CLASS NOTES MATH 527 (SPRING 2011) WEEK 6

Transcription

1 CLASS NOTES MATH 527 (SPRING 2011) WEEK 6 BERTRAND GUILLOU 1. Mon, Feb. 21 Note that since we have C() = X A C (A) and the inclusion A C (A) at time 0 is a coibration, it ollows that the pushout map i : X C() is a coibration. In particular, the quotient C()/X is based homotopy equivalent to C(X i C()). The quotient C()/X can also be identiied with ΣA. So we get a coiber sequence A X i C() ΣA. Proposition 1.1. The coiber o is homotopy equivalent to ΣX, and under this identiication, the map ΣA C( ) ΣX is identiied with the map Σ deined by Σ(x, t) = (x, 1 t). (Draw pictures) So the coiber sequence extends to a long sequence o based maps A X i C() ΣA Σ ΣX Σi.... Theorem 1.2. (Long exact sequence) For any based space Z, mapping the long coiber sequence into Z produces a long exact sequence... [ΣX, Z] [ΣA, Z] [C(), Z] [X, Z] [A, Z]. At the right end o the sequence, these are just pointed sets, but [ΣA, Z] = π1 (Map (A, Z)) and [Σ 2 A, Z] = π2 (Map (A, Z)), so we get groups and eventually abelian groups in this long exact sequence. The proo o this result is completely analogous to the proo o the long exact sequence in homotopy. In act, this result can also be deduced rom the earlier result or the ollowing reason: Proposition 1.3. Let : A X be a based map and let Map () : Map (X, Z) Map (A, Z) be the induced map. Then the homotopy iber F (Map ()) is homeomorphic to the mapping space Map (C(), Z). Proo. This ollows rom the universal property o the pushout C() = X AC (A): Map (C(), Z) = Map (X, Z) Map (A,Z) Map (C (A), Z) = Map (X, Z) Map (A,Z) P Map (A, Z) = F (), where P (Y ) is the space o based paths in Y (ending at the basepoint). Although coiber sequences do not give rise to a long exact sequence in homotopy, they do give rise to a long exact sequence in homology: Theorem 1.4. For any based map : A X, there is a long exact sequence in reduced homology... H n (A) H n (X) H n (C()) H n 1 (A).... 1

2 Proo. Up to homotopy equivalence, we may assume that is a coibration and that the coiber C() is the quotient X/A. This is then the long exact sequence in homology or a good pair. Homotopy pushouts Last time, we said that i : A X is a coibration, then the coiber C() is homotopy equivalent to X/A. This generalizes as ollows: Proposition 1.5. Let j : A X be a coibration and consider a pushout square A j X B P. Then P is homotopy equivalent to the double mapping cylinder B A A I A X. Moreover, suppose given a map o diagrams A j X A j B B in which both j and j are coibrations and the component maps A : A A, B : B B, X : X are homotopy equivalences. Then P P (where P is the pushout o the second diagram). The space P above is called the homotopy pushout o the maps B A X. Proo. A very similar argument to the one rom last time shows that P is homotopy equivalent to the pushout o the diagram A M(j) B. But this pushout is homeomorphic to the double mapping cylinder. For the second part, it is not diicult to see that replacing X or B up to homotopy equivalence results in a homotopy equivalent pushout. That replacing A does not change the homotopy type ollows rom the ollowing lemma: Lemma 1.6. A pushout o a homotopy equivalence along a coibration is a homotopy equivalence. That is, i j is a coibration and is a homotopy equivalence in the square then α is a homotopy equivalence. A j X α A X, Proo. We may as well assume that X is a mapping cylinder M and that j is the inclusion at time 1. Let g be a homotopy inverse or. Up to homeomorphism, we may replace X by the double mapping cylinder M A A I A A and the map α by the inclusion (this is homotopic to the original map). We deine a map β : X M in the other direction by the identity map on the subspace M o X and by using a homotopy h : id A g on the rest. Then β α = id on the nose, and α β id by irst using a homotopy to pull the image o A into A and then using a homotopy g id. 2 X

