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CONFIGURATION SPACES IN ALGEBRAIC : LECTURE 11

BEN KNUDSEN

It remains to produce a criterion guaranteeing that a degreewise weak 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 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 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 |f| 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) × ∂∆ / |skn−1(X)|

  n Nn(X) × ∆ / |skn(X)| 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 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   a skn(X)m = skn−1(X)m q  Nn(X) , [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 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 of the barycenter. Similarly, we cover C by U1 and U2. Clearly, h (Uj) = Uj for j ∈ {0, 1}. Consider the commuting diagrams

' U1 / A U2 o B U1 ∩ U2 A × (D ∩ P )

g h|U1 f h|U2 h|U1∩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 |X| 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 . CONFIGURATION SPACES IN : LECTURE 11 3

Definition. A subspace A ⊆ B is relatively T1 if any open set U ⊆ A may be separated from any b ∈ B \ U by an open set U ⊆ V ⊆ B. An inclusion map is relatively T1 if its 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 of B \ 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 ∈ 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 {xj+1, . . . , xi};

(4) finally, set Uj = colimN Uij. Then Uj ∩ Yi = Uij, so Uj is an open subset the colimit, and, since Yj ⊆ Uj, the collection

{Uj}j∈N is an open cover of colimN Yi. Since K is compact, im(f) is compact, so it is contained N in some finite subcover {Ujr }r=1. But, by construction, Ujr does not contain xi for i > jr, so SN r=1 Ujr does not contain xi for i > max{jr : 1 ≤ r ≤ N}, 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) ∈ B1 ×B2 \U. By S the definition of the product topology, we have U = i∈I Ui1 ×Ui2 for open Uij ⊆ Aj. By our assumption on the inclusions of the Aj, we may find open subsets Uij ⊆ Wij ⊆ Bj for each S i ∈ I such that xj ∈/ Wij. Then U ⊆ W := i∈I Wi1 ×Wi2 is open in B1 ×B2, and (x1, x2) ∈/ 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 ∈ Z \ 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 ∈ 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 z∈ / g(W ), since z∈ / U by assumption. But z ∈ 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 z∈ / Y ; in particular, z = g(b) for a unique b ∈ B. Then −1 −1 b∈ / 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 b∈ / 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 f|W factors through Y as a weak homotopy equivalence, then f|X factors through Z as a weak homotopy equivalence.

Proof. The claim that f|X 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. We claim that the restriction of fn to Nn(X) factors through Nn(Y) as a weak homotopy equivalence for every n ≥ 0. Having established this, it will follow by induction and the first two lemmas above that the induced map |skn(X)| → |skn(Y)| is a weak homotopy equivalence for every n ≥ 0. To establish the claim, we proceed by induction on n, the case following from our assump- tion, since N0(X) = X0 and similarly for Y. For the induction step, we note that the inductive assumption implies that the dashed filler in the diagram

s Nm(X) / Xm / Xn

fm fn

  s  Nm(Y) / Ym / Yn exists and is a weak equivalence for every m < n and every degeneracy s. Thus, the dashed filler in the diagram a Nm(X) / Xn [n][m]6=[n]

fn  a  Nm(Y) / Yn [n][m]6=[n] exists and is a weak equivalence. Since since the righthand map is also a weak equivalence, the claim follows from the last lemma. Now, from the inductive description of skeleta and the lemma on relatively T1 pushouts, it follows that each of the inclusions |skn(X)| → |skn+1(X)| is relatively T1, and similarly for Y. Since CONFIGURATION SPACES IN ALGEBRAIC TOPOLOGY: LECTURE 11 5

m m−1 |X| = colimN |skn(X)|, and likewise for Y, we conclude that any map (D ,S ) → (|Y|, |X|) factors as in the solid commuting diagram Sm−1 / |sk (X)| / |X| : n |f|    m D / |skn(Y)| / |Y|. We have already shown the middle arrow to be a weak homotopy equivalence, so the dashed filler exists making the upper triangle commute and the lower triangle commute up to homotopy m−1 relative to S . Thus, πm(f) = 0 for every m ≥ 0, and the claim follows.  References 1 [DI04] D. Dugger and D. Isaksen, Topological hypercovers and A -realizations, Math. Z. 246 (2004), no. 4, 667–689. [May72] J. P. May, The geometry of iterated loop spaces, Lecture Notes in Math., vol. 271, Springer-Verlag, Berlin, Germany, 1972. [Seg74] G. Segal, Categories and theories, Topology 13 (1974), no. 3, 293–312.