SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS

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1 SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS In this section we will introduce two imortant classes of mas of saces, namely the Hurewicz fibrations and the more general Serre fibrations, which are both obtained by imosing certain homotoy lifting roerties. We will see that u to homotoy equivalence every ma is a Hurewicz fibration. Moreover, associated to a Serre fibration we obtain a long exact sequence in homotoy which relates the homotoy grous of the fibre, the total sace, and the base sace. This sequence secializes to the long exact sequence of a air which we already discussed in the revious lecture. Definition 1. (1) A ma : E X of saces is said to have the right lifting roerty (RLP) with resect to a ma i: A B if for any two mas f : A E and g : B X with f = gi, there exists a ma h: B E with h = g and hi = f: A f E i h B g X (So h at the same time extends f and lifts g.) (2) A ma : E B of saces is a Serre fibration if it has the RLP with resect to all inclusions of the form I n {0} I n I = I n+1, n 0, and a Hurewicz fibration if it has the RLP with resect to all mas of the form A I for any sace A. (So evidently, every Hurewicz fibration is a Serre fibration.) (3) If : E X is a ma of saces (but tyically one of the two kinds of fibrations) and x X, then 1 (x) E is called the fiber of over x. If x = x 0 is a base oint secified earlier, we just say the fiber of for the fiber over x 0. Thus, Hurewicz fibrations are those mas : E X which have the homotoy lifting roerty with resect to all saces: given a homotoy H : A I X of mas with target X and a lift G 0 : A E of H 0 = H(, 0): A X against the fibration : E X then this artial lift can be extended to a lift of the entire homotoy G: A I E, i.e., G satisfies G = H and Gi = G 0 : G 0 E i G A I X By definition, the class of Serre fibrations is given by those mas which have the homotoy lifting roerty with resect to all cubes I n. Examle 2. H (1) Any rojection X F X is a Hurewicz fibration, as you can easily check. 1

2 2 SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS (2) The evaluation ma ɛ = (ɛ 0, ɛ 1 ): X I X X at both end oints is a Hurewicz fibration. Indeed, suose we are given a commutative square Or equivalently, we are given a ma X I i ɛ h A I X X g φ: ( I) (A I {0, 1}) X f which we wish to extend to A I I. But ({0} I) (I {0, 1}) = J 1 I 2 is a retract of I 2, and hence so is A J 1 A I 2. Therefore we can simly recomose φ with the retraction r : A I 2 A J 1 to find the required extension. (3) Let : E X and f : X X be arbitrary mas. Form the fibered roduct or ullback E X X = {(e, x ) (e) = f(x )} toologized as a subsace of the roduct E X. Then if E X is a Hurewicz (or Serre) fibration, so is the induced rojection E X X X. This follows easily from the universal roerty of the ullback (see Exercise 1). (4) If E D X are two Hurewicz (or Serre) fibrations, then so is their comosition E X (see Exercise 2). (5) Let (X, x 0 ) be a ointed sace. The ath sace P (X) (or more recisely, P (X, x 0 ) if necessary) is the subsace of X I (always with the comact-oen toology) given by aths α with α(0) = x 0. The ma ɛ 1 : P (X) X given by evaluation at 1 is a Hurewicz fibration. This follows by combining the revious examles (2) and (3). (6) If f : Y X is any ma, the maing fibration of f is the ma : P (f) X constructed as follows. The sace P (f) is the fibered roduct P (f) = X I X Y = {(α, y) α(1) = f(y)} and the ma is given by (α, y) = α(0). We claim that is a Hurewicz fibration. Indeed, suose we are given a commutative diagram i A I u v P (f) X

