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CONFIGURATION SPACES in ALGEBRAIC TOPOLOGY: LECTURE 11 It Remains to Produce a Criterion Guaranteeing That a Degreewise Weak

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CONFIGURATION SPACES in ALGEBRAIC TOPOLOGY: LECTURE 11 It Remains to Produce a Criterion Guaranteeing That a Degreewise Weak CONFIGURATION SPACES IN ALGEBRAIC TOPOLOGY: LECTURE 11 BEN KNUDSEN It remains to produce a criterion guaranteeing that a degreewise weak homotopy equivalence of simplicial spaces induces a weak homotopy equivalence after geometric realization. The material in this lecture is taken from the appendix to [DI04]. Definition. A simplicial space X is split if there are subspaces Nm(X) ⊆ Xm for each m ≥ 0, called the non-degenerate part in degree n, such that the map a Nm(X) ! Xn [n][m] induced by the degeneracies is a homeomorphism for every n ≥ 0. Our goal is to prove the following result|compare [May72, 11.15] and [Seg74, A.5]. Proposition (Dugger-Isaksen). Let f : X ! Y be a map between split simplicial spaces. If fn : Xn ! Yn is a weak equivalence for every n ≥ 0, then jfj is a weak equivalence. The strategy of the proof is simple. First, we argue that f induces a weak equivalence on geometric realizations of n-skeleta for every n; second, we argue that every element in homotopy of the full realization is captured by some skeleton. In order to put this plan into action, we need to have control over skeleta. Recall that the tensor of a space X with a simplicial space Z is the simplicial space (X ⊗ Z)n = X × Zn, with simplicial structure maps induced by those of Z, together with the identity on X. Lemma. Let X be a split simplicial space. The diagram n Nn(X) × @∆ / jskn−1(X)j n Nn(X) × ∆ / jskn(X)j is a pushout square. Proof. Since geometric realization, as a left adjoint, preserves colimits, it suffices to produce a pushout square in simplicial spaces of the form n Nn(X) ⊗ @∆ / skn−1(X) n Nn(X) ⊗ ∆ / skn(X); where we have indulged in the traditional abuse of using the same notation ∆n for the repre- n sentable simplicial set Hom∆(−; [n]) and its geometric realization, and similarly for @∆ . To Date: 25 September 2017. 1 2 BEN KNUDSEN verify that this diagram is a pushout, it suffices to check in each level. Now, it is direct from the definitions that 0 1 a skn(X)m = skn−1(X)m q @ Nn(X)A ; [m][n] n so skn(X)m is the pushout of skn−1(X)m and Nn(X) × ∆m over a coproduct of copies of Nn(X) n indexed by the set of maps f :[m] ! [n] that fail to be surjective, which is exactly @∆m. This fact will only be useful once we are assured that such pushouts are homotopically well- behaved. With regularity assumptions on the spaces involved, the following type of result is common knowledge, but in fact it holds in complete generality. Lemma. If f : A ! A0 and g : B ! B0 are weak homotopy equivalences, and if the front and back faces in the commuting diagram A × @∆n / B f×id@∆n g x ~ A0 × @∆n / B0 A × ∆n / C f×id@∆n x ~ h A0 × ∆n / C0 are pushout squares, then h : C ! C0 is a weak homotopy equivalence. n Proof. We cover C by two open sets, the first being U1 × A × D, where D ⊆ ∆˚ is a Euclidean ` n neighborhood of the barycenter, and the second U2 = B A×@∆n (A × P ), where P ⊆ ∆ is the 0 0 0 −1 0 complement of the barycenter. Similarly, we cover C by U1 and U2. Clearly, h (Uj) = Uj for j 2 f0; 1g. Consider the commuting diagrams ' U1 / A U2 o B U1 \ U2 A × (D \ P ) g hjU1 f hjU2 hjU1\U2 f×idD\P 0 0 0 0 0 0 ' 0 U1 / A U2 o B U1 \ U2 A × (D \ P ); where the horizontal arrows in the leftmost diagram are the projections onto the first factor, and the horizontal arrows in the middle idagram are induced by the inclusion @∆n ⊆ P . Both horizontal arrows in the leftmost diagram are homotopy equivalences, and f is a weak equivalence by assumption; both horizontal arrows in the middle diagram are inclusions of deformation retracts, and g is a weak equivalence by assumption; and f × idD\P is a weak equivalence by assumption. Thus, by two-out-of-three, all three restrictions of h are weak equivalences, so h itself is a weak equivalence. In verifying that elements in the homotopy groups of jXj are all captured by skeleta, we must be assured that the inclusions among skeleta are not too