3 Example 1.7. Here s an easy example to see that we need to assume one o the maps is a coibration: S n D n+1 D n+1 S n+1 Here we have homotopy equivalences that do not produce a homotopy equivalence on the pushout. 2. Wed, Feb. 23 Last time, we showed that as long as one o the maps B A X is a coibration, then the pushout B A X is equivalent to the double mapping cylinder (the homotopy pushout). Here is another situation in which one can deduce the that the pushout is a homotopy pushout. First recall that a triple (X; A, B) consisting o a space X and subspaces A and B is called an excisive triad i X = inta intb. Theorem 2.1. Let (X; A, B) be an excisive triad and let C = A B. Then the square is a homotopy pushout square. See [May, 10.7] or a proo. Relation o coiber sequences to iber sequences Let : A X be a based map. Then we can orm the coiber sequence Consider the diagram p C A B X A X i C(). F () A X i C() ϕ ΩC() F (i) The let vertical map (ϕ) is speciied by sending (γ, a) to the loop in C() given on the irst hal o the interval by γ 1 and on the second hal by the path t (a, 2t 1). The second vertical map sends a to (δ a, (a)), where δ a is the path t (a, t). The square does not commute on the nose, but it commutes up to homotopy (the pair (δ a γ 1, ) can be contracted to (δ a, (a)) by moving the initial point along γ 1 ). The map ϕ is usually not an equivalence, but we will see later that i A and X are both n- connected, then ϕ induces an isomorphism on π n. We will also see that the analogue o this map in spectra is an equivalence. Example 2.2. Recall the Hop map η : S 3 S 2. In homework 3, you show that F (η) S 1. What is C(η)? Certainly η is not a coibration, so we should irst convert it into a coibration to compute the coiber. Since the coiber is the homotopy pushout o the maps S 3 S 2, by what we have said previously, we may equally well replace the map S 3 by a coibration, namely the coibration S 3 D 4. That is, the coiber is the result o attaching a 4-cell to S 2 via η. This is precisely the CW structure on CP 2 that we mentioned earlier. 3 S n

4 Here is a careul description o this CW structure, rom Hatcher. We usually write CP 2 as the quotient o S 5 by the action o S 1 (complex multiplication). But every point o S 5 C 3 can be identiied, under this action o S 1, with a point (x, y, z) such that the third complex coordinate z is in act real and 0. I z 0, there is a unique such point, whereas i z = 0, then the whole S 1 orbit o (x, y, z) is o this orm. This subspace o S 5 is a hemisphere o S 4 S 5, so it is D 4. Furthermore, we can now write CP 2 as a quotient o D 4 by an action o S 1 on the boundary S 3. This identiication on the boundary is precisely η : S 3 S 2. There is another way to identiy the coiber o η as CP 2, using that excisive triads produce homotopy pushouts. We may replace the diagram S 3 S 2 up to homotopy by the excisive triad diagram V {[0 : 0 : 1]} CP 2 {[0 : 0 : 1]} V, where V CP 2 is the open subset V = {[x : y : z] z 0}. The open subsets V and CP 2 {[0 : 0 : 1]} certainly cover CP 2, so this gives an excisive triad. The subset V is homeomorphic to C 2 (send [x : y : z] to (x/z, y/z)) and is thereore contractible. It also ollows that V {[0 : 0 : 1]} S 3. Finally, the map CP 2 {[0 : 0 : 1]} CP 1 sending [x : y : z] to [x : y] is well-deined since x and y cannot both be zero. The inclusion CP 1 CP 2 {[0 : 0 : 1]} deined by [x : y] [x : y : 0] is a homotopy inverse. So we have identiied the coiber o η with the space CP 2. The comparison map S 1 F (η) ΩC(η) ΩCP 2 cannot be an equivalence: the exact sequence π 5 (S 1 ) π 5 (S 5 ) π 5 (CP 2 )π 4 (S 1 ) shows that π 5 (CP 5 ) = Z, so that π 4 (ΩCP 5 ) = Z, whereas π 4 (S 1 ) = 0. Fibrations & iber sequences We have discussed both (homotopy) ibers and coibers, and we have seen that the coiber o a coibration is just the poin-set quotient. There is an analogue o this statement or the iber o a ibration. The notion o a ibration is completely dual to that o a coibration: Deinition 2.3. A map p : E B is a (Hurewicz) ibration i or every space X, map : X E and homotopy h : X I B with h 0 = p, there is an extension h : X I E. In other words, we get a lit in the diagram X E h p X I h B. We also say that the map p satisies the homotopy liting property (or covering homotopy property). Remark 2.4. There is an important generalization o ibration, called a Serre ibration, in which the map p : E B is only asked to satisy the homotopy liting property with respect to spaces X = I n. We will come back to this later. As or coibrations, there is a universal diagram to check. The pair o maps X E and X h B I that make the diagram commute correspond to a single map X E B B I. This is the 4