3 SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS 3 Denoting by π 1 : P (f) X I and π 2 : P (f) Y the two rojection mas belonging to the ullback P (f), we can first extend π 2 u as in: π2 u Y i ũ A I and then use examle (2) to find a diagonal as in π 1 u X I w A I X X (v,f ũ) ɛ=(ɛ 0,ɛ 1) Then (w, ũ): A I P (f) is a diagonal filling in the original diagram we are looking for. In fact, it lands in P (f) because ɛ 1 w = fũ, and makes the diagram commute because w = ɛ 0 w = v and (w, ũ)i = (π 1 u, π 2 u) = u. Here is a more abstract way of roving this: we showed in examle (2) that ɛ: X I X X is a Hurewicz fibration. Hence, by Exercise 1, so is the ullback Q = (X Y ) (X X) X I in: Q X Y X I ɛ id f X X But then, by examle (1) and Exercise 2, the comosition Q X Y X is a Hurewicz fibration as well. Now note that Q P (f): (α, x, y) (α, y) defines a homeomorhism and that under this homeomorhism the two mas Q X and P (f) X are identified. Thus also the ma P (f) X is a Hurewicz fibration. Exercise 3. Prove that the ma φ in Y φ P (f) f X given by φ(y) = (κ f(y), y) is a homotoy equivalence which makes the diagram commute. Here, as usual, we denote by κ f(y) the constant ath at f(y). This exercise thus shows that every ma is homotoy equivalent to a Hurewicz fibration. Motivated by this, one calls the fiber of over x, 1 (x) = {(α, y) α(1) = f(y), α(0) = x}

4 4 SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS the homotoy fiber of f over x. Note that there is a canonical ma from the fiber of f over x to the homotoy fiber of f over x given by a restriction of φ, namely: f 1 (x) 1 (x): y (κ x, y) Thus, the fiber sits in the homotoy fiber while the homotoy fiber can be thought of as a relaxed version of the fiber: the condition imosed on a oint y Y to lie in the fiber over x is that it has to be maed to x by f, i.e., f(y) = x, while a oint of the homotoy fiber is a air (α, y) consisting of y Y together with a ath α in X witnessing that y lies in the fiber u to homotoy. So far, all our examles are examles of Hurewicz fibrations. However, we will see in the next lecture that the weaker roerty of being a Serre fibration is a local roerty, and hence that all fiber bundles are examles of Serre fibrations. Moreover, this weaker notion suffices to establish the following theorem. Theorem 4. (The long exact sequence of a Serre fibration) Let : (E, e 0 ) (X, x 0 ) be a ma of ointed saces with i: (F, e 0 ) (E, e 0 ) being the fiber. Suose that is a Serre fibration. Then there is a long exact sequence of the form:... π n+1 (X, x 0 ) δ π n (F, e 0 ) i π n (E, e 0 ) π n (X, x 0 ) δ... i π 0 (E, e 0 ) π 0 (X, x 0 ) The connecting homomorhism δ will be constructed exlicitly in the roof. Before turning to the roof, let us deduce an immediate corollary. By considering the homotoy fiber H f instead of the actual fiber, we see that we can obtain a long exact sequence for an arbitrary ma f of ointed saces. Corollary 5. Let f : (Y, y 0 ) (X, x 0 ) be a ma of ointed saces and let H f be its homotoy fiber. Then there is a long exact sequence of the form:... π n+1 (X, x 0 ) π n (H f, ) π n (Y, y 0 ) f π n (X, x 0 )... π 0 (Y, y 0 ) f π 0 (X, x 0 ) Proof. Aly Theorem 4 to the maing fibration (Examle 2.(6)) and use Exercise 3. Before entering in the roof of the theorem, we recall from the revious lecture the definition of the subsace J n I n+1, J n = (I n {0}) ( I n I) I n+1 I n+1. Note that, by flattening the sides of the cube, one can construct a homeomorhism of airs (I n+1, J n ) = (I n+1, I n {0}). Thus, any Serre fibration also has the RLP with resect to the inclusion J n I n+1. We will use this reeatedly in the roof. Proof. (of Theorem 4) The main art of the roof consists in the construction of the oeration δ. Let α: (I n, I n ) (X, x 0 ) reresent an element of π n (X, x 0 ) = π n (X). Let ē 0 : J n 1 E be the constant ma with value e 0. Then the square J n 1 ē 0 E β X I n α

