Math Homotopy Theory Spring 2013 Homework 6 Solutions

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1 Math Homotopy Theory Sprig 2013 Homework 6 Solutios Problem 1. (The Hopf fibratio) Let S 3 C 2 = R 4 be the uit sphere. Stereographic projectio provides a homeomorphism S 2 = CP 1, where the North pole correspods to [0 : 1] CP 1. The composite S 3 C 2 \ {0} CP 1 where the secod map is the atural quotiet map, is called the Hopf map ad is usually deoted by η : S 3 S 2. a. Show that η : S 3 S 2 is a fiber budle with fiber S 1. Solutio. Let us show more geerally that the quotiet map q : S 2+1 CP is a fiber budle with fiber S 1. Recall that the affie patches U i := {[z 0 : z 1 :... : z ] CP z i 0} = C with 0 i form a ope cover of CP. We will show that each U i is a trivializig eighborhood for q. Cosider the preimage q 1 (U i ) = {(z 0, z 1,..., z ) S 2+1 z i 0} ad cosider the (cotiuous) map ϕ: q 1 (U i ) U i S 1 defied by ( ) z i ϕ(z) = q(z),. z i Now cosider the map ψ : U i S 1 p 1 (U i ) defied by ψ ( [z 0 :... : z ], e iθ) ( = e iθ z0,..., z ) zi z i z i z. Oe readily checks that ψ is well defied, cotiuous, ad iverse to ϕ. b. Show that for all 3, the Hopf map iduces a isomorphism η : π (S 3 ) π (S 2 ). Deduce i particular the isomorphism π 3 (S 2 ) Z, where the class [η] π 3 (S 2 ) is a geerator. Solutio. Cosider the log exact sequece of the fibratio S 1 S 3 η S 2. For all 3, the relevat part of the exact sequece is 0 = π (S 1 ) π (S 3 ) η π (S 2 ) π 1 (S 1 ) = 0 which shows that π (S 3 ) η π (S 2 ) is a isomorphism. I particular, for = 3, we obtai the isomorphism π 3 (S 3 ) η π 3 (S 2 ). Give π 3 (S 3 ) = Z with geerator [id S 3], we coclude π 3 (S 2 ) Z with geerator η [id S 3] = [η id S 3] = [η]. 1

2 Problem 2. Let (E, e 0 ) ad (B, b 0 ) be poited spaces ad p: E B a poited map. Deote by F := p 1 (b 0 ) the strict fiber of p, ad F (p) the homotopy fiber of p, defied as F (p) = {(e, γ) E B I γ(0) = p(e), γ(1) = b 0 }. There is a caoical iclusio of the strict fiber ito the homotopy fiber ϕ: F F (p) defied by ϕ(e) = (e, c b0 ) where c b0 : I B is the costat path at b 0. Show that if p: E B is a fibratio, the the map ϕ: F F (p) is a homotopy equivalece. Solutio. Deote by P (p) = {(e, γ) E B I γ(0) = p(e)} = E B B I the path space costructio o p. Sice p is a fibratio, the caoical map (ev 0, p ): E I P (p) admits a sectio s: P (p) E I. Note that the restrictio of s to F (p) P (p) takes values i paths that ed i F. I other words, for (e, γ) F (p), we have p (s(e, γ)(1)) = (p s(e, γ)) (1) = γ(1) = b 0 which meas s(e, γ)(1) p 1 (b 0 ) = F. Now cosider the composite ψ F (p) s ev 1 1 (F ) ev 1 F P (p) s E I ev 1 E. We claim that ψ : F (p) F is homotopy iverse to ϕ: F F (p). a) ψϕ id F. Note that s restricted to F F (p) takes values i paths that live etirely i F. Ideed, the path s(ϕ(e)) projects dow to the costat path p (s(ϕ(e))) = c b0 i B. Cosider the commutative diagram ϕ F F (p) s s ψ F I ev 1 1 (F ) ev 1 F ev 1 F P (p) s E I ev 1 E ad ote that the formula ev t s: F F for t I provides a homotopy from ev 0 s = id F to ev 1 s = ψϕ. 2

3 b) ϕψ id F (p). The family of maps parametrized by t I F (p) F (p) (e, γ) ( ) s(e, γ)(t), γ [t,1] provides a homotopy from the map (e, γ) ( ) s(e, γ)(0), γ [0,1] = (e, γ) to the map (e, γ) ( ) s(e, γ)(1), γ [1,1] = (ψ(e, γ), cb0 ) = ϕψ(e, γ) which proves the relatio ϕψ id F (p). 3

