrct of X. Choose retrction r : X A, nd contrction of X to point which we cn ssume lies in A F : X I X, F = X, F (x) =, x. Consider the composition G : A I i A I F r X I X A. This is homotopy between the identity mp A nd the constnt mp A { }. Htcher, Chp., Problem 4. We denote by c i the number of i-cells. In Figure 2 we hve depicted three cell decompositions of the 2-sphere. The first one hs The second one hs The lst one hs c = = c 2, c =. c = n +, c = n, c 2 =, n >. c = n +, c = n + k, c 2 = k +, k. Any combintion of nonnegtive integers c, c, c 2 such tht c c + c 2 = 2, c, c 2 > belongs to one of the three cses depicted in Figure 2. 2

3 A A A A2 A2 An A A2 A R R k R 2 R An A A 2 Figure 2: Cell decompositions of the 2-sphere. 3

4 Solutions to Homework # 2 Htcher, Chp., Problem 6. Let R := { R = x = (x k ) k ; N : x n =, n N }. n We define topology on R by declring set S R closed if nd only if, n, the intersection S of with the finite dimensionl subspce R n = { (x k ) k ; x k =, k > n }, is closed in the Eucliden topology of R n. For ech x R set ( /2. x := xk) 2 k= S is homeomorphic to the unit sphere in R, S = { x R ; x = }. Observe tht S is deformtion retrct of R \ {} so it suffices to show tht R \ {} is contrctible. Define F : R [, ] R by ( ) ( x, t) F t ( x) = ( t)x, tx + ( t)x, tx + ( t)x 2,... Observe tht F t (R \ {}) R \ {}, t [, ]. Indeed, this is obviously the cse for F nd F. Suppose t (, ), nd F t ( x) =. This mens x =, x k+ = t t x k, k =,, 2,..., so tht x =. We hve thus constructed homotopy F : R \ {} I R \ {} between F = nd S F = S, the shift mp, (x, x, x 2, ) (, x, x, x 2,... ). It is convenient to write this mp s x (, x). Consider now the homotopy G : ( R \ {}) I R \ {} given by G t (, x) = (t, ( t) x). If we first deform R \ {} to R \ {} following F t, nd then to (, ) R following G t, we obtin the desired contrction of R \ {} to point. b b Figure. This CW -complex deformtion retrcts to both the cylinder (yellow) nd the Möbius bnd (grey). Htcher, Chp., Problem 7. (b) Such CW complex is depicted in Figure. For prt () consider continuous mp f : S S. Fix point in S. A cell decomposition See Exmple.B.3 in Htcher s book.

5 2 is depicted in Figure 2. It consists of two vertices, f(), three -cells e, e, t, nd single 2-cell C. The ttching mp of C mps the right verticl side of C onto S = e / e vi f. e t C e f() f e e t f() t f() Figure 2. A cell decomposition of mp f : S S. Htcher, Chp., Problem 22. We investigte ech connected component of the grph seprtely so we my s well ssume tht the grph is connected. We distinguish two cses. Cse.The grph hs vertices on the boundry of the hlf plne. We cn deform the grph inside the hlf-plne so tht ll its vertices lie on the boundry of the hlf-plne (see Figure 3). More precisely, we chieve this by collpsing the edges which connect two different vertices, nd one of them is in the interior of the hlf-plne. Rotting this collpsed grph we obtin closed subset X of R 3 which is finite union of sets of the type R or S s illustrted in Figure 3. More precisely, when n edge connecting different vertices is rotted, we obtin region of type S which is 2-sphere. When loop is rotted, we obtin region of type R, which is 2-sphere with pir of points identified. Two regions obtined by rotting two different edges will intersect in s mny points s the two edges. Thus, two regions of X cn intersect in, or 2 points. Using the rguments in Exmple.8 nd.9 in Htcher we deduce tht X is wedge of S s nd S 2 s. R S Figure 3. Rotting plnr grph.

6 Cse 2. There re no vertices on the boundry. In this cse the grph cn be deformed inside the hlf plne to wedge of circles. By rotting this wedge we obtin spce homotopic to collection of tori piled one on top nother (see Figure 3). Htcher, Chp., Problem 23. Suppose A, B re contrctible subcomplexes of X such tht X = A B, nd A B is lso contrctible. Since B is contrctible we deduce X/B X. The inclusion A X mps A B into B, nd thus defines n injective continuous mp j : A/A B X/B X. Since X = A B, the bove mp is bijection. Note lso tht j mps closed sets to closed sets. From the properties of quotient topology we deduce tht j is homeomorphism. Now observe tht since A B is contrctible we deduce so tht A/A B is contrctible. A A/A B 3

7 w w u u Sec.., Problem 5. () = (b) Suppose we re given mp f : S X. We wnt to prove tht it extends to mp f : D 2 X, given tht f is homotopic to constnt. Consider homotopy F : S I X, F (e iθ, ) = x X, F (e iθ, ) = f(e iθ ), θ [, 2π]. Identify D 2 with the set of complex numbers of norm nd set f(re iθ ) = F (e iθ, r). (b)= (c)suppose f : (S, ) (X, x ) is loop t x X we wnt to show tht [f] = π (X, x ). From (b) we deduce tht there exists f : (D 2, ) (X, x ) such tht the digrm below is commuttive. (S i, ) y (D 2, ) [ [[ f [ [ [] f. (X, x ) We obtin the following commuttive digrm of group morphisms. i π (S, ) π (D 2, ) ' ''' f ') f. π (X, x ) Since π (D 2, ) = {} we deduce tht i is the trivil morphism so tht f = f i must be the trivil morphism s well. The identity mp S : (S, ) (S, ) defines loop on S whose homotopy clss is genertor of π (S, ), nd we hve f ([ S ]) is trivil in π (X, x ). This homotopy clss is precisely the homotopy clss represented by the loop f. (c) = (). Obvious. Sec.., Problem 9. Set Assume the sets A i re open, bounded nd connected. A := A A 2 A 3, V i := vol (A i ). For every unit vector n S 2 nd every t we denote by H + n,t the hlf spce determined by the plne through t n, of norml vector n, nd situted on the sme side of this plne s n. More precisely, if (, ) denotes the Eucliden inner product in R 3, then } H + n,t { x := R 3 ; ( x, n) t.

