K–Theory for Triangulated Categories 3\frac{3}{4}: A Direct Proof

of the Theorem of the Heart ^{∗}^{†}

Amnon Neeman

May 25, 2001

Abstract

Let T be a triangulated category, and assume it admits at least one model. In this article, we define a K–theory for T . The main theorem is that, given any bounded t– structure on T , the K–theory of the heart agrees with the K–theory of T . An immediate consequence tells us that, if two abelian categories occur as hearts of a triangulated category for two different t–structures, then their K–theories must be isomorphic.

The proof was also sketched in previous articles in this series. The virtue of this article is in the careful detail in which it is written down.

Contents

0 Introduction 2

1 Simplicial homotopies 14

2 Model categories 17

3 The comparison with Quillen’s K–theory 21

Step 3.1 : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

∗AMS subject classification: 19D06, 18E30, 18F25 †Key words: Triangulated categories, K–theory, simplicial sets

Step 3.2 : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4 An intermediate simplicial set 52 Step 4.1 : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 Step 4.2 : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

5 The proof (with some points deferred till later) 70 Step 5.1 : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Step 5.2 : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Step 5.3 : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 Step 5.4 : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 Step 5.5 : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

6 Proof of all the Steps but Step 5.5 101

7 An ugly lemma, and the proof of Step 5.5 140

0 Introduction

The introduction to a piece of mathematics ought to be readable by an informed graduate student. The introduction should tell the graduate student, among other things, what preliminaries are assumed. Let me therefore begin by saying something about the prerequisites for the article. In this article, we assume the reader knows about triangulated categories, knows about t–structures in triangulated categories, knows about models for triangulated categories, and knows a little about simplicial sets. But I want to stress here that the required knowledge is very minimal. Chapter 1 of [1] provides quite enough background on triangulated categories and their t–structures. Section 1 of [19] has ample information about models for triangulated categories. And certainly a student who has read Quillen’s [17] is more than adequately prepared for the simplicial techniques of this article.

But since the introduction should have next to no prerequisites at all, let me say here a little about triangulated categories, t–structures and models.

A triangulated category T is an additive category. It is equipped with an invertible endofunctor. There is no consensus in the literature on the name for that functor. If X is an object of T , the functor applied to X goes variously by the names

ΣX, SX, T X, Ω^{−1}X, X[1].

In this article, we will refer to the functor as Σ, and call it the “suspension functor”. (The notation SX is just the Latinised version of ΣX. The notation T X stands for translation. The notation Ω^{−1}X, most common in the group cohomology literature, comes from loop spaces. The notation X[1], fairly standard in algebraic geometry and more particularly in motivic cohomology, is supposed to remind one that this is a “shift operator”.)

Aside from the suspension functor Σ, the category T comes with a collection of “distinguished triangles”

X ⟶ Y ⟶ Z ⟶ ΣX.

These distinguished triangles satisfy certain axioms. An excellent and extensive account of the theory may be found in Verdier’s [21]; for the purpose of this article, the rudiments are quite sufficient. As a matter of terminology, in this article when we say “triangle”, we will mean “distinguished triangle”. For the axiomatic treatment of triangulated categories, I refer the reader elsewhere. But it might be helpful to have here one example.

Example 0.1 Let R be an associative, noetherian ring with unit. The derived category D^{b}(R) is defined as follows. Its objects are bounded chain complexes of finitely generated R–modules. We remind the reader: a chain complex X is a diagram

⋯ ⟶^{∂} X^{i−1} ⟶^{∂} X^{i} ⟶^{∂} X^{i+1} ⟶^{∂} ⋯

where X^{i} are finitely generated left R–modules, and the composites ∂∂ are zero. We assume that all but finitely many of the X^{i}’s vanish. The morphisms are the homotopy equivalence classes of chain maps, with the homology isomorphisms formally inverted.

The suspension functor Σ translates the complex one step to the left; that is

{ΣX}^{i} = X^{i+1}.

It also changes the sign of the differential ∂; that is,

∂_{ΣX} = −∂_{X}.

Finally, we need to describe the distinguished triangles. Given a chain map f : X ⟶ Y of chain complexes, that is a commutative diagram

⋯ ⟶^{∂} X^{i−1}  ⟶^{∂} X^{i}  ⟶^{∂} X^{i+1}  ⟶^{∂} ⋯
f^{i−1} ↓  f^{i} ↓  f^{i+1} ↓
⋯ ⟶^{∂} Y^{i−1}  ⟶^{∂} Y^{i}  ⟶^{∂} Y^{i+1}  ⟶^{∂} ⋯

we can form the mapping cone; it is the chain complex Cone(X ⟶ Y) below

⋯ ⟶ X^{i} ⊕ Y^{i−1} \underrightarrow{( \begin{matrix}−∂ & 0 \\ f^{i} & ∂ \end{matrix} )} X^{i+1} ⊕ Y^{i} \underrightarrow{( \begin{matrix}−∂ & 0 \\ f^{i+1} & ∂ \end{matrix} )} X^{i+2} ⊕ Y^{i+1} ⟶ ⋯

The distinguished triangles in D^{b}(R) are the isomorphs of

X ⟶^{f} Y ⟶ Cone(X → Y) ⟶ ΣX.

Next we should say a little about t–structures. A triangulated category T has a “t–structure” if, for every integer n ∈ ℤ, T has two subcategories denoted T^{≤n} and T^{≥n}. These subcategories are assumed to satisfy certain axioms. The reader can find a wonderful discussion of this in Chapter 1 of [1]. Unlike the other preliminaries, here it is probably worthwhile familiarising oneself thoroughly with Chapter 1 of [1]. As the proof progresses, we will make frequent use of the ideas there. In any case, Chapter 1 of [1] is short and elementary.

One key property of t–structures is the induced truncation functors. Given any n ∈ ℤ, the inclusion T^{≤n} ⊂ T has a right adjoint, while the inclusion T^{≥n} ⊂ T has a left adjoint. If X is an object of T , the right adjoint of the inclusion T^{≤n} ⊂ T , applied to X, will be denoted X^{≤n}. The counit of adjunction will be denoted

X^{≤n} ⟶ X.

The left adjoint of the inclusion T^{≥n} ⊂ T , applied to X, will be denoted X^{≥n}. The unit of adjunction will be denoted

X ⟶ X^{≥n}.

The composite functor

X ↦ {X^{≥0}}^{≤0} = {X^{≤0}}^{≥0}

is a homological functor from the category T to the abelian category T^{≥0} ∩ T^{≤0}. Now it is time for an example.

Example 0.2 Let T be the derived category D^{b}(R) of Example 0.1. Then T has a t–structure. For any integer n, the subcategory T^{≤n} ⊂ T is the full subcategory of all chain complexes

⋯ ⟶^{∂} X^{i−1} ⟶^{∂} X^{i} ⟶^{∂} X^{i+1} ⟶^{∂} ⋯

where X^{i} = 0 for all i > n. The subcategory T^{≥n} ⊂ T is the full subcategory of all chain complexes

⋯ ⟶^{∂} X^{i−1} ⟶^{∂} X^{i} ⟶^{∂} X^{i+1} ⟶^{∂} ⋯

where X^{i} = 0 for all i < n. Given a chain complex X as above, the chain complex X^{≥n} is the complex

⋯ ⟶ 0 ⟶ 0 ⟶ \frac{X^{n}}{Im(X^{n−1} → X^{n})} ⟶^{∂} X^{n+1} ⟶^{∂} X^{n+2} ⟶^{∂} ⋯

The chain complex X^{≤n} is

⋯ ⟶^{∂} X^{n−2} ⟶^{∂} X^{n−1} ⟶^{∂} Ker(X^{n} → X^{n+1}) ⟶ 0 ⟶ 0 ⟶ ⋯

The category T^{≥0} ∩ T^{≤0} is nothing other than the category of chain complexes vanishing outside degree 0, that is the category of finitely generated left R–modules. The functor

X ↦ {X^{≥0}}^{≤0} = {X^{≤0}}^{≥0}

is easily computed to take a chain complex to

\frac{Ker(X^{0} → X^{1})}{Im(X^{−1} → X^{0})}

which is nothing other than the usual homology functor H^{0}.

Finally, we have to talk about models. For a reference, we recommend Section 1 of [19]; the reader may assume that our models are all “biWaldhausen complicial categories”, in Thomason’s terminology. For the benefit of the experts, allow me one paragraph to recall the notation. After that, I will illustrate it with an example. The next paragraph may safely be omitted by the non–experts.

Let T be a triangulated category. In this article, a Waldhausen model T for T , or in short a model category T for T , will be an exact category T, together with a subcategory of weak equivalences, so that if we define the admissible epis to be the fibrations and the admissible monos to be the cofibrations, then both T and T^{op} are Waldhausen categories. We furthermore assume that both suspension functors, the one determined by T and the one determined by T^{op}, are invertible. And finally, inverting the weak equivalences, we must get T . Some more discussion will be given in Section 2. For a thorough discussion the reader is referred to [23] and [19].

Example 0.3 If T is the derived category D^{b}(R) of Example 0.1, then here is one model T. The objects of T are the same as the objects of T ; they are the chain complexes of finitely generated left R–modules. The morphisms in T are the chain maps (NOT homotopy equivalence classes of chain maps). The cofibrations are the chain maps

⋯ ⟶^{∂} X^{i−1} ⟶^{∂} X^{i} ⟶^{∂} X^{i+1} ⟶^{∂} ⋯
↓ ↓ ↓
⋯ ⟶^{∂} Y^{i−1} ⟶^{∂} Y^{i} ⟶^{∂} Y^{i+1} ⟶^{∂} ⋯

so that, for every i, the map X^{i} ⟶ Y^{i} is a split monomorphism. The fibrations are the chain maps so that, for every i, X^{i} ⟶ Y^{i} is a split epimorphism. The weak equivalences are the chain maps inducing homology isomorphisms.

The important relation between T and T is that we obtain T from T by formally inverting the weak equivalences. Other than the assumption that T is an exact category, the only structure we will use is the existence and functoriality of the mapping cone (see Example 0.1), and the dual construction, the mapping fiber.

The reader familiar with Waldhausen’s work will note that this is a very restricted class of Waldhausen models to work with. The theorems remain true if we relax these restrictive hypotheses on the models we are willing to accept, but the arguments become more technical. In this article, we avoid technicalities as much as possible.

Let T be an exact category. A commutative square in T is a diagram

[figure: B′; C; A; B]

When we say the commutative square is bicartesian, we will mean that the sequence

0 ⟶ A ⟶ B ⊕ B′ ⟶ C ⟶ 0

is an admissible short exact sequence. This is perhaps unconventional. It implies that, whenever we embed T as an exact subcategory of any abelian category A, the commutative square is bicartesian in A, in the usual sense of being a pullback and a pushout square. It would perhaps be better to refer to such squares as “stably bicartesian”; they remain bicartesian after any embedding. But since this is the only notion of bicartesian squares we care about, the term bicartesian square will mean, in this article, one of the special ones above.

Now let T be a small triangulated category. Define a bisimplicial set T [figure] as follows:

Definition 0.4 T [figure] is defined to be a bisimplicial set. A (p, q)–simplex in T [figure]

is a diagram in T

[figure: ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0q}; X_{p0}; X_{pq}]

together with a map in T of the form

X_{pq} ⟶ ΣX_{00}.

What is assumed about all this data in T is the following. There must be some model category T for T , and a diagram in T

[figure: ⋮; ⋮; ⋯; ⋯; X̃_{00}; X̃_{0q}; X̃_{p0}; X̃_{pq}]

so that every square in the diagram in T is a bicartesian square. In particular, we have a bicartesian square in T

[figure: X̃_{p0}; X̃_{pq}; X̃_{00}; X̃_{0q}]

and therefore a short exact sequence

0 ⟶ X̃_{00} ⟶ X̃_{0q} ⊕ X̃_{p0} ⟶ X̃_{pq} ⟶ 0.

The short exact sequence tells us that the natural map

Cone(X̃_{00} → X̃_{0q} ⊕ X̃_{p0}) ⟶ X̃_{pq}

is a weak equivalence, hence invertible in T . We may compose the inverse of this map with the natural map

Cone(X̃_{00} → X̃_{0q} ⊕ X̃_{p0}) ⟶ Cone(X̃_{00} → 0) = ΣX̃_{00}.

We deduce a map in T

X̃_{pq} ⟶ ΣX̃_{00}.

We require that the diagrams below and

[figure: ⋮; ⋮; ⋯; ⋯; X̃_{00}; X̃_{0q}; X̃_{p0}; X̃_{pq}]

together with their respective differentials are isomorphic in T . For each i with 0 ≤ i ≤ p and each j with 0 ≤ j ≤ q, there is an isomorphism

X̃_{ij} ⟶ X_{ij}

which yields an isomorphism of diagrams in T .

There are two face maps, one which decreases the integer p and the other decreases the integer q. The one decreasing the integer p is obtained by deleting rows of the diagram, the one decreasing q by deleting columns. The degeneracy maps come from doubling a row or a column.

Now suppose the triangulated category T has a t–structure. Then T has many subcategories, denoted by symbols T^{≤n} or T^{≥m}. These subcategories define the t–structure. We define

Definition 0.5 For any pair of integers m ≤ n, the full subcategory T_{[m,n]} is by definition the intersection

T^{≤n} ∩ T^{≥m}.

Remark 0.6 If T = D^{b}(R) is the derived category of Example 0.1, then T_{[m,n]} is the full subcategory whose objects are the chain complexes

⋯ ⟶ 0 ⟶ X^{m} ⟶ X^{m+1} ⟶ ⋯ ⟶ X^{n−1} ⟶ X^{n} ⟶ 0 ⟶ ⋯

With this, we can define simplicial sets Definition 0.7 For any integers m ≤ n, we define the simplicial set T[figure: [m,n]] to be the

subset of T [figure] where the simplices are diagrams in T_{[m,n]}. In other words, a simplex is a diagram

[figure: ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0q}; X_{p0}; X_{pq}]

where the X_{ij}’s are all assumed to lie in the subcategory T_{[m,n]}. In Section 3 we will prove Lemma 3.4 Suppose that T is a small triangulated category with at least one model. Suppose it has a t–structure. The simplicial set T[figure: [0,0]] is then homotopy equivalent to Quillen’s Q–construction for the abelian category T_{[0,0]}, the heart of T . Perhaps it would be fair to tell the reader that this result is not really new; we will trivially deduce it from work of Jardine. The deep and difficult result of the article is Theorem I.7.1 Suppose T is a small triangulated category with a bounded t–structure. That is, T is the union of all the subcategories T_{[m,n]}. Then the natural inclusion

T[figure: [0,0]]  \overline{}[figure]  T [figure]

induces a homotopy equivalence.

Corollary 0.8 Let T be a small triangulated category with at least one model. It follows that for different bounded t-structures, the K–theory of the hearts are all isomorphic.

The first, obvious reduction of the proof of Theorem I.7.1 is that since T is the union of the

subcategories T_{[m,n]}, it follows that T [figure] is the union of all the simplicial sets T[figure: [m,n]] .

It therefore suffices to prove that for all integer m′ ≤ m ≤ n ≤ n′, the inclusion

T[figure: [m,n]]  \overline{}[figure]  T[figure: [m′,n′]]

induces a homotopy equivalence. Clearly, it suffices to consider the cases where either m′ = m or n = n′. These cases are dual to each other, so we assume n = n′. We may also clearly assume that m = m′ + 1. Finally, by translation, we may assume m′ = 0. Thus the essential case is to prove that the inclusion

T[figure: [1,n]]  \overline{}[figure]  T[figure: [0,n]]

induces a homotopy equivalence. This is as far as we get by trivial reductions. The rest of the article will show this.

Let us give a brief overview of the article. After the introduction come two elementary sections. In Section 1, we remind the reader about simplicial sets and simplicial homotopies. In Section 2, we have a brief reminder about model categories. As we have already mentioned

above, Section 3 is devoted to the proof of Lemma 3.4; we prove that T[figure: [0,0]] is homotopy

equivalent to Quillen’s Q–construction on the heart. The remainder of the article is devoted to the proof of Theorem I.7.1. By the elementary reductions above, this means that we show the map

T[figure: [1,n]]  \overline{}[figure]  T[figure: [0,n]]

a homotopy equivalence.

Conceptually, the proof of Theorem I.7.1 has two parts. We want to prove that some map of topological spaces X ⟶ Y is a homotopy equivalence. The first part is combinatorial. We replace the above by a fairly long string of intermediate spaces and maps

X ⟵ X_{1} ⟶ X_{2} ⟵ X_{3} ⟶ X_{4} ⟵ ⋯ ⟶ X_{n−1} ⟵ X_{n} ⟶ Y

and clearly it suffices to show two things. We must show (1) that all the maps in the string are homotopy equivalences, and (2) that the composite homotopy equivalence agrees with the map induced by X ⟶ Y. This combinatorial part of the article comprises sections 3, 4 and 5. There is not much that one can say to illuminate the argument here. It is a tricky conmbinatorial path, with many intermediate spaces.

Many of the intermediate maps are very easily shown to be homotopy equivalences. But there are nine which are more subtle. We label them as Steps, and leave them till later. Sections 6 and 7 contain the proofs that the nine more subtle maps also induce homotopy equivalences. The interesting feature here is that, in eight of the nine cases, the proofs are all very minor variants on the same theme. The same homotopy works in basically all eight cases. The eight similar cases are found in Section 6, while the different case occupies Section 7.

This article, unlike its predecessors, is written narrowly. It proves Theorem I.7.1, with no mention of adornments or generalisations. The purpose is to make it very checkable. It is, up to some minor permutations of the lemmas, the proof I presented orally to Thomason in Paris in May 1992.

It should be mentioned that the recent work of Vaknin goes a long way towards removing all mention of model categories in this theorem. The reader is referred to Vaknin’s article [20]. The problem is that, despite the illuminating progress, there is still an unsolved problem. It is to be hoped that research over the next few years will lead to substantial improvements and simplifications. At present, this is the best theorem, allowing us to recover the Quillen K–theory of the heart from a triangulated category.

The first written version of this result dates to September 1988. It seems fair to say that at the time, nobody believed the result. In 1990 Thomason became the referee, and unlike the earlier referees, he was determined to find the mistake. Thomason did not believe the result any more than anyone else. He thought that the theorem was probably wrong, and that the proof could not possibly be correct. But this did not stop him from using the next two years trying to find the mistake. Finally, in the summer of 1992, we met in Paris and went through the proof presented here very carefully, in person. After that Thomason became convinced that a proof could be written down.

It seems to me admirable that a referee should take an article such as mine, and spend so long working at it. It must be remembered that in Thomason’s opinion the result was unlikely to be true, the proof was probably wrong, and the exposition was dreadful; vague at crucial points, and much too much was left to the reader. It is clear that there is no way for me to adequately repay Thomason for the time he spent studying my work. Let me thank him for his dedication. Thomason may have set a new standard for excellence in refereeing.

Everything in this paper is an elaboration on the more sketchy outline of the proof to be found in [9], [10], [11], [12], and [13]; especially [13]. This should explain the strange numbering of Theorem I.7.1. It is Theorem 7.1 of K–theory for triangulated categories I; the statement may be found on page 435 of [10]. The reason for writing an elaboration of the proof is the following. When people believe the result is probably incorrect, they are not easily satisfied with the sketch of a proof. When we write an article, we will frequently leave lemmas to the reader, saying the proof is similar to an argument previously seen, in Lemma X. Understandably, referees are very unlikely to tolerate such sloppiness in a result they view as suspect. This article was written to address such skepticism. It is more narrowly focused than the articles cited above. Its virtue is in the careful detail in which the results are checked.

As has already been mentioned, the prerequisites for the article are very minimal. A graduate student should be able to read it, without first having to digest voluminous related work. There is no denying that the article is long, and the proof of the main theorem is difficult. Nevertheless, let me stress once again that the proof, although hard, is elementary. It does not rely on heavy machinery; this is a proof anyone could read.

