SOLUTIONS TO HOMEWORK PROBLEMS

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1 SOLUTIONS TO HOMEWORK PROBLEMS Contents 1. Homework Assignment # Homework Assignment # Homework Assignment # Homework Assignment # Homework Assignment # Homework Assignment # Homework Assignment # Homework Assignment # Homework Assignment # Homework Assignment # Homework Assignment # Homework Assignment # 1 1. Let p: X X be covering map which sends x0 X to x 0 X. Show that the induced map p : π n ( X, x 0 ) π n (X, x 0 ) is injective for n 1. Hint: use the homotopy lifting property for the covering map p. Proof. Since the induced map p is a homomorphism, it suffices to show that the kernel of p is trivial. So let f : (I n, I n ) ( X, x 0 ) be a map of pairs representing an element in the kernel of p. In other words, there is a homotopy H between the map p f : (I n, I n ) (X, x 0 ) and the constant map x 0 : (I n, I n ) (X, x 0 ). More explicitly this means that we have that H is a map H : I n I X with H(s, 0) = p f(s) In particular, the outer square of the diagram for s I n H(s, t) = x 0 for s I n or t = 1. I n 0 I n I H f H X X p commutes. By the homotopy lifting property for the covering space X X, there is a map H making the whole diagram commutative. We claim that H is the desired homotopy 1

2 2 SOLUTIONS TO HOMEWORK PROBLEMS between f : (I n, I n ) ( X, x 0 ) and the constant map x 0 : (I n, I n ) ( X, x 0 ). So we need to show that H(s, t) = x 0 for all s I n and t I. H(s, 1) = x 0 for all s I n and To prove the first property suppose that F I n is one of the (n 1)-dimensional faces that form the boundary of I n. The map H maps all of F I to the base point x 0 X, and hence the lift H maps F I to the fiber p 1 (x 0 ) X. Since F I is connected, H maps all of F I to a connected component of p 1 (x 0 ), that is, to one point of the discrete space p 1 (x 0 ). Since H maps (s, 0) to x 0, it maps every point of p 1 (x 0 ) to x 0. This proves the first property above. The second property follows by the same argument applied to H I n {1}: since H is a lift of H and H maps I n {1} to the basepoint x 0 X, H maps the connected space I n {1} to one point of the fiber p 1 (x 0 ). This point must be x 0, since H maps the boundary of I n {1} to x 0 by the first property. 2. Show that for closed 2-manifolds Σ 1, Σ 2, the Euler characteristic of the connected sum Σ 1 #Σ 2 is given by χ(σ 1 #Σ 2 ) = χ(σ 1 ) + χ(σ 2 ) 2 Proof. Cover both surfaces by patterns of polygons, and let us write V i (resp. E i resp. F i ) for the number of vertices (resp. edges resp. faces) on X i. Let us assume that we chose these pattern such that both contain a face which is a polygon with k vertices and edges for some k. The point of this is that removing that face from both surfaces and identifying the resulting surfaces along the polygon results in the connected sum X 1 #X 2 and the patterns on both surfaces fit together to give a pattern of polygons on X 1 #X 2. Let us determine V (= number of vertices), E (= number of edges) and F (= number of faces) of this pattern on X 1 #X 2 : faces: F = F 1 + F 2 2 (each face on X 1 or X 2 gives a face on X 1 #X 2 except the two polygons that we removed in order to form the connected sum). edges: E = E 1 + E 2 k (each edge on X 1 or X 2 gives an edge on X 1 #X 2 except there are k pairs of edges that get identified with each other when we form the connected sum). vertices: V = V 1 + V 2 k (each vertex on X 1 or X 2 gives a vertex on X 1 #X 2 except there are k pairs of vertices that get identified with each other when we form the connected sum). This implies χ(x 1 #X 2 ) = V E + F = (V 1 + V 2 k) (E 1 + E 2 k) + (F 1 + F 2 2) = V 1 E 1 + F 1 + V 2 E 2 + F 2 2 = χ(x 1 ) + χ(x 2 ) 2 3. a) Calculate the homology groups of the connected sum RP 2 #... #RP 2 of k copies of the real projective plane (we ve calculated the homology groups of the surface of genus g 0

3 SOLUTIONS TO HOMEWORK PROBLEMS 3 in class; after doing this homework problem, by the Classification Theorem for surfaces, we have then calculated the homology groups of all compact connected surfaces). Hint: Use the representation of RP 2 #... #RP 2 as the quotient space of a polygon obtained by identifying pairs of edges. For calculating the homology of the associated chain complex it will be useful to work with a convenient basis for the free Z-module C 1, which is not the obvious basis given by the edges of polygonal pattern we use for the calculation. b) Can the Euler characteristic of a compact connected surface be expressed in terms of its homology groups? Remark: Later we will define the Euler characteristic of more general topological spaces in terms of their homology groups. We will generalize the description of the Euler characteristic as an alternating sum of the number of vertices, edges and faces to an important class of spaces known as CW complexes. Proof. Part a). We recall that the connected sum RP 2 #... #RP 2 of k copies of the projective plane RP 2 can be described as the quotient of the 2k-gon with edge identifications given by the picture a 1 a 1 a k RP 2 #... #RP }{{} 2 k a 2 f a k a 2... (1) The resulting pattern given by this polygon on the surface P #... #P has one vertex v, k edges a 1,..., a k and one face f. Hence the chain complex associated to this pattern looks like Zv 1 Za 1 Za k Since there is only one vertex involved, we have 1 (a i ) = v v = 0 for any edge a i and hence 1 0. To determine 2 (f) we notice that every edge label occurs exactly twice, with arrows pointing in the same direction as the arrow for f. Hence 2 (f) = 2a 1 + 2a a k. It follows that H 0 = Z, H 2 = 0, and 2 Zf H 1 = Za 1 Za k /Z2(a a k ).