8 Set V + 3 ( n, t) := vol (A 3 H + n,t ). Observe tht t V + 3 ( n, t) is continuous, non-incresing function such tht lim V 3 + ( n, t) =, lim V 3 + ( n, t) = V 3. t t The intermedite vlue theorem implies tht the level set { S n = t R; V 3 + ( n, t) = } 2 V 3 is closed nd bounded so it must be compct. t V + ( n, t) is non-incresing we deduce tht S n must be closed, bounded intervl of the rel line. Set t min ( n) := min S n, T mx ( n) := mx S n, s( n) = 2 (t min( n) + T mx ( n)). The numbers t( n), nd T ( n) hve very intuitive menings. Think of the fmily of hyperplnes H t := { x R 3 ; ( x, n) = t} s hyperplne depending on time t, which moves while stying perpendiculr to n. For t the entire region A 3 will be on the side of H t determined by n, while for very lrge t the region A 3 will be on the other side of H t, determined by n. Thus there must exist moments of time when H t divides A into regions of equl volume. t min ( n) is the first such moment, nd T mx ( n) is the lst such moment. Observe tht T mx ( n) = t min ( n), t min ( n) = T mx ( n), s( n) = s( n). Set Observe tht H + n nd H+ n H + n := H+ n,s( n). re complementry hlf-spces. Lemm. S n consists of single point so tht t min ( n) = T mx ( n) = s( η). Lemm 2. The mp S 2 n s( n) R is continuous We will present the proofs of these lemmt fter we hve completed the proof of the clim in problem 9. Set V + i ( n) = vol ( A i H + n ), i =, 2, 3. We need to prove tht there exists n S 2 such tht V + i ( n) = 2 V i, i =, 2, 3. 2

9 Note tht V + 3 ( n) = 2 V 3 so we only need to find n such tht V + i ( n) = 2 V i, i =, 2. Define f : S 2 R 2, f( n) := ( ) V + ( n) + V 2 + ( n), V + ( n). H + n nd H+ n = R3 re complementry hlf spces so tht V + i ( n) + V + i ( n) = vol (A i ), i =, 2, 3. () Lemm 2 implies tht f is continuous, nd using the Borsuk-Ulm theorem we deduce tht there exists n such tht f( n ) = f( n ). The equlity () now implies tht V + ( n ) + V 2 + ( n ) = ( ) vol (A ) + vol (A 2 ), 2 nd V + ( n ) = 2 vol (A ). These equlities imply tht V + 2 ( n ) = 2 vol (A 2). Proof of Lemm. Observe tht since the set A 3 is compct we cn find sufficiently lrge R > such tht A 3 B R (). Set for brevity Observe tht for ech n we hve G n (t) = V + 3 ( n, t). G n (t) =, t R, G n (t) = V 3, t R. We clim tht for every t S n there exists ε t > such tht h (, ε t ) we hve G n (t h) > G n (t) > G n (t + h), which shows tht if S n were n intervl then G n could not hve constnt vlue (V 3 /2) long it. Now observe tht ( ) G n (t h) G n (t) = vol A 3 { x; t h < ( x, n) < t} 3

10 Now observe tht the region A 3 { x; t h < ( x, n) < t} is open. Since A 3 is connected we deduce tht for every h sufficiently smll it must be nonempty nd thus it hs positive volume. The inequlity G n (t) > G n (t + h) is proved in similr fshion. Proof of Lemm 2. We continue to use the sme nottions s bove. Suppose n k n s k. Set G k := G nk, G := G n. Note tht lim G k(t) = G (t), t [ R, R] (2) k On the other hnd G k (t + h) G k (t) = vol (A 3 { x; t ( x, n k ) t + h }) ( vol B R () { x; t ( x, n k ) t + h }) πr 2 h (3) so tht the fmily of functions (G k ) is equicontinuous. Using (2) we deduce from the Arzel- Ascoli theorem tht the sequence of function G k converges uniformly to G on [ R, R]. Observe tht the sequence t min ( n k ) lies [ R, R] so it hs convergent subsequence. Choose such subsequence τ j := t min ( n kj ) t [ R, R]. Since the sequence G kj converges uniformly to G nd G kj (τ j ) = V 3 /2 we deduce G (t ) = V 3 /2, so tht t S n. Since S n consists of single point we deduce tht for every convergent subsequence of t min ( n k ) we hve lim j t min( n kj ) = t min ( n ). This proves the continuity of n s( n) = t min ( n). Sec.., Problem 6. We rgue by contrdiction in ech of the situtions ()-(f). Suppose there exists retrction r : X A. () In this cse r would induce surjection from the trivil group π (R 3, p) to the integers π (S, p). (b) In this cse r would induce surjection from the infinite cyclic group π (S D 2 ) to the direct product of infinite cyclic groups π (S S ). This is not possible since rnk π (S S ) = 2 > = rnk π (S D 2 ). This lso follows directly form (3) without invoking the Arzel-Ascoli theorem. 4

11 w u (c) The inclusion i : A X induces the trivil morphism i : π (A) π (X). Hence π (A) = r i is trivil. This is contrdiction since π (A) is not trivil. (d) Observe first tht S is retrct of S S so tht there exist surjections π (S S ) π (S ). In prticulr π (S S ) is nontrivil so tht there cnnot exist surjections π (D 2 D 2 ) π (S S ). (e) Let p, q be two distinct points on D 2, nd X = D 2 /{p, q}. Denote by x the point in X obtined by identifying p nd q. The chord C connecting p nd q defines circle C on X. C is deformtion retrct of X so tht π (X) = π (C) = Z. To prove tht A is not retrct of X it suffices to show tht π (S S ) is not quotient of Z. We rgue 2 by contrdiction. Suppose π (S S ) is quotient of Z. Since there re surjections π (S S ) Z we deduce tht π (S S ) must be isomorphic to Z. In prticulr there exists exctly two surjections π (S S ) Z. We now show tht in fct there re infinitely mny thus yielding contrction. We denote the two circles entering into S S by C nd C 2. Since C i is deformtion retrct of C C 2 we deduce tht [C i ] is n element of infinite order in π (C C 2 ). Denote by e n : S S the mp θ e iθ. Fix homeomorphisms g i : C i S nd define f n : C C 2 by the composition Define r n : C C 2 S by Observe tht C \ S e \\\] n f n := e n g. f n g S r n C = f n, r n C2 = g 2 (4) r n ([C ]) = [e n ] π (S ), r n ([C 2 ]) = [e ] π (S ) Using the isomorphism Z π (S ), n [e n ] we deduce tht r n r m if n m. 2 We cn chieve this much fster invoking Seifert-vnKmpen theorem. 5

12 (f) Observe first tht π (X) = Z, where the genertor is the core circle C of the Móbus bnd. A is circle so tht π (A) = Z. In terms of these isomorphisms the morphism i : π (A) π (X) induced by i : A X hs the description i (n[a]) = 2n[C]. Clerly there cnnot exist ny surjection f : π (X) π (A) such tht [A] = f i ([A]) = 2k[A], k[a] := f ([C]). Sec.., Problem 7. nottions there we define by R n := g 2 r n. Since We hve lredy constructed these retrction in (4). Using the R n : C C 2 C 2 R n R m, m m we deduce tht these retrctions re pirwise non-homotopic. Sec.., Problem 2. Fix homotopy F : X I X, f s ( ) = F (, s) such tht f = f = X. Denote by g : I X the loop g(t) = f t (x ). Consider nother loop t x, h : (I, I) (X, x ) nd form the mp (see Figure ). H : I t I s X, H(s, t) = F (h(s), t). Set u = g h, u = h g. A homotopy (u t ) rel x connecting u to u is depicted t the bottom of Figure. 6

13 X H s t h h g g x x x x h h g g x x x x u u = h.g u =g.h t= t= s t Figure : g h h g. 7

14 Sec..2, Problem 8. c b R 2 R c b Figure : A cell decomposition The spce in question hs the cell decomposition depicted in Figure. It consists of one -cell, three -cells, b, c nd two 2-cells, R nd R 2. We deduce tht the fundmentl group hs the presenttion genertors:, b, c reltions R = b b =, R 2 = c c =.