There is a similar article, in which I gave a narrow, focused proof of another theorem in this subject. The reader is referred to [14] and [15]. Acknowledgements. As well as Thomason, the author would like to thank the many other referees and editors who went to a great deal of trouble checking the results. In particular let me thank Tony Bak, who handled the article as editor for K–Theory, and showed the patience of an angel.

1 Simplicial homotopies

In this article, our main lemmas all show that certain simplicial maps induce homotopy equivalences. To prove the lemmas, we will need many simpicial homotopies. In this section, we remind the reader of the basic formalism of simplicial maps and homotopies.

Given two simplicial sets, a simplicial map between them is a map taking n–simplices in one to n–simplices in the other, and commuting with the face and degeneracy maps. Let S. and T. be simplicial sets. A homotopy from the realisation of S. to the realisation of T. is a map

|S.| × I ⟶ |T.|

where |S.| is the geometric realisation of S., |T.| is the geometric realisation of T., and I is the unit interval. To do this simplicially, we need to give a simplicial model for I. The model we give is the nerve of the category 1. It is the category

0 ⟶ 1.

The nerve of this category is the simplicial set whose n–simplices are functors

n ⟶ 1

where n is the category

0 ⟶ 1 ⟶ 2 ⟶ ⋯ ⟶ n.

To give a functor n ⟶ 1 is nothing more than giving an integer i with −1 ≤ i ≤ n, and sending the integers {0, 1, ⋯, i} to 0, while the integers {i + 1, i + 2, ⋯, n} map to 1.

If we allow ourselves to confuse the category 1 with its nerve, that is the simplicial set just described, then a simplicial homotopy from S. to T. is just a simplicial map

S. × 1 ⟶ T.

In other words, to any n–simplex in the product S. × 1 we must assign an n–simplex in T.. An n–simplex in S. × 1 is a pair of an n–simplex in S. and an n–simplex in 1. As we have just learned, an n–simplex in 1 is nothing more than an integer i with −1 ≤ i ≤ n. Thus a simplicial homotopy assigns an n–simplex in T. to every pair consisting of an n–simplex in S. and an integer i.

Out of habit, I prefer to think of simplicial homotopies slightly differently. To describe them, I prefer to give their images on slightly degenerate simplices. Let x ∈ S_{n} be an n–simplex. If i is an integer with 0 ≤ i ≤ n, let σ_{i} be the i^{th} degeneracy map. Then σ_{i}x is an n + 1 simplex, which is of course degenerate. The pair (σ_{i}x, i) is an (n + 1)–simplex in S. × 1, and I prefer to describe simplicial homotopies by giving their images on simplices such as these. Any simplex in S. × 1 can be described as a face of one of my favorite simplices; giving the homotopy on those therefore clearly suffices. This way has the appeal that a simplicial homotopy sends an n–simplex x to n + 1 different (n + 1)–simplices in T., namely one for every integer i with 0 ≤ i ≤ n. It is more natural to think of simplicial homotopies this way. A simplicial homotopy takes a simplex x to the image of x × I, and x × I subdivides into n + 1 (n + 1)–simplices.

Next, let us agree to freely confuse a simplicial set with its realisation. When we say that a map of simplicial sets is a homotopy equivalence, we mean that it induces a homotopy equivalence after realisation. Sometimes we will state explicitly that a map is a homotopy equivalence only after realisation, sometimes we will not. There is no mathematical significance to be read into this difference of terminology.

We will frequently be dealing with sets having several commuting simplicial structures. We remind the reader. The category Δ is the category whose objects are n = {0, 1, …, n}. The morphisms in Δ are the order-preserving maps. An r–simplicial set is a functor

{Δ^{op}}^{r} ⟶ Sets.

Thus an r–simplicial set has r commuting simplicial structures. If we realise one of them, we get a functor

{Δ^{op}}^{r−1} ⟶ Spaces.

We can then proceed to realise one of the structures of this (r − 1)–simplicial space, to get a functor

{Δ^{op}}^{r−2} ⟶ Spaces.

We can iterate, and by the time we realise all the simplicial structures, we get a space. It is a classical fact that the resulting space is independent of the order in which we realised the simplicial structures.

We can also do things in one fell swoop. The diagonal inclusion gives a functor

Δ ⟶ Δ^{r}.

Composing an r–simplicial set

{Δ^{op}}^{r} ⟶ Sets

with the diagonal inclusion gives a simplicial set

Δ^{op} ⟶ {Δ^{op}}^{r} ⟶ Sets.

This process is known as diagonalising the r simplicial structures. It is a classical fact that the realisation of the diagonal simplicial set is canonically homotopy equivalent to the iterated realisations of the previous paragraph, for any order of realisation.

This article is a long sequence of lemmas, each of which proves that a map is a homotopy equivalence. Almost in every case, we will study a map from an (m + n)–simplicial set to an n–simplicial set. What we mean by this is of course a map of (m + n)–simplicial sets, where on the target, m of the simplicial structures are degenerate.

The first observation is that the number of simplicial structures is really irrelevant. It is always possible to increase it by introducing some more degenerate structures, or to decrease it by diagonalising some of the structures. We henceforth stop explicitly counting the simplicial structures.

The prototype lemma in this article is therefore:

Prototype Lemma Let S.. be a bisimplicial set, T. a simplicial set. Let f : S.. ⟶ T. be a simplicial map. That is, it is a bisimplicial map, where T. is made into a bisimplicial set with one degenerate structure. The map f induces a homotopy equivalence.

The prototype proof will be: it suffices to realise just one of the structures, and show the map is a homotopy equivalence after realising it. Realise the one that is degenerate on T.; the realisation of T. is then discrete. It then suffices to show that for every point in T. the fiber, which we will refer to as the Segal fiber, is contractible.

The article is therefore a succession of contractibility proofs. The way it has been organised is that first we give the skeleton of the argument, presenting all the statements in logical order, and proving the contractibility statements only if the proof is trivial. We define a proof of contractibility to be trivial if it requires only one simplicial homotopy.

The “difficult” contractibility statements are all given first without proof. The proofs are all presented at the end the article, together.

To make it easy for the reader to see when a statement is being left to be proved later, all such statements are labeled “Step …”. They appear in the table of contents, so the reader can easily check back for the statement of any “Step”, that is check exactly what it was that was being left to prove later. In this way, hopefully the reading of the article will be made a little easier. First the reader gets an overview of the proof, and then come the difficult details.

2 Model categories

While we are still dealing with introductory matter, it is perhaps also good to review what

is meant by a model category. Recall that in Definition 0.4, a simplex in T [figure] was a

diagram

[figure: ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0q}; X_{p0}; X_{pq}]

for which there exists a lifting to some model category for T . What does this mean?

Starting with a category satisfying a long list of axioms, one can construct functorially a triangulated category. There are excellent accounts of this in [23] and [19], as well as a rather skimpy account in [16]. In this section we will assume some familiarity with the notation of these articles.

For us, a model category will be an exact category T. The admissible monos will be the cofibrations, the admissible epis the fibrations. The category T must also have a subcategory of weak equivalences. We assume that both T and its dual are categories with cofibrations and weak equivalences, satisfying the Gluing Lemma, Extension Axiom, Saturation Axiom and Cylinder Axiom. There are two suspension functors, one that comes from the cylinder functor on T, the other from the cylinder functor on the dual T^{op}. The reader should recall that both the category and its dual are assumed to have cylinder functors, which need not have anything to do with each other. We assume that both suspension functors are invertible endofunctors on T.

Given an admissible short exact sequence in T

[figure: A; B; C]

there is a map C ⟶ ΣA in the asssociated triangulated category T so that the sequence

A ⟶ B ⟶ C ⟶ ΣA

is a triangle in T . But the point we wish to make here is that there is a natural choice for C ⟶ ΣA. It is given as follows. We have a pushout square

[figure: 0; C; A; B]

and therefore the induced map on the mapping cones of the rows, that is the map

Cone (A ⟶ B)  ⟶  Cone (0 ⟶ C) ≃ C

must be a weak equivalence in T, hence an isomorphism in T . Now compose the inverse of this isomorphism with the natural projection

Cone (A ⟶ B)  ⟶  Cone (A ⟶ 0) ≃ ΣA

and we have the desired map C ⟶ ΣA.

This construction is clearly natural. Given a map of short exact sequences

[figure: A; B; C; A′; B′; C′]

then, if we choose the maps C ⟶ ΣA and C′ ⟶ ΣA′ as above, the following diagram commutes

A ⟶ B ⟶ C ⟶ ΣA
↓ ↓ ↓ ↓
A′ ⟶ B′ ⟶ C′ ⟶ ΣA′

Now consider a simplex in T [figure] , that is a diagram

[figure: ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0q}; X_{p0}; X_{pq}]

in T which admits a lifting to T. Recall that we required from the lifting

[figure: ⋮; ⋮; ⋯; ⋯; X̃_{00}; X̃_{0q}; X̃_{p0}; X̃_{pq}]

that every square be bicartesian. But then we have an admissible exact sequence

X̃_{00} \underrightarrow{( \begin{matrix}−f \\ g \end{matrix} )} X̃_{p0} ⊕ X̃_{0q} \underrightarrow{( h k )} X̃_{pq}

where f : X̃_{00} ⟶ X̃_{p0}, g : X̃_{00} ⟶ X̃_{0q}, h : X̃_{p0} ⟶ X̃_{pq} and k : X̃_{0q} ⟶ X̃_{pq} are the maps given in the diagram defining the simplex. Note that in passing from a bicartesian square to an exact sequence, there is a choice of signs in the morphisms. Henceforth, our choice of signs is as above; the maps in the sequence

X̃_{00} ⟶ X̃_{p0} ⊕ X̃_{0q} ⟶ X̃_{pq}

all have their natural sign, except f : X̃_{00} ⟶ X̃_{p0}. There is therefore a natural differential

X̃_{pq} ⟶ ΣX̃_{00}

in T , whose sign is determined by our sign choice above. The hypothesis of Definition 0.4, which was left to be made precise in this section, is that the isomorphisms X̃_{ij} ⟶ X_{ij} should send this natural differential to the map X_{pq} ⟶ ΣX_{00} that is part of the structure of the simplex.

Now observe that given any integers i′ with 0 ≤ i′ ≤ p and j′ with 0 ≤ j′ ≤ q, there is a morphism of short exact sequences in T

X̃_{00} ⟶ X̃_{i′0} ⊕ X̃_{0j′} ⟶ X̃_{i′j′}
↓ ↓ ↓
X̃_{00} ⟶ X̃_{p0} ⊕ X̃_{0q} ⟶ X̃_{pq}

and hence, in T we deduce a commutative square

X̃_{i′j′} ⟶ ΣX̃_{00}
↓  ↓
X̃_{pq} ⟶ ΣX̃_{00}

Given any integers i with 0 ≤ i ≤ i′ and j with 0 ≤ j ≤ j′, there is a morphism of short exact sequences in T

X̃_{00} ⟶ X̃_{i′0} ⊕ X̃_{0j′} ⟶ X̃_{i′j′}
↓ ↓ ↓
X̃_{ij} ⟶ X̃_{i′j} ⊕ X̃_{ij′} ⟶ X̃_{i′j′}

and hence, in T we deduce a commutative square

X̃_{i′j′} ⟶ ΣX̃_{00}
↓  ↓
X̃_{i′j′} ⟶ ΣX̃_{ij}

and hence the “natural” differential X̃_{i′j′} ⟶ ΣX̃_{ij} is obtained from X̃_{pq} ⟶ ΣX̃_{00} just as the composite

X̃_{ij} ⟶ X̃_{pq} ⟶ ΣX̃_{00} ⟶ ΣX̃_{ij}

In particular, it is possible to compute what the “natural” differential X̃_{i′j′} ⟶ ΣX̃_{ij} must be, just from the diagram

[figure: ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0q}; X_{p0}; X_{pq}]

and the map X_{pq} ⟶ ΣX_{00}. This establishes that indeed we do have a bisimpicial set. Deleting all rows above i′ and below i, and all columns above j′ and below j, gives a diagram of the same type, still with at least one lifting, as long as we take the map X_{i′j′} ⟶ ΣX_{ij} to be the composite

X_{i′j′} ⟶ X_{pq} ⟶ ΣX_{00} ⟶ ΣX_{ij}.

3 The comparison with Quillen’s K–theory

Let A be an abelian category. Then, following Quillen, one can define higher K–theory for A. The reader is referred to [17] for a treatment.

From Proposition 1.4 of Jardine’s [6], we know that the K–theory space for A is the loop space

[figure: ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0q}; X_{p0}; X_{pq}]

of a bisimplicial set we will denote A [figure] . A simplex in A [figure] is a diagram where the X_{ij}’s are objects of A, and all the squares are assumed bicartesian. As indicated both in the diagram and in the shorthand for the simplicial set, the vertical morphisms are assumed epi, the horizontal ones mono.

To tie this up more with the K–theory of triagulated categories, let T be a triangulated category with a t–structure, and assume that the subcategories T_{[m,n]} are as before, that is

T_{[m,n]} = T^{≤n} ∩ T^{≥m}.

In the special case where m = n = 0, the category T_{[0,0]} is an abelian category, called the heart of T with respect to the t–structure. We remind ourselves of the fact

Reminder 3.1 Let T be a triangulated category with a t–structure. The truncation functor

X ↦ {X^{≤0}}^{≥0}

is a homological functor from T to its heart T_{[0,0]}. We denote this functor

H^{0} : T ⟶ T_{[0,0]}

As is customary, we let H^{n}(X) = H^{0}(Σ^{n}X).

We define

Definition 3.2 A morphism X ⟶ Y in T_{[m,n]} is called mono if the induced map H^{m}(X) ⟶ H^{m}(Y) is mono. A morphism X ⟶ Y in T_{[m,n]} is called epi if the induced map H^{n}(X) ⟶ H^{n}(Y) is epi.

The simplicial set T[figure: [m,n]] is defined to be a subset of T[figure: [m,n]] . A simplex lies in the

subset if the diagram defining it is of the form That is the objects are still in T_{[m,n]}, but the morphisms are now restricted. The vertical morphisms are assumed epi, the horizontal morphisms mono, in the sense of Definition 3.2.

Thus the simplicial set T[figure: [m,n]] is a subset of T [figure] , and the idea of the symbol is

that it should provide the definition of the set. The set consists of simplices in T [figure] , which

are furthermore subject to the restrictions imposed by the symbol. As indicated in the symbol, the objects in a diagram defining a simplex are in the subcategory T_{[m,n]}, while the morphisms satisfy the restrictions indicated by the symbol. The vertical ones are constrained to be epi, the horizontal ones mono.

Now we come to a notational problem. The symbol T[figure: [0,0]] is confusing. If we view T_{[0,0]}

as a subcategory of T , then the definition of T[figure: [0,0]] is that it is a subset of T [figure] . A

simplex in T[figure: [0,0]] is therefore a diagram in T , together with a map X_{pq} ⟶ ΣX_{00}, and the hypothesis is that this diagram can be lifted to at least some model T for T , in such a way that the diagram in T is of bicartesian squares.

But if we view T_{[0,0]} as an abelian category, then T[figure: [0,0]] has also been our notation for the

bisimplicial set of Jardine, homotopy equivalent to Quillen’s Q–construction. We need

Lemma 3.3 Suppose the triangulated category T has at least one model. Then the two constructions above agree. Precisely, given a diagram

[figure: ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0q}; X_{p0}; X_{pq}]

as in Jardine’s model of T[figure: [0,0]] , that is where all the objects are in the abelian category T_{[0,0]}

and all the squares are bicartesian in T_{[0,0]}, then there is a unique way to choose the differential X_{pq} ⟶ ΣX_{00}, and with this (unique) choice, the diagram can be lifted to any model T for T .

Proof. Let us suppose we are given a model category T for T , and a simplex in Jardine’s model of the Q–construction of the abelian category T_{[0,0]}. That is, in the diagram above, all the squares are bicartesian in T_{[0,0]}. Consider the two sequences of morphisms in T

\begin{matrix}X_{00} & ⟶ & X_{01} & ⟶ & ⋯ & ⟶ & X_{0q} \\ X_{00} & ⟶ & X_{10} & ⟶ & ⋯ & ⟶ & X_{p0} \end{matrix}

Choose a lifting of X_{00} to the category T. That is, choose an object X̃_{00} in T, and an isomorphism X̃_{00} ⟶ X_{00} in T . Now we are given a morphism in T

X_{00} ⟶ X_{01},

and by the definition of morphisms in T , that means there is a morphism in T

X̃_{00} ⟶ X̃_{01}

and an isomorphism in T

X̃_{01} ⟶ X_{01}

so that the square

X̃_{00} ⟶ X̃_{01}
↓  ↓
X_{00} ⟶ X_{01}

commutes in T . Continuing this, we can define sequences of objects in T and isomorphisms of sequences in T

X̃_{00} ⟶ X̃_{01} ⟶ ⋯ ⟶ X̃_{0q}
↓ ↓ ↓
X_{00} ⟶ X_{01} ⟶ ⋯ ⟶ X_{0q}

and

X̃_{00} ⟶ X̃_{10} ⟶ ⋯ ⟶ X̃_{p0}
↓ ↓ ↓
X_{00} ⟶ X_{10} ⟶ ⋯ ⟶ X_{p0}

Inductively replacing X̃_{0i} by the mapping cylinder on the map X̃_{0 i−1} ⟶ X̃_{0i}, we may assume that the maps in the row

X̃_{00} ⟶ X̃_{01} ⟶ ⋯ ⟶ X̃_{0q}

are all cofibrations in T. But then we can fill the diagram, by pushing out in T, to

[figure: ⋮; ⋮; ⋯; ⋯; X̃_{00}; X̃_{0q}; X̃_{p0}; X̃_{pq}]

and the point we wish to make is that the resulting diagram is isomorphic in T_{[0,0]} to

[figure: ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0q}; X_{p0}; X_{pq}]

giving us one choice of differential X_{pq} ⟶ ΣX_{00} for which the diagram has a lifting. The uniqueness of the differential is a consequence of Corollaire 1.1.10(ii) on page 24 of [1], where it is shown that for

X_{00} ⟶ X_{p0} ⊕ X_{0q} ⟶ X_{pq}

a short exact sequence in T_{[0,0]}, there is a unique morphism X_{pq} ⟶ ΣX_{00} making the sequence

X_{00} ⟶ X_{p0} ⊕ X_{0q} ⟶ X_{pq} ⟶ ΣX_{00}

a distinguished triangle.

It remains therefore to show that the diagrams

[figure: ⋮; ⋮; ⋯; ⋯; X̃_{00}; X̃_{0q}; X̃_{p0}; X̃_{pq}]

and

[figure: ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0q}; X_{p0}; X_{pq}]

are isomorphic in T . For each i with 0 ≤ i ≤ p and each j with 0 ≤ j ≤ q, we have a triangle in T

X̃_{00} ⟶ X̃_{i0} ⊕ X̃_{0j} ⟶ X̃_{ij} ⟶ ΣX̃_{00}

and hence a long exact sequence in homology. There is an exact sequence in T_{[0,0]}

H^{n−1}(X̃_{ij}) ⟶ H^{n}(X̃_{00}) ⟶ H^{n}(X̃_{i0} ⊕ X̃_{0j}) ⟶ H^{n}(X̃_{ij}) ⟶ H^{n+1}(X̃_{00})

But we know that X̃_{00} ≃ X_{00}, X̃_{i0} ≃ X_{i0} and X̃_{0j} ≃ X_{0j} all lie in T_{[0,0]}; this means that for n ≠ 0

H^{n}(X̃_{00}) = H^{n}(X̃_{i0}) = H^{n}(X̃_{0j}) = 0.

