LECTURE 5: FIBRATIONS AND HOMOTOPY FIBERS

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1 LECTURE 5: FIBRATIONS AND HOMOTOPY FIBERS In ts lecture we wll ntroduce two mortant classes of mas of saces, namely te Hurewcz fbratons and te more general Serre fbratons, wc are bot obtaned by mosng certan omotoy lftng roertes. We wll see tat u to omotoy equvalence every ma s a Hurewcz fbraton. Moreover, assocated to a Serre fbraton we obtan a long exact sequence n omotoy wc relates te omotoy grous of te fbre, te total sace, and te base sace. Ts sequence secalzes to te long exact sequence of a ar wc we already dscussed n te revous lecture. Defnton Fbratons () A ma : E of saces s sad to ave te rgt lftng roerty (RLP) wt resect to a ma : A B f for any two mas f : A E and g : B wt f = g, tere exsts a ma : B E wt = g and = f: A f E B g (So at te same tme extends f and lfts g.) () A ma : E of saces s a Serre fbraton f t as te RLP wt resect to all nclusons of te form I n {0} I n I = I n+1, n 0, and a Hurewcz fbraton f t as te RLP wt resect to all mas of te form for any sace A. (So evdently, every Hurewcz fbraton s a Serre fbraton.) () If : E s a ma of saces (but tycally one of te two knds of fbratons) and x, ten 1 (x) E s called te fber of over x. If x = x 0 s a base ont secfed earler, we just say te fber of for te fber over x 0. Tus, Hurewcz fbratons are tose mas : E wc ave te omotoy lftng roerty wt resect to all saces: gven a omotoy H : of mas wt target and a lft G 0 : A E of H 0 = H(, 0): A aganst te fbraton : E ten ts artal lft can be extended to a lft of te entre omotoy G: E,.e., G satsfes G = H and G = G 0 : 1 G 0 G H E

2 2 LECTURE 5: FIBRATIONS AND HOMOTOPY FIBERS By defnton, te class of Serre fbratons s gven by tose mas wc ave te omotoy lftng roerty wt resect to all cubes I n. Examle 5.2. () Any rojecton F s a Hurewcz fbraton, as you can easly ceck. () Te evaluaton ma ɛ = (ɛ 0, ɛ 1 ): I at bot end onts s a Hurewcz fbraton. Indeed, suose we are gven a commutatve square Or equvalently, we are gven a ma g f I ɛ φ: ( I) ( {0, 1}) wc we ws to extend to I. But ({0} I) (I {0, 1}) = J 1 I 2 s a retract of I 2, and ence so s A J 1 2. Terefore we can smly recomose φ wt te retracton r : 2 A J 1 to fnd te requred extenson. () Let : E and f : be arbtrary mas. Form te fbered roduct or ullback E = {(e, x ) (e) = f(x )} toologzed as a subsace of te roduct E. Ten f E s a Hurewcz (or Serre) fbraton, so s te nduced rojecton E. Ts follows easly from te unversal roerty of te ullback (see Exercse 1). (v) If E D are two Hurewcz (or Serre) fbratons, ten so s ter comoston E (see Exercse 2). (v) Let (, x 0 ) be a onted sace. Te at sace P () (or more recsely, P (, x 0 ) f necessary) s te subsace of I (always wt te comact-oen toology) gven by ats α wt α(0) = x 0. Te ma ɛ 1 : P () gven by evaluaton at 1 s a Hurewcz fbraton. Ts follows by combnng te revous examles () and (). (v) If f : Y s any ma, te mang fbraton of f s te ma : P (f) constructed as follows. Te sace P (f) s te fbered roduct P (f) = I Y = {(α, y) α(1) = f(y)} and te ma s gven by (α, y) = α(0). We clam tat s a Hurewcz fbraton. Indeed, suose we are gven a commutatve dagram u v P (f)