15 Sec..2, Problem. We will first compute the fundmentl group of the complement of b in the cylinder D 2 I (see Figure 2), nd then show tht the loop defined by c defines nontrivil element in this group. b c - + A b y z x c t α c x y z β x t c B b - b + y z t A B The right hnd rule Figure 2: If you cnnot untie it, cut it. Cut the solid torus long the slice D 2 {/2} into two prts A nd B s in Figure 2. We will use the Seifert-vnKmpen theorem for this decomposition of D 2 I. We compute the fundmentl groups π (A, pt), π (B, pt), π (A B, pt), where pt is point situted on the boundry c of the slice. A B is homotopiclly equivlent to the wedge of four circles (see Figure 2), nd thus π (A B, pt) is free group with four genertors x, y, z, t depicted in Figure 2. Wrning: The order in which the elements x, y, z, t re depicted is rther subtle. You should keep in mind tht since the two rcs nd b link then the segment which connects the entrnce nd exit points of b (x nd z) must intersect the segment which connects the entrnce nd exit points of (y nd t); see Figure 2. 2

16 The intersection of b with A consists of three oriented rcs ±, b. Suppose g is one of these rcs. We will denote by l g the loop oriented by the right hnd rule going once round the rc g. (The loop l + is depicted in Figure 2.) - b + - b + - b + - b + Figure 3: Pncking sphere with three solid tori deleted As shown in Figure 3 the complement of these rcs in A is homotopiclly equivlent to disk with three holes bounding the loops l ± nd l. This three-hole disk is homotopiclly equivlent to wedge of three circles nd we deduce tht π (A, pt) is the free group with genertors l ±, l b. We deduce similrly tht π (B, pt) is the free group with genertors l b± nd l. Denote by α the nturl inclusion A B A nd by β the nturl inclusion A B B (see Figure 2). We wnt to compute the induced morphisms α nd β. Upon inspecting 3

17 Figure 2 we deduce 2 the following equlities. α (x) = l b β (x) = l b α (y) = l α (z) = l b, β (y) = l β (z) = l b +. ( ) α (t) = l + β (t) = l Thus the fundmentl group of the complement of b in D 2 I is the group G defined by genertors: l ±, l, l b±, l b, reltions: l b = l b, l = l, l b = l b +, l + = l. It follows tht G is the free group with two genertors l b (= l b± = l b ) nd l (= l ± = l ). Inspecting Figure 2 we deduce tht the loop c defines the element α (xyzt) = ( l b l l b l + ) = ( l b l l b l ) = [lb, l ]. 2 Be very cutions with the right hnd rule. 4

18 Sec..2, Problem. Consider the wedge of two circles (X, x ) = (C, x ) (C 2, x 2 ), x i C i, nd continuous mp f : (X, x ) (X, x ). Consider the mpping torus of f T f := X I/{(x, ) (f(x), )}, nd the loop γ : (I, I) (T f, (x, )), γ(s) = (x, s). We denote by C its imge in T f. Observe tht C is homeomorphic to circle nd the closed set A = X {} C T f is homeomorphic to X C = X S. The complement T f \ A is homeomorphic to X \ {x } (, ) = (C \ x ) (, ) (C }{{} 2 \ x 2 ) (, ) }{{} R R 2. C C R i f(c i) C Figure 4: Attching mps In other words, the complement is the union of two open 2-cells R, R 2, nd thus T f is obtined from A by ttching two 2-cells. The ttching mps re depicted in Figure 4. Thus the fundmentl group of T f hs the presenttion genertors: C, C 2, C reltions R i = Cf (C i )C C i =, i =, 2. 5

19 Sec..2, Problem 4. We define counterclockwise on ech fce using the outer norml convention s in Milnor s little book. For ech fce R of the cube we denote by R the opposite fce, nd by R the counterclockwise rottion by 9 of the fce R. We denote by F, T, S the front, top, nd respectively side fce of the cube s in Figure 5. green c d red red b T green F S d b c d green c red red green green b c b red We mke the identifictions d The -skeleton Figure 5: A 3-dimensionl CW -complex F F, T T, S S. In Figure 5 we lbelled the objects to be identified by identicl symbols or colors. We get CW complex with two -cellls (the green nd red points), four -cells,, b, c, d, three 2-cells, F, T, S, nd one 3-cell, the cube itself. For fundmentl group computtions the 3-cell is irrelevnt. The -skeleton is depicted in Figure 5 nd by collpsing the contrctible subcomplex d to point we deduce tht it is homotopiclly equivlent to wedge of three circles. In other words the fundmentl group of the -skeleton (with bse point the red -cell) is the free group with three genertors α = d, β = b d, γ = c d. Attching the three 2-cells hs the effect of dding three reltions F = c d b = αγ β =, T = bcd = αβ γ =, S = db c = αβγ =. () 6

20 Thus the fundmentl group is isomorphic to the group G with genertors α, β, γ nd reltions (). We deduce from the first reltion Using the third reltion we deduce β = αγ = α(αγ )γ = = α 2 = γ 2. γ = αβ = α 2 = γ 2 = αβγ. Using the second nd third reltion we deduce tht Hence Observe tht α = γ β = γβ = γ 2 = β 2. α 2 = β 2 = γ 2 = αβγ (2) α 2 β = β 2 β = ββ 2 = βα 2, nd similrly α 2 γ = γα 2 so tht the α 2 lies in the center of G. α 2 is n element of order 2, nd the cyclic subgroup α 2 it genertes is norml subgroup. Consider the quotient H := G/ α 2. We deduce tht H hs the presenttion H = α, β, γ α 2 = β 2 = γ 2 = αβγ =, which shows tht H = Z/2Z Z/2Z. It follows tht ord G = 8. Denote by Q the subgroup of nonzero quternions generted by i, j, k. surjective morphism G Q given by We hve α i, β j, γ k. Since ord (G) = ord (Q) we deduce tht this must be n isomorphism. 7

21 w u Sec..3, Problem 9. Suppose f : X S is continuous mp, nd x X. Then f π (X, x ) is finite subgroup of π (S, f(x ) = Z nd thus it must be the trivil subgroup. It follows tht f hs lift f to the universl cover R f f X S exp Since R is contrctible we deduce tht f is nullhomotopic. nullhomotopic s well. Thus f = exp f must be Sec..3, Problem 8. Every norml cover of X hs the form Y := X/G p X where G π (X). In this cse Aut (Y/X) = π (X)/G. We deduce tht the cover X/G X is Abelin iff G contins ll the commuttors in π (X), i.e. Consider the cover. G := [π (X), π (X)] G. p X b := X/G b X. Note tht Aut (X b /X) = Ab (π (X)) cts freely nd trnsitively on X b. We deduce tht for ny Abelin cover of the form X/G we hve n isomorphism of covers X/G = X b /(G/G ) so tht X b is norml covering of X/G. For exmple, when X = S S we hve π (S S ) = Z Z, Ab (Z Z) = Z Z. The universl Abelin cover of S S is homomorphic to the closed set in R 2 } X b = {(x, y) R 2 x Z or y Z. The group Z 2 cts on this set by (x, y) (m, n) := (x + m, y + n) This ction is even nd the quotient is X. The cse S S S cn be nlyzed in similr fshion.