We also know that the map

H^{0}(X̃_{00}) ⟶ H^{0}(X̃_{i0} ⊕ X̃_{0j})

is a monomorphism in T_{[0,0]}. An easy computation with the exact sequence tells us that for n ≠ 0, H^{n}(X̃_{ij}) = 0. Since the t–structure is bounded, it follows that X̃_{ij} ∈ T_{[0,0]}. The long exact sequence in homology reduces to a short exact sequence

0 ⟶ X̃_{00} ⟶ X̃_{i0} ⊕ X̃_{0j} ⟶ X̃_{ij} ⟶ 0

in T_{[0,0]}.

Now choose any integers i, i′ and j, j′, subject to the restrictions 0 ≤ i ≤ i′ ≤ p, 0 ≤ j ≤ j′ ≤ q. The commutative square

X̃_{00} ⟶ X̃_{i0} ⊕ X̃_{0j}
↓  ↓
X_{00} ⟶ X_{i′0} ⊕ X_{0j′}

can be completed, uniquely, to a map of short exact sequences in T_{[0,0]}

0 ⟶ X̃_{00} ⟶ X̃_{i0} ⊕ X̃_{0j} ⟶ X̃_{ij} ⟶ 0
↓ ↓ ↓
0 ⟶ X_{00} ⟶ X_{i′0} ⊕ X_{0j′} ⟶ X_{i′j′} ⟶ 0

If i = i′ and j = j′, the vertical maps in the square

X̃_{00} ⟶ X̃_{i0} ⊕ X̃_{0j}
↓  ↓
X_{00} ⟶ X_{i0} ⊕ X_{0j}

are isomorphisms, and it easily follows that the induced map X̃_{ij} ⟶ X_{ij} is an isomorphism. For any 0 ≤ i ≤ i′ ≤ p, 0 ≤ j ≤ j′ ≤ q, we have the following maps of short exact sequences

0 ⟶ X̃_{00} ⟶ X̃_{i0} ⊕ X̃_{0j} ⟶ X̃_{ij} ⟶ 0
↓ ↓ ↓
0 ⟶ X̃_{00} ⟶ X̃_{i′0} ⊕ X̃_{0j′} ⟶ X̃_{i′j′} ⟶ 0
↓ ↓ ↓
0 ⟶ X_{00} ⟶ X_{i′0} ⊕ X_{0j′} ⟶ X_{i′j′} ⟶ 0

and

0 ⟶ X̃_{00} ⟶ X̃_{i0} ⊕ X̃_{0j} ⟶ X̃_{ij} ⟶ 0
↓ ↓ ↓
0 ⟶ X_{00} ⟶ X_{i0} ⊕ X_{0j} ⟶ X_{ij} ⟶ 0
↓ ↓ ↓
0 ⟶ X_{00} ⟶ X_{i′0} ⊕ X_{0j′} ⟶ X_{i′j′} ⟶ 0

The two composites are the unique map

0 ⟶ X̃_{00} ⟶ X̃_{i0} ⊕ X̃_{0j} ⟶ X̃_{ij} ⟶ 0
↓ ↓ ↓
0 ⟶ X_{00} ⟶ X_{i′0} ⊕ X_{0j′} ⟶ X_{i′j′} ⟶ 0

We deduce that the square

X̃_{ij} ⟶ X̃_{i′j′}
↓  ↓
X_{ij} ⟶ X_{i′j′}

commutes, for any 0 ≤ i ≤ i′ ≤ p, 0 ≤ j ≤ j′ ≤ q. In other words, the isomorphisms X̃_{ij} ⟶ X_{ij} are isomorphisms of the diagrams

[figure: ⋮; ⋮; ⋯; ⋯; X̃_{00}; X̃_{0q}; X̃_{p0}; X̃_{pq}]

and

□

In the introduction, we promised the reader that in this section, the inclusion

T[figure: [0,0]]  \overline{}[figure]  T[figure: [0,0]]

will be shown to be a homotopy equivalence. Just for fun, let us show more: Lemma 3.4 Let m ≤ n be any two integers. Then the inlusion

T[figure: [m,n]]  \overline{}[figure]  T[figure: [m,n]]

induces a homotopy equivalence. Proof. The proof breaks up into two, very similar parts. We factor the inclusion

T[figure: [m,n]]  \overline{}[figure]  T[figure: [m,n]]

as the composite

T[figure: [m,n]]  ϕ  \overline{}[figure]  T[figure: [m,n]]  ψ  \overline{}[figure]  T[figure: [m,n]]

and it clearly suffices to show that the maps ϕ and ψ each induce homotopy equivalences. To make it clear just how nearly identical the proofs are, observe that the inclusion

T[figure: [m,n]]  ψ  \overline{}[figure]  T[figure: [m,n]]

is dual to the inclusion

T[figure: [m,n]]  ψ^{op}  \overline{}[figure]  T[figure: [m,n]]

This means precisely that if we replace the triangulated category T by T^{op}, the inclusions ψ and ψ^{op} are interchanged. It suffices therefore to prove that the inclusions

T[figure: [m,n]]  ϕ  \overline{}[figure]  T[figure: [m,n]]

and

T[figure: [m,n]]  ψ^{op}  \overline{}[figure]  T[figure: [m,n]]

are both homotopy equivalences. Since the proofs are nearly identical, I propose to give the proof for ϕ, and make occasional remarks on how to adapt the argument to also work for ψ^{op}.

We begin by considering the commutative square

T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]]  β  \overline{}[figure]  ^{T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]]} α  γ T[figure: [m,n]] [figure] [figure]  T[figure: [m,n]] [figure] [figure]

We will show that β is a homotopy equivalence, by proving that α and γ are. But before doing any of this, we should explain to the reader what the strange symbols mean.

The simplicial set T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]] , for example, is defined to be a trisimplicial set. An (r, s, t)–

[figure: ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

simplex in it is a diagram Once again, the intention is that the symbol for the trisimplicial set should also be its definition. As suggested by the symbol T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]] , a simplex is made up of two boxes, the X’s and the Y ’s, which for the reader’s convenience I frame below The constraints on the X’s and on the morphisms connecting them are indicated by the symbol

inside the bottom box in T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]] ; in our particular case, the X’s are all constrained to lie

in T_{[m,n]}, the vertical maps between them are assumed epi, and the horizontal maps mono. The constraints on the Y ’s are as indicated in the top square; in our particular case, once again the Y ’s must be in the category T_{[m,n]}, and the vertical maps connecting them are epi, the horizontal mono. The last thing we learn from the symbol is that the vertical morphisms connecting the X’s to the Y ’s are assumed epi. This is indicated by the vertical arrows connecting the bottom box to the top. We assume also that we are given a differential Y_{st} ⟶ ΣX_{00}.

The hypothesis is also that the entire diagram lifts to some model. In other words, there is a model T for T and a diagram of bicartesian squares in T, which is isomorphic in T to the simplex. The isomorphism must respect everything, even the differential. There are three simplicial structures, which vary each of the integers r, s and t.

What is the meaning of the simplicial maps indicated? Take for instance the map α. It is denoted

T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]]  α  \overline{}T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}[figure] [figure]

and, once again, the notation is meant to be self–explanatory. There is a map from the simplicial set T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]] to the simplicial set T[figure: [m,n]] . In fact, there are two maps. There is the one that forgets the X’s, and there is the one that forgets the Y ’s. What should be clear from the symbols is that α is the map forgetting the X’s. They have been explicitly crossed out.

Similarly, the simplicial set T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]] is a trisimplicial set. An (r, s, t)–simplex in it is a diagram

[figure: ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

and once again, the intention is for the symbol to also be the definition of the simplicial set. The restrictions on the X’s are given in the bottom square; the X’s are objects of T_{[m,n]}, the horizontal morphisms among them are mono, while the vertical morphisms are unrestricted. The Y ’s are as given in the top square of the symbol T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]] ; they are objects of T_{[m,n]}, the horizontal morphisms are mono, the vertical ones epi. The arrows connecting the two squares in the symbol

T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]] tell us that there are no restrictions on the vertical maps X ⟶ Y. The map β is the obvious inclusion, while the map γ forgets the bottom square.

Now that we have defined all the simplicial sets and simplicial maps in the diagram

T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]]  β  \overline{}[figure]  ^{T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]]} α  γ T[figure: [m,n]] [figure] [figure]  T[figure: [m,n]] [figure] [figure]

the reader can easily check that it commutes. The first step in the proof that ϕ is a homotopy equivalence is to show that α and γ are. From the diagram above, it then follows that so is β.

Let us consider α. If we are to count simplicial structures, then α is a map from a trisimplicial set to a bisimplicial set. Of course, one can view this either as a trisimplicial map of trisimplicial sets, where for one of the trisimplicial sets, one of the simplicial structures is degenerate. Or one can diagonalise all the structures and consider this a map of 1–simplicial sets. Most of the time this is a frivolous technicality. But here we actually want to make the point precise and view α as a trisimplicial map of trisimplicial sets.

The point is that by a theorem of Segal, (Proposition A.1(ii) on page 308 of [18]), it suffices to show that the map becomes a homotopy equivalence of bisimplicial spaces, after we realise just one of the structures. The structure we wish to realise is the one that is degenerate on the target of the map; that is the one that varies the integer r.

Because the structure is degenerate on the target, the realisation of the target is discrete. We need to show the map of bisimplicial spaces a homotopy equivalence. This means that we will show that the bisimplicial space we get from T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]] by realising just the structure that

varies r gives a space which has contractible components. More precisely, the fiber over each component in the target is contractible.

Such a fiber will henceforth be called the Segal fiber of a map. In our case, for each fixed choice of

[figure: ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}]

we need the contractibility of the fiber over it. (The solid frame around the Y ’s is to indicate that everything inside the box is fixed. The integers s and t are fixed, as are all the Y ’s and the maps among them. Even the differential Y_{st} ⟶ ΣY_{00} is fixed.) The fiber is a simplicial set, and an r–simplex is a diagram and once again, we remind the reader that the Y ’s are fixed by the solid frame around them.

The simplicial set consisting of such diagrams will be denoted  Y^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]] . The letter with no

arrow superscripts in the top box indicates that what is in that box is held fixed. Otherwise, the notational conventions are as before.

We need therefore show that the simplicial set  Y^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]] , which is the Segal fiber of the

map α, is contractible. The contracting homotopy is the “contraction to the terminal object”. Recall that, by Section 1, to give a homotopy from the identity map to a contracting map is to give, for every r–simplex and every integer i with 0 ≤ i ≤ r, an (r + 1)–simplex. Recall that this is slightly non–standard, but in Section 1 we explained how to translate this to the more standard notation. In the case above, we send the simplex

[figure: ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

and the integer i to the simplex

[figure: ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{00}; Y_{0t}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{i0}; X_{it}; \} ( \begin{matrix}i+1 \\ terms \end{matrix} ); \{; \} ( \begin{matrix}r−i+1 \\ terms \end{matrix} ); \{]

More intuitively, the homotopy replaces the X’s by the bottom row of the Y ’s. We have a shorthand for such homotopies. They will be denoted by the curious symbol

Y^{}^{}^{}^{}^{}^{}^{}^{}Y_{S}^{}T[figure: [m,n]]

The rationale for this curious symbol is that the homotopy should be thought of as the South face of the Y ’s, denoted in the symbol by Y_{S}, moving down to sweep across the X’s. This is the reason for the arrow indicating the downward movement.

This homotopy connects the identity on  Y^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]] to a contracting map. Hence α is a

homotopy equivalence.

To prove that the map

T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]]  γ  \overline{}T[figure: [m,n]]^{}^{}^{}^{}[figure] [figure]

induces a homotopy equivalence, one proceeds the same way. The Segal fiber is the simplicial

set  Y^{}^{}^{}^{}T[figure: [m,n]] , and is contracted by the homotopy

Y^{}^{}^{}^{}Y_{S}^{}T[figure: [m,n]]

We now know that α and γ are homotopy equivalences, and hence, from the commutative square

T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]]  β  \overline{}[figure]  ^{T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]]} α  γ T[figure: [m,n]] [figure] [figure]  T[figure: [m,n]] [figure] [figure]

we deduce that so is β. Remark concerning ψ. Recall that the proof of Lemma 3.4 reduced to proving that in the composite

T[figure: [m,n]]  ϕ  \overline{}[figure]  T[figure: [m,n]]  ψ  \overline{}[figure]  T[figure: [m,n]]

both ϕ and ψ are homotopy equivalences. To make the proofs more nearly parallel, we decided to show that ϕ and ψ^{op} are homotopy equivalences, where ψ^{op} is the dual of ψ, that is the map

T[figure: [m,n]]  ψ^{op}  \overline{}[figure]  T[figure: [m,n]]

Finally, I promised the reader a proof that ϕ induces a homotopy equivalence, as well as occasional remarks on how to modify the proof to show that so does ψ^{op}. Now is the time to give such a remark.

To prove that ϕ induces a homotopy equivalence we considered the square

T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]]  β  \overline{}[figure]  ^{T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]]} α  γ T[figure: [m,n]] [figure] [figure]  T[figure: [m,n]] [figure] [figure]

To prove that ψ^{op} is a homotopy equivalence, one just considers instead the square

T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]]  β′  \overline{}[figure]  ^{T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]]} α′  γ′ T[figure: [m,n]] [figure] [figure]  T[figure: [m,n]] [figure] [figure]

Caution 3.5 Maybe it should be noted that the difference between the two commutative squares is that in the first, the simplicial sets all have horizontal monos, in the second the simplicial sets all have free horizontal maps. Since the symbol for the simplicial set is also the definition, one should read carefully the restrictions on the objects and the morphisms. Bökstedt once observed that one needs good eyes to read my articles.

Once again, the Segal fiber of the map α′ is  Y^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]] , and is contracted by the homotopy

Y^{}^{}^{}^{}^{}^{}^{}^{}Y_{S}^{}T[figure: [m,n]]

whereas the Segal fiber of the map γ′ is the simplicial set  Y^{}^{}^{}^{}T[figure: [m,n]] , and the homotopy

Y^{}^{}^{}^{}Y_{S}^{}T[figure: [m,n]]

contracts it away. Thus α′ and γ′ induce homotopy equivalences, and hence so must β′.

To complete the proof that both ϕ and ψ^{op} induce homotopy equivalences, we will now consider the commutative squares

T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]]  β  \overline{}[figure]  ^{T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]]} λ  μ [figure] [figure] T[figure: [m,n]]  ϕ  \overline{}[figure]  [figure] [figure] T[figure: [m,n]] and

47

T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [m,n]]  β′  \overline{}[figure]  ^{T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]]} λ′  μ′ [figure] [figure] T[figure: [m,n]]  ψ^{op}  \overline{}[figure]  [figure] [figure] T[figure: [m,n]]

We know that β and β′ are homotopy equivalences. We want to show that so are ϕ and ψ^{op}. It clearly suffices to establish that λ, λ′, μ and μ′ all induce homotopy equivalences.

For λ and λ′ this is easy. The Segal fiber of λ is the simplicial set T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}X , and is

contracted by the homotopy

T[figure: [m,n]] X_{N}^{}^{}^{}^{}^{}^{}^{}^{}^{}X

Although the notation is meant to be self–explanatory, perhaps we should explain it again at this early stage. The simplicial homotopy above takes a pair consisting of an integer i with 0 ≤ i ≤ s and a simplex

[figure: ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

[figure: ⋮; ⋮; ⋯; ⋯; Y_{i0}; Y_{it}; Y_{s0}; Y_{st}; ⋮; ⋮; ⋯; ⋯; X_{r0}; X_{rt}; X_{r0}; X_{rt}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}; \} ( \begin{matrix}i+1 \\ terms \end{matrix} ); \{; \} ( \begin{matrix}s−i+1 \\ terms \end{matrix} ); \{]

The point of the symbol for the homotopy is that the North face of the X’s is moving up to replace the Y ’s; hence the X_{N} and the arrow pointing up.

Similarly, the Segal fiber of λ′ is the simplicial set T[figure: [m,n]]^{}^{}^{}^{}^{}^{}^{}^{}X . It is contracted by the

homotopy

T[figure: [m,n]] X_{N}^{}^{}^{}^{}^{}^{}^{}^{}^{}X

It now remains only to establish that μ and μ′ induce homotopy equivalences. This is the first time we meet a slightly non–trivial fact, and the proof will be deferred. As I promised, all statements whose proof is being deferred will be labeled “Step”, will appear in the table of contents, and will be highlighted for easy reference.

Step 3.1 :

The natural projection

T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]]  μ \overline{}^{[figure] [figure] T[figure: [m,n]]}

induces a homotopy equivalence.

Step 3.2 :

The natural projection

T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]]  μ′  \overline{}^{[figure] [figure] T[figure: [m,n]]}

induces a homotopy equivalence.

Granting Step 3.1 and Step 3.2, we now know that the inclusion of Quillen’s K–theory

T[figure: [0,0]] into T[figure: [0,0]] is a homotopy equivalence. It remains to show that the inclusion of

T[figure: [1,n]] into T[figure: [0,n]] also is; then Theorem I.7.1 follows formally, as in the introduction.

4 An intermediate simplicial set

We need to show that the inclusion

T[figure: [1,n]]  \overline{}[figure]  T[figure: [0,n]]

induces a homotopy equivalence. To prove this, we will factor it through an intermediate simplicial set. We will produce morphisms

T[figure: [1,n]]  α  \overline{}[figure]  T[figure: [0,n]]  β  \overline{}[figure]  T[figure: [0,n]]

and reduce ourselves to proving that α and β are homotopy equivalences.

To begin with, it seems fair to divulge to the reader what the simplicial set T[figure: [0,n]] might

be. As the notation is intended to suggest, it is a subset of the simplicial set T[figure: [0,n]] , where

the vertical arrows are somehow constrained. They are constrained to be of the type [figure]. A morphism

[figure: X; Y]

is by definition a morphism X ⟶ Y in T_{[0,n]}, such that the induced map H^{0}(X) ⟶ H^{0}(Y) is an isomorphism.

With this definition, it is clear that in fact T[figure: [1,n]] is a subset of T[figure: [0,n]] . For if

H^{0}(X) = 0 = H^{0}(Y), then any map X ⟶ Y induces an H^{0}–isomorphism. We will prove in this section:

Proposition 4.1 The natural inclusion

T[figure: [1,n]]  \overline{}[figure]  T[figure: [0,n]]

induces a homotopy equivalence.

The proof of Proposition 4.1 will break up into a sequence of lemmas, several of which will be Steps left till the end. But before we go much further, we must introduce yet another construction for simplicial sets.

We wish to consider next simplicial sets of the type [figure] 0 ^{}^{}^{}^{}T[figure: [1,n]] . The reader

may well wonder what could possibly be meant by the curious–looking triangle in the symbol.

The simplicial set [figure] 0 ^{}^{}^{}^{}T[figure: [1,n]] is naturally a bisimplicial set. An (s, t)–simplex

is a diagram

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}]

The diagram is assumed to come with many differentials, and it is also assumed to lift to at least one model category T for T . The assumption on the existence of the differentials says that every square in the diagram comes with a differential. In other words, suppose we are given two rows and two columns somewhere in the diagram. If the four corners at which they intersect are all occupied, then they give a square

\begin{matrix}B & ⟶ & B′ \\ ↑ & & ↑ \\ A & ⟶ & A′ \end{matrix}

We assume there is a given map B′ ⟶ ΣA, which we call the differential of the square. If one square is contained in another, the differential of the large square should induce the differential of the small square.

The existence of a lifting means that there is a model T for T and a diagram in T

[figure: ⋮; ⋯; ·; ·; ·; Z̃_{00}; Z̃_{0s}; Z̃_{ss}; ⋮; ⋮; ⋯; ⋯; Ỹ_{00}; Ỹ_{0t}; Ỹ_{s0}; Ỹ_{st}]

in which all squares are bicartesian. This diagram should be isomorphic in T to the diagram in T defining the simplex. In particular, the diagonal terms Z̃_{ii} must be isomorphic in T to zero. The isomorphism must respect the differentials, where for the diagram in T the differentials are the natural ones, defined in Section 2.

