Simplicial Methods

Introduction

This is a minimal introduction to simplicial methods. We just add here whenever something is needed later on. A general reference to this material is perhaps [SimpHom]. An example of the things you can do is the paper by Quillen on Homotopical Algebra, see [Quillen] or the paper on Étale Homotopy by Artin and Mazur, see [ArtinMazur].

The category of finite ordered sets

The category \(\Delta\) is the category with

1. objects \([0], [1], [2], \ldots\) with \([n] = \{0, 1, 2, \ldots, n\}\) and

2. a morphism \([n] \to [m]\) is a nondecreasing map \(\{0, 1, 2, \ldots, n\} \to \{0, 1, 2, \ldots, m\}\) between the corresponding sets.

Here nondecreasing for a map \(\varphi : [n] \to [m]\) means by definition that \(\varphi(i) \geq \varphi(j)\) if \(i \geq j\). In other words, \(\Delta\) is a category equivalent to the “big” category of nonempty finite totally ordered sets and nondecreasing maps. There are exactly \(n + 1\) morphisms \([0] \to [n]\) and there is exactly \(1\) morphism \([n] \to [0]\). There are exactly \((n + 1)(n + 2)/2\) morphisms \([1] \to [n]\) and there are exactly \(n + 2\) morphisms \([n] \to [1]\). And so on and so forth.

Definition

For any integer \(n\geq 1\), and any \(0\leq j \leq n\) we let \(\delta^n_j : [n-1] \to [n]\) denote the injective order preserving map skipping \(j\). For any integer \(n\geq 0\), and any \(0\leq j \leq n\) we denote \(\sigma^n_j : [n + 1] \to [n]\) the surjective order preserving map with \((\sigma^n_j)^{-1}(\{j\}) = \{j, j + 1\}\).

Lemma

Any morphism in \(\Delta\) can be written as a composition of the morphisms \(\delta^n_j\) and \(\sigma^n_j\).

Proof

Let \(\varphi : [n] \to [m]\) be a morphism of \(\Delta\). If \(j \not \in \Im(\varphi)\), then we can write \(\varphi\) as \(\delta^m_j \circ \psi\) for some morphism \(\psi : [n] \to [m - 1]\). If \(\varphi(j) = \varphi(j + 1)\) then we can write \(\varphi\) as \(\psi \circ \sigma^{n - 1}_j\) for some morphism \(\psi : [n - 1] \to [m]\). The result follows because each replacement as above lowers \(n + m\) and hence at some point \(\varphi\) is both injective and surjective, hence an identity morphism.

Lemma

The morphisms \(\delta^n_j\) and \(\sigma^n_j\) satisfy the following relations.

1. If \(0 \leq i < j \leq n + 1\), then \(\delta^{n + 1}_j \circ \delta^n_i = \delta^{n + 1}_i \circ \delta^n_{j - 1}\). In other words the diagram \[\xymatrix{ & [n] \ar[rd]^{\delta^{n + 1}_j} & \\ [n - 1] \ar[ru]^{\delta^n_i} \ar[rd]_{\delta^n_{j - 1}} & & [n + 1] \\ & [n] \ar[ru]_{\delta^{n + 1}_i} & }\] commutes.

2. If \(0 \leq i < j \leq n - 1\), then \(\sigma^{n - 1}_j \circ \delta^n_i = \delta^{n - 1}_i \circ \sigma^{n - 2}_{j - 1}\). In other words the diagram \[\xymatrix{ & [n] \ar[rd]^{\sigma^{n - 1}_j} & \\ [n - 1] \ar[ru]^{\delta^n_i} \ar[rd]_{\sigma^{n - 2}_{j - 1}} & & [n - 1] \\ & [n - 2] \ar[ru]_{\delta^{n - 1}_i} & }\] commutes.

3. If \(0 \leq j \leq n - 1\), then \(\sigma^{n - 1}_j \circ \delta^n_j = \text{id}_{[n - 1]}\) and \(\sigma^{n - 1}_j \circ \delta^n_{j + 1} = \text{id}_{[n - 1]}\). In other words the diagram \[\xymatrix{ & [n] \ar[rd]^{\sigma^{n - 1}_j} & \\ [n - 1] \ar[ru]^{\delta^n_j} \ar[rd]_{\delta^n_{j + 1}} \ar[rr]^{\text{id}_{[n - 1]}} & & [n - 1] \\ & [n] \ar[ru]_{\sigma^{n - 1}_j} & }\] commutes.

4. If \(0 < j + 1 < i \leq n\), then \(\sigma^{n - 1}_j \circ \delta^n_i = \delta^{n - 1}_{i - 1} \circ \sigma^{n - 2}_j\). In other words the diagram \[\xymatrix{ & [n] \ar[rd]^{\sigma^{n - 1}_j} & \\ [n - 1] \ar[ru]^{\delta^n_i} \ar[rd]_{\sigma^{n - 2}_j} & & [n - 1] \\ & [n - 2] \ar[ru]_{\delta^{n - 1}_{i - 1}} & }\] commutes.

5. If \(0 \leq i \leq j \leq n - 1\), then \(\sigma^{n - 1}_j \circ \sigma^n_i = \sigma^{n - 1}_i \circ \sigma^n_{j + 1}\). In other words the diagram \[\xymatrix{ & [n] \ar[rd]^{\sigma^{n - 1}_j} & \\ [n + 1] \ar[ru]^{\sigma^n_i} \ar[rd]_{\sigma^n_{j + 1}} & & [n - 1] \\ & [n] \ar[ru]_{\sigma^{n - 1}_i} & }\] commutes.

Proof

Omitted.

Lemma

The category \(\Delta\) is the universal category with objects \([n]\), \(n \geq 0\) and morphisms \(\delta^n_j\) and \(\sigma^n_j\) such that (a) every morphism is a composition of these morphisms, (b) the relations listed in Lemma 0167 are satisfied, and (c) any relation among the morphisms is a consequence of those relations.

Proof

Omitted.

Simplicial objects

Definition

Let \(\mathcal{C}\) be a category.

1. A simplicial object \(U\) of \(\mathcal{C}\) is a contravariant functor \(U\) from \(\Delta\) to \(\mathcal{C}\), in a formula: \[U : \Delta^{opp} \longrightarrow \mathcal{C}\]

2. If \(\mathcal{C}\) is the category of sets, then we call \(U\) a simplicial set.

3. If \(\mathcal{C}\) is the category of abelian groups, then we call \(U\) a simplicial abelian group.

4. A morphism of simplicial objects \(U \to U'\) is a transformation of functors.

5. The category of simplicial objects of \(\mathcal{C}\) is denoted \(\text{Simp}(\mathcal{C})\).

This means there are objects \(U([0]), U([1]), U([2]), \ldots\) and for \(\varphi\) any nondecreasing map \(\varphi : [m] \to [n]\) a morphism \(U(\varphi) : U([n]) \to U([m])\), satisfying \(U(\varphi \circ \psi) = U(\psi) \circ U(\varphi)\).

In particular there is a unique morphism \(U([0]) \to U([n])\) and there are exactly \(n + 1\) morphisms \(U([n]) \to U([0])\) corresponding to the \(n + 1\) maps \([0] \to [n]\). Obviously we need some more notation to be able to talk intelligently about these simplicial objects. We do this by considering the morphisms we singled out in Section 0164 above.

Lemma

Let \(\mathcal{C}\) be a category.

1. Given a simplicial object \(U\) in \(\mathcal{C}\) we obtain a sequence of objects \(U_n = U([n])\) endowed with the morphisms \(d^n_j = U(\delta^n_j) : U_n \to U_{n-1}\) and \(s^n_j = U(\sigma^n_j) : U_n \to U_{n + 1}\). These morphisms satisfy the opposites of the relations displayed in Lemma 0167, namely

   1. If \(0 \leq i < j \leq n + 1\), then \(d^n_i \circ d^{n + 1}_j = d^n_{j - 1} \circ d^{n + 1}_i\).

   2. If \(0 \leq i < j \leq n - 1\), then \(d^n_i \circ s^{n - 1}_j = s^{n - 2}_{j - 1} \circ d^{n - 1}_i\).

   3. If \(0 \leq j \leq n - 1\), then \(\text{id} = d^n_j \circ s^{n - 1}_j = d^n_{j + 1} \circ s^{n - 1}_j\).

   4. If \(0 < j + 1 < i \leq n\), then \(d^n_i \circ s^{n - 1}_j = s^{n - 2}_j \circ d^{n - 1}_{i - 1}\).

   5. If \(0 \leq i \leq j \leq n - 1\), then \(s^n_i \circ s^{n - 1}_j = s^n_{j + 1} \circ s^{n - 1}_i\).

2. Conversely, given a sequence of objects \(U_n\) and morphisms \(d^n_j\), \(s^n_j\) satisfying (1)(a) – (e) there exists a unique simplicial object \(U\) in \(\mathcal{C}\) such that \(U_n = U([n])\), \(d^n_j = U(\delta^n_j)\), and \(s^n_j = U(\sigma^n_j)\).

3. A morphism between simplicial objects \(U\) and \(U'\) is given by a family of morphisms \(U_n \to U'_n\) commuting with the morphisms \(d^n_j\) and \(s^n_j\).

Proof

This follows from Lemma 0168.

Remark

By abuse of notation we sometimes write \(d_i : U_n \to U_{n - 1}\) instead of \(d^n_i\), and similarly for \(s_i : U_n \to U_{n + 1}\). The relations among the morphisms \(d^n_i\) and \(s^n_i\) may be expressed as follows:

1. If \(i < j\), then \(d_i \circ d_j = d_{j - 1} \circ d_i\).

2. If \(i < j\), then \(d_i \circ s_j = s_{j - 1} \circ d_i\).

3. We have \(\text{id} = d_j \circ s_j = d_{j + 1} \circ s_j\).

4. If \(i > j + 1\), then \(d_i \circ s_j = s_j \circ d_{i - 1}\).

5. If \(i \leq j\), then \(s_i \circ s_j = s_{j + 1} \circ s_i\).

This means that whenever the compositions on both the left and the right are defined then the corresponding equality should hold.

We get a unique morphism \(s^0_0 = U(\sigma^0_0) : U_0 \to U_1\) and two morphisms \(d^1_0 = U(\delta^1_0)\), and \(d^1_1 = U(\delta^1_1)\) which are morphisms \(U_1 \to U_0\). There are two morphisms \(s^1_0 = U(\sigma^1_0)\), \(s^1_1 = U(\sigma^1_1)\) which are morphisms \(U_1 \to U_2\). Three morphisms \(d^2_0 = U(\delta^2_0)\), \(d^2_1 = U(\delta^2_1)\), \(d^2_2 = U(\delta^2_2)\) which are morphisms \(U_2 \to U_1\). And so on.

Pictorially we think of \(U\) as follows: \[\xymatrix{ U_2 \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & U_1 \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & U_0 \ar@<0ex>[l] }\] Here the \(d\)-morphisms are the arrows pointing right and the \(s\)-morphisms are the arrows pointing left.

Example

The simplest example is the constant simplicial object with value \(X \in \Ob(\mathcal{C})\). In other words, \(U_n = X\) and all maps are \(\text{id}_X\).

Example

Suppose that \(Y\to X\) is a morphism of \(\mathcal{C}\) such that all the fibred products \(Y \times_X Y \times_X \ldots \times_X Y\) exist. Then we set \(U_n\) equal to the \((n + 1)\)-fold fibre product, and we let \(\varphi : [n] \to [m]\) correspond to the map (on “coordinates”) \((y_0, \ldots, y_m) \mapsto (y_{\varphi(0)}, \ldots, y_{\varphi(n)})\). In other words, the map \(U_0 = Y \to U_1 = Y \times_X Y\) is the diagonal map. The two maps \(U_1 = Y \times_X Y \to U_0 = Y\) are the projection maps.

Geometrically Example 016E above is an important example. It tells us that it is a good idea to think of the maps \(d^n_j : U_n \to U_{n - 1}\) as projection maps (forgetting the \(j\)th component), and to think of the maps \(s^n_j : U_n \to U_{n + 1}\) as diagonal maps (repeating the \(j\)th coordinate). We will return to this in the sections below.

Lemma

Let \(\mathcal{C}\) be a category. Let \(U\) be a simplicial object of \(\mathcal{C}\). Each of the morphisms \(s^n_i : U_n \to U_{n + 1}\) has a left inverse. In particular \(s^n_i\) is a monomorphism.

Proof

This is true because \(d_i^{n + 1} \circ s^n_i = \text{id}_{U_n}\).

Simplicial objects as presheaves

Another observation is that we may think of a simplicial object of \(\mathcal{C}\) as a presheaf with values in \(\mathcal{C}\) over \(\Delta\). See Sites, Definition 00V3. And in fact, if \(U\), \(U'\) are simplicial objects of \(\mathcal{C}\), then we have [016H]\[\begin{equation} \Mor(U, U') = \Mor_{\textit{PSh}(\Delta)}(U, U'). \end{equation}\] Some of the material below could be replaced by the more general constructions in the chapter on sites. However, it seems a clearer picture arises from the arguments specific to simplicial objects.

Cosimplicial objects

A cosimplicial object of a category \(\mathcal{C}\) could be defined simply as a simplicial object of the opposite category \(\mathcal{C}^{opp}\). This is not really how the human brain works, so we introduce them separately here and point out some simple properties.

Definition

Let \(\mathcal{C}\) be a category.

1. A cosimplicial object \(U\) of \(\mathcal{C}\) is a covariant functor \(U\) from \(\Delta\) to \(\mathcal{C}\), in a formula: \[U : \Delta \longrightarrow \mathcal{C}\]

2. If \(\mathcal{C}\) is the category of sets, then we call \(U\) a cosimplicial set.

3. If \(\mathcal{C}\) is the category of abelian groups, then we call \(U\) a cosimplicial abelian group.

4. A morphism of cosimplicial objects \(U \to U'\) is a transformation of functors.

5. The category of cosimplicial objects of \(\mathcal{C}\) is denoted \(\text{CoSimp}(\mathcal{C})\).

This means there are objects \(U([0]), U([1]), U([2]), \ldots\) and for \(\varphi\) any nondecreasing map \(\varphi : [m] \to [n]\) a morphism \(U(\varphi) : U([m]) \to U([n])\), satisfying \(U(\varphi \circ \psi) = U(\varphi) \circ U(\psi)\).

In particular there is a unique morphism \(U([n]) \to U([0])\) and there are exactly \(n + 1\) morphisms \(U([0]) \to U([n])\) corresponding to the \(n + 1\) maps \([0] \to [n]\). Obviously we need some more notation to be able to talk intelligently about these simplicial objects. We do this by considering the morphisms we singled out in Section 0164 above.

