Algebraic Topology Final

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1 Instituto Superior Técnico Departamento de Matemática Secção de Álgebra e Análise Algebraic Topology Final Solutions 1. Let M be a simply connected manifold with the property that any map f : M M has a fixed point. Show that a covering map p: M B is necessarily a homeomorphism. Answer: Since M is simply connected, p is the universal cover of B. Since M is locally path connected, it follows that the group of deck transformations Aut(p) = π 1 (B) acts transitively on the fibers of p. Since M is connected, the action of Aut(p) on M is free, so the only deck transformation which has a fixed point is the identity. We conclude that the fibers of p have only one element, i.e. that p is injective. It follows that p is a homeomorphism. 2. Let X be the space obtained from CP 2 by attaching a 3-cell to CP 1 by a map of degree 6. (a) Compute H (X). (b) Compute H (X; Z/m) for all m > 1. (c) Compute H (X RP 2 ). (d) Let f : S 1 S 1 be a map of degree k. Determine the homomorphism of graded groups (f id X ) : H (S 1 X) H (S 1 X). Answer: (a) The space X is a cell complex with one cell in each of the dimensions 0, 2, 3 and 4. Therefore its cellular chain complex is given by Ck CW Z if k = 0, 1, 2, 4 (X) = 0 otherwise The space X contains CP 2 as a subcomplex, so C CW (X) contains C CW (CP 2 ) as a sub-chain complex. Thus the boundary maps : C k (X) C k 1 (X) vanish for k 3. For k = 3 the boundary map is multiplication by 6 since the 3-cell is attached by a map of degree 6, We conclude that the homology of the cellular chain complex is (b) By the universal coefficient theorem Z if k = 0, 4 H k (X) = Z/6 if k = 2 H k (X; Z/m) = H k (X) Z/m Tor(H k 1 (X), Z/m) Since Z/6 Z/m = Z/(6, m), Z Z/6 = Z/6, Tor(Z, Z/m) = 0 and Tor(Z/6, Z/m) = Z/(6, m) we have Z/m if k = 0, 4, H k (X; Z/m) = Z/(6, m) if k = 2, 3,

2 (c) By the Künneth formula, H k (X RP 2 ) = k j=0h j (X) H k j (RP 2 ) k 1 j=0 Tor(H j(x), H k 1 j (RP 2 )) Since Z if j = 0, H j (RP 2 ) = Z/2 if j = 1, 0 otherwise, and Z/6 Z/2 = Tor(Z/6, Z/2) = Z/2, we have Z if k = 0, Z/2 if k = 1, Z/6 if k = 2, H k (X RP 2 ) = Z/2 if k = 3, Z Z/2 if k = 4, Z/2 if k = 5. (d) Since the homology of S 1 is free, by the Künneth formula, the cross product gives a natural isomorphism so we have a commutative square n j=0h j (S 1 ) H n j (X) H n (S 1 X) 1 j=0 H j(s 1 ) H n j (X) H n (S 1 X) f id (f id) 1 j=0 H j(s 1 ) H n j (X) H n (S 1 X) 3. Let It follows from the calculation in (a) and the above formula that H n (S 1 X) = H 0 (S 1 ) H n (X) = H n (X) for n even, while H n (S 1 X) = H 1 (S 1 ) H n 1 (X) = H n 1 (X) for n odd. Since a map of degree k induces multiplication by k on H 1 (S 1 ) and the identity on H 0 (S 1 ) we see that (f id X ) is multiplication by k in odd degrees and the identity in even degrees. f 1 f 2 X 1 X2 X3 be a sequence of continuous maps. The telescope of this sequence is the space ( ) T = X n [n, n + 1] / n where is the equivalence relation generated by (x, n + 1) (f n (x), n + 1). ( ) (a) Show that H k (T ) = colim H k (X 1 ) f 1 H k (X 2 ) f 2 H k (X 3 ) for all k.