4 4 SOLUTIONS TO HOMEWORK PROBLEMS To identify this quotient group, we choose a different basis of the free abelian group C 1, namely a 1,..., a k 1, c, with c = a a k. Then we see H 1 = Za 1 Za k 1 Zc/Z2c = Z } {{ Z } Z/2. k 1 Part b). Can the Euler characteristic of a compact connected surface be expressed in terms of its homology groups? Inspection of the Euler characteristic calculation (resp. homology group calculation) for compact connected surfaces X shows that χ(x) = rk H 0 (X) rk H 1 (X) + rk H 2 (X), where rk H q (X) is the rank of the abelian group H q (X). 4. Let Σ be a compact surface, which is not necessarily connected. Then Σ has finitely many connected components Σ 1,..., Σ k each of which is a closed connected surface. a) How can the homology group H q (Σ) be expressed in terms of the homology groups H q (Σ i )? b) What is the group H 0 (Σ)? Proof. Part a). We recall that the homology group H k (Σ) is defined as the k-th homology group of the chain complex (C (Σ, Γ), ). Here Γ is a pattern of polygons on the surface Σ i, equipped with an orientation for all edges and faces. The group C k (Σ i, Γ i ) is the free abelian group generated by the vertices (for k = 0) resp. edges (for k = 1) resp. faces (for k = 2). For an oriented edge e C 1 (Σ) the boundary 1 (e) C 0 (Σ, Γ) is tip(e) tail(e). For an oriented face f C 2 (Σ) the boundary 2 (f) is ±e, where the sum runs over all edges e which are edges of the polygon f; the sign is positive if the orientations of e and f agree, and negative otherwise. Given a pattern of polygons with orientations Γ i on each surface Σ i, the union of all these vertices, oriented edges and faces can be viewed as providing us with a pattern of polygons Γ on Σ (with orientations on edges and faces). Denoting by V i the set of vertices of Γ i, and by V the corresponding sets for the pattern of polygons Γ on Σ whose restriction to Σ i Σ is Γ i. Then the set V is the disjoint union V 1 V k of the sets V i and it follows that C 0 (Σ, Γ) = Z[V ] = Z[V 1 V k ] k k = Z[V i ] = C 0 (Σ i, Γ i ), where the middle isomorphism comes from the map Z[V 1 V k ] = i=1 k Z[V i ] which sends the generator v V j V 1 V k Z[V 1 V k ] to the element (c 1,..., c k ) where c i Z[V i ] is the trivial element for i j and c i = v V i Z[V i ] for i = j. Similarly, we have C 1 (Σ, Γ) k = C 1 (Σ i, Γ i ) and C 2 (Σ, Γ) k = C 2 (Σ i, Γ i ) i=1 for the 1-chains and 2-chains consisting of linear combinations of edges and faces. i=1 i=1 i=1

5 SOLUTIONS TO HOMEWORK PROBLEMS 5 Concerning the boundary homomorphisms n : C n (Σ, Γ) C n 1 (Σ, Γ) for n = 1, 2, if e is an edge or f is a face for the pattern Γ, then e is an edge or f is a face belonging to the pattern Γ i on Σ i for some i, and hence 1 (e) = i 1(e) and 2 (f) = i 2(f) where i n : C n (Σ i, Γ i ) C n 1 (Σ i, Γ i ) is the boundary homomorphism for the pattern Γ i on Σ i. It follows that the diagram C n (Σ, Γ) = k i=1 C n(σ i, Γ i ) k i n C n 1 (Σ, Γ) = k i=1 C n 1(Σ i, Γ i ) is commutative. We summarize this discussion by saying that the chain complex (C (Σ, Γ), ) is isomorphic to the direct sum chain complex i I C n 1(Σ i, Γ i ) i n i I C n(σ i, Γ i ) i n+1 i I C n+1(σ i, Γ i ) Lemma 1. Let (C i, i ), i I be a collection of chain complexes parametrized by some index set I (not necessarily finite). Then the n-th homology group of the direct sum chain complex i n 1 i I Ci n 1 i n i I Ci n i n+1 i I Ci n+1 i n+1 is isomorphic to i I H n(c i, i ). Proof. Let us write Zn i for the n-cycles of the chain complex C, i and Z n for the n-cycles of the direct sum chain complex. Similarly, let Bn, i B n be the n-boundaries of the chain complex C i resp. the direct sum chain complex. Then ( Z n = ker n i : ) Cn i Cn 1 i = ker n i = Zn i i I i I i I i I Similarly, B n = i I Bi n and hence H n (sum chain complex) = Z n B n = Z i n B i n = Z i n B i n = H n (C i ) This finishes the proof of part (a). Part b). By the Classification Theorem for surfaces, any connected closed surface is homeomorphic to either the sphere, or a connected sum of tori or real projective planes. Our calculations show that for all these surfaces the homology group H 0 (Σ) is isomorphic to Z. Part (a) implies that H 0 (Σ) for a not necessarily connected surface is the direct sum of Z s with each copy of Z corresponding to a connected component of Σ.

6 6 SOLUTIONS TO HOMEWORK PROBLEMS 2. Homework Assignment # 2 1. Show that the abelian groups Z/mn and Z/m Z/n are isomorphic if and only if m and n are relatively prime. Proof. We claim that the homomorphism Z/mn Z/m Z/n which sends [k] to ([k], [k]) is an isomorphism if m, n are relatively prime. The condition gcd(m, n) = 1 implies that there are integers a, b such that am + bn = 1. Consider the homomorphism Z/m Z/n Z/mn given by ([k], [l]) [bnk + aml]. This is well-defined since it sends ([m], [0]) to [bnm] = [0] Z/mn and ([0], [n]) to [amn] = [0] Z/mn. It is easy to check that this is an inverse. If m, n are not relatively prime, then the least common multiple of m and n is strictly smaller than mn. Hence multiplication by the least common multiple gives zero for every element in Z/m Z/n, but not Z/mn showing that these groups can t be isomorphic. 2. Show that the singular chain complex of a topological space X is in fact a chain complex; i.e., that q q+1 = 0, where q : C q (X) C q 1 (X) is the boundary map. Proof. It will suffice to show that q q+1 = 0 on the generators of C q+1 since q and q+1 are both homomorphisms. Consider σ : q+1 X in C q+1 (X). q q+1 (σ) = q ( q j=0( 1) j q+1 (σ) [e 0,..., ê j,..., e q+1 ] = ( q q+1 ) ( 1) i ( 1) j σ [e 0,..., ê j,..., e q+1 ] [e 0,..., ê i,..., e q ] i=0 = 0 i<j q+1 + = 0 j<i q+1 0 i<j q+1 =0. 0 j<i q+1 j=0 ) ( 1) i+j σ [e 0,..., ê i,..., ê j,..., e q+1 ] ( 1) i+j+1 σ [e 0,..., ê j,..., ê i,..., e q+1 ] ( 1) i+j σ [e 0,..., ê i,..., ê j,..., e q+1 ] ( 1) i+j σ [e 0,..., ê j,..., ê i,..., e q+1 ] Hence, since q q+1 = 0 on the generators of C q+1 (X), q q+1 = 0, and the singular chain complex of a topological space X is in fact a chain complex. 3. Let X be a topological space with path components X α, α A. Show that H q (X) is isomorphic to α A H q(x α ). Proof. Since the image of a singular simplex q-simplex σ : q X is path connected, its image is contained in exactly one of the path connected components X α. It follows that