Lemma

Let \(\mathcal{C}\) be a category.

1. Given a cosimplicial object \(U\) in \(\mathcal{C}\) we obtain a sequence of objects \(U_n = U([n])\) endowed with the morphisms \(\delta^n_j = U(\delta^n_j) : U_{n - 1} \to U_n\) and \(\sigma^n_j = U(\sigma^n_j) : U_{n + 1} \to U_n\). These morphisms satisfy the relations displayed in Lemma 0167.

2. Conversely, given a sequence of objects \(U_n\) and morphisms \(\delta^n_j\), \(\sigma^n_j\) satisfying these relations there exists a unique cosimplicial object \(U\) in \(\mathcal{C}\) such that \(U_n = U([n])\), \(\delta^n_j = U(\delta^n_j)\), and \(\sigma^n_j = U(\sigma^n_j)\).

3. A morphism between cosimplicial objects \(U\) and \(U'\) is given by a family of morphisms \(U_n \to U'_n\) commuting with the morphisms \(\delta^n_j\) and \(\sigma^n_j\).

Proof

This follows from Lemma 0168.

Remark

By abuse of notation we sometimes write \(\delta_i : U_{n - 1} \to U_n\) instead of \(\delta^n_i\), and similarly for \(\sigma_i : U_{n + 1} \to U_n\). The relations among the morphisms \(\delta^n_i\) and \(\sigma^n_i\) may be expressed as follows:

1. If \(i < j\), then \(\delta_j \circ \delta_i = \delta_i \circ \delta_{j - 1}\).

2. If \(i < j\), then \(\sigma_j \circ \delta_i = \delta_i \circ \sigma_{j - 1}\).

3. We have \(\text{id} = \sigma_j \circ \delta_j = \sigma_j \circ \delta_{j + 1}\).

4. If \(i > j + 1\), then \(\sigma_j \circ \delta_i = \delta_{i - 1} \circ \sigma_j\).

5. If \(i \leq j\), then \(\sigma_j \circ \sigma_i = \sigma_i \circ \sigma_{j + 1}\).

This means that whenever the compositions on both the left and the right are defined then the corresponding equality should hold.

We get a unique morphism \(\sigma^0_0 = U(\sigma^0_0) : U_1 \to U_0\) and two morphisms \(\delta^1_0 = U(\delta^1_0)\), and \(\delta^1_1 = U(\delta^1_1)\) which are morphisms \(U_0 \to U_1\). There are two morphisms \(\sigma^1_0 = U(\sigma^1_0)\), \(\sigma^1_1 = U(\sigma^1_1)\) which are morphisms \(U_2 \to U_1\). Three morphisms \(\delta^2_0 = U(\delta^2_0)\), \(\delta^2_1 = U(\delta^2_1)\), \(\delta^2_2 = U(\delta^2_2)\) which are morphisms \(U_1 \to U_2\). And so on.

Pictorially we think of \(U\) as follows: \[\xymatrix{ U_0 \ar@<1ex>[r] \ar@<-1ex>[r] & U_1 \ar@<0ex>[l] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & U_2 \ar@<1ex>[l] \ar@<-1ex>[l] }\] Here the \(\delta\)-morphisms are the arrows pointing right and the \(\sigma\)-morphisms are the arrows pointing left.

Example

The simplest example is the constant cosimplicial object with value \(X \in \Ob(\mathcal{C})\). In other words, \(U_n = X\) and all maps are \(\text{id}_X\).

Example

Suppose that \(X\to Y\) is a morphism of \(\mathcal{C}\) such that all the pushouts \(Y\amalg_X Y \amalg_X \ldots \amalg_X Y\) exist. Then we set \(U_n\) equal to the \((n + 1)\)-fold pushout, and we let \(\varphi : [n] \to [m]\) correspond to the map \[(y \text{ in }i\text{th component}) \mapsto (y \text{ in }\varphi(i)\text{th component})\] on “coordinates”. In other words, the map \(U_1 = Y \amalg_X Y \to U_0 = Y\) is the identity on each component. The two maps \(U_0 = Y \to U_1 = Y \amalg_X Y\) are the two coprojections.

Example

For every \(n \geq 0\) we denote by \(C[n]\) the cosimplicial set \[\Delta \longrightarrow \textit{Sets},\quad [k] \longmapsto \Mor_{\Delta}([n], [k])\] This example is dual to Example 0176.

Lemma

Let \(\mathcal{C}\) be a category. Let \(U\) be a cosimplicial object of \(\mathcal{C}\). Each of the morphisms \(\delta^n_i : U_{n - 1} \to U_n\) has a left inverse. In particular \(\delta^n_i\) is a monomorphism.

Proof

This is true because \(\sigma_i^{n - 1} \circ \delta^n_i = \text{id}_{U_{n - 1}}\) for \(i < n\); for \(i = n\) use \(\sigma_{n - 1}^{n - 1} \circ \delta^n_n = \text{id}_{U_{n - 1}}\).

Products of simplicial objects

Of course we should define the product of simplicial objects as the product in the category of simplicial objects. This may lead to the potentially confusing situation where the product exists but is not described as below. To avoid this we define the product directly as follows.

Definition

Let \(\mathcal{C}\) be a category. Let \(U\) and \(V\) be simplicial objects of \(\mathcal{C}\). Assume the products \(U_n \times V_n\) exist in \(\mathcal{C}\). The product of \(U\) and \(V\) is the simplicial object \(U \times V\) defined as follows:

1. \((U \times V)_n = U_n \times V_n\),

2. \(d^n_i = (d^n_i, d^n_i)\), and

3. \(s^n_i = (s^n_i, s^n_i)\).

In other words, \(U \times V\) is the product of the presheaves \(U\) and \(V\) on \(\Delta\).

Lemma

If \(U\) and \(V\) are simplicial objects in the category \(\mathcal{C}\), and if \(U \times V\) exists, then we have \[\Mor(W, U \times V) = \Mor(W, U) \times \Mor(W, V)\] for any third simplicial object \(W\) of \(\mathcal{C}\).

Proof

Omitted.

Fibre products of simplicial objects

Of course we should define the fibre product of simplicial objects as the fibre product in the category of simplicial objects. This may lead to the potentially confusing situation where the fibre product exists but is not described as below. To avoid this we define the fibre product directly as follows.

Definition

Let \(\mathcal{C}\) be a category. Let \(U, V, W\) be simplicial objects of \(\mathcal{C}\). Let \(a : V \to U\), \(b : W \to U\) be morphisms. Assume the fibre products \(V_n \times_{U_n} W_n\) exist in \(\mathcal{C}\). The fibre product of \(V\) and \(W\) over \(U\) is the simplicial object \(V \times_U W\) defined as follows:

1. \((V \times_U W)_n = V_n \times_{U_n} W_n\),

2. \(d^n_i = (d^n_i, d^n_i)\), and

3. \(s^n_i = (s^n_i, s^n_i)\).

In other words, \(V \times_U W\) is the fibre product of the presheaves \(V\) and \(W\) over the presheaf \(U\) on \(\Delta\).

Lemma

If \(U, V, W\) are simplicial objects in the category \(\mathcal{C}\), and if \(a : V \to U\), \(b : W \to U\) are morphisms and if \(V \times_U W\) exists, then we have \[\Mor(T, V \times_U W) = \Mor(T, V) \times_{\Mor(T, U)} \Mor(T, W)\] for any fourth simplicial object \(T\) of \(\mathcal{C}\).

Proof

Omitted.

Pushouts of simplicial objects

Of course we should define the pushout of simplicial objects as the pushout in the category of simplicial objects. This may lead to the potentially confusing situation where the pushouts exist but are not as described below. To avoid this we define the pushout directly as follows.

Definition

Let \(\mathcal{C}\) be a category. Let \(U, V, W\) be simplicial objects of \(\mathcal{C}\). Let \(a : U \to V\), \(b : U \to W\) be morphisms. Assume the pushouts \(V_n \amalg_{U_n} W_n\) exist in \(\mathcal{C}\). The pushout of \(V\) and \(W\) over \(U\) is the simplicial object \(V\amalg_U W\) defined as follows:

1. \((V \amalg_U W)_n = V_n \amalg_{U_n} W_n\),

2. \(d^n_i = (d^n_i, d^n_i)\), and

3. \(s^n_i = (s^n_i, s^n_i)\).

In other words, \(V\amalg_U W\) is the pushout of the presheaves \(V\) and \(W\) over the presheaf \(U\) on \(\Delta\).

Lemma

If \(U, V, W\) are simplicial objects in the category \(\mathcal{C}\), and if \(a : U \to V\), \(b : U \to W\) are morphisms and if \(V\amalg_U W\) exists, then we have \[\Mor(V\amalg_U W, T) = \Mor(V, T) \times_{\Mor(U, T)} \Mor(W, T)\] for any fourth simplicial object \(T\) of \(\mathcal{C}\).

Proof

Omitted.

Products of cosimplicial objects

Of course we should define the product of cosimplicial objects as the product in the category of cosimplicial objects. This may lead to the potentially confusing situation where the product exists but is not described as below. To avoid this we define the product directly as follows.

Definition

Let \(\mathcal{C}\) be a category. Let \(U\) and \(V\) be cosimplicial objects of \(\mathcal{C}\). Assume the products \(U_n \times V_n\) exist in \(\mathcal{C}\). The product of \(U\) and \(V\) is the cosimplicial object \(U \times V\) defined as follows:

1. \((U \times V)_n = U_n \times V_n\),

2. for any \(\varphi : [n] \to [m]\) the map \((U \times V)(\varphi) : U_n \times V_n \to U_m \times V_m\) is the product \(U(\varphi) \times V(\varphi)\).

Lemma

If \(U\) and \(V\) are cosimplicial objects in the category \(\mathcal{C}\), and if \(U \times V\) exists, then we have \[\Mor(W, U \times V) = \Mor(W, U) \times \Mor(W, V)\] for any third cosimplicial object \(W\) of \(\mathcal{C}\).

Proof

Omitted.

Fibre products of cosimplicial objects

Of course we should define the fibre product of cosimplicial objects as the fibre product in the category of cosimplicial objects. This may lead to the potentially confusing situation where the product exists but is not described as below. To avoid this we define the fibre product directly as follows.

Definition

Let \(\mathcal{C}\) be a category. Let \(U, V, W\) be cosimplicial objects of \(\mathcal{C}\). Let \(a : V \to U\) and \(b : W \to U\) be morphisms. Assume the fibre products \(V_n \times_{U_n} W_n\) exist in \(\mathcal{C}\). The fibre product of \(V\) and \(W\) over \(U\) is the cosimplicial object \(V \times_U W\) defined as follows:

1. \((V \times_U W)_n = V_n \times_{U_n} W_n\),

2. for any \(\varphi : [n] \to [m]\) the map \((V \times_U W)(\varphi) : V_n \times_{U_n} W_n \to V_m \times_{U_m} W_m\) is the product \(V(\varphi) \times_{U(\varphi)} W(\varphi)\).

Lemma

If \(U, V, W\) are cosimplicial objects in the category \(\mathcal{C}\), and if \(a : V \to U\), \(b : W \to U\) are morphisms and if \(V \times_U W\) exists, then we have \[\Mor(T, V \times_U W) = \Mor(T, V) \times_{\Mor(T, U)} \Mor(T, W)\] for any fourth cosimplicial object \(T\) of \(\mathcal{C}\).

Proof

Omitted.

Simplicial sets

Let \(U\) be a simplicial set. It is a good idea to think of \(U_0\) as the \(0\)-simplices, the set \(U_1\) as the \(1\)-simplices, the set \(U_2\) as the \(2\)-simplices, and so on.

We think of the maps \(s^n_j : U_n \to U_{n + 1}\) as the map that associates to an \(n\)-simplex \(A\) the degenerate \((n + 1)\)-simplex \(B\) whose \((j, j + 1)\)-edge is collapsed to the vertex \(j\) of \(A\). We think of the map \(d^n_j : U_n \to U_{n - 1}\) as the map that associates to an \(n\)-simplex \(A\) one of the faces, namely the face that omits the vertex \(j\). In this way it become possible to visualize the relations among the maps \(s^n_j\) and \(d^n_j\) geometrically.

Definition

Let \(U\) be a simplicial set. We say \(x\) is an \(n\)-simplex of \(U\) to signify that \(x\) is an element of \(U_n\). We say that \(y\) is the \(j\)th face of \(x\) to signify that \(d^n_jx = y\). We say that \(z\) is the \(j\)th degeneracy of \(x\) if \(z = s^n_jx\). A simplex is called degenerate if it is the degeneracy of another simplex.

Here are a few fundamental examples.

Example

For every \(n \geq 0\) we denote \(\Delta[n]\) the simplicial set \[\Delta^{opp} \longrightarrow \textit{Sets},\quad [k] \longmapsto \Mor_{\Delta}([k], [n])\] We leave it to the reader to verify the following statements. Every \(m\)-simplex of \(\Delta[n]\) with \(m > n\) is degenerate. There is a unique nondegenerate \(n\)-simplex of \(\Delta[n]\), namely \(\text{id}_{[n]}\).

Lemma

Let \(U\) be a simplicial set. Let \(n \geq 0\) be an integer. There is a canonical bijection \[\Mor(\Delta[n], U) \longrightarrow U_n\] which maps a morphism \(\varphi\) to the value of \(\varphi\) on the unique nondegenerate \(n\)-simplex of \(\Delta[n]\).

Proof

Omitted.

Example

Consider the category \(\Delta/[n]\) of objects over \([n]\) in \(\Delta\), see Categories, Example 001G. There is a functor \(p : \Delta/[n] \to \Delta\). The fibre category of \(p\) over \([k]\), see Categories, Section 003S, has as objects the set \(\Delta[n]_k\) of \(k\)-simplices in \(\Delta[n]\), and as morphisms only identities. For every morphism \(\varphi : [k] \to [l]\) of \(\Delta\), and every object \(\psi : [l] \to [n]\) in the fibre category over \([l]\) there is a unique object over \([k]\) with a morphism covering \(\varphi\), namely \(\psi \circ \varphi : [k] \to [n]\). Thus \(\Delta/[n]\) is fibred in sets over \(\Delta\). In other words, we may think of \(\Delta/[n]\) as a presheaf of sets over \(\Delta\). See also, Categories, Example 0044. And this presheaf of sets agrees with the simplicial set \(\Delta[n]\). In particular, from Equation (016H) and Lemma 0177 above we get the formula \[\Mor_{\textit{PSh}(\Delta)}(\Delta/[n], U) = U_n\] for any simplicial set \(U\).