3 (b) Let p 1,..., p n,...} be an enumeration of the primes in N. Let X n = S m for all n and let f n be a map of degree p 1 p n. Show that H m (T ) = Q. Answer: (a) Let T j,n+1 T be the image of n X k [k, k + 1] X n+1 n + 1} k=j under the quotient map. Then, for j n, T j,n deformation retracts to T j+1,n via the map T j,n [0, 1] T j,n given by the formula ([(x, t)], s) [(x, t + s(j + 1 t))] if x X j, t [j, j + 1] [(x, t)] otherwise. and composing these deformation retractions we see that T 1,n+1 deformation retracts to T n+1,n+1 = Xn+1. The deformation retraction of T n,n+1 onto T n+1,n+1 = Xn+1 sends [(x, n)] to [(x, n + 1)] = [(f n (x), n + 1)] so the diagram T 1,n T 1,n+1 (1) X n X n+1 f n commutes up to homotopy. There is a continuous map T R defined by [(x, t)] t. If K T is a compact set, it has bounded image under these map and it is therefore contained in T 1,n for some n. By the continuity of homology with respect to direct limits we have H (T ) = colim n H (T 1,n ) and (1) then identifies the previous colimit with ( ) colim H (X 1 ) f 1 H (X 2 ) f 2 H (X 3 ) as required. (b) By (a), H m (T ) is the colimit of the diagram of abelian groups f 1 f 2 G 1 G2 where G i = Z for all i and fi is multiplication by p 1 p i. Let h i : G i Q be the homomorphism that sends 1 to 1/(p i 1 pi 1 2 p i ). Then h i+1 f i = h i and so the h i give a homomorphism h: colim G i Q. This is surjective because clearly any rational number is in the image of some h i. It is injective because any element in colim G i is represented by an element of some G i and all the maps h i are injective. We conclude that h is an isomorphism.

4 4. Compute the cohomology ring H (RP ; Z/6). Answer: The ring map Z/6 Z/2 determines a ring homomorphism H (RP ; Z/6) H (RP ; Z/2) = Z/2[x] where x is a class of degree 1. We can use cellular cohomology to determine the effect of the ring homomorphism: the cellular chain complex of RP has Z in each degree with boundary maps k : C CW k (RP ) C CW k 1 (RP ) given by 0 for k odd and multiplication by 2 for k even. Applying the functor Hom(, Z/6) we obtain C k CW (RP ; Z/6) = Z/6 with coboundary maps C k C k+1 given by 0 for k even and 2 for k odd. Hence H k (RP Z/6 if k = 0 ; Z/6) = Z/2 if k > 0. Moreover the generators in odd degree are represented by homomorphisms that send 1 Z = Ck CW (RP ) to 3 Z/6, while the generators in even degree are represented by homomorphisms which send 1 Z = Ck CW (RP ) to an odd number in Z/6. It follows that the map on cellular cohomology H (RP ; Z/6) H (RP ; Z/2) is an isomorphism in positive degrees (in degree 0 it is the canonical map Z/6 Z/2) and this gives the ring structure with x a class of degree 1. H (RP ; Z/6) = Z/6[x]/2x 5. Let M be a connected manifold of dimension n and p be a point in M. Assuming the Euler characteristic of M is defined, show it is also defined for M \ p and χ(m \ p) = χ(m) + ( 1) n 1 Answer: Consider the long exact sequence of the pair H k+1 (M, M \ p) H k (M \ p) H k (M) H k (M, M \ p) As M is an n-manifold, its local homology groups are Z if k = n H k (M, M \ p) = This implies that the inclusion induced map H k (M \p) H k (M) is injective for k n 1 and surjective for k n (hence an isomorphism for k n 1, n}). The section of the long exact sequence of the pair envolving these degrees is 0 H n (M \ p) H n (M) H n (M, M \ p) H n 1 (M \ p) H n 1 (M) 0 Since M \ p is a non-compact manifold the first group in the sequence is 0. Suppose first that M is closed and orientable. Then the map H n (M) H n (M, M \p) is an isomorphism and so we obtain that H n 1 (M \p) H n 1 (M) is also an isomorphism. Therefore the Betti numbers of M and M \ p agree in all dimensions except n where

5 β n (M) = 1 while β n (M \ p) = 0. We conclude that the Euler characteristic of M \ p is defined and χ(m) = χ(m \ p) + ( 1) n as required. Now suppose M is not closed or not orientable. Then H n (M) = 0 = H n (M \ p) (so the Betti numbers of M and M \ p agree in all dimensions except possibly n 1) and we have a short exact sequence 0 Z H n 1 (M \ p) H n 1 (M) 0. Tensoring with Q, we see that β n 1 (M \ p) = β n 1 (M) + 1 and so again the Euler characteristic of M \ p is defined and which completes the proof. χ(m \ p) = χ(m) + ( 1) n 1 6. Show that if M is a simply connected closed 5-manifold then H 3 (M) is torsion free. Answer: Since M is simply connected, it is orientable. By Poincaré duality H 3 (M) = H 2 (M) and by the universal coefficient theorem H 2 (M) = Hom(H 2 (M), Z) Ext(H 1 (M), Z). Since M is simply connected, H 1 (M) = 0, therefore H 3 (M) = Hom(H 2 (M), Z). For any abelian group G, the abelian group Hom(G, Z) is torsion free (because Z is torsion free). This completes the proof.

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