7 SOLUTIONS TO HOMEWORK PROBLEMS 7 S q (X), the set of singular q-simplices in X, is given by S q (X) = α A S q (X α ), the disjoint union of the sets S q (X α ). As in problem 4 in assignment # 1, this leads to an isomorphism S q (X α )] = Z[S q (X)] = C q (X). α A C q (X α ) = α A Z[S q (X α )] = Z[ α A From the construction it is clear that this isomorphism restricted to C q (X α ) is the map (i α ) : C q (X α ) C q (X) induced by the inclusion i α : X α X. In particular, this map is compatible with boundary maps and hence the map α ) : α A(i C q (X α ) C (X) α A is an isomorphism of chain complexes. By the lemma we proved for problem 4, the homology of the direct sum of chain complexes is isomorphic to the direct sum of the homology groups. This proves the desired isomorphism. 4. Show that the Hurewicz map h : π 1 (X, x 0 ) H 1 (X) given by [γ] [[γ]] is a homomorphism. Proof. Let [γ], [γ ] π 1 (X, x 0 ), and let γγ denote the concatenation of paths γ and γ. To show that h is a homomorphism, we need to verify that h([γ][γ ]) = h([γ]) + h([γ ]); i.e., [[γγ ]] = [[γ]] + [[γ ]]. We will do so by showing that γ + γ γγ B 1 (X) which will then imply that [[γγ ]] = [[γ + γ ]] = [[γ]] + [[γ ]], as desired. Consider the singular 2-simplex, σ : 2 X, defined as the composition 2 [e 0, 1 2 (e 0+e 1 ),e 1 ] 1 γγ X Notice that 2 (σ) = γ γγ + γ, so γ γγ + γ B 1 (X) as desired. Thus, h is indeed a homomorphism. 5. The goal of this problem is to show that the homomorphism h: π ab 1 (X, x 0 ) H 1 (X) induced by the Hurewicz homomorphism h is in fact an isomorphism for path connected space X; here π1 ab (X, x 0 ) is the abelianized fundamental group of X. The idea is to construct an inverse to h as follows. Choose for every point x X a path λ x from x 0 to x. In class we showed that we have a well-defined map Ψ: C 1 (X)/B 1 (X) π ab 1 (X, x 0 ) given by [[γ]] [λ γ(0) γ λ γ(1) ] for any singular 1-simplex γ, also known as path γ : I X. Here λ γ(0) γ λ γ(1) is the concatenation of the path λ γ(0) (from x 0 to γ(1)), the path γ (from γ(0) to γ(1)) and the path λ γ(1) (from γ(1) to x 0, obtained by running the path λ γ(1) from x 0 to γ(1) backwards). Show that the restriction of Ψ to H 1 (X) C 1 (X)/B 1 (X) provides an inverse to the map h. Proof. We need to show: (a) Ψ H1 (X) h = id π ab 1 (X,x 0);

8 8 SOLUTIONS TO HOMEWORK PROBLEMS (b) h Ψ H1 (X) = id H1 (X). To prove (a), let [γ : (I, I) (X, x 0 )] π ab 1 (X, x 0 ). Then Ψ H1 (X) h([γ]) = Ψ H1 (X)([[γ]]) = [λ γ(0) γλ γ(1) ] = [λ x0 γλ x0 ] = [λ x0 ][γ][λ x0 ] = [λ x0 ][λ x0 ][γ] = [γ], which shows Ψ H1 (X) h = id π ab 1 (X,x 0). To prove (b), we will use the following statements: (i) If γ, γ : I X are paths with γ(1) = γ (0), and γγ : I X their concatenation, then γ γγ + γ B 1 (X). (ii) γ + γ B 1 (X). Statement (i) we proved in problem (4) (note that the argument used there does not require γ and γ to be based loops; it is only necessary that γ(1) = γ (0) in order to be able to have the concatenated path γγ ). To deduce the second claim, we apply part (i) for γ = γ which implies γ γ γ + γ B 1 (X). We note that γ γ is a loop based at γ(0) which is homotopic to the constant loop. Hence its image under the Hurewicz map [[γ γ]] H 1 (X) is trivial. In other words, γ γ B 1 (X), which proves (ii). With these preliminaries, we can calculate for any path γ: h Ψ([[γ]]) = h([λ γ(0) γ λ γ(1) ] = [[λ γ(0) γ λ γ(1) ]] (2) = [[λ γ(0) ]] + [[γ]] + [[ λ γ(1) ]] = [[λ γ(0) ]] + [[γ]] [[λ γ(1) ]] = [[γ]] + [[λ( (γ))]], where λ: C 0 (X) C 1 (X) is given on generators by x λ x (and hence λ maps γ = γ(1) γ(0) to λ γ(0) λ γ(1) ). Since formula (2) holds for the generators γ C 1 (X), it hold for every element z C 1 (X); in particular, if z is a cycle, we have h Ψ([[z]]) = [[z]], which is what we wanted to prove. 3. Homework Assignment # 3 1. Calculate the reduced homology groups of the subspace X R 3 which is the union of the sphere S 2 and the x-axis. Proof. The natural idea is to apply the Mayer-Vietoris sequence to X = U V, where U = S 2, and V is the x-axis. Unfortunately, this doesn t work since these aren t open subsets of X as required by the Mayer-Vietoris Theorem. The way around this problem is to work with larger open subsets U, V X that contain U resp. V as deformation retracts. This means that there is a map r U : U U such that r U i U = id U and i U r U id U, where i U : U U is the inclusion map, and similarly for V. There are many possible choices for U, V, for example we can take U := S 2 x-axis without the origin V := x-axis {(x, y, z) S 2 x 0}. The maps r U, r V, the homotopy between i U r U and id U, and the homotopy between i V r V and id V can be written down explicitly in formulas, but it might be more helpful to describe

9 SOLUTIONS TO HOMEWORK PROBLEMS 9 them in a geometric way as follows: the subset {±1} R\{0} is a deformation retract (take the linear homotopy between the identity on R \ {0} and the map x x/ x ). It follows that U = S 2 is a deformation retract of U := S 2 x-axis without the origin. Similarly, V is the union of the x-axis and the two open disks given by the left resp. right hemisphere of S 2. The center of each of these disks is a deformation retract of the disk. These deformation retractions and homotopies fit together to show that V = x-axis is a deformation retract of V. Finally, using both retractions (for the x-axis without 0 and the two disks), we obtain the set U V = {( 1, 0, 0), (1, 0, 0)} as a deformation retract of U V. These homotopy equivalences allow us in particular to calculate the reduced homology groups of U, V and U V as { H k (U ) = H k (U) Z k = 2 = 0 k 2 H k (V ) = H k (V ) = H k (pt) = 0 { H k (U V ) = H k (U V ) Z k = 0 = 0 k 0 These calculations show that in the Mayer-Vietoris sequence for X = U V, there are only two non-trivial reduced homology groups of U, V or U V that contribute. Here are the relevant portions of the Mayer-Vietoris sequence: 0 = H 2 (U V ) H2 (U ) H 2 (V ) H2 (X) H1 (U V ) = 0 The exactness of the sequence implies that the middle map is an isomorphism, and hence H 2 (X) = H 2 (U ) H 2 (V ) = Z. 0 = H 1 (U V ) H1 (X) H0 (U V ) H0 (U ) H 0 (V ) = 0 The exactness of the sequence implies that is an isomorphism, and hence H 1 (X) = H 0 (U V ) = Z. For k 1, 2 the exactness of the Mayer-Vietoris sequence implies that H k (X) = 0. Summarizing our result, we find { H k (X) Z k = 1, 2 = 0 k 1, 2 2. Let x 1,..., x l be distinct points of R n. Calculate the reduced homology groups of the space R n \ {x 1,..., x l }. Hint: Compare the homology groups of R n \ {x 1,..., x l } with those of R n via the Mayer-Vietoris sequence. Proof. Let D 1,..., D l R n be disjoint open disks with center x 1,..., x l. Let H k (U V ) = H k ( l i=1(d i \ {x i }) = U := D 1 D l and V := R n \ {x 1,..., x l }. l H k (D i \ {x i }) = i=1 l H k (S n 1 ) = i=1 { Z l k = 0, n 1 0 k 0, n 1