22 w u u uu w y u w w w w w w w w w u y w w u y w w w w Sec..3, Problem 24. Suppose we re given bsed G-covering (X, x ) p (X, x ) := (X, x )/G. We wnt to clssify the coverings (X, x) p (X, x ) which interpolte between X nd X, i.e. there exists covering mp (X, x ) q (X, x) such tht the digrm below is commuttive. (X, x ) (X, x) [ [[ q [ p p [[ ] u u (X, x ) We will denote such coverings by (X, x; q, p) A morphism between two such covers (X, x ; q, p ) nd (X, x; q, p) is pir of continuous mps f : (X, x) (X, x ), such tht the digrm below is commuttive q (X, x ) (X, x) [ f [ [ [ [^ p q u u p (X, x ) (X, x ) Suppose (X, x; q, p) is such n intermedite cover. Set F i := π (X i, x i ), F := π (X, x). p Since X X is G-covering we obtin short exct sequence p F µ F G Note tht we lso hve commuttive digrm F p " q y F F p which cn be completed to commuttive digrm p y w F y w F G F p µ p q y F y F H := F /q F µ (F ; q, p ) 2

23 w w u u u u w w w y u u y w w w w w w w w u y y u w w w w w w w w w Consider nother such commuttive digrm, p y w F y w F G F p µ p q y F y F H := F /q F µ (F ; q, p ) We define morphism (F ; q, p ) (F ; q, p ) to be group morphism ϕ : F F such tht the digrms below re commuttive ϕ F c[ F [] % p p F, ϕ F c[ F [] % q q F We denote by I the collection of intermedite coverings (X, x ) q (X, x) p (X, x ), nd by D the collection of the digrms of the type (F ; q, p ). We hve constructed mp Ξ : I D which ssocites to covering (X, x; q, p) the digrm Ξ(X, x; q, p) :=(F ; q, p ) D. Moreover if (X, x ; q, p ) I, with ssocited digrm (F ; q, p ), nd f : (X, x;, q, p) (X, x ; q, p ) is morphism of intermedite coverings, then the group morphism f : F F induces morphism of digrms Ξ(f) : Ξ(X, x; q, p) Ξ(X, x ; q, p ). Note tht for every coverings C, C, C I, nd every morphisms C g C f C we hve Ξ( C ) = Ξ(C), Ξ(f g) = Ξ(f) Ξ(g). Thus two coverings C, C I re isomorphic iff the corresponding digrms re isomorphic, Ξ(C) = Ξ(C ). This shows tht we hve n injective correspondence [Ξ] between the collection [I] of isomorphisms clsses of intermedite coverings nd the collection [D] of isomorphism clsses of digrms. Conversely, given digrm D D p y w F y w F G F β µ β α y F y F H := F /αf µ (F ; α, β) we cn form (Y, y;, b) I where (Y, y) := (X, x )/µ β(h), 3

24 w u u w u y w w w w u y w w w w : (X, x ) (Y, y) is the nturl projection, nd b : (Y, y) (X, x ) is the mp (Y, y) z H z G (X, x ), where for z X we hve denoted by z H (resp. z G) the H-orbit (resp the G-orbit) of z. Observe tht the digrm Ξ(Y, y;, b) ssocited to (Y, y;, b) is isomorphic to the initil digrm (F ; α, β). We thus hve bijection [Ξ] : [I] [D]. To complete the solution of the problem it suffices to notice tht the isomorphism clss of the digrm (F ; α, β) is uniquely determined by the subgroup µ β(f ) G. Conversely, to every subgroup H G we cn ssocite the digrm p y w F y w F G µ F p y F y µ (H) H µ (µ (H); p, inclusion) In more modern lnguge, we hve constructed two ctegories I nd D, nd n equivlence of ctegories Ξ : I D. 4

25 Solutions to Homework # 3 Sec. 2., Problem. It is The Möbius bnd; see Figure. V V P b T P b T 2 T b T 2 V V 2 V Figure. The Möbius bnd Sec. 2., Problem 2. For the problem with the Klein bottle the proof is contined in Figure 2, where we view the tetrhedron s the upper hlf-bll in R 3 by rotting the fce [V V V 2 ] bout [V V 2 ] so tht the ngle between the two fces with common edge [V V 2 ] increses until it becomes 8. We now see the Klein bottle sitting t the bottom of this upper hlf-bll. All the other situtions (the torus nd RP 2 ) re delt with similrly. V b V V 2 b V 3 V V V 3 b V2 b Figure 2. A 3-dimensionl -complex which deformtion retrcts to the Klein bottle.

26 2 Sec. 2., Problem 4. V σ c V b V V Figure 3. The homology of prchute. In this cse we hve C n (K) = if n 3 or n, nd C 2 (K) = Z σ, C (K) = Z, b, c, C (K) = Z V nd the boundry opertor is determined by the equlities σ = + b c, = b = c = V =. Then Z 2 (C (K)) =, Z (C (K)) = C (K) = Z, b, c, Z (C (K)) = C (K). H2 ( K ) = (). Moreover B (C (K)) = spn Z ( + b c) Z, b, c Hence so tht H ( K ) = Z, b, c /spn Z ( + b c). The imges of nd b in H ( K ) define bsis of H ( K ). It is cler tht H ( K ) = Z. Sec. 2., Problem 5. v b v U c L v b v Figure 4. The homology of the Klein bottle. nd We hve C 2 = Z U, L, C = Z, b, c, C = Z v. U = + b c, L = c + b, = b = c = v =.

27 If follows tht Z 2 = = H2 ( K ) =, nd H = Z. The first homology group hs the presenttion P Z U, L Z, b, c H where P is the 3 2 mtrix P =. Using the Mple procedure ismith we cn digonlize P over the integers D := 2 = AP B, where A =, B = [ This mens tht by choosing the Z-bsis µ := A, µ 2 := A b, µ 3 = A c in Z, b, c, nd the Z-bsis e := BU, f := BL in Z U, L we cn represent the liner opertor P s the digonl mtrix D. We deduce tht H hs n equivlent presenttion with three genertors µ, µ 2, µ 3 nd two reltions µ =, 2µ 2 =. Thus H = Z 2 µ 2 Z µ 3. Using the MAPLE procedure inverse we find tht A = ] so tht µ 2 is given by the 2nd column of A nd µ 3 is given by the third column of A µ 2 = c b, µ 3 = c. 3

28 Solutions to Homework # 4 Problem 6, 2. We begin by describing the equivlence clsses of k-fces, k =,, 2. Let i [v i vi vi 2 ]. The -fces. We hve so tht [v v ] [v v 2] [v v 2] v v v 2. Denote by v the equivlence clss contining these vertices. Note tht [v v 2] [v v ] = v v, v 2 v [v v ] [v v 2] = v v. Iterting this procedure we deduce tht there exists single equivlence clss of vertices. The -fces. Denote by e the equivlence clss contining the edges of. Then ll the edges [v i vi 2 ] belong to this equivlence clss. We lso hve nother n-equivlence clsses e i contining the pir [v i vi ], [vi vi 2 ]. Observe tht [v i v i 2] e i, i =,, n. The 2-fces. We hve n + equivlence clsses of 2-fces,,,, n. : C 2 C. We hve : C C. We hve nd Z 2 nd H 2. We hve B 2 = nd Thus C 2 = Z,,, C = Z e, e,, e n = e, i = [v i v i ] + [v i v i 2] [v i v i 2] = 2e i e i. C = Z v e i =, i =,,, n. { n Z 2 = x i i ; i= n x i i Z 2 i= We deduce Z 2 = so tht H 2 =. n i= } x i i = x n = x n + 2x n =... x 2 + 2x = x + x = Z nd H. We hve Z = C nd H hs the presenttion e, e,, e n = 2e n e n = 2e e = e. Hence e n = 2e n, e n 2 = 2e n,, e = 2e =