4 Problem 3. Let X be a CW-complex with skeletal filtratio X 0 X 1... X = colim X. Show that for ay k 0, there is a atural isomorphism π k (X) = colim π k (X ). Solutio. Let us show that the atural map ϕ: colim π k (X ) π k (colim X ) = π k (X) is a isomorphism. a) ϕ is surjective. Let [α] π k (X) be represeted by a poited map α: S k X. Sice S k is compact, the image α(s k ) X is compact, ad thus meets oly fiitely may cells of X. Therefore α factors through some fiite skeleto X N X, as illustrated i the diagram: α S k X N X. Therefore [α] is i the image of the composite π k (X N ) colim π k (X ) ϕ π k (X) where the first map is the summad iclusio i the colimitig cocoe. b) ϕ is ijective. Let β ker ϕ colim π k (X ). The coditio ϕ(β ) = 0 π k (X) meas that there is a (poited) ull-homotopy H : S k I + X of ϕ(β ). Agai, S k I + is compact, so that the image H(S k I + ) X meets oly fiitely may cells of X. Therefore H factors through some fiite skeleto X N X, as illustrated i the diagram: H S k I + X N X. Therefore β is the image via the summad iclusio π k (X N ) colim π k (X ) of a elemet which was already 0 i π k (X N ), which proves β = 0. Remark. The same statemet holds for homology H k (X) = colim H k (X ), by compactess of the simplex k. 4

5 Alterate solutio. By cellular approximatio, we have π k (X) = π k (X ) for all k + 1. More precisely, the sequece π k (X k ) π k (X k+1 ) = π k (X k+2 ) = π k (X k+3 ) =... stabilizes, ad skeletal iclusios X X iduce isomorphisms π k (X ) = π k (X) for all k + 1. Therefore the atural map colim π k (X ) π k (X) is a isomorphism. Remark. It may seem that the stroger result obtaied i the alterate solutio makes the first solutio useless. However, the argumet used i the first solutio is much more geeral, ad applies for example to cell complexes X 0 X 1 X 2... X = colim X where cells of arbitrary dimesios are added at each stage. More geerally, it applies to ay sequece of closed embeddigs X 0 X 1 X 2... X = colim X assumig the spaces X are compactly geerated weakly Hausdorff. See May-Poto, Propositio ad Corollary

6 Problem 4. I this problem, feel free to refer to Homework 5 Problem 4. Cosider ifiitedimesioal real projective space RP = colim RP. a. Compute all homotopy groups of RP. Solutio. Usig the stadard cell structure o RP, ote that RP is path-coected: π 0 (RP ) = π 0 (RP 1 ) =. Moreover we have ad for all k 2 we have π 1 (RP ) = π 1 (RP 2 ) = Z/2 π k (RP ) = π k (RP k+1 ) = π k (S k+1 ) = 0. b. Show that RP 2 ad S 2 RP are ot homotopy equivalet. Solutio. The two spaces have very differet homology. For istace, we have { H k (RP 2 Z/2 if 0 k 2 ; Z/2) = 0 otherwise whereas for all k 2, the Küeth theorem yields H k (S 2 RP ; Z/2) = H 0 (S 2 ; Z/2) Z/2 H k (RP ; Z/2) H 2 (S 2 ; Z/2) Z/2 H k 2 (RP ; Z/2) = Z/2 Z/2 Z/2 Z/2 Z/2 Z/2 = Z/2 Z/2 H k (RP 2 ; Z/2). Alterate solutio. The two spaces have a differet π 1 -actio o π 2. By aturality of the π 1 -actio, the actio of o is trivial. π 1 (S 2 RP ) = π 1 (S 2 ) π 1 (RP ) = π 1 (RP ) π 2 (S 2 RP ) = π 2 (S 2 ) π 2 (RP ) = π 2 (S 2 ) O the other had, the actio of π 1 (RP 2 ) = O(1) = { 1, 1} o π 2 (RP 2 ) = π 2 (S 2 ) is iduced by deck trasformatios o the uiversal cover S 2 RP 2. Oe ca show, say, usig the Hurewicz theorem, that the actio of the atipodal map (scalar multiplicatio by 1) o π 2 (S 2 ) = Z is multiplicatio by the degree which is a o-trivial actio. τ : S 2 S 2 deg(τ) = ( 1) 2+1 = 1 6