10 10 SOLUTIONS TO HOMEWORK PROBLEMS H k (U) = H k ( l i=1(d i ) = l H k (D i ) = i=1 { Z l k = 0 0 k 0 We note that that the inclusion map i U : U V U is a bijection on connected components, and hence it induces an isomorphism on H 0. Next we consider the Mayer-Vietoris sequence for the decomposition of R n as the union of the open subsets U and V : Hk+1 (R n ) Hk (U V ) i U i V Hk (U) H k (V ) Hk (R n ) Since R n is contractible, its the reduced homology groups H k (R n ) vanish, and hence the map i U i V is an isomorphism by exactness of the Mayer-Vietoris sequence. For k = 0 the map i U is an isomorphism on the homology group H k as noted above, and hence also on H k. It follows that H 0 (V ) = 0. For k > 0, the vanishing of H k (U) implies { H k (V ) = H k (U V ) Z l k = 0, n 1 = 0 k 0, n 1 So, summarizing, we have H k (R n \ {x 1,..., x l }) = { Z l k = n 1 0 k n 1 3. Let X be a topological space and let ΣX be the suspension of X which is defined as the quotient space X [0, 1]/, where the equivalence relation is generated by (x, 0) (x, 0) and (x, 1) (x, 1) for all x, x X. Show that H k (X) = H k+1 (ΣX) (this is called the suspension isomorphism). Hint: note that the suspension of S n is homeomorphic to S n+1, and think of the suspension isomorphism as a generalization of the isomorphism H q (S n ) = H q+1 (S n+1 ) proved in class. That proof used the decomposition of S n+1 as a union of U = S n+1 \ north pole and V = S n+1 \ south pole. In this more general situation, the subspaces U = {[x, t] Σ 0 t < 1} and V = {[x, t] Σ 0 < t 1} of ΣX play an analogous roll. Proof. We will apply the Mayer-Vietoris sequence for the decomposition of the suspension ΣX as the union of the two open subspaces U and V. We note that the spaces U and V (can be pictured as cones ) are both contractible, i.e., homotopy equivalent to the 1-point space pt. To see this, let i: pt ΣX be the inclusion map that sends pt to the cone point [x, 0] U, and let r : ΣX pt be the unique projection map. Then r i = id pt, and i r : U U sends every point to the cone point, and it remains to show that this map is homotopic to the constant map. A homotopy H : ΣX I ΣX is given by ([x, t], s) [x, st] The argument for V is analogous. We claim that the inclusion map i: X U V, x [x, 1/2], is a homotopy equivalence. To see this, let r : U V X be the projection map given by [x, t] x. Then r i = id X, and i r is homotopic to id U V via the homotopy H : (U V ) I U V H([x, t], s) = [x, (1 s) st]

11 SOLUTIONS TO HOMEWORK PROBLEMS 11 We remark that (1 s) 1 + st is the linear path from 1/2 (for s = 0) to t (for s = 1). Writing 2 down the Mayer-Vietoris sequence for reduced homology groups we have Hk (U) H k (V ) Hk (ΣX) Hk 1 (U V ) H k 1 (U) H k 1 (V ) = r 0 Hk 1 (X) 0 The exactness of the sequence implies that is an isomorphism, and hence the composition r : Hk (ΣX) H k 1 (X) is the desired isomorphism. 4. (a) Let A f B g C α β γ A f B g C α A f B g C β γ A B C f g be a commutative diagram of abelian groups and homomorphisms whose rows are exact at B and B. Moreover, α α = id A, β β = id B, and γ γ = id C. Show that the sequence ker α ker β ker γ is exact (the maps are the restrictions of f resp. g). (b) Show that there is a reduced version of the Mayer-Vietoris long exact sequence, where all homology groups are replaced by the corresponding reduced groups. We recall that the reduced homology group H k (X) of a topological space X is by definition the kernel of the map p : H k (X) H k (pt) induced by the projection map p: X pt. Hint: compare the Mayer-Vietoris sequence for X = U V with the Mayer-Vietoris sequence for the space X = pt considered as the union of the open subsets U = V = pt via the projection map p: X pt. Then apply part (a). Proof. Part (a). For a ker α, its image under g f in ker γ is trivial, since the composition g f is zero by exactness of the top row. Let b ker β and assume that g(b) = 0 ker γ C. Then by exactness of the top row, there is some a A with f(a) = b. In general, α(a) will be non-zero, and so we need to modify a in order to produce an element in the kernel of α. We note that ā := a α α(a) is in the kernel of α, since α(a α α(a)) = α(a) αα α(a) = α(a) α(a) = 0. Here the second equality holds since αα = id A. We check that modifying a by subtracting α α(a) doesn t change its image under f: f(ā) = f(a α α(a)) = f(a) f(α α(a)) = b β f α(a) = b β βf(a) = b β βb = b Here the third (resp. fourth) equation holds due to the commutativity of the left square of the first (resp. second) diagram. The last equality follows from the assumption b ker β.

12 12 SOLUTIONS TO HOMEWORK PROBLEMS Part (b). Consider the following commutative diagram H k (U V ) H k (U) H k (V ) H k (X) H k (pt) H k (pt) H k (pt) H k (pt) where the vertical maps are induced by the projection maps to pt. The rows are exact sequences: the top row is the Mayer-Vietoris sequence for X = U V, while the bottom row is the Mayer-Vietoris sequence for X = U V, where X = U = V = pt. We note that the choice of a point x U V can be interpreted as a map X X compatible with the decompositions X = U V, X = U V, i.e., we have maps U U, V V, U V U V, compatible with the inclusion maps and hence these maps induce homomorphism from the bottom Mayer-Vietoris sequence to the top one which gives a commutative diagram (like the second diagram in part (a)). Finally, the composition pt x U V of the inclusion map and the projection map is the identity map of pt, and hence the analogous statement holds for the homomorphisms they induce in homology. The corresponding statement holds for U V replaced by U, V or X, and hence we can apply part (a) to argue that the sequence given by the kernels of the vertical maps is exact. The kernel of the vertical maps are by definition the reduced homology groups. p pt 4. Homework Assignment # 4 f 1. Let 0 A that the sequence B g C 0 be a short exact sequence of chain complexes. Show... H q (A) f H q (B) g H q (C) H q 1 (A)... is exact at H q (C) and H q (B). Proof. Exactness at H q (C). We need to show im g = ker. Recall that for q Z, the chain map g induces a map g : H q (B) H q (C) where g ([b]) = [g(b)]. Let g ([b]) im g H q (C), where [b] H q (B) (so b Z q (B)), then g ([b]) = ([g(b)]). Recall the construction of : assigns c, which is g(b) in our case, to an a A q 1 for which f(a) = (b ), with b chosen so that c = g(b) = g(b ) for b B q. Since we showed that the choice of such a b is irrelevant, we can then choose b to be b. Hence, f(a) = (b) = 0 since b Z q (B). The map f q is injective (by the exactness of the given sequence), so we have that a = 0, whence g ([b]) = ([g(b)]) = [0]. Therefore, im g ker at H q (C). To show im g ker, let [c] ker H q (C). Using the notation from the construction of, there exists a b B q such that g(b) = c and an a A q 1 for which f(a) = (b), and since [c] ker, a = a for some a A q. We notice that the element b f(a ) B q is a q-cycle since (b f(a )) = b f(a ) = f(a) f a = 0, whence [b f(a )] H q (B). Furthermore, by the exactness of the original sequence, gf = 0, so