pathological. This assurance takes the form of a relative separation axiom. CONFIGURATION SPACES IN ALGEBRAIC TOPOLOGY: LECTURE 11 3 Definition. A subspace A ⊆ B is relatively T1 if any open set U ⊆ A may be separated from any point b 2 B n U by an open set U ⊆ V ⊆ B. An inclusion map is relatively T1 if its image is so. This terminology is motivated by the observation that a space is T1 if and only if each of its points is relatively T1. Since finite intersections of open sets are open, we have the following immediate consequence: Lemma. If A ⊆ B is relatively T1, then any open set U ⊆ A may be separated from any finite subset of B n U by an open set U ⊆ V ⊆ B. The importance of this notion for our purposes is the following result. Proposition. Let Yi ⊆ Yi+1 be a relatively T1 inclusion for i ≥ 1. If K is compact, then any map f : K ! colimN Yi factors through the inclusion of some Yi. Proof. If f does not factor as claimed, then, without loss of generality, we may assume the existence of xi 2 im(f) \ Yi for each i ≥ 1. Recall that a subset of the colimit is open precisely when its intersection with each stage is open; thus, for each j ≥ 1, we may define an open subset Uj ⊆ colimN Yi by the following prescription: (1) for 1 ≤ i < j, set Uij = ?; (2) for i = j, set Uij = Yj; (3) for i > j, take Uij to be an open subset of Yi separating Ui−1;j from the set fxj+1; : : : ; xig; (4) finally, set Uj = colimN Uij. Then Uj \ Yi = Uij, so Uj is an open subset the colimit, and, since Yj ⊆ Uj, the collection fUjgj2N is an open cover of colimN Yi. Since K is compact, im(f) is compact, so it is contained N in some finite subcover fUjr gr=1. But, by construction, Ujr does not contain xi for i > jr, so SN r=1 Ujr does not contain xi for i > maxfjr : 1 ≤ r ≤ Ng, a contradiction. This fact will only be useful once we are able to identify relatively T1 maps, a task that is made easier by the following observation. Lemma. Relatively T1 inclusions are stable under finite products and pushouts along arbitrary continuous maps. Proof. For the first claim, it suffices by induction to show that A1 × A2 ⊆ B1 × B2 is relatively T1 if each Aj ⊆ Bj is so. Fix an open subset U ⊆ A1 ×A2 and a point (x1; x2) 2 B1 ×B2 nU. By S the definition of the product topology, we have U = i2I Ui1 ×Ui2 for open subsets Uij ⊆ Aj. By our assumption on the inclusions of the Aj, we may find open subsets Uij ⊆ Wij ⊆ Bj for each S i 2 I such that xj 2= Wij. Then U ⊆ W := i2I Wi1 ×Wi2 is open in B1 ×B2, and (x1; x2) 2= W , as desired. For the second claim, suppose that the diagram f A / Y i g B / Z is a pushout square and that i is a relatively T1 inclusion. Fix an open subset U ⊆ Y and a point z 2 Z n U (here, in a small abuse, we identity Y with its image in Z, since the pushout of an inclusion is an inclusion). There are two cases to consider. Assume first that z 2 Y . Since f −1(U) is open in A and i is an inclusion, there is an open −1 subset W ⊆ B with W \ A = f (U), and W qf −1(U) U ⊆ Z is open. To see that z is not 4 BEN KNUDSEN contained in this subset, it suffices to show that z2 = g(W ), since z2 = U by assumption. But z 2 Y , and Y \ g(B) = f(A), so Y \ g(W ) = f(W \ A) = U, and the claim follows. On the other hand, suppose that z2 = Y ; in particular, z = g(b) for a unique b 2 B. Then −1 −1 b2 = i(f (U)) and f (U) ⊆ A is open, so, since i is relatively T1, there is an open subset −1 ` i(f (U)) ⊆ W ⊆ B with b2 = W . As before, W f −1(U) U is open in Z and clearly does not contain z. Corollary. For any pushout diagram of the form A × @∆n / Y n n idA×(@∆ ⊆∆ ) i A × ∆n / Z the inclusion Y ! Z is relatively T1. Finally, we will need the following, essentially obvious, observation. Lemma. If f : W q X ! Y q Z is a weak homotopy equivalence such that fjW factors through Y as a weak homotopy equivalence, then fjX factors through Z as a weak homotopy equivalence. Proof. The claim that fjX factors through Z is obvious after applying π0 and considering the analogous claim for bijections of sets, since π0(f) is a bijection. The claim that this factorization is a weak equivalence follows in the same way after applying πn. Proof of proposition. Fix Nn(X) ⊆ Xn and Nn(Y) ⊆ Y witnessing X and Y as split.
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