29 2 so tht H is the cyclic group of order 2 n generted by e n. By generl rguments we hve H = Z. Sec. 2., Problem 7. Consider regulr tetrhedron 3 = [P P P 2 P 3 ], nd fix two opposite edges = [P P ], b = [P 2 P 3 ]. Now glue the fces of this tetrhedron ccording to the prescriptions Type () gluing: [P P P 2 ] [P P P 3 ]. Type (b) gluing: [P P 2 P 3 ] [P P 2 P 3 ]. To see tht the spce obtined by these identifictions is homeomorphic to S 3 we cut the tetrhedron with the plne pssing through the midpoints of the edges of 3 different from nd b (see Figure 2). The solid B b Type (b) gluing P 2 B P b Type () gluing P P 3 The solid A Type () gluing Type (b) gluing Figure. Gluing the fces of tetrhedron to get 3-sphere. We get solid A contining the edge nd solid B contining the edge B. By performing first the type (b) gluing nd then the type () gluing on the solid B we obtin solid torus. Then performing first the type () gluing nd next the type (b) gluing on the solid A we obtin nother solid torus. We obtin in this fshion the stndrd decomposition of S 3 s n union of two solid tori S 3 = D 4 = (D 2 D 2 ) = ( D 2 D 2 ) (D 2 D 2 ).

30 3 Problem 8, 2. Htcher. Denote by [V i V i V i 2 V 3 i ] the i-th 3-simplex. i V 3 V i+ 3 T i Vi Vi V i V i+ 2 2 V i+ i+ T V i+ V i V i V i ~ V 2 i i i V V V ~ V 2 3 i+ i+ i+ V V 3 i+ V V 2 3 Figure 2. Cyclic identifictions of simplices To describe the ssocited chin complex we need to understnd the equivlence clsses of k-fces, k =,, 2, 3. -fces. We deduce V i V i+ contining V i. Similrly V i V i+ i+ i+ i mod n nd we denote by U the equivlence clss nd we denote by U the corresponding equivlence clss. Since V i V i+ we deduce U = U. Now observe tht V2 i V 2 i+ nd we denote by U 2 the corresponding equivlence clss. Similrly the vertices V3 i determine homology clss U 3 nd we deduce from V2 i V 3i + tht U 2 = U 3. Thus we hve only two equivlence clsses of vertices, U nd U 2. The vertices V i, V j belong to U while the vertices V2 i, V j 3 belong to U 2. -fces. The simplex T i hs six -fces (edges) (see Figure 2). A verticl edge v i = [V2 iv 3 i]. A horizontl edge h i = [V iv i]. Two bottom edges: bottom-right br i = [V iv 2 i] nd bottom-left bl i = [V i Two top edges: top-right tr i = [V iv 3 i] nd top-left tl i = [V iv 3 i]. Inspecting Figure 2 we deduce the following equivlence reltions. V i br i bl i+, tr i tl i+, v i v i+, (.) h i h i+, bl i tl i+, br i tr i+. (.2) We denote by v the equivlence clss contining the verticl edges nd by h the equivlence clss contining the horizontl edges. Observe next tht bl i tl i+ tr i, i 2 ].

31 4 so tht bl i tr i for ll i. Denote by e i the equivlence clss contining bl i. Observe tht bl i tr i e i, tl i e i, br i e i+. We thus hve (n + 2) equivlence clsses of edges v, h nd e i, i =,, n. 2-fces. Ech simplex T i hs four 2-fces A bottom fce B i = [V iv iv 2 i]. A top fce τ i = [V iv iv 3 i]. A left fce L i = [V iv 2 iv 3 i]. A right fce R i = [V iv 2 iv 3 i]. We hve the identifictions R i L i+, B i τ i+. We denote by B i the equivlence clss of B i, by L i the equivlence clss of L i nd by R i the equivlence clss of R i. Observe tht R i = L i+, i mod n. There re exctly 2n equivlence clsses of 2-fces. 3-fces. There re exctly n three dimensionl simplices T,, T n. The ssocited chin complex. C = Z U, U 2, C = Z v, h, e i ; i n C 2 = Z B i, R j ; i, j, k n, C 3 = Z T i ; i n. The boundry opertors re defined s follows. : C 3 C 2 T i = R i L i + τ i B i = R i R i + B i B i. : C 2 C B i = h + br i bl i = h + e i+ e i, R i = v tr i + br i = v + e i+ e i, : C C e i = U 2 U, h =, v =. For every sequence of elements x = (x i ) i Z we define its derivtive to be the sequence Using this nottion we cn rewrite The groups of cycles. i x = (x i+ x i ), i Z. T i = i R i B, B i = h + i e, R i = v + i e. Z = { h + bv + i Z = C, k i e i C ;, b, k i Z, = spn Z {v, h, i e; i n}. i } k i = Here we use the elementry fct tht the subgroup of Z n described by the condition x + + x n = is free Abelin group with bsis e 2 e, e 3 e 2,, e n e n, where (e i ) is the cnonicl bsis of Z n

32 5 Suppose c = i x i B i + j y j R j Z 2. Then ( ) = C = x i h + ( ) y j v + i j i (x i + y i ) i e (use Abel s trick 2 ) ( ) = x i h + ( ) y j v i (x + y)e i+. i j i We deduce x i = y j =, i (x + y) = i x + i y =, y. i j The lst condition implies tht (x i + y i ) is constnt α independent of i. Using the first two conditions we deduce = (x i + y i ) = nα i so tht x i = y i, for ll i. This shows { Z 2 = x i (B i R i ); x i Z, To find Z 3 we proceed similrly. Suppose c = x i T i Z 3. i i i } x i =. Then = c = i x i i (R B) = i (R i B i ) i x = ( i x)r i + i ( i x)b i. We deduce i x = for ll i, i.e. x i is independent of i. We conclude tht { Z 3 = xt ; x Z; T = T i} i In prticulr we conclude H 3 = Z. The groups of boundries nd the homology. We hve B = spn Z (U 2 U ) Z U, U 2. We deduce H = Z /B = C /B = Z U, U 2 /spn Z (U 2 U ) = Z. B = spn Z ( B i, R j ; i, j n) Z h, v, e i ; i n. Thus H dmits the presenttion H = Z /B = h, v, i e; h = v = i e, i e = i n i 2 Abel s trick is discrete version of the integrtion-by-prts formul. More precisely if R is commuttive ring, M is n R-module, (x i) i Z is sequence in R, (y i) i Z is sequence in M then we hve n n ( i x) y i = x n+ y n x y x j ( j y). i= j=

33 6 Using the equlity n i e = i= we deduce nh = nv =. This shows H = Z/nZ. Using the fct tht for every sequence x i Z i Z/nZ such tht i x i = there exists sequence y i Z, i Z/nZ such tht Any element c Z 2 hs the form x i = i y, i. c = x i (R i B i ), i where i x i =. Choose y i s bove such tht x i = i y, i mod n. Then c = i y i T i so tht Z 2 = B 2, i.e. H 2 =. Problem, 2., Htcher. Denote by i the cnonicl mp A X. Suppose r : A X is retrction, i.e. r i = A. Then the morphisms induced in homology stisfy r i = Hn (A). This shows tht i is one-to-one since i (u) = i (v) implies u = r i (u) = r (i (u)) = r (i (v)) = r i (v) = v.