13 SOLUTIONS TO HOMEWORK PROBLEMS 13 g(b f(a )) = g(b) gf(a ) = g(b) 0 = g(b) = c. Thus, [c] = [g(b f(a ))] = g ([b f(a )]) im g, and im g ker. Exactness at H q (B). To show im f ker g, let f ([a]) = [f(a)] im f H q (B), where a Z q (A). Then, by the exactness of the original sequence, gf = 0, so g ([f(a)]) = [gf(a)] = [0]. Hence, im f ker g. To show im f ker g, let [b] H q (B) such that g ([b]) = [0]; i.e., g(b) im( : C q+1 C q ). Therefore, c C q+1 such that (c) = g(b). Since g is onto (by the exactness of the given sequence), b B q+1 such that g(b ) = c. The commutativity of the diagram (of the chain complexes, chain maps, and boundary maps) yields: g (b ) = g(b ) = (c) = g(b). Since each g is a homomorphism, we then have g(b (b )) = 0, so b (b ) ker g = im f. Thus, a A q such that f(a) = b (b ). Again, by the commutativity of the diagram, f( a) = f(a) = b b = b = 0, so a = 0 since f is injective (ker f = 0). Thus, a Z q (A) and [a] H q (A). Moreover, since b f(a) = (b ), f ([a]) = [f(a)] = [b] H q (B). Thus, im f ker g. To recap, we now have that: is exact at H q (C) and H q (B).... H q (A) f H q (B) g H q (C) H q 1 (A) Suppose the following diagram of abelian groups and group homomorphisms is commutative with exact rows: q+1 C q+1 Aq f q Bq g q Cq q Aq 1 c q+1 a q b q c q a q 1 C q+1 q+1 A q f q B q g q C q q A q 1 Assuming in addition that the maps c q are isomorphisms show that there is a long exact sequence of the form A q A q B q B q A q 1 A q 1 B q 1 B q 1 First define carefully the homomorphisms in the above sequence. Then prove exactness at each location. Proof. We define the maps in the above sequence as follows: α q : A q A q B q β q : A q B q B q γ q : B q A q 1 a (a q (a), f q (a)) (a, b) f q(a ) b q (b) b q c 1 q g q(b ) Exactness at B q. First we show γ q β q = 0. For (a, b) A q B q we have γ q β q (a, b) = q c 1 q g q(f qa b q b) = q c 1 q g qb q b = q g q b = 0 Here the second equality holds due to g qf q = 0, the third follows from the commutativity of the third square, and the last is due to q g q = 0.

14 14 SOLUTIONS TO HOMEWORK PROBLEMS To show ker γ q imβ q let b B q with γ q b = q c 1 q g qb = 0. By exactness at C q there is an element b B q such that g q b = c 1 q g qb or equivalently g qb = c q g q b = g qb q b, where the second equality follows from commutativity of the third square. It follows that b b q b is in the kernel of g q and hence by exactness at B q, there is an element a A q with f qa = b b q b. This implies β q (a, b) = f qa + b q b = b which shows that b is in the image of β q. Exactness at A q B q. First we show β q α q = 0. For a A q we have β q α q a = β q (a q a, f q a) = f qa q a b q f q a = 0 due to the commutativity of the second square. To show ker β q imα q, let (a, b) A q B q with β q (a, b) = f qa b q b = 0. Then we have c q g q b = g qb q b = g qf qa = 0, where the first equality is due to the commutativity of the third square, and the last is due to exactness at B q. Since c q is an isomorphism, this implies g q b = 0 and hence by exactness at B q, there is an element a A q with f q a = b. If we could show a q a = a, we would be done. However we can only say the following: f q(a q a a ) = f qa q a f qa = b q f q a b q b = 0 where the second equality follows from the commutativity of the second square and our assumption f qa = b q b. Since f q is not necessarily injective, we can t conclude that a q a = a, but thanks to exactness at A q, it implies that there is an element c C q+1 with q+1 c = a q a a. Moreover, since c q+1 is an isomorphism, there is a c C q+1 with c q+1 c = c. Now we modify the element a A q by defining ā := a q+1 c. We calculate f q ā = f q (a q+1 c) = f q a = b a q ā = a q (a q+1 c) = a q a q+1c q+1 c = a q a (a q a a ) = a This shows that α q (ā) = (a, b) as desired. Exactness at A q. First let us show α q γ q+1 = 0. For b B q+1 we have α q γ q+1 b = α q ( q+1 c 1 q+1g q+1b ) = (a q q+1 c 1 q+1g q+1b, f q q+1 c 1 q+1g q+1b ) = ( q+1g q+1b, 0) = (0, 0) since the compositions f q q+1 and q+1g q+1 are zero due to the exactness at A q resp. A q. To show ker α q imγ q+1, let a A q with α q a = (a q a, f q a) = (0, 0). By exactness at A q there is an element c C q+1 with q+1 c = a. Then q+1c q+1 c = a q q+1 c = a q a = 0 and hence by exactness at C q+1, there is an element b B q+1 with g q+1b = c q+1 c. This implies γ q+1 b = q+1 c 1 q+1g q+1b = q+1 c = a,

15 SOLUTIONS TO HOMEWORK PROBLEMS 15 which shows that a is in the image of γ q Prove the Brouwer Fixed Point Theorem: Any continuous map f : D n D n from the closed n-disk to itself has a fixed point, that is, there is some point x 0 D n such that f(x 0 ) = x 0. Hint: Prove by contradiction. More precisely, assuming that f has no fixed point, construct a retraction r : D n S n 1 (i.e., a continuous map whose restriction to S n 1 is the identity), contradicting the fact proved in class that S n 1 is no retract of D n. Proof. Assuming that f has no fixed point, the line through x and f(x) intersects the sphere S n 1 in exactly two points. Let r(x) S n 1 be the intersection point closer to x than to f(x). If x is a point on the sphere, then x is clearly that intersection point; in other words, the map r : D n S n 1 restricted to S n 1 is the identity on S n 1. It remains to show that r is continuous. The idea is to write down a formula for r(x), and to argue that r is a composition of basic functions that we know are continuous from calculus. To derive the formula for r(x), we note that every point of the line through f(x) and x is of the form x + t(x f(x)) for t R. Moreover, the point r(x) = x+t(x f(x)) is characterized by the two conditions r(x) 2 = 1 and t 0. The first condition is a quadratic equation for t. Explicitly, setting v = v(x) = x f(x) we have r(x) 2 = r(x), r(x) = tv + x, tv + x = v 2 t v, x t + x 2 and hence the quadratic formula gives us the following non-negative solution for t: (3) t = b + b 2 4ac 2a where a = v(x) 2, b = 2 v(x), x, c = x 2 1. We note that the functions a(x), b(x), c(x) are continuous maps D n R, since they are expressed as compositions of functions that are well-known to be continuous. This implies that the function t = t(x) defined by equation (3) is a continuous function: the only thing to make sure is that the denominator function a(x) = v(x) 2 is nowhere 0 (this is guaranteed by the assumption that v(x) = x f(x) 0 for all x D n ), and that the expression under the square root is positive (this follows geometrically since we have always two intersection points of the line through x and f(x) and the sphere; this is equivalent to having two solutions of the quadratic equation; this in turn is equivalent to a positive expression under the square root in the quadratic formula). Putting everything together, we conclude that the map r : D n S n 1 given by r(x) = x + t(x)(x f(x)) is continuous since t(x) and f(x) are continuous maps. 4. The open cone CX of a topological space X is the quotient space CX := (X [0, 1))/(X {0}). (a) Show that the cone of S n 1 is homeomorphic to the open n-disk D n. In particular, the cone of S n 1 is a manifold of dimension n.