Note that the 0 in the center of the triangle means that the only conditions imposed on the Z’s are that the diagonal Z’s be zero. Of course, there are some consequences that follow for free. For instance, in the case above, if we consider any pair of integers i ≤ j with 0 ≤ i ≤ j ≤ s, there is a square

\begin{matrix}Z_{jj} & ⟶ & Y_{j0} \\ ↑ & & ↑ \\ Z_{ij} & ⟶ & Y_{i0} \end{matrix}

and since Z_{jj} = 0, it follows that there is a triangle in T

Z_{ij} ⟶ Y_{i0} ⟶ Y_{j0} ⟶ ΣZ_{ij}.

From the long exact sequence in homology and from the fact that Y_{i0} and Y_{j0} are assumed in T_{[1,n]}, it immediately follows that Z_{ij} must be in T_{[1,n+1]}. If we wish to explicitly remind the reader of this, we will write the simplicial set as

[figure] T[figure: [1,n+1]]  T[figure: [1,n]]

The simplicial set [figure] 0 ^{}^{}^{}^{}T[figure: [0,n]] should now be self–explanatory. It is a bisimpli-

cial set, and an (s, t)–simplex is a diagram where now the Y_{ij}’s are assumed in T_{[0,n]}, and the vertical morphisms among them are assumed H^{0}–isomorphisms. The diagram comes with differentials, and with at least one lifting to at least one T. We prove

Lemma 4.2 The natural inclusion

[figure] 0  T[figure: [1,n]]  \overline{}[figure]  [figure] 0  T[figure: [0,n]]

induces a homotopy equivalence.

Proof. In fact, the natual inclusion admits a deformation retraction. There is a homotopy of the

identity on [figure] 0 ^{}^{}^{}^{}T[figure: [0,n]] , which connects the identity to a map factoring through

[figure] 0 ^{}^{}^{}^{}T[figure: [1,n]] . Moreover, the homotopy is relative to the subset.

The homotopy is given by truncation. It sends an integer i with 0 ≤ i ≤ t and an (s, t)– simplex

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}]

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0i}; Y_{s0}; Y_{si}; ⋮; ⋮; ⋯; ⋯; Y_{0i}^{>0}; Y_{0t}^{>0}; Y_{si}^{>0}; Y_{st}^{>0}; _{(i+1)−terms}; _{(t−i+1)−terms}; ; ]

where Y_{ij}^{>0} = Y_{ij}^{≥1} is the t–structure truncation of Y_{ij}. The suggestive shorthand for this homotopy is

[figure] 0  T[figure: [0,n]]  [figure: (T_{[0,n]}^{>0})] \overline{}

Now we need to reason that the homotopy is well–defined.

There is a natural transformation Y ⟶ Y^{>0}, to the truncation functor. Therefore there is no difficulty defining the morphisms in the diagram

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0i}; Y_{s0}; Y_{si}; ⋮; ⋮; ⋯; ⋯; Y_{0i}^{>0}; Y_{0t}^{>0}; Y_{si}^{>0}; Y_{st}^{>0}; _{(i+1)−terms}; _{(t−i+1)−terms}; ; ]

The problem is to define the differentials, and to prove the existence of a lifting to a model T. Given a square of the form

\begin{matrix}Z′ & ⟶ & Y′ \\ ↑ & & ↑ \\ Z & ⟶ & Y \end{matrix}

in the original simplex

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}]

where the Z an Z′ are some Z_{ij}’s and Y and Y′ are some Y_{ij}’s, then we know that we are given a differential Y′ ⟶ ΣZ. But on the other hand, the composite

Y ⟶ Y′ ⟶ ΣZ

must be 0, because

Z′ ⊕ Y ⟶ Y′ ⟶ ΣZ

is part of a triangle. It follows that the map

Y^{≤0} ⟶ Y ⟶ Y′ ⟶ ΣZ

is the zero map. On the other hand, we are assuming that Y and Y′ are in T_{[0,n]}, and that the map Y ⟶ Y′ is an H^{0}–isomorphism. It follows that Y^{≤0} ⟶ (Y′)^{≤0} is an isomorphism. Hence the composite

(Y′)^{≤0} ⟶ Y′ ⟶ ΣZ

is zero. But then Y′ ⟶ ΣZ factors through

Y′ ⟶ (Y′)^{>0} ⟶ ΣZ,

and the factorisation is unique up to a map Σ [(Y′)^{≤0}] ⟶ ΣZ. I assert that any such map must vanish, and hence the factorisation

Y′ ⟶ (Y′)^{>0} ⟶ ΣZ

is unique.

The point is that Σ [(Y′)^{≤0}] is in the category T^{≤−1}, and we will now see that ΣZ must be in the category T^{≥0}. In other words, we will show that Z is in the category T^{≥1}. This follows from the existence of a square

\begin{matrix}Z_{jj} & ⟶ & Y_{j0} \\ ↑ & & ↑ \\ Z_{ij} & ⟶ & Y_{i0} \end{matrix}

for every Z = Z_{ij}. Since Z_{jj} = 0, there is a triangle

Z = Z_{ij} ⟶ Y_{i0} ⟶ Y_{j0} ⟶ ΣZ

and the long exact sequence in homology, coupled with the fact that for k ≤ 0 the map Y_{i0} ⟶ Y_{j0} is an H^{k}–isomorphism, immediately allows us to deduce that H^{k}(Z) = 0 for all k ≤ 0.

Thus we can uniquely define all the differentials (Y′)^{>0} ⟶ ΣZ. But any differential is obtained from one such; any square can be embedded in a square of the form

Z′ ⟶ {Y′}^{>0}
↑  ↑
Z ⟶ Y^{>0}

It therefore only remains to show that the diagram

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0i}; Y_{s0}; Y_{si}; ⋮; ⋮; ⋯; ⋯; Y_{0i}^{>0}; Y_{0t}^{>0}; Y_{si}^{>0}; Y_{st}^{>0}; _{(i+1)−terms}; _{(t−i+1)−terms}; ; ]

has a Waldhausen lifting.

We began with a simplex which, by virtue of being a simplex, must have had a lifting to some Waldhausen model T. Let us choose such a T and a lifting

[figure: ⋮; ⋯; ·; ·; ·; Z̃_{00}; Z̃_{0s}; Z̃_{ss}; ⋮; ⋮; ⋯; ⋯; Ỹ_{00}; Ỹ_{0t}; Ỹ_{s0}; Ỹ_{st}]

We now want to produce a commutative diagram in T

Ỹ_{0} ⟶ Ỹ_{1} ⟶ ⋯ ⟶ Ỹ_{t}
↓ ↓ ↓
Ỹ_{00} ⟶ Ỹ_{01} ⟶ ⋯ ⟶ Ỹ_{0t}

which is isomorphic in T to the diagram

Y_{00}^{≤0} ⟶ Y_{01}^{≤0} ⟶ ⋯ ⟶ Y_{0t}^{≤0}
↓ ↓ ↓
Y_{00} ⟶ Y_{01} ⟶ ⋯ ⟶ Y_{0t}

where the maps Ỹ_{0i} ⟶ Y_{0i} should be the ones already given to us. To do this, we will proceed by descending induction on i. We begin with the case i = t. It is best for the induction to also assume that the maps Ỹ_{i} ⟶ Ỹ_{0i} are fibrations in T.

There is an isomorphism in T

Ỹ_{0t} ⟶ Y_{0t},

and hence a morphism in T

Y_{0t}^{≤0} ⟶ Ỹ_{0t}.

But such a morphisms in T must have a lifting to a morphism in T; there is an object Ỹ_{t} of T, an isomorphism in T

Ỹ_{t} ≃ Y_{0t}^{≤0}

a map in T

Ỹ_{t} ⟶ Ỹ_{0t}

so that the square in T

Ỹ_{t} ⟶^{f} Ỹ_{0t}
↓  ↓
Y_{0t}^{≤0} ⟶ Y_{0t}

commutes. What is more, we may assume that the map f above is a fibration in T; it is an admissible epi in the exact structure.

Suppose now that we have defined Ỹ_{j} for all j ≥ i, together with a diagram

Ỹ_{i} ⟶ Ỹ_{i+1} ⟶ ⋯ ⟶ Ỹ_{t}
↓ ↓ ↓
Ỹ_{0i} ⟶ Ỹ_{0 i+1} ⟶ ⋯ ⟶ Ỹ_{0t}

where the vertical maps are fibrations, and this diagram in T is isomorphic in T to the diagram

Y_{0i}^{≤0} ⟶ Y_{0 i+1}^{≤0} ⟶ ⋯ ⟶ Y_{0t}^{≤0}
↓ ↓ ↓
Y_{0i} ⟶ Y_{0 i+1} ⟶ ⋯ ⟶ Y_{0t}

We wish to extend this to i − 1.

Because the isomorphism has already been established for j ≥ i, we proceed to identify Y_{0j}^{≤0} with Ỹ_{j} for j ≥ i. For all j, we identify Y_{0j} with Ỹ_{0j}. Now we can form the pullback in T, because the map f is a fibration

P ⟶ Ỹ_{0 i−1}
↓  ↓
Ỹ_{i} ⟶^{f}  Ỹ_{0i}

This folds to a triangle in T . We have a triangle in T

P ⟶ Ỹ_{0 i−1} ⊕ Ỹ_{i} ⟶ Ỹ_{0i} ⟶ ΣP.

We also have a commutative square in T

Y_{0 i−1}^{≤0} ⟶ Ỹ_{0 i−1}
↓  ↓
Ỹ_{i}  ⟶^{f}  Ỹ_{0i}

and hence a vanishing composite

Y_{0 i−1}^{≤0} ⟶ Ỹ_{0 i−1} ⊕ Ỹ_{i} ⟶ Ỹ_{0i}.

It follows that in T , the square

Y_{0 i−1}^{≤0} ⟶ Ỹ_{0 i−1}  P ⟶ Ỹ_{0 i−1}
↓  ↓  maps to the square  ↓  ↓
Ỹ_{i}  ⟶^{f}  Ỹ_{0i}  Ỹ_{i} ⟶^{f}  Ỹ_{0i}

This concretely means we have a (non–unique) map Y_{0 i−1}^{≤0} ⟶ P in T , giving a morphism of the commutative squares above. Choose such a map. This map may be lifted to a map Ỹ_{i−1} ⟶ P in T. This means that we have an isomorphism Ỹ_{i−1} ⟶ Y_{0 i−1}^{≤0}, and if we proceed to identify the two via this isomorphism, then Ỹ_{i−1} ⟶ P is a lifting to T of the map Y_{0 i−1}^{≤0} ⟶ P. We may also assume that Ỹ_{i−1} ⟶ P is a fibration.

Since P ⟶ Ỹ_{0 i−1} is the pullback of a fibration, the composite Ỹ_{i−1} ⟶ Ỹ_{0 i−1} is also a fibration. We have therefore established the induction step. We know now that there is a commutative diagram in T

Ỹ_{0} ⟶ Ỹ_{1} ⟶ ⋯ ⟶ Ỹ_{t}
↓ ↓ ↓
Ỹ_{00} ⟶ Ỹ_{01} ⟶ ⋯ ⟶ Ỹ_{0t}

which is isomorphic in T to the diagram

Y_{00}^{≤0} ⟶ Y_{01}^{≤0} ⟶ ⋯ ⟶ Y_{0t}^{≤0}
↓ ↓ ↓
Y_{00} ⟶ Y_{01} ⟶ ⋯ ⟶ Y_{0t}

Using this, we can define a map (in T) from the commutative diagram in T

[figure: ⋮; ⋮; ⋯; ⋯; Ỹ_{00}; Ỹ_{0t}; Ỹ_{s0}; Ỹ_{st}]

to the commutative diagram (in T)

[figure: ⋮; ⋮; ⋯; ⋯; Cone {Ỹ_{0} ⟶ Ỹ_{00}}; Cone {Ỹ_{t} ⟶ Ỹ_{0t}}; Cone {Ỹ_{0} ⟶ Ỹ_{s0}}; Cone {Ỹ_{t} ⟶ Ỹ_{st}}]

where the object

Cone {A ⟶ B}

is the mapping cone on the map A ⟶ B, defined in terms of the cylinder functor in T. Now the diagram

[figure: ⋮; ⋯; ·; ·; ·; Z̃_{00}; Z̃_{0s}; Z̃_{ss}; ⋮; ⋮; ⋯; ⋯; Ỹ_{00}; Ỹ_{0i}; Ỹ_{s0}; Ỹ_{si}; ⋮; ⋮; ⋯; ⋯; Cone{Ỹ_{i}⟶Ỹ_{0i}}; Cone{Ỹ_{t}⟶Ỹ_{0t}}; Cone{Ỹ_{i}⟶Ỹ_{si}}; Cone{Ỹ_{t}⟶Ỹ_{st}}; _{(i+1)−terms}; _{(t−i+1)−terms}; ; ]

is clearly a lifting to T of

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0i}; Y_{s0}; Y_{si}; ⋮; ⋮; ⋯; ⋯; Y_{0i}^{>0}; Y_{0t}^{>0}; Y_{si}^{>0}; Y_{st}^{>0}; _{(i+1)−terms}; _{(t−i+1)−terms}; ; ]

and, being a diagram of bicartesian squares, must come with coherent differentials. Since we already know that these differentials are unique, they must agree with the ones given earlier. □

We know now, by virtue of Lemma 4.2, that the map

[figure] 0  T[figure: [1,n]]  \overline{}[figure]  [figure] 0  T[figure: [0,n]]

induces a homotopy equivalence. Next we wish to consider the commutative square

[figure] 0  T[figure: [1,n]] T[figure: [1,n]]  β  \overline{}[figure]  ^{[figure] 0  T[figure: [0,n]] T[figure: [0,n]]} α  γ [figure] 0  T[figure: [1,n]] [figure] [figure]  δ  \overline{}[figure]  [figure] 0  T[figure: [0,n]] [figure] [figure]

Notation 4.3 We owe the reader a definition of the simplicial sets in this diagram. An (r, s, t)–

simplex in the trisimplicial sets [figure] 0  ^{}^{}^{}^{}T[figure: [1,n]]^{}^{}^{}^{}T[figure: [1,n]] or [figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}T[figure: [0,n]] is a dia-

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

For every pair of rows and every pair of columns, if all four corners in which they meet are occupied, we assume there is a differential for the resulting square. The differentials of large squares induce the differentials of any smaller squares they contain. And the entire diagram admits at least one lifting to at least one model.

There are restrictions on the objects, as well as on the horizontal and vertical morphisms. These are spelt out in the symbol for the simplicial set.

We know that the map δ is a homotopy equivalence. We want to show that so is β. This will follow from the next two easy lemmas, showing that α and γ are homotopy equivalences.

Lemma 4.4 The map

[figure] 0  T[figure: [1,n]] T[figure: [1,n]]  α  \overline{}[figure] 0  T[figure: [1,n]] [figure] [figure]

Proof. The Segal fiber is the simplicial set [figure] Z ^{}^{}^{}^{}Y^{}^{}^{}^{}T[figure: [1,n]] , and is contracted by the homotopy

[figure] Z  Y Y_{S} T[figure: [1,n]]
□

Lemma 4.5 The map

[figure] 0  T[figure: [0,n]] T[figure: [0,n]]  γ  \overline{}[figure] 0  T[figure: [0,n]] [figure] [figure]

induces a homotopy equivalence.

[figure] Z  Y Y_{S} T[figure: [0,n]]
□

Corollary 4.6 Recalling that in the square

[figure] 0  T[figure: [1,n]] T[figure: [1,n]]  β  \overline{}[figure]  ^{[figure] 0  T[figure: [0,n]] T[figure: [0,n]]} α  γ [figure] 0  T[figure: [1,n]] [figure] [figure]  δ  \overline{}[figure]  [figure] 0  T[figure: [0,n]] [figure] [figure]

we have now proved that δ is a homotopy equivalence (Lemma 4.2), that α is a homotopy equivalence (Lemma 4.4) and that γ is a homotopy equivalence (Lemma 4.5), it follows that β is a homotopy equivalence too. □

Next we will study the commutative square

[figure] 0  T[figure: [1,n]] T[figure: [1,n]]  β  \overline{}[figure]  ^{[figure] 0  T[figure: [0,n]] T[figure: [0,n]]} μ  ν [figure] [figure] [figure] [figure] [figure] T[figure: [1,n]]  λ  \overline{}[figure]  [figure] [figure] [figure] [figure] [figure] T[figure: [0,n]]

We already know that β is a homotopy equivalence. Proposition 4.1, which is the proposition this section aims to prove, is that λ induces a homotopy equivalence. It clearly suffices to show that each of μ and ν does. In each case, this is a more subtle argument, which we postpone to the end. As with all arguments we postpone to the end, we carefully highlight these for easy reference.

Step 4.1 :

The natural map

[figure] 0 T[figure: [1,n]] T[figure: [1,n]]  μ  \overline{}[figure] [figure] [figure] [figure] [figure] T[figure: [1,n]]

induces a homotopy equivalence. Step 4.2 : The natural map

[figure] 0 T[figure: [0,n]] T[figure: [0,n]]  ν  \overline{}[figure] [figure] [figure] [figure] [figure] T[figure: [0,n]]

induces a homotopy equivalence.

5 The proof (with some points deferred till later)

Now that the preliminaries are taken care of, we need to establish that the inclusion

T[figure: [0,n]] \overline{}[figure] T[figure: [0,n]]

induces a homotopy equivalence. In this section we will give the proof, up to leaving the more delicate homotopies till later.

To begin with, we consider the commutative diagram

T[figure: [0,n]]  T[figure: [0,n]] [figure] 0  β  \overline{}[figure]  ^{T[figure: [0,n]]  T[figure: [0,n]] [figure] 0} μ  ν T[figure: [0,n]]  [figure] [figure] [figure] [figure] [figure]  λ  \overline{}[figure]  T[figure: [0,n]]  [figure] [figure] [figure] [figure] [figure]

We need to show that the map λ induces a homotopy equivalence. The strategy of the proof will be to show that β, μ and ν do. Of these, μ and ν are technical but straight forward. We leave them till later. We will label them as “Steps” and highlight them. The rest of the section will concern itself with the proof that β induces a homotopy equivalence.

Notation 5.1 In fairness to the reader, we should perhaps explain the meaning of the symbol

T[figure: [0,n]]  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}[figure] 0  ; the other simplicial sets in the diagram are closely analogous. A

simplex in this simplicial set is a diagram

[figure: ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0r}; Y_{t0}; Y_{tr}; ⋮; ⋮; ⋯; ⋯; B_{00}; B_{0s}; B_{t0}; B_{ts}; ⋮; ⋯; 0; A_{s0}; 0; ·; ·; ·]

together with differentials, which has at least one lift to a model category. Perhaps it should be noted that this simplicial set is transpose to the simplicial set

[figure] 0  T[figure: [0,n]] T[figure: [0,n]]

In other words, a simplex in the simplicial set T[figure: [0,n]]  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}[figure] 0  is precisely a dia-

[figure: ⋮; ⋯; ·; ·; ·; 0; A_{s0}; 0; ⋮; ⋮; ⋯; ⋯; B_{00}; B_{t0}; B_{0s}; B_{ts}; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{t0}; Y_{0r}; Y_{tr}]

all we have done is transposed the diagram. Note also that the sign of the differentials must be switched when we transpose.