5 unbased path space construction on the map p, and we will write P (p) or F (p) or this as well. We can see that any liting diagram actors as X P (p) E X I P (p) I B, and inding a lit in the square on the right will give a lit in the big rectangle. 3. Fri, Feb. 25 Proposition 3.1. The map p : E B is a ibration i and only i the natural map k : E I P (p) has a splitting s : P (p) E I so that k s = id P (p). We think o the section s as a path-liting unction or the ibration p. Proposition 3.2. The class o ibrations is closed under (1) composition, (2) pullbacks, (3) products, (4) retracts, and (5) sequential inverse limits. I : E B is any map, we may actor this as E j P () q B. The map q evaluates the path in B at time 1, and j(e) = (e, c (e) ). Proposition 3.3. For any map, the map E j P () is a homotopy equivalence and P () q B is a ibration. Proo. Projection p 1 onto the E actor gives a map p 1 : P () E so that p 1 j = id. The other composition j p 1 id via the homotopy that contracts paths starting at (e) to the constant path at (e). To see that q is a ibration, consider a test diagram (g 1,g 2 ) X P () q X I B. We deine the lit h by the ormula h(x, t) = (g 1 (x), h(x) [0,t] g 2 (x)), h where we renormalize the path h(x) [0,t] g 2 (x) so that we travel along g 2 (x) or time 0 to 1 1+t 1 and along the h(x) [0,t] or time 1+t to 1. Since the initial square commutes, we know that h(x, 0) = g 2 (x)(1), so that it makes sense to compose the paths h(x) [0,t] and g 2 (x)). Our ormula or h(x, t) is a point o P () since the path h(x) [0,t] g 2 (x) begins at g 2 (x). The upper triangle then commutes because concatenating g 2 (x) with the constant path, renormalized to have length zero, just gives g 2 (x) again. The lower triangle commutes because the endpoint o the path h(x) [0,t] g 2 (x) is h(x, t). So any map may be replaced by a ibration by changing the domain up to homotopy equivalence. Examples. (1) For any space X, the map X is a ibration (we say that the space X is ibrant) 5 p

6 (2) Any covering space E B is a ibration. In act, or a covering space the lit h is uniquely determined. (3) Any vector bundle E B is a ibration i the base B is paracompact (essentially means can ind partitions o unity or any cover; any CW complex is paracompact). Recall that a rank n (real) vector bundle over B is a map p : E B such that there exists a cover U o B and a homeomorphism ϕ U : p 1 (U) = U R n compatible with the projections to U. This is subject to the compatibility condition that i U and V are elements o the cover, then the transition unction g U,V deined as the composition (U V ) R n ϕ 1 U p 1 (U V ) ϕ V (U V ) R n restricts, or each x U V to a linear isomorphism R n = R n o ibers. Without the assumption that the base is paracompact, we can still deduce that a vector bundle is a Serre ibration. (4) More generally, any iber bundle in which the base space is paracompact is a ibration. A map p : E B is a iber bundle with iber F i B has a cover U and or each U U a homeomorphism ϕ U : p 1 (U) = U F over U. Without the paracompact hypothesis on the base, we can again say that any bundle is at least a Serre ibration. There is also an intermediate notion o G-bundle, where a group G acts on the iber F, and the transition unctions are assumed to arise rom the action o G on F. Proposition 3.4. A iber bundle p : E B is a Serre ibration. Proo. Let p : E B be a iber bundle, and suppose given a a liting diagram I n E p I n I h B. Let U be a covering or B on which we have trivializations or the bundle. Since I n I is compact, we may divide I n into subcubes C and I into subintervals J, such that each C J is contained in a single h 1 (U). For each C, we build a lit on each C J, starting with the J containting 0, and working our way up in the I coordinate. So by assumption, we have a lit along the initial point o our interval, which we may take to be 0 or simplicity. Moreover, this cube C borders other cubes, and we may have already constructed lits on the other cubes. So we suppose that or some union D o aces o C, we already have a lit along D I. For our ixed C and J, the iber bundle becomes the trivial iber bundle U F U. A lit in the diagram C U F h C J U must be given in the irst coordinate by the map h, so it remains to describe the second coordinate o the lit. By assumption, we already have a lit C D D J U F F. But the space C D D J is a retract o C J, so we may compose the lit we already have with a choice o retraction C J r C D D J. 6