Lemma

Let \(U\), \(V\) be simplicial sets. Let \(a, b \geq 0\) be integers. Assume every \(n\)-simplex of \(U\) is degenerate if \(n > a\). Assume every \(n\)-simplex of \(V\) is degenerate if \(n > b\). Then every \(n\)-simplex of \(U \times V\) is degenerate if \(n > a + b\).

Proof

Suppose \(n > a + b\). Let \((u, v) \in (U \times V)_n = U_n \times V_n\). By assumption, there exists a \(\alpha : [n] \to [a]\) and a \(u' \in U_a\) and a \(\beta : [n] \to [b]\) and a \(v' \in V_b\) such that \(u = U(\alpha)(u')\) and \(v = V(\beta)(v')\). Because \(n > a + b\), there exists an \(0 \leq i \leq a + b\) such that \(\alpha(i) = \alpha(i + 1)\) and \(\beta(i) = \beta(i + 1)\). It follows immediately that \((u, v)\) is in the image of \(s^{n - 1}_i\).

Nerves of categories

Definition

Let \(\mathcal{C}\) be a category. View \([n]\) as the category whose objects are \(0, \ldots, n\) with a unique morphism \(i \to j\) if \(i \leq j\). The nerve of \(\mathcal{C}\) is the simplicial set \(N(\mathcal{C})\) given by \[N(\mathcal{C})_n = \{\text{functors }[n] \longrightarrow \mathcal{C}\}.\] The simplicial operations are given by precomposition. For a simplicial set \(U\) and \(n \geq 2\), set \[P_n(U) = \{(a_1, \ldots, a_n) \in U_1^n \mid d_0(a_i) = d_1(a_{i + 1})\text{ for }1 \leq i < n\}.\] The \(n\)th Segal map is the map \[U_n \longrightarrow P_n(U)\] which sends a simplex to its consecutive edges.

Lemma

The functor \[N : \textit{Cat} \longrightarrow \textit{Simp}(\textit{Sets})\] is fully faithful. A simplicial set \(U\) is isomorphic to the nerve of a category if and only if all its Segal maps are bijective.

Proof

An \(n\)-simplex of \(N(\mathcal{C})\) is the same thing as a string \[x_0 \xrightarrow{a_1} x_1 \xrightarrow{a_2} \ldots \xrightarrow{a_n} x_n\] of composable arrows of \(\mathcal{C}\). Thus all the Segal maps of \(N(\mathcal{C})\) are bijective.

Conversely, suppose the Segal maps of \(U\) are bijective. Define a category \(\mathcal{C}_U\) as follows. Its objects are the elements of \(U_0\), its arrows are the elements of \(U_1\), and the source and target of an arrow \(a\) are \(d_1(a)\) and \(d_0(a)\). The identity of \(x \in U_0\) is \(s_0(x)\). If \(a\) and \(b\) are composable, let \(z \in U_2\) be the unique simplex whose consecutive edges are \(a\) and \(b\), and set \[b \circ a = d_1(z).\] The simplicial identities in degrees at most \(2\) give the identity axioms. The bijectivity of the Segal map in degree \(3\), together with the simplicial identities, gives associativity.

The Segal maps give an isomorphism \(U \to N(\mathcal{C}_U)\). Finally, a map between nerves is determined in degrees \(0\) and \(1\). Compatibility with the face and degeneracy maps says that it preserves sources, targets, and identities, while compatibility in degree \(2\) says that it preserves composition. Hence it comes from a unique functor. This proves full faithfulness and the description of the essential image.

Remark

Let \(\mathcal{A}\) be a category with finite limits. The same construction works internally. Namely, a simplicial object \(U\) of \(\mathcal{A}\) is the nerve of a category object if and only if the Segal morphisms \[U_n \longrightarrow U_1 \times_{U_0} \ldots \times_{U_0} U_1\] are isomorphisms for all \(n \geq 2\). The source, target, and identity are \(d_1\), \(d_0\), and \(s_0\). Composition is obtained from \(d_1 : U_2 \to U_1\) after inverting the Segal morphism in degree \(2\). The proof of the preceding lemma uses only finite limits and commutative diagrams, so it applies without change.

Truncated simplicial objects and skeleton functors

Let \(\Delta_{\leq n}\) denote the full subcategory of \(\Delta\) with objects \([0], [1], [2], \ldots, [n]\). Let \(\mathcal{C}\) be a category.

Definition

An \(n\)-truncated simplicial object of \(\mathcal{C}\) is a contravariant functor from \(\Delta_{\leq n}\) to \(\mathcal{C}\). A morphism of \(n\)-truncated simplicial objects is a transformation of functors. We denote the category of \(n\)-truncated simplicial objects of \(\mathcal{C}\) by the symbol \(\text{Simp}_n(\mathcal{C})\).

Given a simplicial object \(U\) of \(\mathcal{C}\) the truncation \(\text{sk}_n U\) is the restriction of \(U\) to the subcategory \(\Delta_{\leq n}\). This defines a skeleton functor \[\text{sk}_n : \text{Simp}(\mathcal{C}) \longrightarrow \text{Simp}_n(\mathcal{C})\] from the category of simplicial objects of \(\mathcal{C}\) to the category of \(n\)-truncated simplicial objects of \(\mathcal{C}\). See Remark 018Q to avoid possible confusion with other functors in the literature.

Products with simplicial sets

Let \(\mathcal{C}\) be a category. Let \(U\) be a simplicial set. Let \(V\) be a simplicial object of \(\mathcal{C}\). We can consider the covariant functor which associates to a simplicial object \(W\) of \(\mathcal{C}\) the set [017B]\[\begin{equation} \left\{ (f_{n, u} : V_n \to W_n)_{n \geq 0, u \in U_n} \text{ such that } \begin{matrix} \forall \varphi : [m] \to [n] \\ f_{m, U(\varphi)(u)} \circ V(\varphi) = W(\varphi) \circ f_{n, u} \end{matrix} \right\} \end{equation}\] If this functor is of the form \(\Mor_{\text{Simp}(\mathcal{C})}(Q, -)\) then we can think of \(Q\) as the product of \(U\) with \(V\). Instead of formalizing this in this way we just directly define the product as follows.

Definition

Let \(\mathcal{C}\) be a category such that the coproduct of any two objects of \(\mathcal{C}\) exists. Let \(U\) be a simplicial set. Let \(V\) be a simplicial object of \(\mathcal{C}\). Assume that each \(U_n\) is finite nonempty. In this case we define the product \(U \times V\) of \(U\) and \(V\) to be the simplicial object of \(\mathcal{C}\) whose \(n\)th term is the object \[(U \times V)_n = \coprod\nolimits_{u\in U_n} V_n\] with maps for \(\varphi : [m] \to [n]\) given by the morphism \[\coprod\nolimits_{u\in U_n} V_n \longrightarrow \coprod\nolimits_{u'\in U_m} V_m\] which maps the component \(V_n\) corresponding to \(u\) to the component \(V_m\) corresponding to \(u' = U(\varphi)(u)\) via the morphism \(V(\varphi)\). More loosely, if all of the coproducts displayed above exist (without assuming anything about \(\mathcal{C}\)) we will say that the product \(U \times V\) exists.

Lemma

Let \(\mathcal{C}\) be a category such that the coproduct of any two objects of \(\mathcal{C}\) exists. Let \(U\) be a simplicial set. Let \(V\) be a simplicial object of \(\mathcal{C}\). Assume that each \(U_n\) is finite nonempty. The functor \(W \mapsto \Mor_{\text{Simp}(\mathcal{C})}(U \times V, W)\) is canonically isomorphic to the functor which maps \(W\) to the set in Equation (017B).

Proof

Omitted.

Lemma

Let \(\mathcal{C}\) be a category such that the coproduct of any two objects of \(\mathcal{C}\) exists. Let us temporarily denote \(\textit{FSSets}\) the category of simplicial sets all of whose components are finite nonempty.

1. The rule \((U, V) \mapsto U \times V\) defines a functor \(\textit{FSSets} \times \text{Simp}(\mathcal{C}) \to \text{Simp}(\mathcal{C})\).

2. For every \(U\), \(V\) as above there is a canonical map of simplicial objects \[U \times V \longrightarrow V\] defined by taking the identity on each component of \((U \times V)_n = \coprod_u V_n\).

Proof

Omitted.

We briefly study a special case of the construction above. Let \(\mathcal{C}\) be a category. Let \(X\) be an object of \(\mathcal{C}\). Let \(k \geq 0\) be an integer. If all coproducts \(X \amalg \ldots \amalg X\) exist then according to the definition above the product \[X \times \Delta[k]\] exists, where we think of \(X\) as the corresponding constant simplicial object.

Lemma

With \(X\) and \(k\) as above. For any simplicial object \(V\) of \(\mathcal{C}\) we have the following canonical bijection \[\Mor_{\text{Simp}(\mathcal{C})}(X \times \Delta[k], V) \longrightarrow \Mor_\mathcal{C}(X, V_k).\] which maps \(\gamma\) to the restriction of the morphism \(\gamma_k\) to the component corresponding to \(\text{id}_{[k]}\). Similarly, for any \(n \geq k\), if \(W\) is an \(n\)-truncated simplicial object of \(\mathcal{C}\), then we have \[\Mor_{\text{Simp}_n(\mathcal{C})}(\text{sk}_n(X \times \Delta[k]), W) = \Mor_\mathcal{C}(X, W_k).\]

Proof

A morphism \(\gamma : X \times \Delta[k] \to V\) is given by a family of morphisms \(\gamma_\alpha : X \to V_n\) where \(\alpha : [n] \to [k]\). The morphisms have to satisfy the rules that for all \(\varphi : [m] \to [n]\) the diagrams \[\xymatrix{ X \ar[r]^{\gamma_\alpha} \ar[d]^{\text{id}_X} & V_n \ar[d]^{V(\varphi)} \\ X \ar[r]^{\gamma_{\alpha \circ \varphi}} & V_m }\] commute. Taking \(\alpha = \text{id}_{[k]}\), we see that for any \(\varphi : [m] \to [k]\) we have \(\gamma_\varphi = V(\varphi) \circ \gamma_{\text{id}_{[k]}}\). Thus the morphism \(\gamma\) is determined by the value of \(\gamma\) on the component corresponding to \(\text{id}_{[k]}\). Conversely, given such a morphism \(f : X \to V_k\) we easily construct a morphism \(\gamma\) by putting \(\gamma_\alpha = V(\alpha) \circ f\).

The truncated case is similar, and left to the reader.

A particular example of this is the case \(k = 0\). In this case the formula of the lemma just says that \[\Mor_\mathcal{C}(X, V_0) = \Mor_{\text{Simp}(\mathcal{C})}(X, V)\] where on the right hand side \(X\) indicates the constant simplicial object with value \(X\). We will use this formula without further mention in the following.

Hom from simplicial sets into cosimplicial objects

Let \(\mathcal{C}\) be a category. Let \(U\) be a simplicial object of \(\mathcal{C}\), and let \(V\) be a cosimplicial object of \(\mathcal{C}\). Then we get a cosimplicial set \(\Hom_\mathcal{C}(U, V)\) as follows:

1. we set \(\Hom_\mathcal{C}(U, V)_n = \Mor_\mathcal{C}(U_n, V_n)\), and

2. for \(\varphi : [m] \to [n]\) we take the map \(\Hom_\mathcal{C}(U, V)_m \to \Hom_\mathcal{C}(U, V)_n\) given by \(f \mapsto V(\varphi) \circ f \circ U(\varphi)\).

This is our motivation for the following definition.

Definition

Let \(\mathcal{C}\) be a category with finite products. Let \(V\) be a cosimplicial object of \(\mathcal{C}\). Let \(U\) be a simplicial set such that each \(U_n\) is finite nonempty. We define \(\Hom(U, V)\) to be the cosimplicial object of \(\mathcal{C}\) defined as follows:

1. we set \(\Hom(U, V)_n = \prod_{u \in U_n} V_n\), in other words the unique object of \(\mathcal{C}\) such that its \(X\)-valued points satisfy \[\Mor_\mathcal{C}(X, \Hom(U, V)_n) = \text{Map}(U_n, \Mor_\mathcal{C}(X, V_n))\] and

2. for \(\varphi : [m] \to [n]\) we take the map \(\Hom(U, V)_m \to \Hom(U, V)_n\) given by \(f \mapsto V(\varphi) \circ f \circ U(\varphi)\) on \(X\)-valued points as above.

We leave it to the reader to spell out the definition in terms of maps between products. We also point out that the construction is functorial in both \(U\) (contravariantly) and \(V\) (covariantly), exactly as in Lemma 017E in the case of products of simplicial sets with simplicial objects.

Hom from cosimplicial sets into simplicial objects

Let \(\mathcal{C}\) be a category. Let \(U\) be a cosimplicial object of \(\mathcal{C}\), and let \(V\) be a simplicial object of \(\mathcal{C}\). Then we get a simplicial set \(\Hom_\mathcal{C}(U, V)\) as follows:

1. we set \(\Hom_\mathcal{C}(U, V)_n = \Mor_\mathcal{C}(U_n, V_n)\), and

2. for \(\varphi : [m] \to [n]\) we take the map \(\Hom_\mathcal{C}(U, V)_n \to \Hom_\mathcal{C}(U, V)_m\) given by \(f \mapsto V(\varphi) \circ f \circ U(\varphi)\).

This is our motivation for the following definition.

Definition

Let \(\mathcal{C}\) be a category with finite products. Let \(V\) be a simplicial object of \(\mathcal{C}\). Let \(U\) be a cosimplicial set such that each \(U_n\) is finite nonempty. We define \(\Hom(U, V)\) to be the simplicial object of \(\mathcal{C}\) defined as follows:

1. we set \(\Hom(U, V)_n = \prod_{u \in U_n} V_n\), in other words the unique object of \(\mathcal{C}\) such that its \(X\)-valued points satisfy \[\Mor_\mathcal{C}(X, \Hom(U, V)_n) = \text{Map}(U_n, \Mor_\mathcal{C}(X, V_n))\] and

2. for \(\varphi : [m] \to [n]\) we take the map \(\Hom(U, V)_n \to \Hom(U, V)_m\) given by \(f \mapsto V(\varphi) \circ f \circ U(\varphi)\) on \(X\)-valued points as above.

We leave it to the reader to spell out the definition in terms of maps between products. We also point out that the construction is functorial in both \(U\) (contravariantly) and \(V\) (covariantly), exactly as in Lemma 017E in the case of products of simplicial sets with simplicial objects.

