SECTION 12: HOMOTOPY EXTENSION AND LIFTING PROPERTY

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1 SECTION 12: HOMOTOPY EXTENSION AND LIFTING PROPERTY In the revious section, we exloited the interlay between (relative) CW comlexes and fibrations to construct the Postnikov and Whitehead towers aroximating a given sace. In this context, K(π, n)-saces naturally came u. In this lecture, we ll give a few of the most elementary examles of K(π, n)-saces, and study the interlay between relative CW comlexes and fibrations. Remark 1. Recall from the revious lecture that a K(π, n)-sace, or an Eilenberg-MacLane sace of tye (π, n), is a sace (X, x 0 ) such that π i (X, x 0 ) = for all i n together with an isomorhism π n (X, x 0 ) = π. Here π can be a ointed set if n = 0, a grou is n = 1, or an abelian grou if n 2. It can be shown that for such a π, a K(π, n)-sace always exists, and is unique u to homotoy. We will not give the general construction in this lecture, but restrict ourselves to the discussion of some examles. Examle 2. (Examles of K(π, n)-saces) (1) The circle S 1 is a K(Z, 1)-sace. Indeed, it is a connected sace with fundamental grou Z, and one way to see that the higher homotoy grous vanish is to consider the universal covering sace R S 1. This is a fiber bundle with discrete fiber F and contractible total sace, so the long exact sequence gives us isomorhisms 0 = π i (F ) = π i+1 (S 1 ) for i > 0. (2) The same argument alies to wedges of sheres. Consider for examle the figure eight S 1 S 1. Its fundamental grou is the free grou on two generators Z Z. The universal cover of S 1 S 1 can be exlicitly described in terms of the grid in the lane, G = (Z R) (R Z) R 2. The ma w : G S 1 S 1 can be described by wraing each edge of length 1 in the grid around one of the circles (in a way resecting orientations): say the vertical edges to the left hand circle and the horizontal edges to the right hand one. The universal cover E of S 1 S 1 is the sace of homotoy classes of aths in G which start in the origin, and E S 1 S 1 is the comosition E ɛ1 G w S 1 S 1 (where ɛ 1 is evaluation at the endoint). The fiber of ɛ 1 : E G over a given grid oint (n, m) with n, m Z is the set of combinatorial aths from (0, 0) to (n, m): a sequence of alternating decisions: go left or go right, go u or go down, where successions of u-down and left-right cancel each other. Since each homotoy class of aths in E has a unique such combinatorial descrition, the sace E is clearly contractible. (3) Recall that RP n, the real rojective sace of dimension n, is the sace of lines in R n+1. It can be constructed as S n / Z 2 where the grou Z 2 = {0, 1} acts by the antiodal ma on the unit shere S n = {(x 0,..., x n ) R n+1 x x 2 n = 1}. The embedding S n S n+1 sending (x 0,..., x n ) to (x 0,..., x n, 0) sends S n to the equator inside S n+1, and is comatible with this antiodal action so that we get a commutative 1

2 2 SECTION 12: HOMOTOPY EXTENSION AND LIFTING PROPERTY diagram S 0 S 1 S 2... RP 0 RP 1 RP 2... There is a natural CW decomosition of S n+1, given inductively by a CW decomosition of S n with two (n + 1)-cells attached to it: thorthern and the southern hemisheres. This makes S n into a CW comlex with exactly two k-cells in each dimension k n. One can also take the union along the uer row of the diagram (with the weak toology) to obtain the infinite-dimensional shere S = n S n, a CW comlex with exactly two n-cells in each dimension n. Note that since π i (S n ) = 0 for i < k we also obtain π i (S ) = 0 for all i 0. In other words, S is a weakly contractible CW comlex, and hence by Whitehead s theorem is contractible. In a similar way, we can take the union along the lower row in the above diagram to obtain RP = n RP n, a CW comlex, the infinite-dimensional real rojective sace, with exactly on-cell in each dimension n. The long exact sequence of the covering rojection S n RP n with discrete fiber Z 2 shows that π i (RP n ) = 0, 1 < i < n, or i = 0, π 1 (RP n ) = Z 2, and by assing to the limit, one concludes that RP is a K(Z 2, 1)-sace. (Alternatively, one can show that S RP is still a covering rojection with fiber Z 2 to conclude that RP is a K(Z 2, 1).) (4) Recall that CP n, the comlex rojective sace of (comlex) dimension n, is the sace of (comlex) lines in C n+1. It can be constructed as (C n+1 {0})/ C where C = C {0} acts by multilication; or, by choosing oints on the line of norm 1, as the quotient of the unit shere in C n+1, CP n = S 2n+1 /S 1, where S 1 C again acts by multilication. The quotient S 2n+1 CP n has enough local sections (check this!), hence is a fiber bundle with fiber S 1. The embedding induces mas C n+1 C n+2 : (z 0,..., z n ) (z 0,..., z n, 0) S 2n+1 S 2n+3 CP n CP n+1 and one can again take the union, to obtain a ma S CP with CP the infinitedimensional comlex rojective sace. The sace CP is a quotient of S by S 1, and the ma is again a fiber bundle. The saces CP n have comatible CW comlex structures,