Step 5.1 : The simplicial map

T[figure: [0,n]]  T[figure: [0,n]] [figure] 0  μ  \overline{}T[figure: [0,n]]  [figure] [figure] [figure] [figure] [figure]

induces a homotopy equivalence. Step 5.2 : The simplicial map

T[figure: [0,n]]  T[figure: [0,n]] [figure] 0  ν  \overline{}T[figure: [0,n]]  [figure] [figure] [figure] [figure] [figure]

induces a homotopy equivalence.

Modulo the various Steps left for later, we are therefore reduced to showing that the map β above, that is

T[figure: [0,n]]  T[figure: [0,n]] [figure] 0  β  \overline{}[figure]  T[figure: [0,n]]  T[figure: [0,n]] [figure] 0

induces a homotopy equivalence. To do this, we make use of the following few lemmas.

Lemma 5.2 The natural projection

T[figure: [0,n]]  T[figure: [0,n]] [figure] 0  \overline{}[figure] [figure] T[figure: [0,n]] [figure] 0

Proof. The Segal fiber is the simplicial set T[figure: [0,n]]  ^{}^{}^{}^{}B^{}^{}^{}^{}[figure] A  , and the following homo-

topy contracts it away

T[figure: [0,n]]  B_{W} \overline{}B [figure] A
□

Next, let us observe that the simplicial set T[figure: [0,n]]^{}^{}^{}^{}[figure] 0 by definition consists of diagrams

[figure: ⋮; ⋮; ⋯; ⋯; B_{00}; B_{0s}; B_{t0}; B_{ts}; ⋮; ⋯; 0; A_{s0}; 0; ·; ·; ·]

and the meaning of the zero inside the triangle is that the only restriction placed on the A’s is that the diagonal ones be zero. But of course there may be other restrictions that follow as a consequence. In particular, for any s ≥ i ≥ j ≥ 0, there is a square

B_{0j} ⟶ B_{0i}
↑  ↑
A_{ij} ⟶ A_{ii}

and since A_{ii} = 0, this gives a triangle

A_{ij} ⟶ B_{0j} ⟶ B_{0i} ⟶ ΣA_{ij}

We know that B_{0j} and B_{0i} are objects in T_{[0,n]}, and that the map B_{0j} ⟶ B_{0i} is epi. This means, by Definition 3.2, that the map H^{n}(B_{0j}) ⟶ H^{n}(B_{0i}) is epi. By the long exact sequence for H, applied to the above triangle, we get that A_{ij} must be in T_{[0,n]}.

But more turns out to be true. Given any s ≥ i ≥ j′ ≥ j ≥ 0, there is a square

A_{ij} ⟶ A_{ij′}
↑  ↑
A_{j′j} ⟶ A_{j′j′}

and as A_{j′j′} = 0, we deduce a triangle

A_{j′j} ⟶ A_{ij} ⟶ A_{ij′} ⟶ ΣA_{j′j}

where every object is in T_{[0,n]}. But then the long exact sequence in homology tells us that H^{n}(A_{ij}) ⟶ H^{n}(A_{ij′}) must be epi. Thus the horizontal maps A_{ij} ⟶ A_{ij′} are epi in T_{[0,n]}. It follows that the following simplicial sets are equal:

T[figure: [0,n]] [figure] 0  T[figure: [0,n]] [figure] T[figure: [0,n]]

Now we are ready for the next lemma.

Lemma 5.3 The simplicial map

T[figure: [0,n]] [figure] T[figure: [0,n]]  \overline{}[figure] [figure] [figure] T[figure: [0,n]]

Proof. The Segal fiber is the simplicial set T[figure: [0,n]]^{}^{}^{}^{}[figure] A , and it is contracted by the homotopy

T[figure: [0,n]] A_{N} [figure] A
□

Corollary 5.4 The map

T[figure: [0,n]]  T[figure: [0,n]] [figure] T[figure: [0,n]]  π  \overline{}[figure] [figure] [figure] [figure] [figure] T[figure: [0,n]]

induces a homotopy equivalence. This is just putting together Lemmas 5.2 and 5.3. We need to show that the map

T[figure: [0,n]]  T[figure: [0,n]] [figure] 0  β  \overline{}[figure]  T[figure: [0,n]]  T[figure: [0,n]] [figure] 0

induces a homotopy equivalence. By Corollary 5.4 the map π certainly does. It therefore suffices to show that the composite π ∘ β induces a homotopy equivalence. This is the map

T[figure: [0,n]]  T[figure: [0,n]] [figure] T[figure: [0,n]]  π ∘ β \overline{}[figure] [figure] [figure] [figure] [figure] T[figure: [0,n]]

What we will show, which clearly suffices, is that the Segal fiber is contractible. The rest of the proof amount to a study of this Segal fiber,

T[figure: [0,n]]  T[figure: [0,n]] [figure] A

We begin with the commutative square

[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  T[figure: [0,n]] [figure] A  α  \overline{}^{T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  T[figure: [0,n]] [figure] A} β  γ [figure] [figure] [figure] [figure] [figure] [figure] [figure] T[figure: [0,n]]  T[figure: [0,n]] [figure] A  [figure] [figure] [figure] [figure] T[figure: [0,n]]  T[figure: [0,n]] [figure] A

We want to show that α is a homotopy equivalence. We will prove it by showing that β and γ are. For γ, this is a trivial lemma.

Lemma 5.5 The map

T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  T[figure: [0,n]] [figure] A  γ  \overline{}[figure] [figure] [figure] [figure] T[figure: [0,n]]  T[figure: [0,n]] [figure] A

induces a homotopy equivalence. Proof. The Segal fiber is contracted by the homotopy

T[figure: [0,n]]  T[figure: [0,n]] Y_{N}  B_{N} Y  B [figure] A
□

Not so trivial is the next Step, left till later:

Step 5.3 : The natural map

[figure] 0 T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  T[figure: [0,n]] [figure] A  β \overline{}[figure] [figure] [figure] [figure] [figure] [figure] [figure] T[figure: [0,n]]  T[figure: [0,n]] [figure] A

induces a homotopy equivalence.

Modulo the Step, we then know that the map α is a homotopy equivalence. Next we consider the square

[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  T[figure: [0,n]] [figure] A  α  \overline{}^{T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  T[figure: [0,n]] [figure] A} β  γ [figure] 0  T[figure: [0,n]]  T[figure: [0,n]] [figure] [figure] [figure] [figure] [figure] A  δ  \overline{}T[figure: [0,n]]  T[figure: [0,n]] [figure] [figure] [figure] [figure] [figure] A

We know that α is a homotopy equivalence. We wish to know that so is δ. This follows from the two trivial lemmas:

Lemma 5.6 The natural map

[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  T[figure: [0,n]] [figure] A  β  \overline{}[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] [figure] [figure] [figure] [figure] [figure] A

induces a homotopy equivalence. Proof. The Segal fiber is contracted by the homotopy

[figure] Z  Y  B Y_{S}  B_{S} T[figure: [0,n]]  T[figure: [0,n]] [figure] A
□

Lemma 5.7 The natural map

T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  T[figure: [0,n]] [figure] A  γ  \overline{}T[figure: [0,n]]  T[figure: [0,n]] [figure] [figure] [figure] [figure] [figure] A

induces a homotopy equivalence. Proof. The Segal fiber is contracted by the homotopy

Y  B Y_{S}  B_{S} T[figure: [0,n]]  T[figure: [0,n]] [figure] A
□

Now we know that the map δ is a homotopy equivalence. In other words, we have learned so far Corollary 5.8 The natural projection

[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] [figure] A  \overline{}T[figure: [0,n]]  T[figure: [0,n]] [figure] A

induces a homotopy equivalence. Let us remind the reader that what remains to be proved, modulo the various Steps, is the

contractibility of T[figure: [0,n]]  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}[figure] A  .

The next few Lemmas will prove that the map

[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A  \overline{}[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] [figure] [figure]  [figure] A

induces a homotopy equivalence. But first maybe we should explain the meaning of the simplicial

set [figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}[figure] A  . A simplex is a diagram

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0t}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0r}; Y_{t0}; Y_{tr}; ⋮; ⋮; ⋯; ⋯; B_{00}; B_{0s}; B_{t0}; B_{ts}; ⋮; ⋯; 0; A_{s0}; 0; ·; ·; ·; ⋯; X_{0}; X_{r}]

In other words, the meaning of the line in the symbol for the simplicial set is that there is a row of X’s. The hypotheses on the X’s are as indicated; they lie in the category T_{[0,n]}, and the morphisms between them (which are all horizontal) are unrestricted.

We begin by considering the square

[figure] 0  [figure] [figure] T[figure: [0,n]]  T[figure: [0,n]] [figure] [figure]  T[figure: [0,n]]  [figure] A  α [figure]^{[figure] 0  T[figure: [0,n]]  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  T[figure: [0,n]]  [figure] A} β [figure] 0  [figure] [figure] T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A  γ [figure]^{[figure] 0  T[figure: [0,n]]  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A}

We will show that the simplicial map β is a homotopy equivalence, by establishing the two easy lemmas, namely that α and γ are. Lemma 5.9 The map α in the diagram above induces a homotopy equivalence. Proof. The Segal fiber is contracted by the homotopy

[figure] Z  T[figure: [0,n]]  Y_{W} \overline{}Y  B T[figure: [0,n]]  X_{W}  X  [figure] A

□

Lemma 5.10 The map γ in the diagram above induces a homotopy equivalence. Proof. The Segal fiber is contracted by the homotopy

[figure] Z  T[figure: [0,n]]  Y_{W} \overline{}Y  B X  [figure] A

□

Now we turn our attention to the commutative square

[figure] 0  T[figure: [0,n]]  [figure] [figure] T[figure: [0,n]] T[figure: [0,n]]  [figure] [figure]  [figure] A  μ [figure]^{[figure] 0  T[figure: [0,n]]  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  T[figure: [0,n]]  [figure] A} λ  β [figure] 0  T[figure: [0,n]]  [figure] [figure] T[figure: [0,n]] [figure] [figure]  [figure] A  ν [figure]^{[figure] 0  T[figure: [0,n]]  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A}

We know that β induces a homotopy equivalence. We want to know that so does λ. To prove this, we will establish the trivial fact that μ is a homotopy equivalence, and leave till later the Step that so does ν.

Lemma 5.11 In the diagram above, the map μ is a homotopy equivalence.

Proof. The Segal fiber is contracted by the homotopy

[figure] Z  T[figure: [0,n]] Y_{E} \overline{}Y  B T[figure: [0,n]] X_{E} X  [figure] A
□

Not quite so trivial is the Step

Step 5.4 : The natural map

[figure] 0 T[figure: [0,n]]  [figure] [figure] T[figure: [0,n]] [figure] [figure]  [figure] A  ν [figure]^{[figure] 0 T[figure: [0,n]]  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A}

induces a homotopy equivalence.

Lemmas 5.9, 5.10 and 5.11, as well as Step 5.4, combine to allow us to deduce that Corollary 5.12 The natural projection

[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A  \overline{}[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] [figure] [figure]  [figure] A

induces a homotopy equivalence.

By Corollary 5.8 we already know that the map

[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] [figure] A  \overline{}T[figure: [0,n]]  T[figure: [0,n]] [figure] A

induces a homotopy equivalence. We deduce that so does the composite. We want to show

that T[figure: [0,n]]  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}[figure] A  is contractible. Equivalently, we can show that any of the other,

homotopy equivalent sets is contractible. What we will actually show is that the homotopy equivalence above factors through a contractible set. Precisely,

Step 5.5 : The map

[figure] 0 T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A  ϕ \overline{}T[figure: [0,n]]  T[figure: [0,n]] [figure] A

which we know to be a homotopy equivalence by Corollaries 5.8 and 5.12, factors up to homotopy through the simplicial set

[figure] 0 ^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}[figure] A  .

The remainder of the proof is essentially all trivial. We will show that

[figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}[figure] A  is contractible, and show it very concretely. We proceed

in two lemmas.

Lemma 5.13 The projection

[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A  \overline{}[figure] 0  [figure] [figure] T[figure: [0,n]] [figure] [figure]  [figure] A

Proof. The Segal fiber is the simplicial set [figure] Z  ^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}^{}^{}B^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}[figure] A  , and it is

contracted by a homotopy which I would like to denote

Z_{E} \overline{}^{[figure] Z  T[figure: [0,n]]  B T[figure: [0,n]]  [figure] A}

but this symbol requires a little explanation. Recall that a homotopy sends a pair, consisting of an integer i and a simplex, to a simplex. Take the simplex

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0t}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0r}; Y_{t0}; Y_{tr}; ⋮; ⋮; ⋯; ⋯; B_{00}; B_{0s}; B_{t0}; B_{ts}; ⋮; ⋯; 0; A_{s0}; 0; ·; ·; ·; ⋯; X_{0}; X_{r}]

Perhaps we should explain the meaning of the various arrows. An arrow [figure] is one which

is simultaneously  [figure] and  [figure]. In other words, it is an H^{0} isomorphism, and an

H^{n}–epimorphism. An arrow labeled [figure] is simultaneously [figure] and [figure]. It is simultaneously an H^{0}–monomorphism and an H^{n}–epimorphism.

By virtue of being a simplex, we know that there is at least one lifting to at least one model T for T . Write this diagram

[figure: ⋮; ⋯; ·; ·; ·; Z̃_{00}; Z̃_{0t}; Z̃_{tt}; ⋮; ⋮; ⋯; ⋯; Ỹ_{00}; Ỹ_{0r}; Ỹ_{t0}; Ỹ_{tr}; ⋮; ⋮; ⋯; ⋯; B̃_{00}; B̃_{0s}; B̃_{t0}; B̃_{ts}; ⋮; ⋯; Ã_{00}; Ã_{s0}; Ã_{ss}; ·; ·; ·; ⋯; X̃_{0}; X̃_{r}]

There is no loss in assuming the map Ã_{ss} ⟶ B̃_{0s} a fibration in the category T; this may be done by the Cylinder Axiom for the dual of T. Hence we may form the pullbacks

[figure: ⋮; ⋯; ·; ·; ·; Z̃_{00}; Z̃_{0t}; Z̃_{tt}; ⋮; ⋮; ⋯; ⋯; Ỹ_{00}; Ỹ_{0r}; Ỹ_{t0}; Ỹ_{tr}; ⋮; ⋮; ⋯; ⋯; B̃_{00}; B̃_{0s}; B̃_{t0}; B̃_{ts}; ⋮; ⋯; Ã_{00}; Ã_{s0}; Ã_{ss}; ·; ·; ·; ⋯; R̃_{0}; R̃_{t}; ⋯; X̃_{0}; X̃_{r}]

where the R’s are fixed. In fact, the natural differential for the square

\begin{matrix}Z̃_{ii} & ⟶ & B̃_{is} \\ ↑ & & ↑ \\ R̃_{i} & ⟶ & Ã_{ss} \end{matrix}

establishes an isomorphism in the category T

B̃_{is} ≃ ΣR̃_{i}.

So not only are the R̃_{i} independent of the integer r and of the non–fixed part of the data, but in T the pullback, together with the maps into it, give no new data. The homotopy we have in mind takes the integer i with 0 ≤ i ≤ r and the cell

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0t}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0r}; Y_{t0}; Y_{tr}; ⋮; ⋮; ⋯; ⋯; B_{00}; B_{0s}; B_{t0}; B_{ts}; ⋮; ⋯; 0; A_{s0}; 0; ·; ·; ·; ⋯; X_{0}; X_{r}] to the cell

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0t}; 0; ⋮; ⋮; ⋯; ⋯; Z_{0t}; Z_{0t}; Z_{tt}; Z_{tt}; ⋮; ⋮; ⋯; ⋯; Y_{0i}; Y_{0r}; Y_{ti}; Y_{tr}; ⋮; ⋮; ⋯; ⋯; B_{00}; B_{0s}; B_{t0}; B_{ts}; ⋮; ⋯; 0; A_{s0}; 0; ·; ·; ·; ; ; ⋯; Σ^{−1}B_{ts}; Σ^{−1}B_{ts}; ⋯; X_{i}; X_{r}; _{(i+1)−terms}; _{(r−i+1)−terms}]

This diagram has a Waldhausen lifting, namely

[figure: ⋮; ⋯; ·; ·; ·; Z̃_{00}; Z̃_{0t}; Z̃_{tt}; ⋮; ⋮; ⋯; ⋯; Z̃_{0t}; Z̃_{0t}; Z̃_{tt}; Z̃_{tt}; ⋮; ⋮; ⋯; ⋯; Ỹ_{0i}; Ỹ_{0r}; Ỹ_{ti}; Ỹ_{tr}; ⋮; ⋮; ⋯; ⋯; B̃_{00}; B̃_{0s}; B̃_{t0}; B̃_{ts}; ⋮; ⋯; Ã_{00}; Ã_{s0}; Ã_{ss}; ·; ·; ·; ; ; ⋯; R̃_{t}; R̃_{t}; ⋯; X̃_{i}; X̃_{r}; _{(i+1)−terms}; _{(r−i+1)−terms}]

What needs a little discussion is why the homotopy is well–defined. In particular, we need to show that the cell just given is in the simplicial set. This requires us to show that the objects R̃_{t} and Z_{kt} are in T_{[0,n]}, and that the morphisms among them satisfy the hypotheses. The hypotheses which would qualify the diagram to belong to our simplicial set are that all the vertical maps be epi, and that the the vertical maps Z_{kt} ⟶ Z_{lt} are H^{0}–isomorphisms, while R̃_{t} ⟶ Z_{kt} are H^{0}–monomorphisms.

Whatever happens, given any object A which is either R̃_{j} or Z_{kj}, we can find a square

\begin{matrix}Z_{jj} & ⟶ & B′ \\ ↑ & & ↑ \\ A & ⟶ & B \end{matrix}

where A is R̃_{j} or Z_{kj}, and B and B′ are either X’s or Y ’s. That is, B′ can be taken to be Y_{j0}, while B will be either X_{0} or Y_{k0}. In any case, the map B ⟶ B′ is certainly epi, and certainly mono. (If both B and B′ are Y ’s, then we know more, namely that the map is an H^{0}–isomorphism. We know this, but we do not care.)

Because Z_{jj} is isomorphic to 0, there is a long exact sequence in cohomology for the triangle

A ⟶ B ⟶ B′ ⟶ ΣA

and since we know that B and B′ are in T_{[0,n]} and the map B ⟶ B′ is both mono and epi, it follows immediately that A is in T_{[1,n]}. Thus H^{0}(A) = 0, and we immediately deduce that the hypotheses on H^{0} of the vertical maps R_{t} ⟶ Z_{kt} and Z_{kt} ⟶ Z_{lt} are true. The identity map on the zero object is both mono and an isomorphism.

It remains to show that the vertical maps are epi. For this, observe that given a vertical map B ⟶ B′, where B is R̃_{t} or Z_{kt} and B′ is Z_{lt}, then there is a square

\begin{matrix}Z_{ll} & ⟶ & B′ \\ ↑ & & ↑ \\ A & ⟶ & B \end{matrix}

where A is again either an R or a Z. But Z_{ll} = 0, and we already know that A, B and B′ lie in T_{[1,n]}. From the long exact sequence for the triangle

A ⟶ B ⟶ B′ ⟶ ΣA

we immediately deduce that B ⟶ B′ is an H^{n}–epimorphism. □

Lemma 5.14 The simplicial set [figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}[figure] A  is contractible.