16 16 SOLUTIONS TO HOMEWORK PROBLEMS (b) Show that the cone of X is not a manifold of dimension n if the homology groups of X are not isomorphic to those of S n 1. Hint: Relate the local homology groups of CX at the cone point (the image of X {0} under the projection map X [0, 1) CX) to the homology groups of X. Proof. Part (a). Consider the map f : CS n 1 D n given by [x, t] xt. The map f is a continuous, since its composition with the projection map S n 1 [0, t) CS n 1 is given by (x, t) xt, and hence continuous as the restriction of the continuous multiplication map R n R R n, (x, t) xt. It is straightforward to check that f is a bijection, and hence it remains to show that its inverse is a bijection. We note that the domain is not compact, but the closed cone (X [0, 1])/(X {0}) on X = S n 1 is compact (as quotient of the product of the spaces S n 1 and [0, 1] which are compact by Heine-Borel). The map f extends in an obvious way to a continuous bijection from the closed cone to the closed disk D n. This extension then is a homeomorphism, since its domain is compact and its codomain Hausdorff. Restricting this homeomorphism to the open cone gives the desired homeomorphism. Part (b). Let c CX be the cone point. To calculate the local homology groups H q (CX, CX \ c) we will use the long exact homology sequence of the pair (CX, CX \ c): Hq (CX) H q (CX, CX \ c) Hq 1 (CX \ c) Hq 1 (CX) We observe that the cone CX deformation retracts onto the cone point, and hence the reduced homology groups of CX vanish. This implies that the boundary map provides an isomorphism between the q-th local homology group of CX at c and H q 1 (X). In particular, if CX is a manifold of dimension n, then { H q 1 (X) = H q (CX, CX \ c) Z q = n = 0 q n, and hence the homology groups of X are isomorphic to those of S n Homework Assignment # 5 1. For a polynomial p(z) = a n z n + + a 0 of degree n with complex coefficients a i, let p: CP 1 CP 1 be given by [z 0, z 1 ] [a n z n 0 + a n 1 z n 1 0 z a 0 z n 1, z n 1 ]. Using the fact that CP 1 is homeomorphic to S 2, we can interpret p as a map p: S 2 S 2 which has a degree deg( p) Z. (a) Show that deg( p) = n for the polynomial p(z) = z n. (b) Show that deg( q) = n for any polynomial q(z) of degree n. Hint: construct a homotopy between q and p.

17 SOLUTIONS TO HOMEWORK PROBLEMS 17 Proof. Part (a). We will calculate the degree of p: CP 1 CP 1 as the sum of the local degrees of p at the preimage points of a suitable point y CP 1. For determining p 1 (y) it is useful to note that the diagram C i CP 1 p p is commutative, where the vertical map i is the embedding given by mapping z C to [z, 1] CP 1. If we chose y = 1 C CP 1, then p 1 (y) = p 1 (1) is the set {1, ζ, ζ 2,..., ζ n 1 } of n-th roots of unity with ζ := e 2πi/n. To calculate the local degree deg(p, z) at a point z C, we will argue that the derivative Dp z : R 2 R 2 is invertible and we use part (b) of problem # 3 in this assignment according to which deg(p, z) = sign det(dp z ). We observe that the map p(z) = z n is holomorphic and hence its derivative Dp z is the endomorphism of R 2 = C given by multiplication by the complex number p = z nzn 1. This is a non-zero complex number for any z 0. To calculate det(dp z ), we write nz n 1 = a + ib for a, b R. Using the basis {1, i} for C as a real vector space, we have Dp z (1) = a + ib Dp z (i) = i(a + ib) = b + ia and hence the matrix representing the linear map Dp z : R 2 R 2 is ( ) a b b a. It follows that det(dp z ) = a 2 + b 2 > 0 for z 0. In particular, deg(p, z) = 1 for any 0 z C. Hence C i CP 1 n 1 deg(p) = deg(p, ζ k ) = n. k=0 Part (b). Let q(z) be the polynomial q(z) = a n z n + + a 0, a n 0. For t [0, 1] let q t (z) be the polynomial q t (z) = a n z n + t(a n 1 z n a 0 ). We note that for any t this is again a polynomial of degree n. This guarantees that we can extend q t to a well-defined map q t : CP 1 CP 1 by the formula above (note that it is essential to assume that q t has degree n, since if the coefficient of z n is zero, then q t is not well-defined since it would map [1, 0] CP 1 to [0, 0] which is not a point of CP 1!). This shows that q = q 1 is homotopic to q 0 and hence these maps have the same degree. It remains to show that there is a homotopy of polynomials of degree n between q 0 (z) = a n z n and p(z) = z n to conclude that deg( q 0 ) = deg( p) = n. Such a homotopy is given by p t (z) = e tw z n, where w C with e w = a n C and t [0, 1]. 2. Prove the following statements for the local degree. (a) Show that if f : R n R n is a linear isometry (i.e., f belongs to the orthogonal group), then deg(f, 0) = deg(f S n 1). Hint: apply the statement deg(σg) = deg(g) (which we proved in class) to g = f S n 1, and apply the theorem expressing the degree of a map as a sum of local degrees to Σg. (b) Show that if f n : R n R n is the reflection map (x 1,..., x n ) ( x 1, x 2,..., x n ), then deg(f n, 0) = 1. Hint: argue by induction over n, first using part (a) to show the statement for n = 1. For the inductive step use part (a) again to argue that deg(f n+1, 0) = deg((f n+1 ) S n, e n+1 ), where e n+1 = (0,..., 0, 1) S n. Then show that deg((f n+1 ) S n, e n+1 ) = deg(f n, 0) by showing that (f n+1 ) S n near e n+1 S n corresponds