We spell out the construction above in a special case. Let \(X\) be an object of a category \(\mathcal{C}\). Assume that self products \(X \times \ldots \times X\) exist. Let \(k\) be an integer. Consider the simplicial object \(U\) with terms \[U_n = \prod\nolimits_{\alpha \in \Mor([k], [n])} X\] and maps given \(\varphi : [m] \to [n]\) \[U(\varphi) : \prod\nolimits_{\alpha \in \Mor([k], [n])} X \longrightarrow \prod\nolimits_{\alpha' \in \Mor([k], [m])} X, \quad (f_{\alpha})_{\alpha} \longmapsto (f_{\varphi \circ \alpha'})_{\alpha'}\] In terms of “coordinates”, the element \((x_\alpha)_\alpha\) is mapped to the element \((x_{\varphi \circ \alpha'})_{\alpha'}\). We claim this object is equal to \(\Hom(C[k], X)\) where we think of \(X\) as the constant simplicial object \(X\) and where \(C[k]\) is the cosimplicial set from Example 0B13.

Lemma

With \(X\), \(k\) and \(U\) as above.

1. For any simplicial object \(V\) of \(\mathcal{C}\) we have the following canonical bijection \[\Mor_{\text{Simp}(\mathcal{C})}(V, U) \longrightarrow \Mor_\mathcal{C}(V_k, X).\] which maps \(\gamma\) to the morphism \(\gamma_k\) composed with the projection onto the factor corresponding to \(\text{id}_{[k]}\).

2. Similarly, if \(W\) is an \(k\)-truncated simplicial object of \(\mathcal{C}\), then we have \[\Mor_{\text{Simp}_k(\mathcal{C})}(W, \text{sk}_k U) = \Mor_\mathcal{C}(W_k, X).\]

3. The object \(U\) constructed above is an incarnation of \(\Hom(C[k], X)\) where \(C[k]\) is the cosimplicial set from Example 0B13.

Proof

We first prove (1). Suppose that \(\gamma : V \to U\) is a morphism. This is given by a family of morphisms \(\gamma_{\alpha} : V_n \to X\) for \(\alpha : [k] \to [n]\). The morphisms have to satisfy the rules that for all \(\varphi : [m] \to [n]\) the diagrams \[\xymatrix{ X \ar[d]^{\text{id}_X} & V_n \ar[d]^{V(\varphi)} \ar[l]^{\gamma_{\varphi \circ \alpha'}} \\ X & V_m \ar[l]_{\gamma_{\alpha'}} }\] commute for all \(\alpha' : [k] \to [m]\). Taking \(\alpha' = \text{id}_{[k]}\), we see that for any \(\varphi : [k] \to [n]\) we have \(\gamma_\varphi = \gamma_{\text{id}_{[k]}} \circ V(\varphi)\). Thus the morphism \(\gamma\) is determined by the component of \(\gamma_k\) corresponding to \(\text{id}_{[k]}\). Conversely, given such a morphism \(f : V_k \to X\) we easily construct a morphism \(\gamma\) by putting \(\gamma_\alpha = f \circ V(\alpha)\).

The truncated case is similar, and left to the reader.

Part (3) is immediate from the construction of \(U\) and the fact that \(C[k]_n = \Mor([k], [n])\) which are the index sets used in the construction of \(U_n\).

Internal Hom

Let \(\mathcal{C}\) be a category with finite nonempty products. Let \(U\), \(V\) be simplicial objects of \(\mathcal{C}\). In some cases the functor \[\text{Simp}(\mathcal{C})^{opp} \longrightarrow \textit{Sets}, \quad W \longmapsto \Mor_{\text{Simp}(\mathcal{C})}(W \times V, U)\] is representable. In this case we denote \(\SheafHom(V, U)\) the resulting simplicial object of \(\mathcal{C}\), and we say that the internal hom of \(V\) into \(U\) exists. Moreover, in this case, given \(X\) in \(\mathcal{C}\), we would have \[\begin{align*} \Mor_\mathcal{C}(X, \SheafHom(V, U)_n) & = \Mor_{\text{Simp}(\mathcal{C})}(X \times \Delta[n], \SheafHom(V, U)) \\ & = \Mor_{\text{Simp}(\mathcal{C})}(X \times \Delta[n]\times V, U) \\ & = \Mor_{\text{Simp}(\mathcal{C})}(X, \SheafHom(\Delta[n] \times V, U)) \\ & = \Mor_\mathcal{C}(X, \SheafHom(\Delta[n] \times V, U)_0) \end{align*}\] provided that \(\SheafHom(\Delta[n] \times V, U)\) exists also. The first and last equalities follow from Lemma 017F.

The lesson we learn from this is that, given \(U\) and \(V\), if we want to construct the internal hom then we should try to construct the objects \[\SheafHom(\Delta[n] \times V, U)_0\] because these should be the \(n\)th term of \(\SheafHom(V, U)\). In the next section we study a construction of simplicial objects “\(\Hom(\Delta[n], U)\)”.

Hom from simplicial sets into simplicial objects

Motivated by the discussion on internal hom we define what should be the simplicial object classifying morphisms from a simplicial set into a given simplicial object of the category \(\mathcal{C}\).

Definition

Let \(\mathcal{C}\) be a category such that the coproduct of any two objects exists. Let \(U\) be a simplicial set, with \(U_n\) finite nonempty for all \(n \geq 0\). Let \(V\) be a simplicial object of \(\mathcal{C}\). We denote \(\Hom(U, V)\) any simplicial object of \(\mathcal{C}\) such that \[\Mor_{\text{Simp}(\mathcal{C})}(W, \Hom(U, V)) = \Mor_{\text{Simp}(\mathcal{C})}(W \times U, V)\] functorially in the simplicial object \(W\) of \(\mathcal{C}\).

Of course \(\Hom(U, V)\) need not exist. Also, by the discussion in Section 017G we expect that if it does exist, then \(\Hom(U, V)_n = \Hom(U \times \Delta[n], V)_0\). We do not use the italic notation for these Hom objects since \(\Hom(U, V)\) is not an internal hom.

Lemma

Assume the category \(\mathcal{C}\) has coproducts of any two objects and countable limits. Let \(U\) be a simplicial set, with \(U_n\) finite nonempty for all \(n \geq 0\). Let \(V\) be a simplicial object of \(\mathcal{C}\). Then the functor \[\begin{eqnarray*} \mathcal{C}^{opp} & \longrightarrow & \textit{Sets} \\ X & \longmapsto & \Mor_{\text{Simp}(\mathcal{C})}(X \times U, V) \end{eqnarray*}\] is representable.

Proof

A morphism from \(X \times U\) into \(V\) is given by a collection of morphisms \(f_u : X \to V_n\) with \(n \geq 0\) and \(u \in U_n\). And such a collection actually defines a morphism if and only if for all \(\varphi : [m] \to [n]\) all the diagrams \[\xymatrix{ X \ar[r]^{f_u} \ar[d]_{\text{id}_X} & V_n \ar[d]^{V(\varphi)} \\ X \ar[r]^{f_{U(\varphi)(u)}} & V_m }\] commute. Thus it is natural to introduce a category \(\mathcal{U}\) and a functor \(\mathcal{V} : \mathcal{U}^{opp} \to \mathcal{C}\) as follows:

1. The set of objects of \(\mathcal{U}\) is \(\coprod_{n \geq 0} U_n\),

2. a morphism from \(u' \in U_m\) to \(u \in U_n\) is a \(\varphi : [m] \to [n]\) such that \(U(\varphi)(u) = u'\)

3. for \(u \in U_n\) we set \(\mathcal{V}(u) = V_n\), and

4. for \(\varphi : [m] \to [n]\) such that \(U(\varphi)(u) = u'\) we set \(\mathcal{V}(\varphi) = V(\varphi) : V_n \to V_m\).

At this point it is clear that our functor is nothing but the functor defining \[\lim_{\mathcal{U}^{opp}} \mathcal{V}\] Thus if \(\mathcal{C}\) has countable limits then this limit and hence an object representing the functor of the lemma exist.

Lemma

Assume the category \(\mathcal{C}\) has coproducts of any two objects and finite limits. Let \(U\) be a simplicial set, with \(U_n\) finite nonempty for all \(n \geq 0\). Assume that all \(n\)-simplices of \(U\) are degenerate for all \(n \gg 0\). Let \(V\) be a simplicial object of \(\mathcal{C}\). Then the functor \[\begin{eqnarray*} \mathcal{C}^{opp} & \longrightarrow & \textit{Sets} \\ X & \longmapsto & \Mor_{\text{Simp}(\mathcal{C})}(X \times U, V) \end{eqnarray*}\] is representable.

Proof

We have to show that the category \(\mathcal{U}\) described in the proof of Lemma 017J has a finite subcategory \(\mathcal{U}'\) such that the limit of \(\mathcal{V}\) over \(\mathcal{U}'\) is the same as the limit of \(\mathcal{V}\) over \(\mathcal{U}\). We will use Categories, Lemma 002R. For \(m > 0\) let \(\mathcal{U}_{\leq m}\) denote the full subcategory with objects \(\coprod_{0 \leq n \leq m} U_m\). Let \(m_0\) be an integer such that every \(n\)-simplex of the simplicial set \(U\) is degenerate if \(n > m_0\). For any \(m \geq m_0\) large enough, the subcategory \(\mathcal{U}_{\leq m}\) satisfies property (1) of Categories, Definition 09WP.

Suppose that \(u \in U_n\) and \(u' \in U_{n'}\) with \(n, n' \leq m_0\) and suppose that \(\varphi : [k] \to [n]\), \(\varphi' : [k] \to [n']\) are morphisms such that \(U(\varphi)(u) = U(\varphi')(u')\). A simple combinatorial argument shows that if \(k > 2m_0\), then there exists an index \(0 \leq i \leq 2m_0\) such that \(\varphi(i) =\varphi(i + 1)\) and \(\varphi'(i) = \varphi'(i + 1)\). (The pigeon hole principle would tell you this works if \(k > m_0^2\) which is good enough for the argument below anyways.) Hence, if \(k > 2m_0\), we may write \(\varphi = \psi \circ \sigma^{k - 1}_i\) and \(\varphi' = \psi' \circ \sigma^{k - 1}_i\) for some \(\psi : [k - 1] \to [n]\) and some \(\psi' : [k - 1] \to [n']\). Since \(s^{k - 1}_i : U_{k - 1} \to U_k\) is injective, see Lemma 016F, we conclude that \(U(\psi)(u) = U(\psi')(u')\) also. Continuing in this fashion we conclude that given morphisms \(u \to z\) and \(u' \to z\) of \(\mathcal{U}\) with \(u, u' \in \mathcal{U}_{\leq m_0}\), there exists a commutative diagram \[\xymatrix{ u \ar[rd] \ar[rrd] & & \\ & a \ar[r] & z \\ u' \ar[ru] \ar[rru] }\] with \(a \in \mathcal{U}_{\leq 2m_0}\).

It is easy to deduce from this that the finite subcategory \(\mathcal{U}_{\leq 2m_0}\) works. Namely, suppose given \(x' \in U_n\) and \(x'' \in U_{n'}\) with \(n, n' \leq 2m_0\) as well as morphisms \(x' \to x\) and \(x'' \to x\) of \(\mathcal{U}\) with the same target. By our choice of \(m_0\) we can find objects \(u, u'\) of \(\mathcal{U}_{\leq m_0}\) and morphisms \(u \to x'\), \(u' \to x''\). By the above we can find \(a \in \mathcal{U}_{\leq 2m_0}\) and morphisms \(u \to a\), \(u' \to a\) such that \[\xymatrix{ u \ar[rd] \ar[rrd] \ar[r] & x' \ar[rd] & \\ & a \ar[r] & x \\ u' \ar[ru] \ar[rru] \ar[r] & x'' \ar[ru] & }\] is commutative. Turning this diagram 90 degrees clockwise we get the desired diagram as in (2) of Categories, Definition 09WP.

Lemma

Assume the category \(\mathcal{C}\) has coproducts of any two objects and finite limits. Let \(U\) be a simplicial set, with \(U_n\) finite nonempty for all \(n \geq 0\). Assume that all \(n\)-simplices of \(U\) are degenerate for all \(n \gg 0\). Let \(V\) be a simplicial object of \(\mathcal{C}\). Then \(\Hom(U, V)\) exists, moreover we have the expected equalities \[\Hom(U, V)_n = \Hom(U \times \Delta[n], V)_0.\]

Proof

We construct this simplicial object as follows. For \(n \geq 0\) let \(\Hom(U, V)_n\) denote the object of \(\mathcal{C}\) representing the functor \[X \longmapsto \Mor_{\text{Simp}(\mathcal{C})}(X \times U \times \Delta[n], V)\] This exists by Lemma 017K because \(U \times \Delta[n]\) is a simplicial set with finite sets of simplices and no nondegenerate simplices in high enough degree, see Lemma 0179. For \(\varphi : [m] \to [n]\) we obtain an induced map of simplicial sets \(\varphi : \Delta[m] \to \Delta[n]\). Hence we obtain a morphism \(X \times U \times \Delta[m] \to X \times U \times \Delta[n]\) functorial in \(X\), and hence a transformation of functors, which in turn gives \[\Hom(U, V)(\varphi) : \Hom(U, V)_n \longrightarrow \Hom(U, V)_m.\] Clearly this defines a contravariant functor \(\Hom(U, V)\) from \(\Delta\) into the category \(\mathcal{C}\). In other words, we have a simplicial object of \(\mathcal{C}\).