Proof. The contracting homotopy takes an integer i with 0 ≤ i ≤ t and a simplex

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0t}; 0; ⋮; ⋮; ⋯; ⋯; B_{00}; B_{0s}; B_{t0}; B_{ts}; ⋮; ⋯; 0; A_{s0}; 0; ·; ·; ·]

to the simplex

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{it}; 0; ⋮; ⋮; ⋯; ⋯; B_{i0}; B_{is}; B_{t0}; B_{ts}; ⋮; ⋯; ·; ·; ·; 0; 0; 0; ⋮; ⋮; ⋯; ⋯; Σ^{−1}B_{is}; Σ^{−1}B_{ts}; Σ^{−1}B_{is}; Σ^{−1}B_{ts}; ⋮; ⋮; ⋯; ⋯; A_{s0}; A_{ss}; A_{s0}; A_{ss}; ⋮; ⋯; 0; A_{s0}; 0; ·; ·; ·; \} ( \begin{matrix}(i+1) \\ times \end{matrix} ); \} ( \begin{matrix}(t−i+1) \\ terms \end{matrix} )]

and the fact that a Waldhausen lift exists is as in Lemma 5.13, where the lifts for the Σ^{−1}B_{js} were the R̃_{j}. □

6 Proof of all the Steps but Step 5.5

The proof is now finished, modulo the Steps which we left as technical details, to be proved at the end. To complete the article we now need to prove them. Of these Steps, the first 8 are nearly identical, and will be dealt with in this section. Step 5.5 is somewhat different and its proof will be the next section.

It is perhaps best to start with one of the Steps, give a complete proof of it, and then explain why the other Steps follow by essentially the same argument. Let us therefore prove Step 4.1. It is:

Step 4.1: The natural map

[figure] 0 T[figure: [1,n]] T[figure: [1,n]]  μ  \overline{}[figure] [figure] [figure] [figure] [figure] T[figure: [1,n]]

induces a homotopy equivalence.

Proof of Step 4.1. It clearly suffices to prove the contractibility of the Segal fiber

[figure] 0  ^{}^{}^{}^{}T[figure: [1,n]]^{}^{}^{}^{}X  . To contract it, we will use two simplicial homotopies. The first one

is really the key homotopy in this entire theory. We will now discuss it carefully.

Any homotopy takes an s–simplex and an integer i with 0 ≤ i ≤ s to a simplex. Let us consider the homotopy taking i and the s–simplex

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{is}; 0; ⋮; ⋮; ⋯; ⋯; Y_{i0}; Y_{it}; Y_{s0}; Y_{st}; ⋮; ⋯; ·; ·; ·; 0; Z_{0i}; 0; ⋮; ⋮; ⋯; ⋯; Z_{0i}⊕X_{r0}; Z_{0s}⊕X_{r0}; Z_{ii}⊕X_{r0}; Z_{is}⊕X_{r0}; ⋮; ⋮; ⋯; ⋯; Y_{00}⊕X_{r0}; Y_{00}⊕X_{rt}; Y_{i0}⊕X_{r0}; Y_{i0}⊕X_{rt}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

to the diagram To explain this cell, it is perhaps simplest and clearest if we do this at the level of Waldhausen liftings. Recall that the s–simplex we began with, namely is a simplex, and therefore has a Waldhausen lifting. There is some model category T for T , and a diagram of bicartesian squares in T

[figure: ⋮; ⋯; ·; ·; ·; Z̃_{00}; Z̃_{0s}; Z̃_{ss}; ⋮; ⋮; ⋯; ⋯; Ỹ_{00}; Ỹ_{0t}; Ỹ_{s0}; Ỹ_{st}; ⋮; ⋮; ⋯; ⋯; X̃_{00}; X̃_{0t}; X̃_{r0}; X̃_{rt}]

[figure: ⋮; ⋯; ·; ·; ·; Z̃_{ii}; Z̃_{is}; Z̃_{ss}; ⋮; ⋮; ⋯; ⋯; Ỹ_{i0}; Ỹ_{it}; Ỹ_{s0}; Ỹ_{st}; ⋮; ⋯; ·; ·; ·; Z̃_{00}; Z̃_{0i}; Z̃_{ii}; ⋮; ⋮; ⋯; ⋯; Z̃_{0i}⊕X̃_{r0}; Z̃_{0s}⊕X̃_{r0}; Z̃_{ii}⊕X̃_{r0}; Z̃_{is}⊕X̃_{r0}; ⋮; ⋮; ⋯; ⋯; Ỹ_{00}⊕X̃_{r0}; Ỹ_{00}⊕X̃_{rt}; Ỹ_{i0}⊕X̃_{r0}; Ỹ_{i0}⊕X̃_{rt}; ⋮; ⋮; ⋯; ⋯; X̃_{00}; X̃_{0t}; X̃_{r0}; X̃_{rt}]

lifting the simplex. Now consider the diagram where the maps are as follows. We divide the diagram into regions, by framing each one with a dashed box.

Within each region, the maps are the obvious diagonals. There are only two “interesting” regions in the diagram; the two where the objects are sums of two objects in the original simplex. Maps connecting two “uninteresting” regions are defined to be just the maps from the original simplex. It remains to give the maps connecting uninteresting to interesting regions, and the maps between the two interesting regions.

The maps X̃ ⟶ Ỹ ⊕ X̃′ in the diagram above are the matrices

( \begin{matrix}0 \\ α \end{matrix} )

where α : X̃ ⟶ X̃′ is the natural map of the given simplex. The map Z̃ ⟶ Ỹ ⊕ X̃′ is the matrix

( \begin{matrix}α \\ 0 \end{matrix} )

where α : Z̃ ⟶ Ỹ is the natural map. The map Z̃ ⊕ X̃_{r0} ⟶ Ỹ ⊕ X̃′ is the matrix

( \begin{matrix}α & −β \\ 0 & γ \end{matrix} )

where α : Z̃ ⟶ Ỹ, β : X̃_{r0} ⟶ Ỹ and γ : X̃_{r0} ⟶ X̃′ are the maps of the simplex we began with.

Thus far, we have described all the maps into the interesting region containing objects of the form Ỹ_{k0} ⊕ X̃_{rl}. The map Ỹ ⊕ X̃′ ⟶ Ỹ′ is the matrix

( α β )

where α : Ỹ ⟶ Ỹ′ and β : X̃′ ⟶ Ỹ′ are the maps of the original simplex. We have now defined all the maps in and out of Ỹ ⊕ X̃′.

The map Z̃ ⟶ Z̃′ ⊕ X̃_{r0} is just

( \begin{matrix}α \\ 0 \end{matrix} )

where α : Z̃ ⟶ Z̃′ is a map of the original simplex. The maps Z̃′⊕X̃_{r0} ⟶ Z̃ and Z̃′⊕X̃_{r0} ⟶ Ỹ are all

( α 0 )

where α : Z̃′ ⟶ Z̃ or α : Z̃′ ⟶ Ỹ is the given map of the original simplex.

Now that I have given all the maps in this diagram, it is easy to verify that the diagram commutes, and all squares in it are bicartesian. It suffices to check this on the small squares that one obtains from two adjacent rows and two adjacent columns. For those, the argument divides into a small number of trivial cases.

It follows that the above diagram in T gives a simplex; there are coherent differentials, namely the natural ones. And clearly the diagram has a lifting. What needs to be checked is that the resulting diagram in T is independent of the the choice of T and of the lifting. In other words, we will now show that the coherent differentials that one obtains from the lifting are independent of the lifting.

It clearly suffices to consider “large” squares in the diagram; that is, ones that contain every other square. There are three cases. Case 1. Suppose the square has the form

\begin{matrix}Ỹ & ⟶ & Ỹ′ \\ ↑ & & ↑ \\ X̃ & ⟶ & X̃′ \end{matrix}

There is clearly nothing to do. The square lies entirely in the uninteresting region of the diagram; we have a bicartesian square in T which lies entirely in the old simplex. The natural differential for it is therefore the differential given as part of the structure of that simplex. Case 2. Suppose that the square has the form

Z̃_{i′j}  ⟶  Ỹ_{i′j′}
↑  ↑
Z̃_{ij} ⊕ X̃_{r0} ⟶ Ỹ_{i0} ⊕ X̃_{rj′}

Then this square maps to the following bicartesian square in T

Z̃_{i′j} ⟶ Ỹ_{i′j′}
↑  ↑
Z̃_{ij} ⟶ Ỹ_{ij′}

It follows that we get a commutative square in T

Ỹ_{i′j′} ⟶ Σ (Z̃_{ij} ⊕ X̃_{r0})
↓  ↓
Ỹ_{i′j′} ⟶  ΣZ̃_{ij}

Now recall that the square

Z̃_{i′j}  ⟶  Ỹ_{i′j′}
↑  ↑
Z̃_{ij} ⊕ X̃_{r0} ⟶ Ỹ_{i0} ⊕ X̃_{rj′}

corresponds to a short exact sequence

Z̃_{ij} ⊕ X̃_{r0}  ⟶  Z̃_{i′j} ⊕ Ỹ_{i0} ⊕ X̃_{rj′}  ⟶  Ỹ_{i′j′}

and it is this short exact sequence which determines the differential. Of course, this is also the short exact sequence of a bicartesian square of the form

Z̃_{i′j} ⊕ Ỹ_{i0} ⟶ Ỹ_{i′j′}
↑  ↑
Z̃_{ij} ⊕ X̃_{r0} ⟶ X̃_{rj′}

and it may easily be computed that the interesting maps in this bicartesian square are:

( 0 α ) : Z̃_{ij} ⊕ X̃_{r0} ⟶ X̃_{rj′}
( β β′ ) : Z̃_{i′j} ⊕ Ỹ_{i0} ⟶ Ỹ_{i′j′}

and

( \begin{matrix}γ & 0 \\ −γ′ & γ′′ \end{matrix} ) : Z̃_{ij} ⊕ X̃_{r0} ⟶ Z̃_{i′j} ⊕ Ỹ_{i0}

where α, β, β′, γ, γ′ and γ′′ are all the given maps of the original lifting of the simplex. Now the bicartesian square

Z̃_{i′j} ⊕ Ỹ_{i0} ⟶ Ỹ_{i′j′}
↑  ↑
Z̃_{ij} ⊕ X̃_{r0} ⟶ X̃_{rj′}

clearly maps to the bicartesian square

Ỹ_{i′0} ⟶ Ỹ_{i′j′}
↑  ↑
X̃_{r0} ⟶ X̃_{rj′}

It follows that we get a commutative square in T

Ỹ_{i′j′} ⟶ Σ (Z̃_{ij} ⊕ X̃_{r0})
↓  ↓
Ỹ_{i′j′} ⟶  ΣX̃_{r0}

We deduce that the composites of the natural differential Ỹ_{i′j′} ⟶ Σ (Z̃_{ij} ⊕ X̃_{r0}) with each of the two projections Σ (Z̃_{ij} ⊕ X̃_{r0}) ⟶ ΣZ̃_{ij} and Σ (Z̃_{ij} ⊕ X̃_{r0}) ⟶ ΣX̃_{r0} is a given map in the original simplex. Case 3. Suppose we are looking at the square

Z̃′ ⟶ Ỹ′ ⊕ X̃
↑  ↑
Z̃ ⟶ Ỹ ⊕ X̃

then this maps to the square

\begin{matrix}Z̃′ & ⟶ & Ỹ′ \\ ↑ & & ↑ \\ Z̃ & ⟶ & Ỹ \end{matrix}

and in T we deduce a commutative diagram, with the natural choices of differentials

Ỹ ⊕ X̃ ⟶ ΣZ̃
↓  ↓
Ỹ  ⟶ ΣZ̃

and it follows immediately that the differential Ỹ ⊕ X̃ ⟶ ΣZ̃ is the composite of the natural projection Ỹ ⊕ X̃ ⟶ Ỹ with the given differential Ỹ ⟶ ΣZ̃.

Cases 1, 2 and 3 suffice. Every square can be embedded in one of the ones dealt with in Cases 1, 2 or 3. Since we know that the differentials are coherent, this tells us that the differentials of any smaller square will be determined by a larger one. Since the large differentials have already been shown to depend only on the original simplex and not on the choice of T or of the lifting, we are done.

The homotopy above is all–important. For this reason we introduce a special symbol for it. The homotopy is denoted by

T[figure: [1,n]] [figure] 0  T[figure: [1,n]] X

The intention of the symbol, with the curved arrow, is to represent the fact that in the homotopy, the West face of the Y ’s and the North face of the X’s are used to fill in the cell. The map is not quite the contraction to an initial object, but is close in the sense that most of the Y ’s, that is all but the West face, are forgotten.

The homotopy connects the identity to a map taking the simplex

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00} ⊕ X_{r0}; Y_{00} ⊕ X_{rt}; Y_{s0} ⊕ X_{r0}; Y_{s0} ⊕ X_{rt}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

This map is in turn homotopic to the null map, by a homotopy taking the simplex above and the integer i to

[figure: ⋮; ⋯; ·; ·; ·; 0; 0; 0; ⋮; ⋮; ⋯; ⋯; X_{r0}; X_{rt}; X_{r0}; X_{rt}; ⋮; ⋯; ·; ·; ·; 0; Z_{0i}; 0; ⋮; ⋮; ⋯; ⋯; Z_{0i}⊕X_{r0}; Z_{0s}⊕X_{r0}; Z_{ii}⊕X_{r0}; Z_{is}⊕X_{r0}; ⋮; ⋮; ⋯; ⋯; Y_{00}⊕X_{r0}; Y_{00}⊕X_{rt}; Y_{i0}⊕X_{r0}; Y_{i0}⊕X_{rt}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

□

We have now completed the proof of Step 4.1. Next we will dicuss how the other Steps follow, by nearly identical homotopies. Let us begin with the Steps that are really essentially identical to Step 4.1. Consider in particular

Step 5.1: The simplicial map

T[figure: [0,n]]  T[figure: [0,n]] [figure] 0  μ  \overline{}T[figure: [0,n]]  [figure] [figure] [figure] [figure] [figure]

induces a homotopy equivalence.

Step 5.2: The simplicial map

T[figure: [0,n]]  T[figure: [0,n]] [figure] 0  ν  \overline{}T[figure: [0,n]]  [figure] [figure] [figure] [figure] [figure]

induces a homotopy equivalence.

Proof of Step 5.1 and Step 5.2. The Segal fibers are, respectively, the simplicial sets

X  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}[figure] 0  and  X  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}[figure] 0  . We will prove these Segal fibers con-

tractible. Observe first that the transposes of the Segal fibers are nearly the same as what we saw in the proof of Step 4.1. There are two differences. One is that the objects now lie in the category T_{[0,n]} instead of T_{[1,n]}. This is clearly minor. More serious is the fact that the morphisms are

restricted. Let us transpose the fibers. In [figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}X  the vertical morphisms in

the Y box are assumed epi. In [figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}X  , there are also restrictions on the hori-

zontal morphisms. The horizontal morphisms in the X box are H^{0}–isomorphisms, the horizontal morphisms in the Y box are mono.

We contract the two Segal fibers with the pair of homotopies of the proof of Step 4.1, and what is required is to check that the homotopies thus defined take the smaller simplicial sets, with restricted morphisms, to themselves. Consider the s–simplex x_{s}

x_{s}  =  ( [figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}] )

Here, to be appropriately vague and cover both cases at once, we view x_{s} as either a simplex

in [figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}X  or in [figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}X  . In [figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}X  , there

are hypotheses on the horizontal maps.

Our homotopies contracting the Segal fibers took an integer i with 0 ≤ i ≤ s and the s– simplex x_{s} first to the simplex

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{is}; 0; ⋮; ⋮; ⋯; ⋯; Y_{i0}; Y_{it}; Y_{s0}; Y_{st}; ⋮; ⋯; ·; ·; ·; 0; Z_{0i}; 0; ⋮; ⋮; ⋯; ⋯; Z_{0i}⊕X_{r0}; Z_{0s}⊕X_{r0}; Z_{ii}⊕X_{r0}; Z_{is}⊕X_{r0}; ⋮; ⋮; ⋯; ⋯; Y_{00}⊕X_{r0}; Y_{00}⊕X_{rt}; Y_{i0}⊕X_{r0}; Y_{i0}⊕X_{rt}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

and then to the simplex It is trivial to check that the hypotheses on the horizontal morphisms (if we are in the case

[figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}X  ) are respected by these two homotopies. The only thing that really

needs checking is that in the first homotopy, the vertical maps stay epi. In particular, why is the map

Y_{i0} ⊕ X_{rj} ⟶ Y_{ij}

necessarily epi?

The point is that there is a square in the original simplex

Y_{i0} ⟶ Y_{ij}
↑  ↑
X_{r0} ⟶ X_{rj}

and the resulting long exact sequence in cohomology gives in particular

H^{n}(Y_{i0} ⊕ X_{rj}) ⟶ H^{n}(Y_{ij}) ⟶ H^{n+1}(X_{r0})

and since X_{r0} ∈ T_{[0,n]}, it follows that H^{n+1}(X_{r0}) = 0, and hence

Y_{i0} ⊕ X_{rj} ⟶ Y_{ij}

is indeed epi, as required. □ Even easier are the proofs of Step 3.1 and Step 3.2. Recall

Step 3.1: The natural projection

T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]]  μ \overline{}^{[figure] [figure] T[figure: [m,n]]}

induces a homotopy equivalence.

Step 3.2: The natural projection

T[figure: [m,n]]^{}^{}^{}^{}T[figure: [m,n]]  μ′  \overline{}^{[figure] [figure] T[figure: [m,n]]}

induces a homotopy equivalence.

Proof of Step 3.1 and Step 3.2. In the above cases, the Segal fibers are even simpler. Take

the case of Step 3.2. The Segal fiber is T[figure: [m,n]]^{}^{}^{}^{}X , which is obtained from the Segal fiber of

Step 5.2 by deleting the triangle of Z’s. The homotopy contracting the Segal fiber of Step 5.2 will also contract the Segal fiber of Step 3.2, simply by deleting the triangle of Z’s.

The Segal fiber of Step 3.1 is T[figure: [m,n]]^{}^{}^{}^{}X . It is a subset of a Segal fiber of Step 3.2, where the

horizontal morphisms are assumed mono. As we already observed in the proof of Step 5.1, the contracting homotopy can easily be verified to respect horizontal monos, and hence the inclusion of the subset. Hence, it contracts also the subset. □

The argument we have just given needs to be modified to prove the next two Steps, Step 4.2 and Step 5.3. Recall

Step 4.2: The natural map

[figure] 0 T[figure: [0,n]] T[figure: [0,n]]  ν  \overline{}[figure] [figure] [figure] [figure] [figure] T[figure: [0,n]]

Step 5.3: The natural map

[figure] 0 T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  T[figure: [0,n]] [figure] A  β \overline{}[figure] [figure] [figure] [figure] [figure] [figure] [figure] T[figure: [0,n]]  T[figure: [0,n]] [figure] A

induces a homotopy equivalence.

In each case, we will prove the contractibility of the Segal fiber. The arguments are essentially identical. I will give the argument for Step 4.2, and ask the reader to provide it for Step 5.3.

Proof of Step 4.2. The Segal fiber is the simplicial set [figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}X  , and we

would like to contract it. This will be done by means of two simplicial homotopies, the first of which is closely related to our interesting one.

A homotopy should start with an s–simplex and an integer i with 0 ≤ i ≤ s, and produce a new simplex. Let i be the integer, and let us be given an s–simplex

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

Then, motivated by the proof of Step 4.1, we might study the diagram Our problem is that this is not a simplex in our simplicial set. It has a Waldhausen lifting, by the argument of Step 4.1. But the vertical maps fail to satisfy the restrictions. The vertical maps X ⟶ X′ ⊕ Y and X′ ⊕ Y ⟶ Y′ need not, in general, be H^{0} isomorphisms.

Consider now the same diagram, where the interesting region has now been highlighted for the reader’s convenience.

There is a map from X_{r0} into the interesting region. It is given as

( \begin{matrix}0 \\ 1 \end{matrix} ) : X_{r0} ⟶ Z_{ij} ⊕ X_{r0}

and

( \begin{matrix}−α \\ β \end{matrix} ) : X_{r0} ⟶ Y_{i0} ⊕ X_{rj}

As we will soon see, this map is always an H^{0}–monomorphism. The idea is to replace everything in the interesting region by its quotient by H^{0}(X_{r0}).