18 18 SOLUTIONS TO HOMEWORK PROBLEMS to f n near 0 R n via a homeomorphism from a neighborhood of e n+1 to a neighborhood of 0. Proof. Part (a). Following the hint, deg(f S n 1) = deg(g) = deg(σg) = deg(σg, c ), where Σg : ΣS n 1 ΣS n 1 is the suspension of g, and c = [x, 0] ΣS n 1 is the bottom cone point (I m thinking of the suspension ΣS n 1 as the union of the two cones C = {[x, t] ΣS n 1 t [0, 1/2]} with cone point c and C = {[x, t] ΣS n 1 t [1/2, 1]} with cone point c + = [x, 1]). The last equality follows from the theorem expressing the degree of the map Σg as the sum of the local degree of Σg at the points of (Σg) 1 (c ) = {c }. To show that deg(σg, c ) = deg(f, 0) we will construct a homeomorphism h: C D n which maps c to 0 such that the diagram C Σg C C h h D n D n f D n commutes. The induced diagram of local homology groups H n (C, C \ c ) (Σg C ) H n (C, C \ c ) h = h = H n (D n, D n \ 0) (f D n) H n (D n, D n \ 0) then implies deg(σg, c ) = deg(f, 0). We define the map h: C D n by h([x, t]) = 2tx This is evidently a continuous bijection. Since C is compact (as quotient of the product of the compact spaces S n 1 and [0, 1/2]) and D n is Hausdorff, it follows that h is a homeomorphism. Part (b). We prove the statement deg(f n, 0) = 1 by induction over n. For n = 1 we have deg(f 1, 0) = deg((f 1 ) S 0) by part (a). To calculate the degree of g = (f 1 ) S 0 we need to determine the map it induces on H 0 (S 0 ) = ker(h 0 (S 0 ) H 0 (pt)). An element of H 0 (X) for any space X is represented by a finite linear combination i n ix i of points x i X with coefficients n i Z. Such a sum belongs to H 0 (X) if and only if i n i = 0. In particular, X = S 0 consists of the two points x 1 := 1 and x 2 := 1, and the reduced homology group H 0 (S 0 ) is the infinity cyclic group generated by the homology class [x 1 x 2 ] H 0 (S 0 ). Since the map g permutes the two points x 1, x 2 we have which proves that g has degree 1. we note that S 0 = {±1} R. g ([x 1 x 2 ]) = [g(x 1 ) g(x 2 )] = [x 2 x 1 ] = [x 1 x 2 ],

19 SOLUTIONS TO HOMEWORK PROBLEMS 19 For the inductive step, let g : S n S n be the restriction of f n+1 to S n R n+1. Then we have deg(f n+1, 0) = deg(g) = deg(g, e n+1 ) = deg(f n, 0). Here the first equation is a consequence of part (a), the second equation follows again from the result expressing the degree as a sum of local degrees at the points belonging to the preimage g 1 (e n+1 ) = {e n+1 }, e n+1 = (0,..., 0, 1) R n+1. The argument for the last equation is as in part (a): we show that the map g restricted to a neighborhood of e n+1 S n is isomorphic to the map f n restricted to a neighborhood of 0 R n, namely the homeomorphism h from the upper hemisphere D n + S n to the disk D n given by (v 1,..., v n, v n+1 ) (v 1,..., v n ) (with inverse v = (v 1,..., v n ) (v 1,..., v n, 1 v 2 )) maps e n+1 to 0 and makes the diagram D n + g D n + h = h = D n D n f n commutative. The induced diagram of local homology groups then implies deg(g, e n+1 ) = deg(f n, 0). 3. Prove the following further statements for the local degree. (a) Show that if f : R n R n is a linear isomorphism, then deg(f, 0) = sign det(f). Hint: First show that the statement holds when f is reflection at the hypersurface v for any non-zero v R n. Then use the fact that any linear isometry f : R n R n can be written as a composition of such reflections to argue that the statement holds for such f. Finally show that the statement holds for any isomorphism f by finding a path of isomorphisms f t connecting f and a linear isometry. (b) Let f : R n R n be a continuous map which is differentiable at the point x 0 R n. Let Df x0 be the derivative at x 0 (which is a linear map Df x0 : R n R n ; the corresponding matrix is the Jacobian of f at the point x 0 ). Show that if Df x0 is invertible, then deg(f, x 0 ) = sign det(df x0 ). Hint: Use the assumption that f is differentiable at x 0 to write f(x) in the form f(x) = f(x 0 ) + Df x0 (x x 0 ) + e(x x 0 ) where the error term e(h) is o(h) for h 0. Use the homotopy f t (x) := f(x 0 ) + Df x0 (x x 0 ) + te(x x 0 ) to argue (carefully!) that the local degrees of f 0 and f 1 = f at x 0 agree. Proof. Part (a). As suggested by the hint, we first show deg(f, 0) = 1 if f is the reflection R v at the hyperplane v orthogonal to a non-zero vector v R n. By part (b) the statement holds for v = e 1 = (1, 0,..., 0). We recall that the reflection R v : R n R n is given explicitly by R v (w) = w 2 w, v v 2 v. This shows that if v(t), t I is a path of non-zero vectors, then R v(t) is a homotopy between R v(0) and R v(1) and hence their local degrees at 0 agree. For any non-zero v R n, n > 1,

20 20 SOLUTIONS TO HOMEWORK PROBLEMS with v e 1 the linear path (1 t)e 1 + tv does not go through zero (since v, e 1 are linearly independent), and hence deg(r v, 0) = deg(r e1, 0) = 1. For a general element of the orthogonal group O(n) consisting of linear isometries f : R n R n, we use the fact that it can be written as a composition of reflections R v1 R vk. Using the fact that the degree of a composition is the product of the degrees we have deg(f, 0) = deg(r v1, 0) deg(r vk, 0) = ( 1) k. For the determinant we have the analogous equation det(f) = det(r v1 ) det(r vk ) = ( 1) k, which implies that deg(f, 0) = det(f) for any f O(n). To prove the result for a general isomorphism f : R n R n, it suffices to show that there is a path f(t) in the group GL n (R) of isomorphism of R n which connects f = f(0) with an element f(1) O(n) GL n (R). We note that the sign of det(f(t)) is independent of t, since it depends continuously on t, and det(f(t)) 0 for all t [0, 1]. To construct the path, we identify linear maps R n R n with n n matrices in the usual way. We will write a matrix A in the form A = (a 1, a 2,..., a n ), where a i R n are the column vectors of the matrix. We recall that A belongs to GL n (R) if and only if the vectors a 1,..., a n are linearly independent; A O(n) if and only if the vectors a i are unit vectors which are mutually perpendicular. Let p i : R n R n be the orthogonal projection onto the subspace spanned by a 1,..., a i, and let p i : R n R n be the orthogonal projection onto the orthogonal complement of that subspace; in particular, we have v = p i v + p i v for all v R n. Given a matrix A, inspired by the Gram-Schmidt procedure, let us define A t = (v 1, p 1 v 2 + tp 1 v 2,..., p i 1v i + tp i 1 v i + + p n 1v n + tp n 1 v n ) We note that det(a t ) is independent of t, since the determinant depends linearly on each column vector and this determinant is zero if the vectors are linearly dependent. In particular, if A = A 1 invertible, then so is A t. The column vectors of B = A 0 are mutually perpendicular, but not necessarily of unit length. Now for B = (w 1,..., w n ) we define B t = ((1 t)w 1 + t w 1 w 1,..., (1 t)w n + t w n w n ), which is a path connecting B = B 0 with B 1 O(n). Concatenating these paths, we obtain the desired path from A to B 0 O(n). Part(b). To prove part (b), we note that the assumption that f is differentiable at x 0 means that f(x) can be written in the form f(x) = f(x 0 ) + Df x0 (x x 0 ) + e(x x 0 ), where the error term e(h) is o(h) for h 0, which means (4) lim h 0 e(h) h We define and want to argue that f t is a map of pairs = 0. f(x) t := f(x 0 ) + Df x0 (x x 0 ) + te(x x 0 ), f t : (B ɛ (x 0 ), B ɛ (x 0 ) \ {x 0 }) (R n, R n \ f(x 0 )),