We have to show that \(\Hom(U, V)\) satisfies the desired universal property \[\Mor_{\text{Simp}(\mathcal{C})}(W, \Hom(U, V)) = \Mor_{\text{Simp}(\mathcal{C})}(W \times U, V)\] To see this, let \(f : W \to \Hom(U, V)\) be given. We want to construct the element \(f' : W \times U \to V\) of the right hand side. By construction, each \(f_n : W_n \to \Hom(U, V)_n\) corresponds to a morphism \(f_n : W_n \times U \times \Delta[n] \to V\). Further, for every morphism \(\varphi : [m] \to [n]\) the diagram \[\xymatrix{ W_n \times U \times \Delta[m] \ar[rr]_{W(\varphi)\times \text{id} \times \text{id}} \ar[d]_{\text{id} \times \text{id} \times \varphi} & & W_m \times U \times \Delta[m] \ar[d]^{f_m} \\ W_n \times U \times \Delta[n] \ar[rr]^{f_n} & & V }\] is commutative. For \(\psi : [n] \to [k]\) in \((\Delta[n])_k\) we denote \((f_n)_{k, \psi} : W_n \times U_k \to V_k\) the component of \((f_n)_k\) corresponding to the element \(\psi\). We define \(f'_n : W_n \times U_n \to V_n\) as \(f'_n = (f_n)_{n, \text{id}}\), in other words, as the restriction of \((f_n)_n : W_n \times U_n \times (\Delta[n])_n \to V_n\) to \(W_n \times U_n \times \text{id}_{[n]}\). To see that the collection \((f'_n)\) defines a morphism of simplicial objects, we have to show for any \(\varphi : [m] \to [n]\) that \(V(\varphi) \circ f'_n = f'_m \circ W(\varphi) \times U(\varphi)\). The commutative diagram above says that \((f_n)_{m, \varphi} : W_n \times U_m \to V_m\) is equal to \((f_m)_{m, \text{id}} \circ W(\varphi) : W_n \times U_m \to V_m\). But then the fact that \(f_n\) is a morphism of simplicial objects implies that the diagram \[\xymatrix{ W_n \times U_n \times (\Delta[n])_n \ar[r]_-{(f_n)_n} \ar[d]_{\text{id} \times U(\varphi) \times \varphi} & V_n \ar[d]^{V(\varphi)} \\ W_n \times U_m \times (\Delta[n])_m \ar[r]^-{(f_n)_m} & V_m }\] is commutative. And this implies that \((f_n)_{m, \varphi} \circ U(\varphi)\) is equal to \(V(\varphi) \circ (f_n)_{n, \text{id}}\). Altogether we obtain \(V(\varphi) \circ (f_n)_{n, \text{id}} = (f_n)_{m, \varphi} \circ U(\varphi) = (f_m)_{m, \text{id}} \circ W(\varphi)\circ U(\varphi) = (f_m)_{m, \text{id}} \circ W(\varphi)\times U(\varphi)\) as desired.

On the other hand, given a morphism \(f' : W \times U \to V\) we define a morphism \(f : W \to \Hom(U, V)\) as follows. By Lemma 017F the morphisms \(\text{id} : W_n \to W_n\) corresponds to a unique morphism \(c_n : W_n \times \Delta[n] \to W\). Hence we can consider the composition \[W_n \times \Delta[n] \times U \xrightarrow{c_n} W \times U \xrightarrow{f'} V.\] By construction this corresponds to a unique morphism \(f_n : W_n \to \Hom(U, V)_n\). We leave it to the reader to see that these define a morphism of simplicial sets as desired.

We also leave it to the reader to see that \(f \mapsto f'\) and \(f' \mapsto f\) are mutually inverse operations.

Lemma

Assume the category \(\mathcal{C}\) has coproducts of any two objects and finite limits. Let \(a : U \to V\), \(b : U \to W\) be morphisms of simplicial sets. Assume \(U_n, V_n, W_n\) finite nonempty for all \(n \geq 0\). Assume that all \(n\)-simplices of \(U, V, W\) are degenerate for all \(n \gg 0\). Let \(T\) be a simplicial object of \(\mathcal{C}\). Then \[\Hom(V, T) \times_{\Hom(U, T)} \Hom(W, T) = \Hom(V \amalg_U W, T)\] In other words, the fibre product on the left hand side is represented by the Hom object on the right hand side.

Proof

By Lemma 017L all the required \(\Hom\) objects exist and satisfy the correct functorial properties. Now we can identify the \(n\)th term on the left hand side as the object representing the functor that associates to \(X\) the first set of the following sequence of functorial equalities \[\begin{align*} & \Mor(X \times \Delta[n], \Hom(V, T) \times_{\Hom(U, T)} \Hom(W, T)) \\ & = \Mor(X \times \Delta[n], \Hom(V, T)) \times_{\Mor(X \times \Delta[n], \Hom(U, T))} \Mor(X \times \Delta[n], \Hom(W, T)) \\ & = \Mor(X \times \Delta[n] \times V, T) \times_{\Mor(X \times \Delta[n] \times U, T)} \Mor(X \times \Delta[n] \times W, T) \\ & = \Mor(X \times \Delta[n] \times (V \amalg_U W), T) \end{align*}\] Here we have used the fact that \[(X \times \Delta[n] \times V) \amalg_{X \times \Delta[n] \times U} (X \times \Delta[n] \times W) = X \times \Delta[n] \times (V \amalg_U W)\] which is easy to verify term by term. The result of the lemma follows as the last term in the displayed sequence of equalities corresponds to \(\Hom(V \amalg_U W, T)_n\).

Splitting simplicial objects

A subobject \(N\) of an object \(X\) of the category \(\mathcal{C}\) is an object \(N\) of \(\mathcal{C}\) together with a monomorphism \(N \to X\). Of course we say (by abuse of notation) that the subobjects \(N\), \(N'\) are equal if there exists an isomorphism \(N \to N'\) compatible with the morphisms to \(X\). The collection of subobjects forms a partially ordered set. (Because of our conventions on categories; not true for category of spaces up to homotopy for example.)

Definition

Let \(\mathcal{C}\) be a category which admits finite nonempty coproducts. We say a simplicial object \(U\) of \(\mathcal{C}\) is split if there exist subobjects \(N(U_m)\) of \(U_m\), \(m \geq 0\) with the property that [017Q]\[\begin{equation} \coprod\nolimits_{\varphi : [n] \to [m]\text{ surjective}} N(U_m) \longrightarrow U_n \end{equation}\] is an isomorphism for all \(n \geq 0\). If \(U\) is an \(r\)-truncated simplicial object of \(\mathcal{C}\) then we say \(U\) is split if there exist subobjects \(N(U_m)\) of \(U_m\), \(r \geq m \geq 0\) with the property that (017Q) is an isomorphism for \(r \geq n \geq 0\).

If this is the case, then \(N(U_0) = U_0\). Next, we have \(U_1 = U_0 \amalg N(U_1)\). Second we have \[U_2 = U_0 \amalg N(U_1) \amalg N(U_1) \amalg N(U_2).\] It turns out that in many categories \(\mathcal{C}\) every simplicial object is split.

Lemma

Let \(U\) be a simplicial set. Then \(U\) has a unique splitting with \(N(U_m)\) equal to the set of nondegenerate \(m\)-simplices.

Proof

From the definition it follows immediately, that if there is a splitting then \(N(U_m)\) has to be the set of nondegenerate simplices. Let \(x \in U_n\). Suppose that there are surjections \(\varphi : [n] \to [k]\) and \(\psi : [n] \to [l]\) and nondegenerate simplices \(y \in U_k\), \(z \in U_l\) such that \(x = U(\varphi)(y)\) and \(x = U(\psi)(z)\). Choose a right inverse \(\xi : [l] \to [n]\) of \(\psi\), i.e., \(\psi \circ \xi = \text{id}_{[l]}\). Then \(z = U(\xi)(x)\). Hence \(z = U(\xi)(x) = U(\varphi \circ \xi)(y)\). Since \(z\) is nondegenerate we conclude that \(\varphi \circ \xi : [l] \to [k]\) is surjective, and hence \(l \geq k\). Similarly \(k \geq l\). Hence we see that \(\varphi \circ \xi : [l] \to [k]\) has to be the identity map for any choice of right inverse \(\xi\) of \(\psi\). This easily implies that \(\psi = \varphi\).

Of course it can happen that a map of simplicial sets maps a nondegenerate \(n\)-simplex to a degenerate \(n\)-simplex. Thus the splitting of Lemma 017R is not functorial. Here is a case where it is functorial.

Lemma

Let \(f : U \to V\) be a morphism of simplicial sets. Suppose that (a) the image of every nondegenerate simplex of \(U\) is a nondegenerate simplex of \(V\) and (b) the restriction of \(f\) to a map from the set of nondegenerate simplices of \(U\) to the set of nondegenerate simplices of \(V\) is injective. Then \(f_n\) is injective for all \(n\). Same holds with “injective” replaced by “surjective” or “bijective”.

Proof

Under hypothesis (a) we see that the map \(f\) preserves the disjoint union decompositions of the splitting of Lemma 017R, in other words that we get commutative diagrams \[\xymatrix{ \coprod\nolimits_{\varphi : [n] \to [m]\text{ surjective}} N(U_m) \ar[r] \ar[d] & U_n \ar[d] \\ \coprod\nolimits_{\varphi : [n] \to [m]\text{ surjective}} N(V_m) \ar[r] & V_n. }\] And then (b) clearly shows that the left vertical arrow is injective (resp. surjective, resp. bijective).

Lemma

Let \(U\) be a simplicial set. Let \(n \geq 0\) be an integer. The rule \[U'_m = \bigcup\nolimits_{\varphi : [m] \to [i], \ i\leq n} \Im(U(\varphi))\] defines a sub simplicial set \(U' \subset U\) with \(U'_i = U_i\) for \(i \leq n\). Moreover, all \(m\)-simplices of \(U'\) are degenerate for all \(m > n\).

Proof

If \(x \in U_m\) and \(x = U(\varphi)(y)\) for some \(y \in U_i\), \(i \leq n\) and some \(\varphi : [m] \to [i]\) then any image \(U(\psi)(x)\) for any \(\psi : [m'] \to [m]\) is equal to \(U(\varphi \circ \psi)(y)\) and \(\varphi \circ \psi : [m'] \to [i]\). Hence \(U'\) is a simplicial set. By construction all simplices in dimension \(n + 1\) and higher are degenerate.

Lemma

Let \(U\) be a simplicial abelian group. Then \(U\) has a splitting obtained by taking \(N(U_0) = U_0\) and for \(m \geq 1\) taking \[N(U_m) = \bigcap\nolimits_{i = 0}^{m - 1} \Ker(d^m_i).\] Moreover, this splitting is functorial on the category of simplicial abelian groups.

Proof

By induction on \(n\) we will show that the choice of \(N(U_m)\) in the lemma guarantees that (017Q) is an isomorphism for \(m \leq n\). This is clear for \(n = 0\). In the rest of this proof we are going to drop the superscripts from the maps \(d_i\) and \(s_i\) in order to improve readability. We will also repeatedly use the relations from Remark 016C.

First we make a general remark. For \(0 \leq i \leq m\) and \(z \in U_m\) we have \(d_i(s_i(z)) = z\). Hence we can write any \(x \in U_{m + 1}\) uniquely as \(x = x' + x''\) with \(d_i(x') = 0\) and \(x'' \in \Im(s_i)\) by taking \(x' = (x - s_i(d_i(x)))\) and \(x'' = s_i(d_i(x))\). Moreover, the element \(z \in U_m\) such that \(x'' = s_i(z)\) is unique because \(s_i\) is injective.

Here is a procedure for decomposing any \(x \in U_{n + 1}\). First, write \(x = x_0 + s_0(z_0)\) with \(d_0(x_0) = 0\). Next, write \(x_0 = x_1 + s_1(z_1)\) with \(d_1(x_1) = 0\). Continue like this to get \[\begin{eqnarray*} x & = & x_0 + s_0(z_0), \\ x_0 & = & x_1 + s_1(z_1), \\ x_1 & = & x_2 + s_2(z_2), \\ \ldots & \ldots & \ldots \\ x_{n - 1} & = & x_n + s_n(z_n) \end{eqnarray*}\] where \(d_i(x_i) = 0\) for all \(i = n, \ldots, 0\). By our general remark above all of the \(x_i\) and \(z_i\) are determined uniquely by \(x\). We claim that \(x_i \in \Ker(d_0) \cap \Ker(d_1) \cap \ldots \cap \Ker(d_i)\) and \(z_i \in \Ker(d_0) \cap \ldots \cap \Ker(d_{i - 1})\) for \(i = n, \ldots, 0\). Here and in the following an empty intersection of kernels indicates the whole space; i.e., the notation \(z_0 \in \Ker(d_0) \cap \ldots \cap \Ker(d_{i - 1})\) when \(i = 0\) means \(z_0 \in U_n\) with no restriction.

We prove this by ascending induction on \(i\). It is clear for \(i = 0\) by construction of \(x_0\) and \(z_0\). Let us prove it for \(0 < i \leq n\) assuming the result for \(i - 1\). First of all we have \(d_i(x_i) = 0\) by construction. So pick a \(j\) with \(0 \leq j < i\). We have \(d_j(x_{i - 1}) = 0\) by induction. Hence \[0 = d_j(x_{i - 1}) = d_j(x_i) + d_j(s_i(z_i)) = d_j(x_i) + s_{i - 1}(d_j(z_i)).\] The last equality by the relations of Remark 016C. These relations also imply that \(d_{i - 1}(d_j(x_i)) = d_j(d_i(x_i)) = 0\) because \(d_i(x_i)= 0\) by construction. Then the uniqueness in the general remark above shows the equality \(0 = x' + x'' = d_j(x_i) + s_{i - 1}(d_j(z_i))\) can only hold if both terms are zero. We conclude that \(d_j(x_i) = 0\) and by injectivity of \(s_{i - 1}\) we also conclude that \(d_j(z_i) = 0\). This proves the claim.

The claim implies we can uniquely write \[x = s_0(z_0) + s_1(z_1) + \ldots + s_n(z_n) + x_n\] with \(x_n \in N(U_{n + 1})\) and \(z_i \in \Ker(d_0) \cap \ldots \cap \Ker(d_{i - 1})\). We can reformulate this as saying that we have found a direct sum decomposition \[U_{n + 1} = N(U_{n + 1}) \oplus \bigoplus\nolimits_{i = 0}^{i = n} s_i\Big(\Ker(d_0) \cap \ldots \cap \Ker(d_{i - 1})\Big)\] with the property that \[\Ker(d_0) \cap \ldots \cap \Ker(d_j) = N(U_{n + 1}) \oplus \bigoplus\nolimits_{i = j + 1}^{i = n} s_i\Big(\Ker(d_0) \cap \ldots \cap \Ker(d_{i - 1})\Big)\] for \(j = 0, \ldots, n\). The result follows from this statement as follows. Each of the \(z_i\) in the expression for \(x\) can be written uniquely as \[z_i = s_i(z'_{i, i}) + \ldots + s_{n - 1}(z'_{i, n - 1}) + z_{i, 0}\] with \(z_{i, 0} \in N(U_n)\) and \(z'_{i, j} \in \Ker(d_0) \cap \ldots \cap \Ker(d_{j - 1})\). The first few steps in the decomposition of \(z_i\) are zero because \(z_i\) already is in the kernel of \(d_0, \ldots, d_i\). This in turn uniquely gives \[x = x_n + s_0(z_{0, 0}) + s_1(z_{1, 0}) + \ldots + s_n(z_{n, 0}) + \sum\nolimits_{0 \leq i \leq j \leq n - 1} s_i(s_j(z'_{i, j})).\] Continuing in this fashion we see that we in the end obtain a decomposition of \(x\) as a sum of terms of the form \[s_{i_1} s_{i_2} \ldots s_{i_k} (z)\] with \(0 \leq i_1 \leq i_2 \leq \ldots \leq i_k \leq n - k + 1\) and \(z \in N(U_{n + 1 - k})\). This is exactly the required decomposition, because any surjective map \([n + 1] \to [n + 1 - k]\) can be uniquely expressed in the form \[\sigma^{n - k}_{i_k} \ldots \sigma^{n - 1}_{i_2} \sigma^n_{i_1}\] with \(0 \leq i_1 \leq i_2 \leq \ldots \leq i_k \leq n - k + 1\).