Perhaps this will be clearer if we do it on the level of Waldhausen liftings. Let

[figure: ⋮; ⋯; ·; ·; ·; Z̃_{00}; Z̃_{0s}; Z̃_{ss}; ⋮; ⋮; ⋯; ⋯; Ỹ_{00}; Ỹ_{0t}; Ỹ_{s0}; Ỹ_{st}; ⋮; ⋮; ⋯; ⋯; X̃_{00}; X̃_{0t}; X̃_{r0}; X̃_{rt}]

be a lifting for the simplex to a model T for T . Let X_{r0}^{<1} be the t–structure truncation of X_{r0}. Let X_{r0}^{<1} ⟶ X_{r0} be the natural map. Then it is possible to lift it to the category T. There is an object A in T, and a map A ⟶ X̃_{r0} in T, which is isomorphic to the map X_{r0}^{<1} ⟶ X_{r0}. That is, there is an isomorphism A ⟶ X_{r0}^{<1} in T , so that the square

A ⟶ X̃_{r0}
↓  ↓
X_{r0}^{<1} ⟶ X_{r0}

commutes, where X̃_{r0} ⟶ X_{r0} is the isomorphism that comes from saying that the diagram in T is a lifting. Now form

[figure: ⋮; ⋯; ·; ·; ·; Z̃_{ii}; Z̃_{is}; Z̃_{ss}; ⋮; ⋮; ⋯; ⋯; Ỹ_{i0}; Ỹ_{it}; Ỹ_{s0}; Ỹ_{st}; ⋮; ⋯; ·; ·; ·; Z̃_{00}; Z̃_{0i}; Z̃_{ii}; ⋮; ⋮; ⋯; ⋯; \frac{Z̃_{0i}⊕X̃_{r0}}{A}; \frac{Z̃_{0s}⊕X̃_{r0}}{A}; \frac{Z̃_{ii}⊕X̃_{r0}}{A}; \frac{Z̃_{is}⊕X̃_{r0}}{A}; ⋮; ⋮; ⋯; ⋯; \frac{Ỹ_{00}⊕X̃_{r0}}{A}; \frac{Ỹ_{00}⊕X̃_{rt}}{A}; \frac{Ỹ_{i0}⊕X̃_{r0}}{A}; \frac{Ỹ_{i0}⊕X̃_{rt}}{A}; ⋮; ⋮; ⋯; ⋯; X̃_{00}; X̃_{0t}; X̃_{r0}; X̃_{rt}]

The reader should be warned of the notation. In the diagram above, when we write B/A, we do not mean the quotient of B by A; we mean the homotopy quotient. There are many quotients B/A in the diagram, where there is no guarantee that A ⟶ B is a cofibration in T. To make sense of the quotients, define, for the purpose of the diagram above,

B/A = Cone (A ⟶ B)

to be the mapping cone on the map A ⟶ B.

It is trivial to see that the diagram above is a commutative diagram of bicartesian squares in T. We need to show that it satisfies the hypothesis that all the objects in the right column are in T_{[0,n]}, and that the vertical maps X ⟶ \frac{Y ⊕X′}{A}, \frac{Y ⊕X}{A} ⟶ \frac{Y ⊕X′}{A} and \frac{Y ⊕X′}{A} ⟶ Y′ are H^{0}–isomorphisms. Then it must also be checked that the maps in the diagram are independent of the choice of lifting.

We begin with checking that the objects \frac{Y ⊕X}{A} lie in T_{[0,n]}, and that the vertical morphisms \frac{Y ⊕X}{A} ⟶ Y′ are H^{0}–isomorphisms. Note that it then follows formally that X ⟶ \frac{Y ⊕X′}{A} and \frac{Y ⊕X}{A} ⟶ \frac{Y ⊕X′}{A} are H^{0}–isomorphisms. The point is that in the commutative triangles

X  ⟶  \frac{Y ⊕X′}{A}  \frac{Y ⊕X}{A}  ⟶  \frac{Y ⊕X′}{A}
↘  ↙  ↘  ↙
Y′  Y′

the slanted arrows are H^{0}–isomorphisms, forcing the horizontal ones to also be.

So we need to show that the objects \frac{Y ⊕X}{A} lie in T_{[0,n]}, and that the vertical morphisms \frac{Y ⊕X}{A} ⟶ Y′ are H^{0}–isomorphisms. For both, it helps to consider the square

Y_{i0} ⟶ Y_{ij}
↑  ↑
X_{r0} ⟶ X_{rj}

Recalling that the vertical maps are H^{0}–isomorphisms by hypothesis, the H^{0} of this diagram is a square with vertical isomorphisms

H^{0}(Y_{i0}) ⟶ H^{0}(Y_{ij})
↑  ↑
H^{0}(X_{r0}) ⟶ H^{0}(X_{rj})

and we immediately deduce that H^{0}(X_{r0}) injects into H^{0}(Y_{i0} ⊕ X_{rj}), with quotient H^{0}(Y_{ij}). But this precisely means that, if A = X_{0r}^{<1}, the map A ⟶ Y_{i0} ⊕ X_{rj} is mono, that is the quotient is in T_{[0,n]}. Furthermore, the map

H^{0} (\frac{Y_{i0} ⊕ X_{rj}}{A} ) ⟶ H^{0}(Y′)

is an isomorphism.

Now we need to show that all the morphisms in the diagram are independent of the choice of lifting. In the interesting region, we have replaced every C by C/A, where the map A ⟶ C is a mono of an object in T_{[0,0]} into an object of T^{≥0}. There are two types of new morphisms in the diagram. There are the morphisms of the type B ⟶ C/A, and the morphisms of the type C/A ⟶ B. The morphisms of the type B ⟶ C/A are easy. In the original diagram we were given a map B ⟶ C, and the morphism B ⟶ C/A must just be the composite

B ⟶ C ⟶ C/A.

More subtle is what to do with morphisms of the type C/A ⟶ B.

For the uniqueness of the morphisms, observe that any morphism C ⟶ Y of C into an object Y of T^{≥0}, for which the composite

A ⟶ C ⟶ Y

vanishes, factors uniquely through C/A. The ambiguity is given by a map ΣA ⟶ Y, and as ΣA ∈ T^{≤−1} and Y ∈ T^{≥0}, the ambiguity must vanish. All objects in the diagram above lie in T^{≥0}, so this might tempt the reader to conclude that we have just proved the uniqueness (and hence the independence of lifting) of the morphisms. But here there is a caution. The differentials require some extra care. The differentials are maps of the form C ⟶ ΣB. Even if B lies in T^{≥0}, ΣB in general does not.

Clearly, it suffices to consider the “large” differentials, i.e. differentials of rectangles with the property that any rectangle can be embedded in one of them. In the diagram above, all rectangles can be embedded in ones of the form

Y ⟶ Y′  Z′ ⟶ Y′  Z′ ⟶ \frac{Y′⊕X}{A}
↑  ↑  ,  ↑  ↑  or  ↑  ↑
X ⟶ X′  \frac{Z⊕X}{A} ⟶ \frac{Y ⊕X}{A}  Z ⟶ \frac{Y ⊕X}{A}

The first two present no difficulties; the differentials involved are maps Y′ ⟶ ΣX (given), and Y′ ⟶ Σ [\frac{Z⊕X}{A} ] (the easy map). We need to reflect a little about the map \frac{Y′⊕X}{A} ⟶ ΣZ.

The reason this map is unique is that ΣZ lies in T^{≥0}. In other words, Z lies in T^{≥1}. In fact, we have already proved (see page 59) that Z lies in T_{[1,n+1]}.

Thus, because the maps are unambiguous, the diagram above is a Waldhausen lifting for a well–defined cell

[figure: ⋮; ⋯; ·; ·; ·; Z_{ii}; Z_{is}; Z_{ss}; ⋮; ⋮; ⋯; ⋯; Y_{i0}; Y_{it}; Y_{s0}; Y_{st}; ⋮; ⋯; ·; ·; ·; Z_{00}; Z_{0i}; Z_{ii}; ⋮; ⋮; ⋯; ⋯; \frac{Z_{0i}⊕X_{r0}}{X_{r0}^{<1}}; \frac{Z_{0s}⊕X_{r0}}{X_{r0}^{<1}}; \frac{Z_{ii}⊕X_{r0}}{X_{r0}^{<1}}; \frac{Z_{is}⊕X_{r0}}{X_{r0}^{<1}}; ⋮; ⋮; ⋯; ⋯; \frac{Y_{00}⊕X_{r0}}{X_{r0}^{<1}}; \frac{Y_{00}⊕X_{rt}}{X_{r0}^{<1}}; \frac{Y_{i0}⊕X_{r0}}{X_{r0}^{<1}}; \frac{Y_{i0}⊕X_{rt}}{X_{r0}^{<1}}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

and this defines a homotopy. The homotopy connects the identity to a map sending the simplex

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

to the simplex

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; \frac{Y_{00}⊕X_{r0}}{X_{r0}^{<1}}; \frac{Y_{00}⊕X_{rt}}{X_{r0}^{<1}}; \frac{Y_{s0}⊕X_{r0}}{X_{r0}^{<1}}; \frac{Y_{s0}⊕X_{rt}}{X_{r0}^{<1}}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

This map is, in its turn, connected to the null map by a homotopy having the cells

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{is}; 0; ⋮; ⋮; ⋯; ⋯; \frac{Y_{i0}⊕X_{r0}}{X_{r0}^{<1}}; \frac{Y_{i0}⊕X_{rt}}{X_{r0}^{<1}}; \frac{Y_{s0}⊕X_{r0}}{X_{r0}^{<1}}; \frac{Y_{s0}⊕X_{rt}}{X_{r0}^{<1}}; ⋮; ⋯; ·; ·; ·; 0; 0; 0; ⋮; ⋮; ⋯; ⋯; Σ^{−1}Y_{i0}^{>0}; Σ^{−1}Y_{s0}^{>0}; Σ^{−1}Y_{i0}^{>0}; Σ^{−1}Y_{s0}^{>0}; ⋮; ⋮; ⋯; ⋯; \frac{X_{r0}^{<1}⊕X_{r0}}{X_{r0}^{<1}}; \frac{X_{r0}^{<1}⊕X_{rt}}{X_{r0}^{<1}}; \frac{X_{r0}^{<1}⊕X_{r0}}{X_{r0}^{<1}}; \frac{X_{r0}^{<1}⊕X_{rt}}{X_{r0}^{<1}}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{is}; 0; ⋮; ⋮; ⋯; ⋯; Y_{i0}⊕X_{r0}; Y_{i0}⊕X_{rt}; Y_{s0}⊕X_{r0}; Y_{s0}⊕X_{rt}; ⋮; ⋯; ·; ·; ·; 0; 0; 0; ⋮; ⋮; ⋯; ⋯; Σ^{−1}Y_{i0}^{>0}; Σ^{−1}Y_{s0}^{>0}; Σ^{−1}Y_{i0}^{>0}; Σ^{−1}Y_{s0}^{>0}; ⋮; ⋮; ⋯; ⋯; X_{r0}^{<1}⊕X_{r0}; X_{r0}^{<1}⊕X_{rt}; X_{r0}^{<1}⊕X_{r0}; X_{r0}^{<1}⊕X_{rt}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

Once again, it might help the reader to look at this cell before we divide by A. One gets the simpler cell and to obtain the homotopy we want, the objects in the dashed box are all divided by A. To show that such a cell exists, observe that since the vertical maps are H^{0}–isomorphisms, A ≃ X_{r0}^{<1} is isomorphic to Y_{j0}^{<1}, and therefore there is a triangle

A ⟶ Y_{j0} ⟶ Y_{j0}^{>0} ⟶ ΣA

By Corollaire 1.1.10 on page 24 of [1], we know that this triangle is unique up to canonical isomorphism, and in particular the map Y_{j0}^{>0} ⟶ ΣA is unique. Suppose we now choose A so that the map A ⟶ Ỹ_{00} is a fibration. To do this we may need to modify also the choice of X̃_{r0}, to guarantee that X̃_{r0} ⟶ Ỹ_{00} is a fibration, and then choose A ⟶ X̃_{r0} to also be a fibration. Then complete by pulling back to get the objects R_{j} in the diagram

[figure: ⋮; ⋯; ·; ·; ·; Z̃_{ii}; Z̃_{is}; Z̃_{ss}; ⋮; ⋮; ⋯; ⋯; Ỹ_{i0}⊕X̃_{r0}; Ỹ_{i0}⊕X̃_{rt}; Ỹ_{s0}⊕X̃_{r0}; Ỹ_{s0}⊕X̃_{rt}; ⋮; ⋯; ·; ·; ·; 0; 0; 0; ⋮; ⋮; ⋯; ⋯; R_{i}; R_{s}; R_{i}; R_{s}; ⋮; ⋮; ⋯; ⋯; A⊕X̃_{r0}; A⊕X̃_{rt}; A⊕X̃_{r0}; A⊕X̃_{rt}; ⋮; ⋮; ⋯; ⋯; X̃_{00}; X̃_{0t}; X̃_{r0}; X̃_{rt}]

This is a diagram with bicartesian squares. The squares

Z̃_{jj} ⟶ Ỹ_{j0} ⊕ X̃_{rk}
↑  ↑
R̃_{j} ⟶ A ⊕ X̃_{rk}

coupled with the fact that Z̃_{jj} is isomorphic to 0 in T and the existence and uniqueness of a triangle

A ⟶ Y_{j0} ⟶ Y_{j0}^{>0} ⟶ ΣA

allow us to identify R_{j} with Σ^{−1}Y_{j0}^{>0}. We leave it to the reader to check that all the morphisms in the diagram are uniquely determined.

As already mentioned, the homotopy cell is obtained from the above by dividing everything inside the dashed box by A. Once again, it needs to be checked that the maps are unique. The differentials are the ones requiring some subtlety. Note that the differential \frac{Y ⊕X′}{A} ⟶ ΣX is given, since the square

\frac{Y′⊕X′}{A} ⟶ \frac{Y ⊕X′}{A}
↑  ↑
X ⟶ X′

was contained in the diagram for the previous homotopy. The other potentially non–vanishing large differentials come from the square

Z  ⟶ \frac{Y′⊕X′}{A}
↑  ↑
Σ^{−1}Y^{>0} ⟶ \frac{A⊕X′}{A}

This differential is a map

\frac{Y′ ⊕ X′}{A} ⟶ Σ [Σ^{−1}Y^{>0}]

and this is unique because ΣΣ^{−1}Y^{>0} is clearly in T^{≥0}, and hence the map

\frac{Y′ ⊕ X′}{A} ⟶ Σ [Σ^{−1}Y^{>0}]

is uniquely determined by the given map

Y′ ⊕ X′ ⟶ Σ [Σ^{−1}Y^{>0}]

Taken together, these homotopies connect the identity to the map taking the simplex

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0s}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0t}; Y_{s0}; Y_{st}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

[figure: ⋮; ⋯; ·; ·; ·; 0; 0; 0; ⋮; ⋮; ⋯; ⋯; \frac{A⊕X_{r0}}{A}; \frac{A⊕X_{rt}}{A}; \frac{A⊕X_{r0}}{A}; \frac{A⊕X_{rt}}{A}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

which is easily identified with

[figure: ⋮; ⋯; ·; ·; ·; 0; 0; 0; ⋮; ⋮; ⋯; ⋯; X_{r0}; X_{rt}; X_{r0}; X_{rt}; ⋮; ⋮; ⋯; ⋯; X_{00}; X_{0t}; X_{r0}; X_{rt}]

Thus the map is a contraction, and hence the contractibility of the Segal fiber. □ It remains only to prove Step 5.4 and Step 5.5. As has already been said, Step 5.5 is rather different from the others, and is therefore being postponed to a section by itself. In this Section, we will still prove

Step 5.4: The natural map

[figure] 0 T[figure: [0,n]]  [figure] [figure] T[figure: [0,n]] [figure] [figure]  [figure] A  ν [figure]^{[figure] 0 T[figure: [0,n]]  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A}

induces a homotopy equivalence.

Proof of Step 5.4. We will contract the Segal fiber

[figure] Z  Y  T[figure: [0,n]]  B T[figure: [0,n]]  [figure] A

The homotopy is actually very straight forward; the awkward thing about writing it down is the enormous size of the simplices. Since the Z’s are fixed, I will permit myself to omit them in the notation. Thus, a typical simplex is a diagram

[figure: ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0r}; Y_{t0}; Y_{tr}; ⋮; ⋮; ⋯; ⋯; P_{00}; P_{0q}; P_{t0}; P_{tq}; ⋮; ⋮; ⋯; ⋯; B_{00}; B_{0s}; B_{t0}; B_{ts}; ⋮; ⋯; 0; A_{s0}; 0; ·; ·; ·; ⋯; X_{0}; X_{q}]

Once again, the arrows to the right of the Y ’s are to remind us that although the Z’s are not included in the above diagram, they are definitely part of a simplex. They are only omitted to save space.

In the diagram, everything is fixed except the integer q, the P ’s and the X’s. To contract this simplicial set, we use two homotopies. We will give what they do only on the rectangle of P ’s and X’s. The first homotopy is a transposed version of the key homotopy, the one we met in the proof of Step 4.1, and with a large part of the diagram deleted. It takes the integer i and the q–simplex above to

[figure: Y; ⋮; ⋮; ⋯; ⋯; Y_{0r} ⊕ P_{00}; Y_{0r} ⊕ P_{0i}; Y_{tr} ⊕ P_{00}; Y_{tr} ⊕ P_{0i}; ⋮; ⋮; ⋯; ⋯; P_{0i}; P_{0s}; P_{ti}; P_{ts}; B; A; ⋯; Y_{0r} ⊕ X_{0}; Y_{0r} ⊕ X_{i}; ⋯; X_{i}; X_{s}]

Perhaps the homotopy will become more understandable if we write down its shorthand. The shorthand for this homotopy has the symbol

[figure] Z  Y  \overline{}T[figure: [0,n]]  T[figure: [0,n]]  B T[figure: [0,n]]⊕Y_{0r} T[figure: [0,n]]  [figure] A

In other words, the homotopy really is a version of what we have been seeing all this section. It takes the East face of the Y ’s and the South face of the P ’s, and out of these, constructs the new P rectangle. The X’s are largely unaffected by the homotopy. All that happens to them is that a Y_{0r} is added to each.

The homotopy takes the identity to a map which has forgotten most of the P ’s; all that remains are the maps X_{j} ⟶ P_{0j}. To contract, we follow with the homotopy that replaces the maps X_{j} ⟶ P_{0j} by A_{s0} ⟶ B_{00}. The homotopy takes the s–simplex we began with and the integer i to the simplex

[figure: Y; ⋮; ⋮; ⋯; ⋯; Y_{0r}⊕P_{00}; Y_{0r}⊕P_{0i}; Y_{tr}⊕P_{00}; Y_{tr}⊕P_{0i}; ⋮; ⋮; ⋯; ⋯; Y_{0r}⊕B_{00}; Y_{0r}⊕B_{00}; Y_{tr}⊕B_{00}; Y_{tr}⊕B_{00}; B; A; ⋯; Y_{0r}⊕X_{0}; Y_{0r}⊕X_{i}; ⋯; Y_{0r}⊕A_{s0}; Y_{0r}⊕A_{s0}]

Together, the two homotopies contract. □

7 An ugly lemma, and the proof of Step 5.5

To complete the proof of Theorem I.7.1, and hence also this article, it remains only to establish Step 5.5. For the reader’s convenience, we recall what must be proved:

Step 5.5: The map

[figure] 0 T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A  ϕ \overline{}T[figure: [0,n]]  T[figure: [0,n]] [figure] A

which we know to be a homotopy equivalence by Corollaries 5.8 and 5.12, factors up to homotopy through the simplicial set

[figure] 0 ^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}[figure] A  .