34 Solutions to Homework # 5 Problem 7, 2., Htcher. Denote by A n set consisting of n distinct points in X. The long exct sequence of the triple (X, A n, A n ) is H k (A n, A n ) H k (X, A n ) H k (X, A n ) H k (A n, A n ) We deduce tht for k 2 we hve isomorphisms H k (X, A n ) H k (X, A n ). Thus for every k 2 nd every n we hve n isomorphism For k = we hve n exct sequence H k (X) = H k (X) = H k (X, A ) H k (X, A n ). (5.) H (X, A n ) H (X, A n ) H (A n, A n ) jn H (X, A n ) Since H (A n, A n ) is free Abelin group ker j n is free Abelin nd we hve H (X, A n ) = H (X, A n ) ker j n. Assume X is pth connected CW -complex. H (X, A n ) =. Hence Then X/A n is pth connected so tht H (X, A n ) = H (X, A n ) H (A n, A n ) = H (X, A n ) H (A n /A n ) = H (X, A n ) Z. Hence H (X, A n ) = H (X, A ) Z n = H (X) Z n. (5.2) Finlly ssuming the pth connectivity of X s bove we deduce H (X, A n ) = H (X/A n ) =. (5.3) Now pply (5.)-(5.3) using the informtion H (S 2 ) = H (S S ) = Z, H (S 2 ) =, H (S S ) = Z Z, H 2 (S 2 ) = H 2 (S S ) = Z. A X Figure. The cycle A is seprting while B is non-seprting (b) Denote by à collr round A nd by B collr round B. Then à deformtion retrcts onto A while B deformtion retrcts onto B. Then H (X, A) excision = H (X, Ã) = H (X A, à A). Cn you visulize the isomorphisms in (5.2)? B

35 2 The spce X A hs two connected components Y, Y 2 both homeomorphic to torus with disk removed. Then à A consists of two collrs round the boundries of Y j so tht H (X A, à A) = H (Y, Y ) H (Y 2, Y 2 ). We now use the following simple observtion. Suppose Σ is surfce, S is finite set of points in Σ, nd DS is set of disjoint disks centered t the points in S. By homotopy invrince we hve H (Σ, S) = H (Σ, D S ). Denote by Σ S the mnifold with boundry obtined by removing the disks D S. Using excision gin we deduce H (Σ, D S ) = H (Σ S, Σ S ) so tht H (Σ S, Σ S ) = H (Σ, S) (5.4) Note tht the groups on the right hnd side were computed in prt (). We deduce tht H (X, A) = H (torus, pt) H (torus, pt). Observe tht X B is torus with two disks removed so tht H (X, B) = H (torus, { pt, pt 2 }). Problem 2, 2. () Consider the cone over X CX = I X/{} X. We will regrd X s subspce of CX vi the inclusion X = {} X CX. Then CX is contrctible nd we deduce H (CX) =. (CX, X) is good pir, nd SX = CX/Xso tht H (SX) = H (CX, X). From the long exct sequence of the pir (CX, X) we deduce Thus for k we hve so tht H k+ (CX) H k+ (CX, X) H k (X) H k (CX) (5.5) Using k = in (5.5) we deduce H k (CX) = H k+ (CX) = H k+ (SX) = H k+ (CX, X) = H k (X). H (CX, X) H (X) H (CX) The inclusion induced morphism H (X) H (CX) is onto so tht H (SX) = H (CX, X) = ker(h (X) H (CX)) = H (X). (b) Denote by S n X the spce obtined by ttching n-cones over X long their bses using the tutologicl mps (see Figure 2).

36 3 X Figure 2. Stcking-up severl cones We see copy of X inside S n X. It hs n open neighborhood U which deformtion retrcts onto this copy of X nd such tht its complement is homeomorphic to disjoint union of n cones on X. The Myer-Vietoris sequence of the decomposition S n X = S n X X CX is H k (X) H k (S n X) H k (CX) H k (S n X) H k (X). For k > we hve H k (CX) =. Moreover, the inclusion induced morphism H k (X) H k (S n X) is trivil since ny cycle in X bounds inside 2 S n X. Hence we get short exct sequence H k (S n X) H k (S n X) H k (X) H k (S n X). For k > we hve H k (X) ( ) = ker H k (X) H k (S n X) while for k = we hve H k (X) ( ) = ker H k (X) H k (S n X). Thus, for every k we hve the short exct sequence H k (S n X) H k (S n X) H k (X). (5.6) Now observe tht there exists nturl retrction r : S n X S n X. To describe it consider first the obvious retrction from the disjoint union of n cones to the disjoint union of (n ) cones { (j, p) if j < n r : {,, n} CX {,, n } CX, r(j, p) = (, p) if j = n Now observe tht 2 The cone on z bounds z. r({,, n} X) = {,, n } X

37 u w u w u w u w u w u 4 nd S n X = {,, n} CX/{,, n} X, S n X = {,, n } CX/{,, n } X so tht r descends to retrction r : S n X S n X. This shows tht the sequence (5.6) splits so tht H k (S n X) = H k (S n X) H k (X) = inductively = n j= H k (X). Problem 27, 2. () We hve the following commuttive digrm H n+ (A) H n+ (X) H n+ (X, A) H n (A) H n (X) f f f f f H n+ (B) w H n+ (Y ) w H n+ (Y, B) w H n (B) w H n (Y ) The rows re exct. The morphisms induced on bsolute homology re isomorphisms so the five lemm implies tht the middle verticl morphism between reltive homology groups is n isomorphism s well. (b) We rgue by contrdiction. Suppose there exists mp g : (D n, D n \ ) (D n, D n ) such tht g f is homotopic s mps of pirs with (D n, D n ). If x D n \ then, g(tx) D n, t (, ]. We deduce tht g() = lim t g(tx) D n. Hence g(d n ) D n so we cn regrd g s mp D n D n. Note tht g D n D n. Equivlently, if we denote by i the nturl inclusion D n D n then we hve g i D n, so tht for every k we get commuttive digrm i H k ( D n ) Hk (D n ) = ' '' Hk ( D n ) ' '') g = H k ( D n ) In prticulr for k = n we hve H n ( D n ) = Z nd we reched contrdiction. Problem 28, 2. The cone on the -skeleton of 3 is depicted in Figure 3. Before we proceed with the proof let us introduce bit of terminology. The cone X is linerly embedded in R 3 so tht it is equipped with metric induced by the Eucliden metric. For every point x X we set B r (x ) := {x X; x x r}.