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(U\) be a simplicial object in \(\mathcal{A}\). Then \(U\) has a splitting obtained by taking \(N(U_0) = U_0\) and for \(m \geq 1\) taking \[N(U_m) = \bigcap\nolimits_{i = 0}^{m - 1} \Ker(d^m_i).\] Moreover, this splitting is functorial on the category of simplicial objects of \(\mathcal{A}\).

Proof

For any object \(A\) of \(\mathcal{A}\) we obtain a simplicial abelian group \(\Mor_\mathcal{A}(A, U)\). Each of these are canonically split by Lemma 017U. Moreover, \[N(\Mor_\mathcal{A}(A, U_m)) = \bigcap\nolimits_{i = 0}^{m - 1} \Ker(d^m_i) = \Mor_\mathcal{A}(A, N(U_m)).\] Hence we see that the morphism (017Q) becomes an isomorphism after applying the functor \(\Mor_\mathcal{A}(A, -)\) for any object of \(\mathcal{A}\). Hence it is an isomorphism by the Yoneda lemma.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(f : U \to V\) be a morphism of simplicial objects of \(\mathcal{A}\). If the induced morphisms \(N(f)_i : N(U)_i \to N(V)_i\) are injective for all \(i\), then \(f_i\) is injective for all \(i\). Same holds with “injective” replaced with “surjective”, or “isomorphism”.

Proof

This is clear from Lemma 017V and the definition of a splitting.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(U\) be a simplicial object in \(\mathcal{A}\). Let \(N(U_m)\) as in Lemma 017V above. Then \(d^m_m(N(U_m)) \subset N(U_{m - 1})\).

Proof

For \(j = 0, \ldots, m - 2\) we have \(d^{m - 1}_j d^m_m = d^{m - 1}_{m - 1} d^m_j\) by the relations in Remark 016C. The result follows.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(U\) be a simplicial object of \(\mathcal{A}\). Let \(n \geq 0\) be an integer. The rule \[U'_m = \sum\nolimits_{\varphi : [m] \to [i], \ i\leq n} \Im(U(\varphi))\] defines a sub simplicial object \(U' \subset U\) with \(U'_i = U_i\) for \(i \leq n\). Moreover, \(N(U'_m) = 0\) for all \(m > n\).

Proof

Pick \(m\), \(i \leq n\) and some \(\varphi : [m] \to [i]\). The image under \(U(\psi)\) of \(\Im(U(\varphi))\) for any \(\psi : [m'] \to [m]\) is equal to the image of \(U(\varphi \circ \psi)\) and \(\varphi \circ \psi : [m'] \to [i]\). Hence \(U'\) is a simplicial object. Pick \(m > n\). We have to show \(N(U'_m) = 0\). By definition of \(N(U_m)\) and \(N(U'_m)\) we have \(N(U'_m) = U'_m \cap N(U_m)\) (intersection of subobjects). Since \(U\) is split by Lemma 017V, it suffices to show that \(U'_m\) is contained in the sum \[\sum\nolimits_{\varphi : [m] \to [m']\text{ surjective}, \ m' < m} \Im(U(\varphi)|_{N(U_{m'})}).\] By the splitting each \(U_{m'}\) is the sum of images of \(N(U_{m''})\) via \(U(\psi)\) for surjective maps \(\psi : [m'] \to [m'']\). Hence the displayed sum above is the same as \[\sum\nolimits_{\varphi : [m] \to [m']\text{ surjective}, \ m' < m} \Im(U(\varphi)).\] Clearly \(U'_m\) is contained in this by the simple fact that any \(\varphi : [m] \to [i]\), \(i \leq n\) occurring in the definition of \(U'_m\) may be factored as \([m] \to [m'] \to [i]\) with \([m] \to [m']\) surjective and \(m' < m\) as in the last displayed sum above.

Coskeleton functors

Let \(\mathcal{C}\) be a category. The coskeleton functor (if it exists) is a functor \[\text{cosk}_n : \text{Simp}_n(\mathcal{C}) \longrightarrow \text{Simp}(\mathcal{C})\] which is right adjoint to the skeleton functor. In a formula [0181]\[\begin{equation} \Mor_{\text{Simp}(\mathcal{C})}(U, \text{cosk}_n V) = \Mor_{\text{Simp}_n(\mathcal{C})}(\text{sk}_n U, V) \end{equation}\] Given a \(n\)-truncated simplicial object \(V\) we say that \(\text{cosk}_nV\) exists if there exists a \(\text{cosk}_nV \in \Ob(\text{Simp}(\mathcal{C}))\) and a morphism \(\text{sk}_n \text{cosk}_n V \to V\) such that the displayed formula holds, in other words if the functor \(U \mapsto \Mor_{\text{Simp}_n(\mathcal{C})}(\text{sk}_n U, V)\) is representable. If it exists it is unique up to unique isomorphism by the Yoneda lemma. See Categories, Section 001L.

Example

Suppose the category \(\mathcal{C}\) has finite nonempty self products. A \(0\)-truncated simplicial object of \(\mathcal{C}\) is the same as an object \(X\) of \(\mathcal{C}\). In this case we claim that \(\text{cosk}_0(X)\) is the simplicial object \(U\) with \(U_n = X^{n + 1}\) the \((n + 1)\)-fold self product of \(X\), and structure of simplicial object as in Example 016E. Namely, a morphism \(V \to U\) where \(V\) is a simplicial object is given by morphisms \(V_n \to X^{n + 1}\), such that all the diagrams \[\xymatrix{ V_n \ar[r] \ar[d]_{V([0] \to [n], 0 \mapsto i)} & X^{n + 1} \ar[d]^{\text{pr}_i} \\ V_0 \ar[r] & X }\] commute. Clearly this means that the map determines and is determined by a unique morphism \(V_0 \to X\). This proves that formula (0181) holds.

Recall the category \(\Delta/[n]\), see Example 0178. We let \((\Delta/[n])_{\leq m}\) denote the full subcategory of \(\Delta/[n]\) consisting of objects \([k] \to [n]\) of \(\Delta/[n]\) with \(k \leq m\). In other words we have the following commutative diagram of categories and functors \[\xymatrix{ (\Delta/[n])_{\leq m} \ar[r] \ar[d] & \Delta/[n] \ar[d] \\ \Delta_{\leq m} \ar[r] & \Delta }\] Given a \(m\)-truncated simplicial object \(U\) of \(\mathcal{C}\) we define a functor \[U(n) : (\Delta/[n])_{\leq m}^{opp} \longrightarrow \mathcal{C}\] by the rules \[\begin{eqnarray*} ([k] \to [n]) & \longmapsto & U_k \\ \psi : ([k'] \to [n]) \to ([k] \to [n]) & \longmapsto & U(\psi) : U_k \to U_{k'} \end{eqnarray*}\] For a given morphism \(\varphi : [n] \to [n']\) of \(\Delta\) we have an associated functor \[\overline{\varphi} : (\Delta/[n])_{\leq m} \longrightarrow (\Delta/[n'])_{\leq m}\] which maps \(\alpha : [k] \to [n]\) to \(\varphi \circ \alpha : [k] \to [n']\). The composition \(U(n') \circ \overline{\varphi}\) is equal to the functor \(U(n)\).

Lemma

If the category \(\mathcal{C}\) has finite limits, then \(\text{cosk}_m\) functors exist for all \(m\). Moreover, for any \(m\)-truncated simplicial object \(U\) the simplicial object \(\text{cosk}_mU\) is described by the formula \[(\text{cosk}_mU)_n = \lim_{(\Delta/[n])_{\leq m}^{opp}} U(n)\] and for \(\varphi : [n] \to [n']\) the map \(\text{cosk}_mU(\varphi)\) comes from the identification \(U(n') \circ \overline{\varphi} = U(n)\) above via Categories, Lemma 002L.

Proof

During the proof of this lemma we denote \(\text{cosk}_mU\) the simplicial object with \((\text{cosk}_mU)_n\) equal to \(\lim_{(\Delta/[n])_{\leq m}^{opp}} U(n)\). We will conclude at the end of the proof that it does satisfy the required mapping property.

Suppose that \(V\) is a simplicial object. A morphism \(\gamma : V \to \text{cosk}_mU\) is given by a sequence of morphisms \(\gamma_n : V_n \to (\text{cosk}_mU)_n\). By definition of a limit, this is given by a collection of morphisms \(\gamma(\alpha) : V_n \to U_k\) where \(\alpha\) ranges over all \(\alpha : [k] \to [n]\) with \(k \leq m\). These morphisms then also satisfy the rules that \[\xymatrix{ V_n \ar[r]_{\gamma(\alpha)} & U_k \\ V_{n'} \ar[r]^{\gamma(\alpha')} \ar[u]^{V(\varphi)} & U_{k'} \ar[u]_{U(\psi)} }\] are commutative, given any \(0 \leq k, k' \leq m\), \(0 \leq n, n'\) and any \(\psi : [k] \to [k']\), \(\varphi : [n] \to [n']\), \(\alpha : [k] \to [n]\) and \(\alpha' : [k'] \to [n']\) in \(\Delta\) such that \(\varphi \circ \alpha = \alpha' \circ \psi\). Taking \(n = k = k'\), \(\varphi = \alpha'\), and \(\alpha = \psi = \text{id}_{[k]}\) we deduce that \(\gamma(\alpha') = \gamma(\text{id}_{[k]}) \circ V(\alpha')\). In other words, the morphisms \(\gamma(\text{id}_{[k]})\), \(k \leq m\) determine the morphism \(\gamma\). And it is easy to see that these morphisms form a morphism \(\text{sk}_m V \to U\).

Conversely, given a morphism \(\gamma : \text{sk}_m V \to U\), we obtain a family of morphisms \(\gamma(\alpha)\) where \(\alpha\) ranges over all \(\alpha : [k] \to [n]\) with \(k \leq m\) by setting \(\gamma(\alpha) = \gamma(\text{id}_{[k]}) \circ V(\alpha)\). These morphisms satisfy all the displayed commutativity restraints pictured above, and hence give rise to a morphism \(V \to \text{cosk}_m U\).

Lemma

Let \(\mathcal{C}\) be a category. Let \(U\) be an \(m\)-truncated simplicial object of \(\mathcal{C}\). For \(n \leq m\) the limit \(\lim_{(\Delta/[n])_{\leq m}^{opp}} U(n)\) exists and is canonically isomorphic to \(U_n\).

Proof

This is true because the category \((\Delta/[n])_{\leq m}\) has a final object in this case, namely the identity map \([n] \to [n]\).

Lemma

Let \(\mathcal{C}\) be a category with finite limits. Let \(U\) be an \(n\)-truncated simplicial object of \(\mathcal{C}\). The morphism \(\text{sk}_n \text{cosk}_n U \to U\) is an isomorphism.

Proof

Combine Lemmas 0183 and 0184.

Let us describe a particular instance of the coskeleton functor in more detail. By abuse of notation we will denote \(\text{sk}_n\) also the restriction functor \(\text{Simp}_{n'}(\mathcal{C}) \to \text{Simp}_n(\mathcal{C})\) for any \(n' \geq n\). We are going to describe a right adjoint of the functor \(\text{sk}_n : \text{Simp}_{n + 1}(\mathcal{C}) \to \text{Simp}_n(\mathcal{C})\). For \(n \geq 1\), \(0 \leq i < j \leq n + 1\) define \(\delta^{n + 1}_{i, j} : [n - 1] \to [n + 1]\) to be the increasing map omitting \(i\) and \(j\). Note that \(\delta^{n + 1}_{i, j} = \delta^{n + 1}_j \circ \delta^n_i = \delta^{n + 1}_i \circ \delta^n_{j - 1}\), see Lemma 0167. This motivates the following lemma.

Lemma

Let \(n\) be an integer \(\geq 1\). Let \(U\) be a \(n\)-truncated simplicial object of \(\mathcal{C}\). Consider the contravariant functor from \(\mathcal{C}\) to \(\textit{Sets}\) which associates to an object \(T\) the set \[\{ (f_0, \ldots, f_{n + 1}) \in \Mor_\mathcal{C}(T, U_n) \mid d^n_{j - 1} \circ f_i = d^n_i \circ f_j \ \forall\ 0\leq i < j\leq n + 1\}\] If this functor is representable by some object \(U_{n + 1}\) of \(\mathcal{C}\), then \[U_{n + 1} = \lim_{(\Delta/[n + 1])_{\leq n}^{opp}} U(n)\]

Proof

The limit, if it exists, represents the functor that associates to an object \(T\) the set \[\{ (f_\alpha)_{\alpha : [k] \to [n + 1], k \leq n} \mid f_{\alpha \circ \psi} = U(\psi) \circ f_\alpha\ \forall \ \psi : [k'] \to [k], \alpha : [k] \to [n + 1] \}.\] In fact we will show this functor is isomorphic to the one displayed in the lemma. The map in one direction is given by the rule \[(f_\alpha)_{\alpha} \longmapsto (f_{\delta^{n + 1}_0}, \ldots, f_{\delta^{n + 1}_{n + 1}}).\] This satisfies the conditions of the lemma because \[d^n_{j - 1} \circ f_{\delta^{n + 1}_i} = f_{\delta^{n + 1}_i \circ \delta^n_{j - 1}} = f_{\delta^{n + 1}_j \circ \delta^n_i} = d^n_i \circ f_{\delta^{n + 1}_j}\] by the relations we recalled above the lemma. To construct a map in the other direction we have to associate to a system \((f_0, \ldots, f_{n + 1})\) as in the displayed formula of the lemma a system of maps \(f_\alpha\). Let \(\alpha : [k] \to [n + 1]\) be given. Since \(k \leq n\) the map \(\alpha\) is not surjective. Hence we can write \(\alpha = \delta^{n + 1}_i \circ \psi\) for some \(0 \leq i \leq n + 1\) and some \(\psi : [k] \to [n]\). We have no choice but to define \[f_\alpha = U(\psi) \circ f_i.\] Of course we have to check that this is independent of the choice of the pair \((i, \psi)\). First, observe that given \(i\) there is a unique \(\psi\) which works. Second, suppose that \((j, \phi)\) is another pair. Then \(i \not = j\) and we may assume \(i < j\). Since both \(i, j\) are not in the image of \(\alpha\) we may actually write \(\alpha = \delta^{n + 1}_{i, j} \circ \xi\) and then we see that \(\psi = \delta^n_{j - 1} \circ \xi\) and \(\phi = \delta^n_i \circ \xi\). Thus \[\begin{eqnarray*} U(\psi) \circ f_i & = & U(\delta^n_{j - 1} \circ \xi) \circ f_i \\ & = & U(\xi) \circ d^n_{j - 1} \circ f_i \\ & = & U(\xi) \circ d^n_i \circ f_j \\ & = & U(\delta^n_i \circ \xi) \circ f_j \\ & = & U(\phi) \circ f_j \end{eqnarray*}\] as desired. We still have to verify that the maps \(f_\alpha\) so defined satisfy the rules of a system of maps \((f_\alpha)_\alpha\). To see this suppose that \(\psi : [k'] \to [k]\), \(\alpha : [k] \to [n + 1]\) with \(k, k' \leq n\). Set \(\alpha' = \alpha \circ \psi\). Choose \(i\) not in the image of \(\alpha\). Then clearly \(i\) is not in the image of \(\alpha'\) also. Write \(\alpha = \delta^{n + 1}_i \circ \phi\) (we cannot use the letter \(\psi\) here because we’ve already used it). Then obviously \(\alpha' = \delta^{n + 1}_i \circ \phi \circ \psi\). By construction above we then have \[U(\psi) \circ f_\alpha = U(\psi) \circ U(\phi) \circ f_i = U(\phi \circ \psi) \circ f_i = f_{\alpha \circ \psi} = f_{\alpha'}\] as desired. We leave to the reader the pleasant task of verifying that our constructions are mutually inverse bijections, and are functorial in \(T\).

