ALGEBRAIC TOPOLOGY MASTERMATH (FALL 2014) Written exam, 21/01/2015, 3 hours Outline of solutions

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1 ALGERAIC TOPOLOGY MASTERMATH FALL 014) Written exam, 1/01/015, 3 hours Outline of solutions Exercise 1. i) There are various definitions in the literature. ased on the discussion on. 5 of Lecture 3, as well as the fact that all saces and mas should be taken ointed, the correct answer should at least be an examle of the following definition: Definition. An H-sace is a sace X equied with a multilication ma µ: X X X and a unit element e: X such that the mas µe, ): X X µ, e): X X are homotoic to the identity mas on X, via a homotoy that fixes the unit e so that µe, e) = e). There are variants asking for associativity u to homotoy or the existence of strict units. All these variants are counted as a correct answer. Note that exercise iii) gives an extra hint for taking X, e) as a ointed sace, so that all homotoies should fix the unit e. ii) For [α] π 1 X) and [β] π n X), take reresentatives α: I X and β : I n X sending I and I n to the baseoint. Consider the ma I n {0} I n I X given by β on I n {0} and by α π on I n I. It has an extension to a ma H : I n I X, whose restriction to I n {1} resents the element [α] [β]. For n = 1, the ma H can be ictured as α α β This clearly shows that the remaining face reresents [α][β][α] 1. See. 6-7 of Lecture 4. iii) π 1 X) is abelian by the Eckman-Hilton trick. 3 of Lecture 3). It also follows from the argument for generic n, by taking n = 1. If the unit is a strict unit, one can take the extension H from art ii) to be given by Hx, t) = µβx), αt)) Since α1) = e, one sees that Hx, 1) = βx), showing that the action is trivial. If the e is only the unit u to ointed homotoy, observe that Hx, 1) = µβx), e) is homotoic to βx) via the homotoy witnessing unitality. ut the restriction of Hx, 1) to I n is constant with value e and the unitality homotoy fixes e. It follows that H, 1) β via a homotoy that fixes the boundary I n. Exercise. i) False, π 3 S ) 0 see. 4 of Lecture 3 or Exercise 6 from sheet 6). ii) True, by cellular aroximation see. 3 of Lecture 9). iii) True, to check the defining right lifting roerty one roduces a lifting in two stes. iv) True, to check the defining right lifting roerty one roduces a lifting in two stes. v) False in general, f should land in X n 1). 1

2 ALGERAIC TOPOLOGY vi) True, f necessarily lands in X n 1). vii) False. As was mentioned in the lectures, there are many mas KA, n) K, m) which are not nullhomotoic, even when n m. Such mas always induce the zero ma on homotoy grous when n m. However, this remark does not aear in the lecture notes and therefore exercise vii) has not been marked. A concrete counterexamle to vii) is given by the quotient ma S 1 S 1 S 1 S 1 /S 1 S 1 S. However, to rove that this ma is indeed not nullhomotoic requires material beyond the scoe of this course the ma induces an isomorhism between the second homology grous). Exercise 3. i) In general, if q 1 c 0 ) q D c 0 C φ A realizes q as the ullback of a Serre fibration, then q is a Serre fibration as well. Furthermore, the fiber q 1 c 0 ) can be realized by the dotted ullback square. y the asting lemma for ullbacks, the total square is a ullback as well, which show that q 1 c 0 ) is homeomorhic to 1 φc 0 )). Since we know that the ma ɛ 0, ɛ 1 ): X I X X is a Serre fibration see. of Lecture 5), this gives i). Alternatively, one can construct an exlicit homeomorhism between the fiber of Y h X Z Y Z and ΩX, x 0). ii) The long exact sequence of the fibration Y h X Z Y Z gives an exact sequence π n ΩX, x 0 )) π n Y h X Z) π n Y Z) π n 1 ΩX, x 0 )) π 0 ΩX, x 0 )) π 0 Y h X Z) π 0 Y Z) Using that π n X, x 0 ) π n 1 ΩX, x 0 )) for n 1 and that π n Y Z) π n Y ) π n Z), one obtains a long exact sequence of the desired form. The ma π 1 Y ) π 1 Z) π 1 X) corresonds to the connecting homomorhism of the long exact sequence for Y h X Z Y Z. It is constructed as follows: let α: I Y and β : I Z resent an element in π 1 Y ) π 1 Z). Pick a lift {0} I κ x0,y 0,z 0) h α,β) Y h X Z X I ɛ 0,ɛ 1) Y Z fg X X. Then h1) is a oint in the fiber over y 0, z 0 ) which was ΩX, x 0 ) whose homotoy class rovides the desired element in π 0 ΩX, x 0 )). Now the comosite I Y h X Z XI determines a ma H : I I X with the roerty that H, 0) = f α H, 1) = g β H0, ) = κ x0 In other words, H determines a square of the form f α κ x0 γ g β