21 SOLUTIONS TO HOMEWORK PROBLEMS 21 for sufficiently small ɛ > 0, where B ɛ (x 0 ) is the ball of radius ɛ around x 0. In other words, we want to argue that ft 1 (f(x 0 )) B ɛ (x 0 ) = {x 0 }, or equivalently, that Df x0 (h) + te(h) 0 for all h with 0 < h < ɛ. The idea is to show that for 0 < h < ɛ the norm of Df x0 (h) is large compared to the norm of te(h). To make this precise, let m := min h S n 1 Df x0 (h). We note that m > 0, since m = Df x0 (h 0 ) for some h 0 S n 1, and Df x0 (h 0 ) 0 due to our assumption that Df x0 is invertible. Now the statement (4) allows us to choose ɛ > 0 such that e(h) < m for h h < ɛ. This implies that for 0 < h < ɛ we have te(h) e(t) < m h Df x0 ( h h ) h = Df x 0 (h) and hence Df x0 (h) + te(h) 0 as desired. We conclude that f = f 1 is homotopic to g = f 0 as maps from (B ɛ (x 0 ), B ɛ (x 0 ) \ {x 0 }) to (R n, R n \ f(x 0 )) and hence deg(f, x 0 ) = deg(g, x 0 ). Finally, we want to compare deg(g, x 0 ) and deg(d, 0), where D = Df x0. We note that by construction of g the following diagram is commutative B ɛ (0) Dfx 0 R n T x0 B ɛ (x 0 ) g where T x0, T f(x0 ) are translation maps defined for v R n by T v : R n R n, w w + v. The corresponding diagram of local homology groups then implies deg(df x0 ) = deg(g, 0). 4. Let (X, V, A) be a triple of topological spaces (i.e., A V X). Show that there is a long exact sequence of homology groups T f(x0 ) R n... H q (V, A) H q (X, A) H q (X, V ) H q 1 (V, A)... Hint: use the algebraic fact that a short exact sequence of chain complexes leads to a long exact sequence of homology groups. Proof. Let i: (V, A) (X, A) and j : (X, A) (X, V ) be the maps of pairs induced by the inclusion map V X resp. the identity on X. The induced maps on singular q-chains C q (V, A) i q Cq (X, A) j q Cq (X, V ) C q (V )/C q (A) C q (X)/C q (A) C q (X)/C q (V ) form a short exact sequence since j q is an epimorphism whose kernel is equal to C q (V )/C q (A) C q (X)/C q (A). This implies that i j C (V, A) C (X, A) C (X, V ) is a short exact sequence of chain complexes which implies the desired long exact sequence of homology groups.

22 22 SOLUTIONS TO HOMEWORK PROBLEMS 6. Homework Assignment # 6 1. Prove the following statement which is known as the 5-lemma. Suppose we have a commutative diagram of abelian groups and group homomorphisms f 1 f A 1 2 f A 2 3 f A 3 4 A 4 A 5 h 1 h 2 B 1 g 1 B2 g 2 B3 g 3 B4 g 4 B5 h 3 h 4 h 5 such that the rows are exact sequences. Show that if the vertical maps h 1, h 2, h 4, and h 5 are isomorphisms, then also the middle map h 3 is an isomorphism. Remark: the assumptions that the maps h 1, h 2, h 4, h 5 are all isomorphisms are slightly stronger than needed for the proof. What weaker assumptions will do? Proof. We first prove that h 3 is injective. Let a 3 A 3 such that h 3 a 3 = 0 B 3. Then by the commutativity of the diagram, we have h 4 f 3 a 3 = g 3 h 3 a 3 = g 3 (0) = 0, so f 3 (a 3 ) ker h 4 = 0 since h 4 is injective. Hence, a 3 ker f 3 = im f 2, so there is an a 2 A 2 such that f 2 (a 2 ) = a 3. Again, by the commutativity of the diagram, we have g 2 h 2 (a 2 ) = h 3 f 2 (a 2 ) = h 3 (a 3 ) = 0, so h 2 (a 2 ) ker g 2 = im g 1. Thus, there is an element b 1 B 1 such that g 1 (b 1 ) = h 2 (a 2 ). Moreover, since h 1 is surjective, there is some a 1 A 1 with h 1 a 1 = b 1. It follows that h 2 f 1 a 1 = g 1 h 1 a 1 = g 1 b 1 = h 2 a 2. Thus, since h 2 is injective, a 2 = f 1 a 1, so a 3 = f 2 a 2 = f 2 f 1 a 1 = 0. Therefore, h 3 is injectiv. To show surjectivity of h 3, let b 3 B 3. Since h 4 is surjective, there is some a 4 A 4 with h 4 a 4 = g 3 b 3 A 4 and by commutativity, h 5 f 4 a 4 = g 4 h 4 a 4 = g 4 g 3 b 3 = 0. Since h 5 is injective, this implies f 4 a 4 = 0. By exactness of the top row at A 4, there is an a 3 A 3 such that f 3 (a 3 ) = a 4. Hence, g 3 (h 3 (a 3 ) b 3 ) = h 4 f 3 a 3 g 3 b 3 = h 4 a 4 g 3 b 3 = 0. By the exactness of the lower row at B 3, this implies that there exists b 2 B 2 such that g 2 (b 2 ) = h 3 (a 3 ) b 3. Since h 2 is surjective, there is some a 2 A 2 with h 2 a 2 = b 2 and hence h 3 (a 3 f 2 a 2 ) = h 3 a 3 h 3 f 2 a 2 = h 3 a 3 g 2 h 2 a 2 = b 3, which shows that b 3 is in the image of h 3. Since b 3 was arbitrary, this shows that h 3 is surjective. We see that we ve used the assumptions that h 4, h 2 are injective, and that h 1 is surjective to show injectivity of h 3. Our proof that h 3 is surjective required the assumptions that h 2, h 4 are surjective, and that h 5 is injective. So it is sufficient to assume that h 2 and h 4 are isomorphisms, that h 1 is an epimorphism, and that h 5 is a monomorphism. 2. Show that the complex projective space CP n is a CW complex with one cell of dimension 2i for 0 i n. Proof. It suffices to show that CP n is obtained from CP n 1 by attaching a cell of dimension 2n. Define Φ: D 2n CP n z = (z 0,..., z n 1 ) [z 0, z 1,..., z n 1, 1 z 2 ],