Lemma

Let \(n\) be an integer \(\geq 1\). Let \(U\) be a \(n\)-truncated simplicial object of \(\mathcal{C}\). Consider the contravariant functor from \(\mathcal{C}\) to \(\textit{Sets}\) which associates to an object \(T\) the set \[\{ (f_0, \ldots, f_{n + 1}) \in \Mor_\mathcal{C}(T, U_n) \mid d^n_{j - 1} \circ f_i = d^n_i \circ f_j \ \forall\ 0\leq i < j\leq n + 1\}\] If this functor is representable by some object \(U_{n + 1}\) of \(\mathcal{C}\), then there exists an \((n + 1)\)-truncated simplicial object \(\tilde U\), with \(\text{sk}_n \tilde U = U\) and \(\tilde U_{n + 1} = U_{n + 1}\) such that the following adjointness holds \[\Mor_{\text{Simp}_{n + 1}(\mathcal{C})}(V, \tilde U) = \Mor_{\text{Simp}_n(\mathcal{C})}(\text{sk}_nV, U)\]

Proof

By Lemma 0184 there are identifications \[U_i = \lim_{(\Delta/[i])_{\leq n}^{opp}} U(i)\] for \(0 \leq i \leq n\). By Lemma 0186 we have \[U_{n + 1} = \lim_{(\Delta/[n + 1])_{\leq n}^{opp}} U(n).\] Thus we may define for any \(\varphi : [i] \to [j]\) with \(i, j \leq n + 1\) the corresponding map \(\tilde U(\varphi) : \tilde U_j \to \tilde U_i\) exactly as in Lemma 0183. This defines an \((n + 1)\)-truncated simplicial object \(\tilde U\) with \(\text{sk}_n \tilde U = U\).

To see the adjointness we argue as follows. Given any element \(\gamma : \text{sk}_n V \to U\) of the right hand side of the formula consider the morphisms \(f_i = \gamma_n \circ d^{n + 1}_i : V_{n + 1} \to V_n \to U_n\). These clearly satisfy the relations \(d^n_{j - 1} \circ f_i = d^n_i \circ f_j\) and hence define a unique morphism \(V_{n + 1} \to U_{n + 1}\) by our choice of \(U_{n + 1}\). Conversely, given a morphism \(\gamma' : V \to \tilde U\) of the left hand side we can simply restrict to \(\Delta_{\leq n}\) to get an element of the right hand side. We leave it to the reader to show these are mutually inverse constructions.

Remark

Let \(U\), and \(U_{n + 1}\) be as in Lemma 0187. On \(T\)-valued points we can easily describe the face and degeneracy maps of \(\tilde U\). Explicitly, the maps \(d^{n + 1}_i : U_{n + 1} \to U_n\) are given by \[(f_0, \ldots, f_{n + 1}) \longmapsto f_i.\] And the maps \(s^n_j : U_n \to U_{n + 1}\) are given by \[\begin{eqnarray*} f & \longmapsto & ( s^{n - 1}_{j - 1} \circ d^{n - 1}_0 \circ f, \\ & & s^{n - 1}_{j - 1} \circ d^{n - 1}_1 \circ f, \\ & & \ldots\\ & & s^{n - 1}_{j - 1} \circ d^{n - 1}_{j - 1} \circ f, \\ & & f, \\ & & f, \\ & & s^{n - 1}_j \circ d^{n - 1}_{j + 1} \circ f, \\ & & s^{n - 1}_j \circ d^{n - 1}_{j + 2} \circ f, \\ & & \ldots\\ & & s^{n - 1}_j \circ d^{n - 1}_n \circ f ) \end{eqnarray*}\] where we leave it to the reader to verify that the RHS is an element of the displayed set of Lemma 0187. For \(n = 0\) there is one map, namely \(f \mapsto (f, f)\). For \(n = 1\) there are two maps, namely \(f \mapsto (f, f, s_0d_1f)\) and \(f \mapsto (s_0d_0f, f, f)\). For \(n = 2\) there are three maps, namely \(f \mapsto (f, f, s_0d_1f, s_0d_2f)\), \(f \mapsto (s_0d_0f, f, f, s_1d_2f)\), and \(f \mapsto (s_1d_0f, s_1d_1f, f, f)\). And so on and so forth.

Remark

The construction of Lemma 0187 above in the case of simplicial sets is the following. Given an \(n\)-truncated simplicial set \(U\), we make a canonical \((n + 1)\)-truncated simplicial set \(\tilde U\) as follows. We add a set of \((n + 1)\)-simplices \(U_{n + 1}\) by the formula of the lemma. Namely, an element of \(U_{n + 1}\) is a numbered collection of \((f_0, \ldots, f_{n + 1})\) of \(n\)-simplices, with the property that they glue as they would in a \((n + 1)\)-simplex. In other words, the \(i\)th face of \(f_j\) is the \((j-1)\)st face of \(f_i\) for \(i < j\). Geometrically it is obvious how to define the face and degeneracy maps for \(\tilde U\). If \(V\) is an \((n + 1)\)-truncated simplicial set, then its \((n + 1)\)-simplices give rise to compatible collections of \(n\)-simplices \((f_0, \ldots, f_{n + 1})\) with \(f_i \in V_n\). Hence there is a natural map \(\Mor(\text{sk}_nV, U) \to \Mor(V, \tilde U)\) which is inverse to the canonical restriction mapping the other way.

Also, it is enough to do the combinatorics of the construction in the case of truncated simplicial sets. Namely, for any object \(T\) of the category \(\mathcal{C}\), and any \(n\)-truncated simplicial object \(U\) of \(\mathcal{C}\) we can consider the \(n\)-truncated simplicial set \(\Mor(T, U)\). We may apply the construction to this, and take its set of \((n + 1)\)-simplices, and require this to be representable. This is a good way to think about the result of Lemma 0187.

Remark

Inductive construction of coskeleta. Suppose that \(\mathcal{C}\) is a category with finite limits. Suppose that \(U\) is an \(m\)-truncated simplicial object in \(\mathcal{C}\). Then we can inductively construct \(n\)-truncated objects \(U^n\) as follows:

1. To start, set \(U^m = U\).

2. Given \(U^n\) for \(n \geq m\) set \(U^{n + 1} = \tilde U^n\), where \(\tilde U^n\) is constructed from \(U^n\) as in Lemma 0187.

Since the construction of Lemma 0187 has the property that it leaves the \(n\)-skeleton of \(U^n\) unchanged, we can then define \(\text{cosk}_m U\) to be the simplicial object with \((\text{cosk}_m U)_n = U^n_n = U^{n + 1}_n = \ldots\). And it follows formally from Lemma 0187 that \(U^n\) satisfies the formula \[\Mor_{\text{Simp}_n(\mathcal{C})}(V, U^n) = \Mor_{\text{Simp}_m(\mathcal{C})}(\text{sk}_mV, U)\] for all \(n \geq m\). It also then follows formally from this that \[\Mor_{\text{Simp}(\mathcal{C})}(V, \text{cosk}_mU) = \Mor_{\text{Simp}_m(\mathcal{C})}(\text{sk}_mV, U)\] with \(\text{cosk}_mU\) chosen as above.

Lemma

Let \(\mathcal{C}\) be a category which has finite limits.

1. For every \(n\) the functor \(\text{sk}_n : \text{Simp}(\mathcal{C}) \to \text{Simp}_n(\mathcal{C})\) has a right adjoint \(\text{cosk}_n\).

2. For every \(n' \geq n\) the functor \(\text{sk}_n : \text{Simp}_{n'}(\mathcal{C}) \to \text{Simp}_n(\mathcal{C})\) has a right adjoint, namely \(\text{sk}_{n'}\text{cosk}_n\).

3. For every \(m \geq n \geq 0\) and every \(n\)-truncated simplicial object \(U\) of \(\mathcal{C}\) we have \(\text{cosk}_m \text{sk}_m \text{cosk}_n U = \text{cosk}_n U\).

4. If \(U\) is a simplicial object of \(\mathcal{C}\) such that the canonical map \(U \to \text{cosk}_n \text{sk}_nU\) is an isomorphism for some \(n \geq 0\), then the canonical map \(U \to \text{cosk}_m \text{sk}_mU\) is an isomorphism for all \(m \geq n\).

Proof

The existence in (1) follows from Lemma 0183 above. Parts (2) and (3) follow from the discussion in Remark 018A. After this (4) is obvious.

Remark

We do not need all finite limits in order to be able to define the coskeleton functors. Here are some remarks

1. We have seen in Example 0182 that if \(\mathcal{C}\) has products of pairs of objects then \(\text{cosk}_0\) exists.

2. For \(k > 0\) the functor \(\text{cosk}_k\) exists if \(\mathcal{C}\) has finite connected limits.

This is clear from the inductive procedure of constructing coskeleta (Remarks 0189 and 018A) but it also follows from the fact that the categories \((\Delta/[n])_{\leq k}\) for \(k \geq 1\) and \(n \geq k + 1\) used in Lemma 0183 are connected. Observe that we do not need the categories for \(n \leq k\) by Lemma 0184 or Lemma 0185. (As \(k\) gets higher the categories \((\Delta/[n])_{\leq k}\) for \(k \geq 1\) and \(n \geq k + 1\) are more and more connected in a topological sense.)

Lemma

Let \(U\), \(V\) be \(n\)-truncated simplicial objects of a category \(\mathcal{C}\). Then \[\text{cosk}_n (U \times V) = \text{cosk}_nU \times \text{cosk}_nV\] whenever the left and right hand sides exist.

Proof

Let \(W\) be a simplicial object. We have \[\begin{eqnarray*} \Mor(W, \text{cosk}_n (U \times V)) & = & \Mor(\text{sk}_n W, U \times V) \\ & = & \Mor(\text{sk}_n W, U) \times \Mor(\text{sk}_nW, V) \\ & = & \Mor(W, \text{cosk}_n U) \times \Mor(W, \text{cosk}_n V) \\ & = & \Mor(W, \text{cosk}_n U \times \text{cosk}_n V) \end{eqnarray*}\] The lemma follows.

Lemma

Assume \(\mathcal{C}\) has fibre products. Let \(U \to V\) and \(W \to V\) be morphisms of \(n\)-truncated simplicial objects of the category \(\mathcal{C}\). Then \[\text{cosk}_n (U \times_V W) = \text{cosk}_nU \times_{\text{cosk}_n V} \text{cosk}_nW\] whenever the left and right hand side exist.

Proof

Omitted, but very similar to the proof of Lemma 018C above.

Lemma

Let \(\mathcal{C}\) be a category with finite limits. Let \(X \in \Ob(\mathcal{C})\). The functor \(\mathcal{C}/X \to \mathcal{C}\) commutes with the coskeleton functors \(\text{cosk}_k\) for \(k \geq 1\).

Proof

The statement means that if \(U\) is a simplicial object of \(\mathcal{C}/X\) which we can think of as a simplicial object of \(\mathcal{C}\) with a morphism towards the constant simplicial object \(X\), then \(\text{cosk}_k U\) computed in \(\mathcal{C}/X\) is the same as computed in \(\mathcal{C}\). This follows for example from Categories, Lemma 002T because the categories \((\Delta/[n])_{\leq k}\) for \(k \geq 1\) and \(n \geq k + 1\) used in Lemma 0183 are connected. Observe that we do not need the categories for \(n \leq k\) by Lemma 0184 or Lemma 0185.

Lemma

The canonical map \(\Delta[n] \to \text{cosk}_1 \text{sk}_1 \Delta[n]\) is an isomorphism.

Proof

Consider a simplicial set \(U\) and a morphism \(f : U \to \Delta[n]\). This is a rule that associates to each \(u \in U_i\) a map \(f_u : [i] \to [n]\) in \(\Delta\). Furthermore, these maps should have the property that \(f_u \circ \varphi = f_{U(\varphi)(u)}\) for any \(\varphi : [j] \to [i]\). Denote \(\epsilon^i_j : [0] \to [i]\) the map which maps \(0\) to \(j\). Denote \(F : U_0 \to [n]\) the map \(u \mapsto f_u(0)\). Then we see that \[f_u(j) = F(\epsilon^i_j(u))\] for all \(0 \leq j \leq i\) and \(u \in U_i\). In particular, if we know the function \(F\) then we know the maps \(f_u\) for all \(u\in U_i\) all \(i\). Conversely, given a map \(F : U_0 \to [n]\), we can set for any \(i\), and any \(u \in U_i\) and any \(0 \leq j \leq i\) \[f_u(j) = F(\epsilon^i_j(u))\] This does not in general define a morphism \(f\) of simplicial sets as above. Namely, the condition is that all the maps \(f_u\) are nondecreasing. This clearly is equivalent to the condition that \(F(\epsilon^i_j(u)) \leq F(\epsilon^i_{j'}(u))\) whenever \(0 \leq j \leq j' \leq i\) and \(u \in U_i\). But in this case the morphisms \[\epsilon^i_j, \epsilon^i_{j'} : [0] \to [i]\] both factor through the map \(\epsilon^i_{j, j'} : [1] \to [i]\) defined by the rules \(0 \mapsto j\), \(1 \mapsto j'\). In other words, it is enough to check the inequalities for \(i = 1\) and \(u \in U_1\). In other words, we have \[\Mor(U, \Delta[n]) = \Mor(\text{sk}_1 U, \text{sk}_1 \Delta[n])\] as desired.