3 LECTURE 5: FIBRATIONS AND HOMOTOPY FIBERS 3 Denotng by π 1 : P (f) I and π 2 : P (f) Y te two rojecton mas belongng to te ullback P (f), we can frst extend π 2 u as n: π2 u Y and ten use examle () to fnd a dagonal as n π 1 u w (v,f ũ) ũ I ɛ=(ɛ 0,ɛ 1) Ten (w, ũ): P (f) s a dagonal fllng n te orgnal dagram we are lookng for. In fact, t lands n P (f) because ɛ 1 w = fũ, and makes te dagram commute because w = ɛ 0 w = v and (w, ũ) = (π 1 u, π 2 u) = u. Here s a more abstract way of rovng ts: we sowed n () tat ɛ: I s a Hurewcz fbraton. Hence, by (), so s te ullback Q = ( Y ) ( ) I n: Q Y I ɛ d f But ten, by () and (v), te comoston Q Y s a Hurewcz fbraton as well. Now note tat Q P (f): (α, x, y) (α, y) defnes a omeomorsm and tat under ts omeomorsm te two mas Q and P (f) are dentfed. Tus also te ma P (f) s a Hurewcz fbraton. Exercse 5.3. Prove tat te ma φ n Y φ f P (f) gven by φ(y) = (κ f(y), y) s a omotoy equvalence wc makes te dagram commute. Here, as usual, we denote by κ f(y) te constant at at f(y). Ts exercse tus sows tat every ma s omotoy equvalent to a Hurewcz fbraton. Motvated by ts, one calls te fber of over x, 1 (x) = {(α, y) α(1) = f(y), α(0) = x} te omotoy fber of f over x. Note tat tere s a canoncal ma from te fber of f over x to te omotoy fber of f over x gven by a restrcton of φ, namely: f 1 (x) 1 (x): y (κ x, y)

4 4 LECTURE 5: FIBRATIONS AND HOMOTOPY FIBERS Tus, te fber sts n te omotoy fber wle te omotoy fber can be tougt of as a relaxed verson of te fber: te condton mosed on a ont y Y to le n te fber over x s tat t as to be maed to x by f,.e., f(y) = x, wle a ont of te omotoy fber s a ar (α, y) consstng of y Y togeter wt a at α n wtnessng tat y les n te fber u to omotoy. 2. Te long exact sequence assocated to a fbraton So far, all our examles are examles of Hurewcz fbratons. However, we wll see n te next lecture tat te weaker roerty of beng a Serre fbraton s a local roerty, and ence tat all fber bundles are examles of Serre fbratons. Moreover, ts weaker noton suffces to establs te followng teorem. Teorem 5.4. (Te long exact sequence of a Serre fbraton) Let : (E, e 0 ) (, x 0 ) be a ma of onted saces wt : (F, e 0 ) (E, e 0 ) beng te fber. Suose tat s a Serre fbraton. Ten tere s a long exact sequence of te form:... π n+1 (, x 0 ) δ π n (F, e 0 ) π n (E, e 0 ) π n (, x 0 ) δ... π 0 (E, e 0 ) π 0 (, x 0 ) Te connectng omomorsm δ wll be constructed exlctly n te roof. Before turnng to te roof, let us deduce an mmedate corollary. By consderng te omotoy fber H f nstead of te actual fber, we see tat we can obtan a long exact sequence for an arbtrary ma f of onted saces. Corollary 5.5. Let f : (Y, y 0 ) (, x 0 ) be a ma of onted saces and let H f be ts omotoy fber. Ten tere s a long exact sequence of te form:... π n+1 (, x 0 ) π n (H f, ) π n (Y, y 0 ) f π n (, x 0 )... π 0 (Y, y 0 ) f π 0 (, x 0 ) Proof. Aly Teorem 5.4 to te mang fbraton (Examle 5.2(v)) and use Exercse 3. Before enterng n te roof of te teorem, we recall from te revous lecture te defnton of te subsace J n I n+1, J n = (I n {0}) ( I n I) I n+1 I n+1. Note tat, by flattenng te sdes of te cube, one can construct a omeomorsm of ars (I n+1, J n ) = (I n+1, I n {0}). Tus, any Serre fbraton also as te RLP wt resect to te ncluson J n I n+1. We wll use ts reeatedly n te roof. Proof of Teorem 5.4. Te man art of te roof conssts n te constructon of te oeraton δ. Let α: (I n, I n ) (, x 0 ) reresent an element of π n (, x 0 ) = π n (). Let ē 0 : J n 1 E be te constant ma wt value e 0. Ten te square J n 1 ē 0 E I n commutes, so by te defnton of a Serre fbraton we fnd a dagonal β. Ten δ[α] s te element of π n 1 (F ) reresented by te ma α β β(, 1): I n 1 F, t β(t, 1).