38 5 O V V V 3 V 2 Figure 3. A cone over the -skeleton of tetrhedron. By excising X B r (x ), < r we deduce H (X, X x ) = H (B r (x ), B r (x ) x ). Now observe tht B r (x ) deformtion retrcts onto L r (x ), the link of x in X, L r (x ) = {x X; x x = r. Hence H (X, X x ) = H (B r (x ), L r (x )) = H (B r (x )/L r (x )). We now discuss seprtely vrious cses (see Figure 4). (i) (ii) (iii) (iv) Figure 4. The links of vrious points on X. (i) x is in the interior of 2-fce. In this cse B r (x )/L r (x ) = S 2 for ll r so tht H (X, X x ) = H (S 2 ). (ii)x is inside one of the edges [V i V j ]. In this cse B r (x ) is the upper hlf-disk, nd the link is the upper hlf-circle. H (X, X x ) =.

39 6 (iii) x is inside one of the edges [OV i ]. In this cse B r consists of three hlf-disks glued long their dimeters. The link consists of three rcs with identicl initil points nd finl points. Then B r (x )/L r (x ) S 2 S 2 so tht H (X, X x ) = H (S 2 S 2 ) = H (S 2 ) H (S 2 ). (iv) x is one of the vertices V i. In this cse B r consists of three circulr sectors with common edge. The link is the wedge of three rcs. In this cse B r /L r is contrctible so tht H (X, X x ) = (v) x = O. In this cse B r = X nd the link coincides with the -skeleton of 3. We denote this -skeleton by Y. Using the long exct sequence of the pir (X, Y ) nd the contrctibility of X we obtin isomorphisms { H n (X, Y ) = H n (Y ) = if n 2 Z 3 if n = 2. We deduce tht the boundry points re the points in (ii) nd (iv). These re precisely the points situted on Y. To understnd the invrint sets of homeomorphism f of X note first tht H (X, X x) = H (X, X f(x)). In prticulr ny homeomorphism of X induces by restriction homeomorphism of Y. By nlyzing in similr fshion the vrious locl homology groups H (Y, Y y) we deduce tht ny homeomorphism of Y mps vertices to vertices so it must permute them. Any homeomorphism f of X mps the vertex O to itself. Also, it mps ny point on one of the edges [OV i ] to point on n edge [OV j ]. Thus ny homeomorphism permutes the edges [OV i ]. We deduce tht the nonempty subsets of X left invrint by ll the homeomorphisms of X re obtined from the following sets {O}, {V, V, V 2, V 3 }, Y, [OV ] [OV 3 ], X. vi the bsic set theoretic opertions,, \.

40 Homework. We denote by Z[t] the ring of polynomils with integer coefficients in one vrible t. If A, B Z[t], we sy tht A domintes B, nd we write this A B, if there exists polynomil Q Z[t], with nonnegtive coefficients such tht A(t) = B(t) + ( + t)q(t). () Show tht if A B, A B nd C then A + A B + B nd CA CB. (b) Suppose A(t) = + t + n t n Z[t], B = b + b t + + b m t m. Show tht A B if nd only if, for every k we hve ( ) i j ( ) i b j, (M ) i+j=k i+j=k j ( ) j j = ( ) j b j. (M = ) k (c) We define grded Abelin group to be sequence of Abelin groups C := (C n ) n. We sy tht C is of finite type if rnk C n <. The Poincré polynomil of grded group C of finite type is defined s n The Euler chrcteristic of C is the integer P C (t) = n (rnk C n )t n. χ(c ) = P C ( ) = n ( ) n rnk C n. A short exct sequence of grded groups (A ), (B ), (C ) is sequence of short exct sequences A n B n C n, n. Prove tht if A B C is short exct sequence of grded Abelin groups of finite type, then P B (t) = P A (t) + P C (t). (2) (d)(morse inequlities. Prt ) Suppose C n Cn C C is chin complex such tht the grde group C is of finite type. We denote by H n the n-th homology group of this complex nd we form the corresponding grded group H = (H n ) n. Show tht H is of finite type nd P C (t) P H (t) nd χ(c ) = χ(h ). (e) (Morse inequlities. Prt 2 ) Suppose we re given three finite type grded groups A, B nd C which re prt of long exct sequence Show tht A k i k Bk j k Ck k Ak A B C. P A (t) + P C (t) P B (t),

41 2 nd Proof. () We hve χ(b ) = χ(a ) + χ(c ). A (t) = B (t) + ( + t)q (t), A (t) = B (t) + ( + t)q (t) so tht A (t) + A (t) = B (t) + B (t) + ( + t) ( Q (t) + Q (t) ). Note tht if Q nd Q hve nonnegtive integrl coefficients, so does Q + Q. Next observe tht CA = CB + ( + t)cq. If C nd Q hve nonnegtive integrl coefficients, so does CQ. (b) Use the identity ( + t) = k ( ) k t k. Then Hence A B = ( + t)q Q(t) = ( + t) ( A(t) B(t) ) q n = i+j=n( ) i ( j b j ), where Q = q n t n. n q n, n i+j=n ( ) i j i+j=n ( ) i b j. This proves (M ). The equlity (M = ) is nother wy of writing the equlity A( ) = B( ). (c) Set n = rnk A n, b n = rnk B n, c n = rnk C n. If is short exct sequence then A n B n C n, n. b n = n + c n = n which is exctly (2). (d) Observe tht we hve short exct sequences We set b n t n = n n t n + n c n t n., Z n (C) C n Bn (C), (3) B n (C) Z n (C) H n (C). (4) z n := rnk Z n (C), b n = rnk B n (C), h n = rnk H n (C), c n = rnk C n. From (3) we deduce c n = z n + b n, n, where we hve B (C) =. Hence On the other hnd, the sequence (4) implies P C (t) = P Z (t) + tp B (t). P Z = P B + P H.

42 3 Hence P C = P H + ( + t)p B = P C P H. The equlity χ(c) = χ(h) follows from (M = ). (e) Set k := rnk A k, b k := rnk B k, c k = rnk C k, Then α k = rnk ker i k, β k = rnk ker j k, γ k = rnk ker k. = k k = α k + β k b k = β k + γ k = k b k + c k = α k + α k c k = γ k + α k ( k b k + c k )t k = k t k (α k + α k ) = P A (t) P B (t) + P C (t) = ( + t)q(t), Q(t) = k α k t k. Htcher, 2., Problem 4. We will use the identifiction { } Z n = i/n Q/Z; i Z. () Consider the injection j : Z 4 Z 8 Z 2, /4 (/4, /2). Then (/8, ) is n element of order 4 in (Z 8 Z 2 )/j(z 4 ) so tht we hve short exct sequence Z 4 Z 8 Z 2 Z 4. (b) Suppose we hve short exct sequence Z p m j A π Z p n. (5) Then A is n Abelin group of order p m+n so tht it hs direct sum decomposition A k = Z p ν i, ν ν 2 ν k, ν i = m + n. (6) i= On the other hnd A must hve n element of order p m, nd n element of order p n so tht ν mx(m, n). Fix n element A which projects onto genertor of Z p n, nd denote by A the imge of genertor in Z p m. Then A is generted by nd so the number k of summnds in (6) is t most 2. Hence A = A α,β := Z p α Z p β, α mx(m, n, β), α + β = m + n. (7) We clim tht ny group A α,β s in (7) fits in n exct sequence of the type (5). To prove this we need to find n inclusion j : Z p n A α,β such tht the group A α,β /j(z p m) hs n element of order p n. Observe first tht β min(m, n) becuse β = (m + n) α = min(m, n) + (mx(m, n) α) min(m, n). }{{} i