5 SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS 5 commutes, so by the definition of a Serre fibration we find a diagonal β. Then δ[α] is the element of π n 1 (F ) reresented by the ma β(, 1): I n 1 F, t β(t, 1). Note that this indeed reresents an element of π n 1 (F ), because the boundary of I n 1 {1} is contained in J n 1, and β mas the to face I n 1 {1} into F since β = α mas it to x 0. The first thing to check is that δ is well defined on homotoy classes. Suose [α 0 ] = [α 1 ], as witnessed by a homotoy h: I n I X from α 0 to α 1. Suose also that we have chosen liftings β 0 and β 1 of α 0 and α 1 as above. Then we can define a ma k making the solid square J n k E l I n I X commute. Here Jn is the union of all the faces of I n+1 excet {t n = 1}. (It is like J n excet that we have interchanged the roles of t n and t n+1.) On I n {0} and I n {1} the ma k is defined to be β 0 and β 1 resectively. On the faces {t i = 0}, {t i = 1} (i < n) and {t n = 0} the ma k has constant value e 0. Now a diagonal l restricted to I n 1 {1} I gives a homotoy from β 0 (, 1) to β 1 (, 1), and lies entirely in the fiver over x 0 because h is a homotoy relative to I n. This roves that β 0 (, 1) and β 1 (, 1) define the same element of π n 1 (F ). It also roves that δ[α] thus defined does not deend on the choice of the filling β (Why?). With these details about δ being well-defined out of the way, it is quite easy to rove that the sequence of the theorem is an exact sequence of ointed sets. (We write ointed sets here because we haven t roved yet that δ is a homomorhism of grous for n > 1. We leave this to you as Exercise 7.) Exactness at π n (E). Clearly i = 0 because i is constant, so im(i ) ker( ). For the reverse inclusion, suose α: I n E reresents an element of π n (E) with [α] = [ α] = 0. Let h: I n I X be a homotoy rel I n from α to the constant ma on x 0. Choose a lift l in h J n k E l I n I X where k I n {0}= α and k is constant e 0 on the other faces. Then γ = l I n {1} mas entirely into F, so reresents an element [γ] π n (F ) with i [γ] = [iγ] = [α] (by the homotoy l). Exactness at π n (X). If β : I n E reresents an element of π n (E) then for α = β we can take the same β as the diagonal filling in the construction of δ[α]. So δ [β] = β I n 1 {1} which is constant e 0. Thus δ = 0, or im( ) ker(δ). For the converse inclusion, suose α: I n X reresents an element of π n (X) with δ[α] = 0. Then for a lift β as in h J n 1 ē 0 E β X I n α

6 6 SECTION 5: FIBRATIONS AND HOMOTOPY FIBERS we have that β(, 1) is homotoic to the constant ma by a homotoy h relative to I n 1 which mas into the fiber F. But then, stacking this homotoy h on to of β (i.e., by forming h n β), we obtain a ma reresenting an element β of π n (E). The image β is obviously homotoic to β = α because h is constant, showing that [α] lies in the image of. Exactness at π n 1 (F ). For α: I n X reresenting an element of π n (X), the ma β in the construction of δ[α] = [β(, 1)] shows that β(, 1) β(, 0) = ē 0 in E, so i δ[α] = 0. For the other inclusion, suose γ : I n 1 F reresents an element of π n 1 (F ) with i [γ] = 0, as reresented by a homotoy h: I n 1 I E with h(, 1) = γ and h(, 0) = ē 0. Then α = h reresents an element of π n (X), and in the construction of δ[α] we can choose the diagonal filling β to be identical to h, in which case δ[α] is reresented by γ. This shows that ker(i ) im(δ), and comletes the roof of the theorem. Exercise 6. Show that the long exact sequence of a ointed air (X, A), constructed in the revious lecture, can be obtained from this long exact sequence, by considering the maing fibration of the inclusion A X (see also the last exercise sheet). Exercise 7. Prove that the connecting homomorhism δ : π n (X, x 0 ) π n 1 (F, e 0 ) is a homomorhism of grous for n 2. Exercise 8. Let : E X be a Hurewicz fibration, and let α: I X be a ath from x to y. Use the lifting roerty of E X with resect to 1 (x) {0} 1 (x) I to show that α induces a ma α : 1 (x) 1 (y). Show that the homotoy class of α only deends on the homotoy class of α, and that this construction in fact defines a functor on the fundamental grouoid, π(x) Ho(To). This last exercise shows in articular that the homotoy tye of the fiber of a Hurewicz fibration is constant on ath comonents. More recisely, if : E X is a Hurewicz fibration, then any ath α: I X induces a homotoy equivalence between the fiber over α(0) and the fiber over α(1).