5 LECTURE 5: FIBRATIONS AND HOMOTOPY FIBERS 5 Note tat ts ndeed reresents an element of π n 1 (F ), because te boundary of I n 1 {1} s contaned n J n 1, and β mas te to face I n 1 {1} nto F snce β = α mas t to x 0. Te frst tng to ceck s tat δ s well defned on omotoy classes. Suose [α 0 ] = [α 1 ], as wtnessed by a omotoy : I n I from α 0 to α 1. Suose also tat we ave cosen lftngs β 0 and β 1 of α 0 and α 1 as above. Ten we can defne a ma k makng te sold square J n I n I k l E commute. Here J n s te unon of all te faces of I n+1 excet {t n = 1}. (It s lke J n excet tat we ave ntercanged te roles of t n and t n+1.) On I n {0} and I n {1} te ma k s defned to be β 0 and β 1 resectvely. On te faces {t = 0}, {t = 1} ( < n) and {t n = 0} te ma k as constant value e 0. Now a dagonal l restrcted to I n 1 {1} I gves a omotoy from β 0 (, 1) to β 1 (, 1), and les entrely n te fber over x 0 because s a omotoy relatve to I n. Ts roves tat β 0 (, 1) and β 1 (, 1) defne te same element of π n 1 (F ). It also roves tat δ[α] tus defned does not deend on te coce of te fllng β (Wy?). Wt tese detals about δ beng well-defned out of te way, t s qute easy to rove tat te sequence of te teorem s an exact sequence of onted sets. (We wrte onted sets ere because we aven t roved yet tat δ s a omomorsm of grous for n > 1. We leave ts to you as Exercse 5.7.) Exactness at π n (E). Clearly = 0 because s constant, so m( ) ker( ). For te reverse ncluson, suose α: I n E reresents an element of π n (E) wt [α] = [ α] = 0. Let : I n I be a omotoy rel I n from α to te constant ma on x 0. Coose a lft l n J n k E I n I were k In {0}= α and k s constant e 0 on te oter faces. Ten γ = l In {1} mas entrely nto F, so reresents an element [γ] π n (F ) wt [γ] = [γ] = [α] (by te omotoy l). Exactness at π n (). If β : I n E reresents an element of π n (E) ten for α = β we can take te same β as te dagonal fllng n te constructon of δ[α]. So δ [β] = β I n 1 {1} wc s constant e 0. Tus δ = 0, or m( ) ker(δ). For te converse ncluson, suose α: I n reresents an element of π n () wt δ[α] = 0. Ten for a lft β as n l J n 1 ē 0 E I n α we ave tat β(, 1) s omotoc to te constant ma by a omotoy relatve to I n 1 wc mas nto te fber F. But ten, stackng ts omotoy on to of β (.e., by formng n β), we obtan a ma reresentng an element β of π n (E). Te mage β s obvously omotoc to β = α because s constant, sowng tat [α] les n te mage of. β

6 6 LECTURE 5: FIBRATIONS AND HOMOTOPY FIBERS Exactness at π n 1 (F ). For α: I n reresentng an element of π n (), te ma β n te constructon of δ[α] = [β(, 1)] sows tat β(, 1) β(, 0) = ē 0 n E, so δ[α] = 0. For te oter ncluson, suose γ : I n 1 F reresents an element of π n 1 (F ) wt [γ] = 0, as reresented by a omotoy : I n 1 I E wt (, 1) = γ and (, 0) = ē 0. Ten α = reresents an element of π n (), and n te constructon of δ[α] we can coose te dagonal fllng β to be dentcal to, n wc case δ[α] s reresented by γ. Ts sows tat ker( ) m(δ), and comletes te roof of te teorem. Exercse 5.6. Sow tat te long exact sequence of a onted ar (, A), constructed n te revous lecture, can be obtaned from ts long exact sequence, by consderng te mang fbraton of te ncluson A (see also te last exercse seet). Exercse 5.7. Prove tat te connectng omomorsm δ : π n (, x 0 ) π n 1 (F, e 0 ) s a omomorsm of grous for n 2. Exercse 5.8. Let : E be a Hurewcz fbraton, and let α: I be a at from x to y. Use te lftng roerty of E wt resect to 1 (x) {0} 1 (x) I to sow tat α nduces a ma α : 1 (x) 1 (y). Sow tat te omotoy class of α only deends on te omotoy class of α, and tat ts constructon n fact defnes a functor on te fundamental grouod, π() Ho(To). Ts last exercse sows n artcular tat te omotoy tye of te fber of a Hurewcz fbraton s constant on at comonents. More recsely, f : E s a Hurewcz fbraton, ten any at α: I nduces a omotoy equvalence between te fber over α(0) and te fber over α(1).