All the other Steps were proved by relying on the same basic homotopy. Unlike all of them, Step 5.5 will be proved by using a homotopy based entirely on the t–structure truncation. Let us state the slightly revolting lemma which will do the truncation for us.

Lemma 7.1 Let us be given a model category T for the triangulated category T , and in T a diagram of bicartesian squares

[figure: ⋮; ⋯; ·; ·; ·; Z̃_{i+1i+1}; Z̃_{i+1t}; Z̃_{tt}; ⋮; ⋮; ⋯; ⋯; Ỹ_{i+10}; Ỹ_{i+1r}; Ỹ_{t0}; Ỹ_{tr}; ⋮; ⋮; ⋯; ⋯; B̃_{i+10}; B̃_{i+1s}; B̃_{t0}; B̃_{ts}; ⋮; ⋮; ⋯; ⋯; Z̃_{0i+1}; Z̃_{0t}; Z̃_{ii+1}; Z̃_{it}; ⋮; ⋮; ⋯; ⋯; Ỹ_{00}; Ỹ_{0r}; Ỹ_{i0}; Ỹ_{ir}; ⋮; ⋮; ⋯; ⋯; B̃_{00}; B̃_{0s}; B̃_{i0}; B̃_{is}; ⋮; ⋯; R_{00}; R_{i0}; R_{ii}; ·; ·; ·; ⋮; ⋯; Ã_{00}; Ã_{s0}; Ã_{ss}; ·; ·; ·; ⋯; X̃_{0}; X̃_{r}]

Suppose all the objects in the diagram are such that their image in T lies in T_{[0,n]}. Choose an integer j with −1 ≤ j < i. Then R_{ij} is an object of T_{[0,n]}. If j = −1 we define R_{ij} to be B_{is}.

Assume that for all k ≥ i + 1, B_{ks} ∈ T_{[0,n−1]}. Assume further that for our choice of j with −1 ≤ j < i, we have that for all j′ > j, R_{ij′} ∈ T_{[0,n−1]}. Then we will show that we can truncate to achieve R_{ij} ∈ T_{[0,n−1]}. There is a new diagram

[figure: ⋮; ⋯; ·; ·; ·; Z̃_{i+1i+1}; Z̃_{i+1t}; Z̃_{tt}; ⋮; ⋮; ⋯; ⋯; Ỹ_{i+10}; Ỹ_{i+1r}; Ỹ_{t0}; Ỹ_{tr}; ⋮; ⋮; ⋯; ⋯; B̃_{i+10}; B̃_{i+1s}; B̃_{t0}; B̃_{ts}; ⋮; ⋮; ⋯; ⋯; Z̃′_{0i+1}; Z̃′_{0t}; Z̃′_{ii+1}; Z̃′_{it}; ⋮; ⋮; ⋯; ⋯; Ỹ′_{00}; Ỹ′_{0r}; Ỹ′_{i0}; Ỹ′_{ir}; ⋮; ⋮; ⋯; ⋯; B̃′_{00}; B̃′_{0s}; B̃′_{i0}; B̃′_{is}; ⋮; ⋯; R′_{00}; R′_{i0}; R′_{ii}; ·; ·; ·; ⋮; ⋯; Ã_{00}; Ã_{s0}; Ã_{ss}; ·; ·; ·; ⋯; X̃_{0}; X̃_{r}]

and a map in T from the new diagram to the old. The new diagram is a diagram of bicartesian squares. In it, R′_{ij} ∈ T_{[0,n−1]}. Furthermore, the new diagram is universal in T with respect to this property. Any diagram of the same shape, of squares with differentials in T_{[0,n]}, mapping to the old diagram and satisfying R′_{ij} ∈ T_{[0,n−1]}, must factor uniquely through the universal one we will construct.

Furthermore, in the category T we get a three dimesional diagram, describing the map from the new diagram to the old. In this three dimensional diagram, not only are there two planes of diagrams of bicartesian squares (the original diagram and the new), but more is true. If we delete the Z’s, Z′’s, R’s and R′’s, then every horizontal plane is also a diagram of bicartesian squares. That is, if Y ⟶ B is a horizontal map from a Y to a B in the old diagram, then the square

\begin{matrix}Y & ⟶ & B \\ ↑ & & ↑ \\ Y′ & ⟶ & B′ \end{matrix}

describing the map from the new to the old diagram, is also bicartesian.

Finally, in the modified diagram, the A’s and X’s are unaltered.

Proof. The Lemma is quite a mouthful to state, but very easy to prove. Let us choose a lifting to T of the natural map R_{ij} ⟶ R_{ij}^{>n−1} in T . That is, we choose a map R_{ij} ⟶ Q in T which is isomorphic in T to the natural map. The rule for forming the R′’s, Y′’s X′’s A′’s and B′’s is very simple. Everything of the form W to the left and below R_{ij} gets replaced by the mapping fiber Fiber(W ⟶ Q), unless it is also to the left and below R_{jj}, in which case it gets replaced by the mapping fiber Fiber(W ⟶ R_{jj}). By the fiber we mean the mapping cone in the dual Waldhausen category T^{op}.

It is clear that we end up with a diagram of bicartesian squares in T. It is also clear that it admits a map to the original diagram, and that if we delete the R’s and Z’s then we have a three dimensional diagram whose horizontal planes have bicartesian squares. Equally evident is that R′_{ij} is just R_{ij}^{<n}, which certainly lies in T_{[0,n−1]}. What is not so immediately clear is that every object in this diagram is in T_{[0,n]}.

If W is below and to the left of R_{jj}, then it is replaced by W′ = Fiber(W ⟶ R_{jj}). But since R_{jj} is an object of T isomorphic to 0 in T , W′ is isomorphic in T to W and lies in T_{[0,n]}. We need therefore only consider W ’s not to the left and below R_{jj}. This means they must lie above R_{jj}. Choose such a W.

Then W fits in a square

\begin{matrix}W & ⟶ & R_{kj} \\ ↑ & & ↑ \\ S & ⟶ & R_{jj} \end{matrix}

Recalling that R_{jj} is isomorphic in T to 0, we get a triangle

S ⟶ W ⟶ R_{kj} ⟶ ΣS

and because we know that S ∈ T_{[0,n]}, the long exact sequence in cohomology immediately implies that H^{n}(W) ⟶ H^{n}(R_{kj}) is surjective. On the other hand, W lies below R_{ij} and to its left. This means that k ≤ i. There is then a square

R_{ij} ⟶ R_{ik}
↑  ↑
R_{kj} ⟶ R_{kk}

and hence, recalling that R_{kk} = 0, a triangle

R_{kj} ⟶ R_{ij} ⟶ R_{ik} ⟶ ΣR_{kj}

Now recall that for k > j we assumed that R_{ik} ∈ T_{[0,n−1]}. From the long exact sequence in cohomology we therefore deduce that H^{n}(R_{kj}) ⟶ H^{n}(R_{ij}) is epi. That makes the composite map

H^{n}(W) ⟶ H^{n}(R_{kj}) ⟶ H^{n}(R_{ij})

also epi. Thus the map W ⟶ Q is an epimorphism in H^{n}, and from the long exact sequence it immediately follows that the mapping fiber is in T_{[0,n]}.

The universality of the diagram is just the observation that if a map X ⟶ W with X ∈ T^{≤n} is given, and the composite

X ⟶ W ⟶ Q

vanishes, then the map factors through W′. We have a triangle in T

W′ ⟶ W ⟶ Q ⟶ ΣW′.

Not only do we know that the map X ⟶ W must factor as X ⟶ W′ ⟶ W, but we also know that two factorisations will differ by a map

X ⟶ Σ^{−1}Q ⟶ W′.

But the map X ⟶ Σ^{−1}Q must vanish, since X ∈ T^{≤n} and Σ^{−1}Q ∈ T^{≥n+1}.

As always, the universality becomes a little ticklish when one considers differentials. The problem, as always, is that differentials are maps Σ^{−1}B ⟶ C, and even if B lies in T_{[0,n]}, Σ^{−1}B may not. So some extra caution is in order as regards differentials.

It clearly suffices to consider the “large” differentials. These fall into several cases. Case 1. The differentials coming from squares

\begin{matrix}Y & ⟶ & B \\ ↑ & & ↑ \\ X & ⟶ & A \end{matrix}

where B′ is some B_{kℓ}, Y is either a Y_{kℓ} or a B_{kℓ}, X is either an X_{k} or an A_{kℓ} and A is some A_{kℓ}. Hence we are looking at a differential B ⟶ ΣX. By hypothesis, the truncation does not change X. After truncation, B may be replaced by B′ with a map B′ ⟶ B. In any case, the differential B′ ⟶ B ⟶ ΣX is clearly unique and universal. Case 2. The differentials coming from squares

\begin{matrix}\overline{Z} & ⟶ & \overline{B} \\ ↑ & & ↑ \\ Z & ⟶ & B \end{matrix}

where the Z’s are Z_{kℓ}’s and the B’s are B_{kℓ}’s. Because we are only considering large squares, we may embed this in a square where \overline{B} is B_{ks} with k ≥ i + 1. But then, by hypothesis, B_{ks} ∈ T [0, n − 1]. The differential Σ^{−1}\overline{B} ⟶ Z′ is therefore a map from Σ^{−1}B_{ks} ∈ T^{≤n} into Z′ ⟶ Z, which may be truncated. The uniqueness argument above then applies. Case 3. The differentials coming from squares

\begin{matrix}\overline{Z} & ⟶ & \overline{R} \\ ↑ & & ↑ \\ Z & ⟶ & R \end{matrix}

where the Z’s are Z_{kℓ}’s and the R’s are R_{kℓ}’s. Because we are only considering large squares, we may embed this in a square where \overline{R} is R_{ik}. Now there are two cases to consider. If k ≤ j then Z may be taken to be Z_{k i+1}, which is unchanged by the truncation I claim universal. Hence the differential into ΣZ′ = ΣZ presents no problem. If k > j, then by hypothesis R_{ik} ∈ T_{[0,n−1]}. Thus Σ^{−1}\overline{R} is an object of T^{≤n}, and hence the uniqueness argument above applies to the map Σ^{−1}\overline{R} ⟶ Z′, even if Z′ ⟶ Z is a truncation. □

This completed the proof of the Lemma. Now we proceed to apply the Lemma successively to construct our homotopy. Let us recall again the statement of the Step:

Step 5.5: The map

[figure] 0 T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A  ϕ \overline{}T[figure: [0,n]]  T[figure: [0,n]] [figure] A

which we know to be a homotopy equivalence by Corollaries 5.8 and 5.12, factors up to homotopy through the simplicial set

[figure] 0 ^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}[figure] A  .

Proof of Step 5.5. We will produce a simplicial map

[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A  θ  \overline{}[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A

so that, if i is the natural inclusion

[figure] 0  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A  i  \overline{}[figure]  [figure] 0  T[figure: [0,n]]  T[figure: [0,n]] T[figure: [0,n]]  [figure] A

then ϕ ∘ i ∘ θ is homotopic to ϕ. In this sense, ϕ factors, up to homotopy, through

[figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}[figure] A  . Let us begin by describing the simplicial map θ. It

might be clearest if, once again, we rely on liftings of simplices to models.

A simplex in [figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}[figure] A  is a diagram in T

[figure: ⋮; ⋯; ·; ·; ·; 0; Z_{0t}; 0; ⋮; ⋮; ⋯; ⋯; Y_{00}; Y_{0r}; Y_{t0}; Y_{tr}; ⋮; ⋮; ⋯; ⋯; B_{00}; B_{0s}; B_{t0}; B_{ts}; ⋮; ⋯; 0; A_{s0}; 0; ·; ·; ·; ⋯; X_{0}; X_{r}]

[figure: ⋮; ⋯; ·; ·; ·; 0; Z′_{0t}; 0; ⋮; ⋮; ⋯; ⋯; Y′_{00}; Y′_{0r}; Y′_{t0}; Y′_{tr}; ⋮; ⋮; ⋯; ⋯; B′_{00}; B′_{0s}; B′_{t0}; B′_{ts}; ⋮; ⋯; 0; A_{s0}; 0; ·; ·; ·; ⋯; X_{0}; X_{r}]

Now, we propose to find a diagram of the same shape, but where all the objects B_{is} lie in T_{[0,n−1]}. We will find a map from the new diagram to the old diagram, which will be universal. That is, we propose to find a diagram where the last column of the B’s satisfy H^{n}(B′_{is}) = 0, maps of the from the W′ ⟶ W for every W in the diagram, that is W can be B, Y or Z. The maps W′ ⟶ W must be morphisms of diagrams. The universality we will show is that any other diagram of this type with the B′_{is} ∈ T_{[0,n−1]} with a map to the original factors through the one we will construct.

Because the diagram we began with is a simplex, it has at least one lifting to at least one model T. Choose such a T and a lifting, that is a diagram

[figure: ⋮; ⋯; ·; ·; ·; 0; Z̃_{0t}; 0; ⋮; ⋮; ⋯; ⋯; Ỹ_{00}; Ỹ_{0r}; Ỹ_{t0}; Ỹ_{tr}; ⋮; ⋮; ⋯; ⋯; B̃_{00}; B̃_{0s}; B̃_{t0}; B̃_{ts}; ⋮; ⋯; 0; Ã_{s0}; 0; ·; ·; ·; ⋯; X̃_{0}; X̃_{r}]

Suppose that for some i we know that j > i implies the vanishing of H^{n}(B_{js}). We will show how to improve this to a diagram where H^{n}(B_{is}) also vanishes. Then, descending induction on i will provide the proof.

To begin with, complete to a diagram [figure: ⋮; ⋯; ·; ·; ·; Z̃_{i+1i+1}; Z̃_{i+1t}; Z̃_{tt}; ⋮; ⋮; ⋯; ⋯; Ỹ_{i+10}; Ỹ_{i+1r}; Ỹ_{t0}; Ỹ_{tr}; ⋮; ⋮; ⋯; ⋯; B̃_{i+10}; B̃_{i+1s}; B̃_{t0}; B̃_{ts}; ⋮; ⋮; ⋯; ⋯; Z̃_{0i+1}; Z̃_{0t}; Z̃_{ii+1}; Z̃_{it}; ⋮; ⋮; ⋯; ⋯; Ỹ_{00}; Ỹ_{0r}; Ỹ_{i0}; Ỹ_{ir}; ⋮; ⋮; ⋯; ⋯; B̃_{00}; B̃_{0s}; B̃_{i0}; B̃_{is}; ⋮; ⋯; R_{00}; R_{i0}; R_{ii}; ·; ·; ·; ⋮; ⋯; Ã_{00}; Ã_{s0}; Ã_{ss}; ·; ·; ·; ⋯; X̃_{0}; X̃_{r}]

In other words, choose a cofibration B̃_{0s} ⟶ R_{00} with R_{00} an object of T isomorphic to 0 in T . Then push out to obtain the first column of R’s. Choose a cofibration R_{10} ⟶ R_{11} with R_{11} isomorphic to 0 in T , and push out to obtain the next column of R’s, and so on. Note that we have deleted a triangle of Z̃’s, namely Z̃_{jk} with j ≤ k ≤ i. In T they would be superfluous, because the square

Z_{kk} ⟶ R_{kj}
↑  ↑
Z_{jk} ⟶ R_{jj}

gives an isomorphism R_{kj} ⟶ ΣZ_{jk}, and hence with this isomorphism R_{kj} and ΣZ_{jk} determine each other, and all the maps in and out of each other.

Now we would like to apply Lemma 7.1. By hypothesis, we know that for k > i, B_{ks} ∈ T_{[0,n−1]}. But to apply the lemma we must first show that every object in the diagram is in T_{[0,n]}. This is given to us for the A’s, B’s, X’s and Y ’s; however, it was never guaranteed for the Z’s or R’s.

To prove that the R’s are in T_{[0,n]}, consider the square

\begin{matrix}B′ & ⟶ & R_{jk} \\ ↑ & & ↑ \\ B & ⟶ & R_{kk} \end{matrix}

where B = B_{k0} and B′ = B_{j0}. The morphism B ⟶ B′ is by hypothesis mono. The long exact sequence for the triangle

B ⟶ B′ ⟶ R_{jk} ⟶ ΣB

immediately allows us to establish that R_{jk} ∈ T_{[0,n]}.

What about the Z’s? For any k > i and any j ≤ k, consider the square

Z_{kk} ⟶ B_{ks}
↑  ↑
Z_{jk} ⟶ B_{js}

In any case, we know that the B’s are in T_{[0,n]}. Our induction hypothesis was that for k > i, H^{n}(B_{ks}) = 0. Now the long exact sequence for the triangle

Z_{jk} ⟶ B_{js} ⟶ B_{ks} ⟶ ΣZ_{jk}

permits us to easily deduce that Z_{jk} lies in T_{[0,n]}.

Now we can successively apply Lemma 7.1. If for some j with −1 ≤ j < i we know that H^{n}(R_{ik}) = 0 for all k > j, then we get a universal diagram mapping to the given one where H^{n}(R_{ij}) has also been made 0. Working down through the j’s, we get all the way down to j = −1; we can even make H^{n}(B_{is}) = 0.

By induction, we now know that there is a universal diagram

where B′_{is} ∈ T_{[0,n−1]}. I assert that this diagram is a simplex in [figure] 0  ^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}^{}^{}T[figure: [0,n]]^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}^{}T[figure: [0,n]]  ^{}^{}[figure] A  .

In other words, I assert that all the vertical maps X ⟶ Y, Y ⟶ Y′, B ⟶ B′ and A_{sj} ⟶ B_{ij} are epi and mono. To prove the map epi, recall that it fits in a square

\begin{matrix}W′ & ⟶ & B_{is} \\ ↑ & & ↑ \\ W & ⟶ & F \end{matrix}

and we know that H^{n}(F) = H^{n}(B_{is}) = 0. Now the long exact sequence in cohomology immediately tells us that H^{n}(W) ⟶ H^{n}(W′) is epi.

It remains to show that the vertical maps X_{j} ⟶ Y_{ij} and A_{sj} ⟶ B_{ij} are mono. The other vertical maps we already know to be mono. Now let W be either X_{j} or A_{sj}, and let W ⟶ W′ be a vertical map. Consider the square

\begin{matrix}W′ & ⟶ & B_{is} \\ ↑ & & ↑ \\ W & ⟶ & A_{ss} \end{matrix}

Since A_{ss} = 0, we get a triangle

W ⟶ W′ ⟶ B_{is} ⟶ ΣW

and as B_{is} ∈ T_{[0,n]}, it immediately follows from the long exact cohomology sequence that W ⟶ W′ is mono.

Since by construction the diagram above clearly has a lifting to T, it qualifies as a simplex. Now we need to show that the map assigning to the starting simplex the simplex with the dashes is a simplicial map. This is not so clear by construction. To construct the truncations we used the entire simplex inductively. If we apply a face map, much of the simplex disappears, and the truncation goes much faster. It is not á priori clear that we end up with the same.

Here is where we use the universality of the truncated simplex. By the universality, there is a natural map from the face of a truncation of a simplex to the truncation of the face. This map is easily shown to be a cohomology isomorphism. Thus the truncation of a face is canonically isomorphic to the face of the truncation.

Therefore, for all zero simplices, choose and fix a truncation together with its structural map to the original simplex. For any 1-simplex, choose and fix the truncation of the corresponding Z and the map Z′ ⟶ Z. For larger simplices, define the truncation so that its 0–faces are equal to our choices, and its 1–faces have the Z′ we chose above. Having thus rigidified the construction, we get a simplicial map which we call θ.

It only remains to show that ϕ∘i∘θ is homotopic to ϕ. But the natural map of the truncated simplex to the original simplex provides the homotopy, once we recall that in Lemma 7.1 we proved that this three dimensional diagram gives horizontal planes of bicartesian squares.

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