3 ALGERAIC TOPOLOGY 3 for some γ : I X with γ0) = γ1) = x 0. It follows that [γ] f [α] = g [β] [κ x0 ] in π 1 X), so that [γ] = g [β] f [α] 1. y construction, the h1) is given by γ, y 0, z 0 ). It follows that the connecting homomorhism sends α, β) to [γ] = g [β] f [α] 1 in π 0 ΩX, x 0 )) π 1 X). Remark: the ma π n Y ) π n Z) π n X) sends α, β) to g β) f α) for all n. This either follows from a similar argument as the one for n = 1, or one can reduce to the case n = 1 as follows: alying Ω n to ullback square??), we find a ullback square Ω n Y h X Z) Ω n X ) I Ω n Y Ω n Z Ω n X Ω n X in which the vertical mas are the n-fold looings of the original Serre fibrations. If : E X is a Serre fibration with fiber F, then Ω n : Ω n E Ω n X is a fiber sequence with fiber Ω n F. Under the isomorhism π 0 Ω n X) π n X), the long exact sequence of Ω n corresonds to the art of the long exact sequence of that sits in dimensions n. The exercise then shows that the ma π n+1 Y ) π n+1 Z) π 1 Ω n Y Ω n Z) sends α, β) to g β f α) 1. π 1 Ω n X) π n+1 X) Exercise 4. i) See Lecture 11. A correct answer should include: iteratively attaching cells along mas e n X resenting nontrivial elements in π n 1 X) and relacing the resulting sequence of relative CW-comlexes by a homotoy equivalent sequence of fibrations. ii) See Lecture 11. The fibration ψ n 1 induces isomorhisms of homotoy grous in dimensions n: in dimensions > n the homotoy grous of P n X) and P n 1 X) are both trivial and in dimensions k < n there is a commuting diagram π k Pn X) ) ψ n 1 π k X) π k Pn 1 X) ) f n 1 so that ψ n 1 induces isomorhisms of homotoy grous in dimensions k < n. Furthermore, the ma ψ n 1 induces the zero ma on the n-th homotoy grou. Insection of the long exact sequence of the fibration ψ n 1 now shows that the fiber of ψ n 1 is a Kπ, n), where π = π n X). Exercise 5. i) One ossible CW-decomosition is given by 1 c This has three 0-cells,,, five 1-cells and two -cells. The inclusion of the central cirlce is the inclusion of a subcomlex we take to be the unique 0-cell of the central circle).

4 4 ALGERAIC TOPOLOGY If we give the boundary circle the CW-decomosition with a unique 0-cell and one 1-cell, then the inclusion of the boundary becomes cellular since it sends the 0-, res. 1- skeleton of the circle to the 0-, res. 1-skeleton of M). For exercise iii), it is useful to ick a different CW-structure for the Möbius stri, for which both the central circle and the boundary circle are inclusions of subcomlexes: 1 c This decomosition has two 0-cells, three 1-cells the central circle, the boundary circle 1 and the segment between and ) and one -cell. ii) The most obvious retraction is given by : M S 1 ; [s, t] = [s, 0] It is immediate that c is the identity on S 1 I/ I. Furthermore, c is homotoic to the identity on M via H : M I M; H [s, t], τ ) = [s, t τ]. The main oint is that the comosite S 1 M S 1 wras the boundary circle twice around the central circle, so it induces multilication by two on π 1 S 1 ) = Z. The ma is a homotoy equivalence by construction, while the ma is a cofibration since it is the inclusion of a subcomlex. iii) Using the second CW-structure from i), we have that c and are both inclusions of subcomlexes. In general, if A and A C are inclusions of subcomlexes giving the same CW-structure on A!), then the ushout A C is the inclusion of a subcomlex. In articular, A C is itself a CW-comlex. Indeed, define A C) n) as the ushout A n) C n) n) A C) n). Then A C) n+1) is obtained from A C) n) by adding the n + 1)-cells of and the n + 1)-cells of C identifying the n + 1)-cells of A). Alying this inductively to the sequence of M n) shows that M n) is a CW-comlex and that both M n 1) M n) and the inclusion of the central circle c n) : S 1 M n) are inclusions of subcomlexes; the latter allows one to roceed inductively. Phrased differently, one obtains a CW-structure on M n) by 1) taking the CW-structure on M n 1) ) adding a 0-cell the 0-cell of the new central circle) 3) adding two 1-cells: add the new central circle to the added oint, and add a 1-cell between the newly added oint and the 0-cell in the old central circle of M n 1).

5 ALGERAIC TOPOLOGY 5 4) finally, adding a -cell according to the CW-structure of the Möbius stri. The result is a CW-comlex since we only attach n-cells to the n 1)-skeleton. Furthermore, it is immediate that M n 1) is a CW subcomlex. For the case n =, the CW-structure is given by taking the images of the 0-, 1- and -cells of each M n). iv) Any finite subcomlex of M ) is contained in some M n) this holds in general for the colimit of a sequence of subcomlex inclusions). This imlies that any ma S k M ) takes values in some M n). ut it is given that π k M n) ) is zero when k 1, so any ma S k M ) is ointed) homotoic to the constant ma when k 1. This shows that M ) is a KG, 1). In articular, it is ath-connected, so we just have to look at the fundamental grou at some baseoint that lies in M 1) M ). For any sequence of subcomlex inclusions X 0 X 1..., one has that π 1 colim X n ) = colim n π 1 X n ) It this case, we know that each π 1 M n) ) Z and that each ma π 1 M n) ) π 1 M n+1) is given by multilication by. We thus have that π 1 M ) ) is the colimit of the sequence Z Z The colimit of this sequence is Z[ 1 ], i.e. the additive) grou of fractions q ower of. where q is a Alternatively, one can exlicitly construct an isomorhism between Z[ 1 ] and π 1M ) ) as follows: send the element 1 Z[ 1 ] to the image of the generating element 1 π 1M 1) ) in π 1 M ) ). Similarly, send 1 to the image in π n 1 M ) ) of the generating element of π 1 M n+1) ). This gives a well-defined grou homomorhism because the generator of π 1 M n) ) is exactly identified with twice the generator of π 1 M n+1) ). The resulting grou homomorhism is surjective: this follows from the fact that any ma S 1 M ) takes values in some M n+1) and therefore is given by for some n and n. The resulting grou homomorhism is injective: if and q n are sent to the m same element in π 1 M ) ), then the reresenting loos inside M n), res. M m), become homotoic in some M N) for N large enough. ut this means recisely that n = N n N = N m q N = q m.