43 4 Consider the inclusion Z p m A α,β = Z p α Z p β, /p m (/p m, /p β ). Then the element g = (/p α, ) hs order p n in the quotient A α,β /j(z p m). To prove this observe first tht the order of g is power p ν of p, ν n. Since p ν g j(z p m), there exists x Z, < x < p m, such tht Hence We cn now write x = x p β, so tht p ν g = (/p α ν, ) = x (/p m, /p β ) mod Z. p β x, p α+m (p m+ν xp α ). p n+m (x p α+β p m+ν ). Since α + β = m + n we deduce p n+m p m+ν so tht n ν. (c) Consider short exct sequence Z f A g Z n. We will construct group morphism χ : Z n Q/Z s follows. For every x Z n there exists ˆx A such tht g(ˆx) = x. Then g(n ˆx) = nx = so tht Hence there exists k Z such tht Set 2 n ˆx ker g = f(z). f(k) = n ˆx. χ(x) := k mod Z. n The definition of χ(x) is independent of the choice ˆx. Indeed if ˆx A is different element of A such tht g(ˆx ) = x then ˆx ˆx ker g so there exists s Z such tht Then ˆx ˆx = f(s). nˆx = nˆx f(ns) = f(k ns) so tht k n = k ns n mod Z. Now define mp ( f (n) ) h : A Q Z n,, g(). n Observe tht h is injective. Its imge consists of pirs (q, x) Q Z n such tht q = χ(x) mod Z. We deduce tht A is isomorphic to Z Im (χ). The imge of χ is cyclic group whose order is divisor of n. Conversely,given group morphism λ : Z n Q/Z, we denote by C λ Q/Z its imge, nd we form the group A λ := { (q, c) Q Z n ; q = λ(c) mod Z }. Observe tht A = Z C λ, nd C λ is finite cyclic group whose order is divisor of n. A group morphism G Q/Z is clled chrcter of the group. 2 Less rigorously χ(x) = f (ng (x)) n mod Z.

44 We hve nturl injection f : Z Q A λ, nturl surjection A λ Q Z n Z n, nd the sequence Z A λ Z n is exct. Given ny divisor m of n, we consider k λ m : Z n Q/Z, n mod Z k mod Z. m Its imge is cyclic group of order m. We hve thus shown tht there exists short exct sequences Z A Z n if nd only if A = Z Z m, m n. 5

45 Homework # 7 Definition 7.. A spce X is sid to be of finite type if it stisfies the following conditions. () N > such tht H n (X) =, n > N. (b) rnk H k (X) <, k.. () Suppose A, B re open subsets of the spce X such tht X = A B. Assume A, B nd A B re of finite type. Prove tht X is of finite type nd χ(x) = χ(a) + χ(b) χ(a B). (b) Suppose X is spce of finite type. Prove tht χ(s X) =. (c) Suppose we re given structure of finite -complex on spce X. We denote by c k the number of equivlence clsses of k-fces. Prove tht χ(x) = c c + c (d) Let us define grph to be connected, -dimensionl, finite -complex. (A grph is llowed to hve loops, i.e., edges originting nd ending t the sme vertex, see Figure.) Figure. A grph with loops. Suppose G is grph with vertex set V. For simplicity, we ssume tht it is embedded in the Eucliden spce R 3. We denote by c (G) the number of vertices, nd by c (G) the number of edges, nd by χ(g) the Euler chrcteristic of G. We set Prove tht l(v) := rnk H (G, G \ {v}), d(v) = + l(v). c (G) = d(v), χ(g) = ( ) l(v). 2 2 v V Proof. () From the Myer-Vietoris sequence... H n (A B) H n (A) H n (B) H n (X) H n (A B) v V

46 2 tht X is of finite type. Using prt (e) of Problem in Homework # 6 for the bove long exct sequence we deduce (b) View S s the round circle in the plne χ(a) + χ(b) = χ(a B) + χ(x). S = { (x, y) R 2 ; x 2 + y 2 = }. Denote by p + the North pole p + = (, ), nd by p the South pole, p = (, ). We set A ± = (S \ {p ± }) X. Then A ± re open subsets of S X nd S A + A. Ech of them is homeomorphic to (, ) X, nd thus homotopic with X nd therefore The overlp χ(a ± ) = χ(x). A = A + A = (S \ {p +, p }) X, hs two connected components, ech homeomorphic to (, ) X, nd thus homotopic with X so tht χ(a ) = 2χ(X). From prt () we deduce tht χ(x) = χ(a + ) + χ(a ) χ(a ) =. (c) The homology of X cn be computed using the -complex structure. Thus, the homology groups H k (X) re the homology groups of chin complex n (X) n (X), where rnk n (X) = c n. The desired conclusion now follows from prt (d) of Problem in Homework # 6. (d) For every v V we denote by B r (v) the closed bll of rdius r centered t x, nd we set G r (v) := B r (v) G. For r sufficiently smll G r (x) is contrctible. We ssume r is such. Using excision, we deduce H (G, G \ {v}) = H (G r (v), G r (x) \ {v}). We set G r(x) := G r (v) \ {x}. Using the long exct sequence of the pir (G r (v), G r(v)) we obtin the exct sequence Hence = H (G r (x)) H (G r (v), G r(x) ) H ( G r(v) ) i H ( G r (v) ) = Z. l(x) = rnk ker i = rnk H (G r(v) ) = d(v) = rnk H (G r(x) ). In other words, d(v) is the number of components of G r(v), when r is very smll. Equivlently, d(v) is the number of edges originting /nd/or ending t v, where ech loop is to be counted twice. This is clled the degree of the vertex x. For exmple, the degree of the top vertex of the grph depicted in Figure is 8, becuse there re 3 loops nd 2 regulr edges t tht vertex. The equlity d(v) = 2c (G), v V

47 is now cler, becuse in the bove sum ech edge is counted twice. From prt (c) we deduce so tht χ(g) = c (G) c (G) χ(g) = d(v) = ( l(v) ) 2 2 v V v V v V v V = ( ) + l(v). 2 v V 2. Consider connected plnr grph G situted in hlf plne H, such tht the boundry of the hlf plne intersects G in nonempty set of vertices. Denote by ν the number of such vertices, nd by χ G the Euler chrcteristic of G. Let S be the spce obtined by rotting G bout the y xis. () Compute the Betti numbers of S. (b) Determine these Betti numbers in the specil cse when G is the grph depicted in Figure 2, where the red dotted line is the boundry of the hlf plne. 3 n n- m- m Figure 2. Rotting plnr grph. Proof. For every grph Γ, we denote by c (Γ) (respectively (c (Γ)) the number of vertices (respectively edges) of Γ. As in Homework # 2, we cn deform the grph G inside the hlfplne, by collpsing one by one the edges which hve t lest one vertex not situted on the y-xis. We obtin new plnr grph G, tht is homotopic to G, nd hs exctly ν vertices, ll situted on the xis of rottion. From the equlity χ G = χ(g ), we deduce χ G = c (G ) c (G ) = ν c (G ) = c (G ) = ν χ G. Denote by S the spce obtined by rotting G bout the y-xis. Then S is homotopic with S, nd the result you proved in Homework 2 shows tht S is wedge of number n circles, nd number n 2 of spheres. Using Corollry 2.25 of your textbook we deduce