3 SECTION 12: HOMOTOPY EXTENSION AND LIFTING PROPERTY 3 given by exactly one k-cell in each dimension k n. One way to see this is to reresent a line in C n+1 by a oint z = (z 0,..., z n ), z n R, z n 0, and z = z z 2 n = 1. There is a unique way of doing this. Then the last coordinate t = z n is uniquely determined by z = (z 0,..., z n 1 ) (since t = 1 z ), and these (z 0,..., z n 1 ) form a disk of dimension 2n. The boundary of this disk is given by z = 1, in other words t = 0, and this is exactly the art already in CP n 1. In any case, either of the two arguments at the end of the revious examle shows that CP is a K(Z, 2)-sace. Next, we will discuss the relation between Serre fibrations and relative CW comlexes in terms of lifting roerties. Recall that, by definition, a Serre fibration is a ma having the right-liftingroerty (RLP) with resect to inclusions of the form I n {0} I n+1 ; or equivalently, with resect to the inclusion J n = I n {0} I n I I n+1. Each of these two kinds of inclusions are relative CW comlexes as well as homotoy equivalences. (In fact, all the saces involved are contractible.) We will rove a much more general statement in Theorem 4 below. But first, we state the following lemma, where HELP stands for the homotoy extension and lifting roerty. Lemma 3. (HELP) Let : E X be a Serre fibration. A B, the ma has the RLP with resect to the inclusion A I B {0} B I. Proof. Consider a commutative square of solid arrows A I B {0} E γ B I X β α Then for any relative CW comlex in which we wish to find a diagonal γ, as indicated. One can write B = n B(n) where A = B ( 1) and B (n) is obtained from B (n 1) by attaching n-cells. It thus suffices to construct larger and larger liftings γ (n) : B (n) I E making the aroriate diagram (n 1) γ α B (n 1) I B {0} E γ (n) B (n) I X commute. Suose γ (n 1) has been constructed. Then we can find γ (n) by defining it on each of th-cells searately, making sure it agrees with γ (n 1) on the boundary of that cell.this reduces the roblem to finding a lift in a diagram of the form β I {0} E I X.

4 4 SECTION 12: HOMOTOPY EXTENSION AND LIFTING PROPERTY But such a lift exists because the left hand inclusion is essentially (i.e., u to homeomorhism) the same as J n I n+1. Theorem 4. A ma : E X is a Serre fibration if and only if it has the RLP with resect to any inclusion A B which is at the same time a relative CW comlex and a weak equivalence. Proof. One imlication is immediate from the definition of Serre fibrations. Notext that, by Whitehead s theorem, such an inclusion i: A B is a strong deformation retract, i.e., there is a retraction r : B A (meaning ri = id A ) and a homotoy h: B I B from ir to id B relative to A. Now suose we are given a commutative diagram A α E i B β X. Let γ : B E be the comosition γ = αr. Then γ is not yet a diagonal filler, because one of the triangles γi = α commutes, but the other doesn t, in general, since γ = αr = βir is only homotoic to β (via the homotoy βh). We can correct this by invoking HELP, and finding a lift k in the commutative (!) square απ 1 γ A I B {0} E k B I X. Then γ = k(, 1) is the required ma filling the first diagram, since γ = β while γ i = α as one easily checks. Theorem 4 states that in a square of the form βh A α E i B β X. in which is a Serre fibration while i is a relative CW comlex, one can find a diagonal in case i is a weak homotoy equivalence. In fact, the situation is symmetric in the sense that one can also ask to be a weak homotoy equivalence. Theorem 5. Let : E X be a Serre fibration and a weak homotoy equivalence. Then has the RLP with resect to any relative CW comlex A B. For the roof, we use the following easy lemma. Lemma 6. Let Y X E E π 1 Y f X

5 SECTION 12: HOMOTOPY EXTENSION AND LIFTING PROPERTY 5 be a ullback square. If : E X is a Serre fibration and a weak homotoy equivalence then so is its ullback π 1 along f. Proof. We already know from an earlier lecture that the ullback of a Serre fibration is again one. The fact that π 1 is a weak homotoy equivalence whenever is now easily follows from the long exact sequence of a fibration. Indeed, is a weak homotoy equivalence if and only its fiber F is weakly contractible and if is surjective on ath comonents. Using that the fiber of π 1 is homeomorhic to that of one easily concludes. Proof. (of Theorem 5) Suose : E X is a Serre fibration and a weak homotoy equivalence. Given a commutative square A α E B β X. where A B is a relative CW comlex, one can try to construct a diagonal filler by induction over the cells, exactly as in the roof of HELP, and this reduces the roblem to the case where A B is the inclusion i: of the boundary of a cell. So, suose we are given commutative square of the form α E i X. Finding a diagonal lift in this diagram is the same as finding one in the square on the left β (i,α) X E i π 1 id β π 2 E X where the right hand square is the ullback (check this!). Write : E for π 1 : X E, so that the square on the left looks like α i = with α = (i, α). Then α reresents an element of π n 1 (E ) which must be zero because is a weak homotoy equivalence (by the lemma) and is contractible. So α extends to a ma γ : E extends in the sense that γi = α, but we do not know whether γ = id. On the other hand, γ is homotoic to the identity by a homotoy relative to the boundary, which can be described exlicitly as the convex combination E h(x, t) = tx + (1 t)γ (x).

6 6 SECTION 12: HOMOTOPY EXTENSION AND LIFTING PROPERTY Wow invoke HELP again, exactly as in the roof of Theorem 4: choose a lift in and let γ = k(, 1). diagram. I {0} I h α π 1 γ k Then γ i = α and γ = id, so γ is the required lift in the revious Remark 7. Theorem 4 and Theorem 5 reresent the most imortant art of Quillen s famous axioms for a (closed) model category, about which we will say more in one of the remaining lectures. E