Augmentations

Definition

Let \(\mathcal{C}\) be a category. Let \(U\) be a simplicial object of \(\mathcal{C}\). An augmentation \(\epsilon : U \to X\) of \(U\) towards an object \(X\) of \(\mathcal{C}\) is a morphism from \(U\) into the constant simplicial object \(X\).

Lemma

Let \(\mathcal{C}\) be a category. Let \(X \in \Ob(\mathcal{C})\). Let \(U\) be a simplicial object of \(\mathcal{C}\). To give an augmentation of \(U\) towards \(X\) is the same as giving a morphism \(\epsilon_0 : U_0 \to X\) such that \(\epsilon_0 \circ d^1_0 = \epsilon_0 \circ d^1_1\).

Proof

Given a morphism \(\epsilon : U \to X\) we certainly obtain an \(\epsilon_0\) as in the lemma. Conversely, given \(\epsilon_0\) as in the lemma, define \(\epsilon_n : U_n \to X\) by choosing any morphism \(\alpha : [0] \to [n]\) and taking \(\epsilon_n = \epsilon_0 \circ U(\alpha)\). Namely, if \(\beta : [0] \to [n]\) is another choice, then there exists a morphism \(\gamma : [1] \to [n]\) such that \(\alpha\) and \(\beta\) both factor as \([0] \to [1] \to [n]\). Hence the condition on \(\epsilon_0\) shows that \(\epsilon_n\) is well defined. Then it is easy to show that \((\epsilon_n) : U \to X\) is a morphism of simplicial objects.

Lemma

Let \(\mathcal{C}\) be a category with fibred products. Let \(f : Y\to X\) be a morphism of \(\mathcal{C}\). Let \(U\) be the simplicial object of \(\mathcal{C}\) whose \(n\)th term is the \((n + 1)\)fold fibred product \(Y \times_X Y \times_X \ldots \times_X Y\). See Example 016E. For any simplicial object \(V\) of \(\mathcal{C}\) we have \[\begin{align*} \Mor_{\text{Simp}(\mathcal{C})}(V, U) & = \Mor_{\text{Simp}_1(\mathcal{C})}(\text{sk}_1 V, \text{sk}_1 U) \\ & = \{g_0 : V_0 \to Y \mid f \circ g_0 \circ d^1_0 = f \circ g_0 \circ d^1_1\} \end{align*}\] In particular we have \(U = \text{cosk}_1 \text{sk}_1 U\).

Proof

Suppose that \(g : \text{sk}_1V \to \text{sk}_1U\) is a morphism of \(1\)-truncated simplicial objects. Then the diagram \[\xymatrix{ V_1 \ar@<1ex>[r]^{d^1_0} \ar@<-1ex>[r]_{d^1_1} \ar[d]_{g_1} & V_0 \ar[d]^{g_0} \\ Y \times_X Y \ar@<1ex>[r]^{pr_1} \ar@<-1ex>[r]_{pr_0} & Y \ar[r] & X }\] is commutative, which proves that the relation shown in the lemma holds. We have to show that, conversely, given a morphism \(g_0\) satisfying the relation \(f \circ g_0 \circ d^1_0 = f \circ g_0 \circ d^1_1\) we get a unique morphism of simplicial objects \(g : V \to U\). This is done as follows. For any \(n \geq 1\) let \(g_{n, i} = g_0 \circ V([0] \to [n], 0 \mapsto i) : V_n \to Y\). The equality above implies that \(f \circ g_{n, i} = f \circ g_{n, i + 1}\) because of the commutative diagram \[\xymatrix{ [0] \ar[rd]_{0 \mapsto 0} \ar[rrrrrd]^{0 \mapsto i} \\ & [1] \ar[rrrr]^{0 \mapsto i, 1\mapsto i + 1} & & & & [n] \\ [0] \ar[ru]^{0 \mapsto 1} \ar[rrrrru]_{0 \mapsto i + 1} }\] Hence we get \((g_{n, 0}, \ldots, g_{n, n}) : V_n \to Y \times_X\ldots \times_X Y = U_n\). We leave it to the reader to see that this is a morphism of simplicial objects. The last assertion of the lemma is equivalent to the first equality in the displayed formula of the lemma.

Remark

Let \(\mathcal{C}\) be a category with fibre products. Let \(V\) be a simplicial object. Let \(\epsilon : V \to X\) be an augmentation. Let \(U\) be the simplicial object whose \(n\)th term is the \((n + 1)\)st fibred product of \(V_0\) over \(X\). By a simple combination of Lemmas 018H and 018I we obtain a canonical morphism \(V \to U\).

Left adjoints to the skeleton functors

In this section we construct a left adjoint \(i_{m!}\) of the skeleton functor \(\text{sk}_m\) in certain cases. The adjointness formula is \[\Mor_{\text{Simp}_m(\mathcal{C})}(U, \text{sk}_mV) = \Mor_{\text{Simp}(\mathcal{C})}(i_{m!}U, V).\] It turns out that this left adjoint exists when the category \(\mathcal{C}\) has finite colimits.

We use a similar construction as in Section 017Z. Recall the category \([n]/\Delta\) of objects under \([n]\), see Categories, Example 001H. Its objects are morphisms \(\alpha : [n] \to [k]\) and its morphisms are commutative triangles. We let \(([n]/\Delta)_{\leq m}\) denote the full subcategory of \([n]/\Delta\) consisting of objects \([n] \to [k]\) with \(k \leq m\). Given a \(m\)-truncated simplicial object \(U\) of \(\mathcal{C}\) we define a functor \[U(n) : ([n]/\Delta)_{\leq m}^{opp} \longrightarrow \mathcal{C}\] by the rules \[\begin{eqnarray*} ([n] \to [k]) & \longmapsto & U_k \\ \psi : ([n] \to [k']) \to ([n] \to [k]) & \longmapsto & U(\psi) : U_k \to U_{k'} \end{eqnarray*}\] For a given morphism \(\varphi : [n] \to [n']\) of \(\Delta\) we have an associated functor \[\underline{\varphi} : ([n']/\Delta)_{\leq m} \longrightarrow ([n]/\Delta)_{\leq m}\] which maps \(\alpha : [n'] \to [k]\) to \(\varphi \circ \alpha : [n] \to [k]\). The composition \(U(n) \circ \underline{\varphi}\) is equal to the functor \(U(n')\).

Lemma

Let \(\mathcal{C}\) be a category which has finite colimits. The functors \(i_{m!}\) exist for all \(m\). Let \(U\) be an \(m\)-truncated simplicial object of \(\mathcal{C}\). The simplicial object \(i_{m!}U\) is described by the formula \[(i_{m!}U)_n = \colim_{([n]/\Delta)_{\leq m}^{opp}} U(n)\] and for \(\varphi : [n] \to [n']\) the map \(i_{m!}U(\varphi)\) comes from the identification \(U(n) \circ \underline{\varphi} = U(n')\) above via Categories, Lemma 002K.

Proof

In this proof we denote \(i_{m!}U\) the simplicial object whose \(n\)th term is given by the displayed formula of the lemma. We will show it satisfies the adjointness property.

Let \(V\) be a simplicial object of \(\mathcal{C}\). Let \(\gamma : U \to \text{sk}_mV\) be given. A morphism \[\colim_{([n]/\Delta)_{\leq m}^{opp}} U(n) \to T\] is given by a compatible system of morphisms \(f_\alpha : U_k \to T\) where \(\alpha : [n] \to [k]\) with \(k \leq m\). Certainly, we have such a system of morphisms by taking the compositions \[U_k \xrightarrow{\gamma_k} V_k \xrightarrow{V(\alpha)} V_n.\] Hence we get an induced morphism \((i_{m!}U)_n \to V_n\). We leave it to the reader to see that these form a morphism of simplicial objects \(\gamma' : i_{m!}U \to V\).

Conversely, given a morphism \(\gamma' : i_{m!}U \to V\) we obtain a morphism \(\gamma : U \to \text{sk}_m V\) by setting \(\gamma_i : U_i \to V_i\) equal to the composition \[U_i \xrightarrow{\text{id}_{[i]}} \colim_{([i]/\Delta)_{\leq m}^{opp}} U(i) \xrightarrow{\gamma'_i} V_i\] for \(0 \leq i \leq n\). We leave it to the reader to see that this is the inverse of the construction above.

Lemma

Let \(\mathcal{C}\) be a category. Let \(U\) be an \(m\)-truncated simplicial object of \(\mathcal{C}\). For any \(n \leq m\) the colimit \[\colim_{([n]/\Delta)_{\leq m}^{opp}} U(n)\] exists and is equal to \(U_n\).

Proof

This is so because the category \(([n]/\Delta)_{\leq m}\) has an initial object, namely \(\text{id} : [n] \to [n]\).

Lemma

Let \(\mathcal{C}\) be a category which has finite colimits. Let \(U\) be an \(m\)-truncated simplicial object of \(\mathcal{C}\). The map \(U \to \text{sk}_m i_{m!}U\) is an isomorphism.

Proof

Combine Lemmas 018L and 018M.

Lemma

If \(U\) is an \(m\)-truncated simplicial set and \(n > m\) then all \(n\)-simplices of \(i_{m!}U\) are degenerate.

Proof

This can be seen from the construction of \(i_{m!}U\) in Lemma 018L, but we can also argue directly as follows. Write \(V = i_{m!}U\). Let \(V' \subset V\) be the simplicial subset with \(V'_i = V_i\) for \(i \leq m\) and all \(i\) simplices degenerate for \(i > m\), see Lemma 017T. By the adjunction formula, since \(\text{sk}_m V' = U\), there is an inverse to the injection \(V' \to V\). Hence \(V' = V\).

Lemma

Let \(U\) be a simplicial set. Let \(n \geq 0\) be an integer. The morphism \(i_{n!} \text{sk}_n U \to U\) identifies \(i_{n!} \text{sk}_n U\) with the simplicial set \(U' \subset U\) defined in Lemma 017T.

Proof

By Lemma 018O the only nondegenerate simplices of \(i_{n!} \text{sk}_n U\) are in degrees \(\leq n\). The map \(i_{n!} \text{sk}_n U \to U\) is an isomorphism in degrees \(\leq n\). Combined we conclude that the map \(i_{n!} \text{sk}_n U \to U\) maps nondegenerate simplices to nondegenerate simplices and no two nondegenerate simplices have the same image. Hence Lemma 017S applies. Thus \(i_{n!} \text{sk}_n U \to U\) is injective. The result follows easily from this.

Remark

In some texts the composite functor \[\text{Simp}(\mathcal{C}) \xrightarrow{\text{sk}_m} \text{Simp}_m(\mathcal{C}) \xrightarrow{i_{m!}} \text{Simp}(\mathcal{C})\] is denoted \(\text{sk}_m\). This makes sense for simplicial sets, because then Lemma 018P says that \(i_{m!} \text{sk}_m V\) is just the sub simplicial set of \(V\) consisting of all \(i\)-simplices of \(V\), \(i \leq m\) and their degeneracies. In those texts it is also customary to denote the composition \[\text{Simp}(\mathcal{C}) \xrightarrow{\text{sk}_m} \text{Simp}_m(\mathcal{C}) \xrightarrow{\text{cosk}_m} \text{Simp}(\mathcal{C})\] by \(\text{cosk}_m\).

Lemma

Let \(U \subset V\) be simplicial sets. Suppose \(n \geq 0\) and \(x \in V_n\), \(x \not \in U_n\) are such that

1. \(V_i = U_i\) for \(i < n\),

2. \(V_n = U_n \cup \{x\}\),

3. any \(z \in V_j\), \(z \not \in U_j\) for \(j > n\) is degenerate.

Let \(\Delta[n] \to V\) be the unique morphism mapping the nondegenerate \(n\)-simplex of \(\Delta[n]\) to \(x\). In this case the diagram \[\xymatrix{ \Delta[n] \ar[r] & V \\ i_{(n - 1)!} \text{sk}_{n - 1} \Delta[n] \ar[r] \ar[u] & U \ar[u] }\] is a pushout diagram.

Proof

Let us denote \(\partial \Delta[n] = i_{(n - 1)!} \text{sk}_{n - 1} \Delta[n]\) for convenience. There is a natural map \(U \amalg_{\partial \Delta[n]} \Delta[n] \to V\). We have to show that it is bijective in degree \(j\) for all \(j\). This is clear for \(j \leq n\). Let \(j > n\). The third condition means that any \(z \in V_j\), \(z \not \in U_j\) is a degenerate simplex, say \(z = s^{j - 1}_i(z')\). Of course \(z' \not \in U_{j - 1}\). By induction it follows that \(z'\) is a degeneracy of \(x\). Thus we conclude that all \(j\)-simplices of \(V\) are either in \(U\) or degeneracies of \(x\). This implies that the map \(U \amalg_{\partial \Delta[n]} \Delta[n] \to V\) is surjective. Note that a nondegenerate simplex of \(U \amalg_{\partial \Delta[n]} \Delta[n]\) is either the image of a nondegenerate simplex of \(U\), or the image of the (unique) nondegenerate \(n\)-simplex of \(\Delta[n]\). Since clearly \(x\) is nondegenerate we deduce that \(U \amalg_{\partial \Delta[n]} \Delta[n] \to V\) maps nondegenerate simplices to nondegenerate simplices and is injective on nondegenerate simplices. Hence it is injective, by Lemma 017S.

Lemma

Let \(U \subset V\) be simplicial sets, with \(U_n, V_n\) finite nonempty for all \(n\). Assume that \(U\) and \(V\) have finitely many nondegenerate simplices. Then there exists a sequence of sub simplicial sets \[U = W^0 \subset W^1 \subset W^2 \subset \ldots W^r = V\] such that Lemma 018R applies to each of the inclusions \(W^i \subset W^{i + 1}\).

Proof

Let \(n\) be the smallest integer such that \(V\) has a nondegenerate simplex that does not belong to \(U\). Let \(x \in V_n\), \(x\not \in U_n\) be such a nondegenerate simplex. Let \(W \subset V\) be the set of elements which are either in \(U\), or are a (repeated) degeneracy of \(x\) (in other words, are of the form \(V(\varphi)(x)\) with \(\varphi : [m] \to [n]\) surjective). It is easy to see that \(W\) is a simplicial set. The inclusion \(U \subset W\) satisfies the conditions of Lemma 018R. Moreover the number of nondegenerate simplices of \(V\) which are not contained in \(W\) is exactly one less than the number of nondegenerate simplices of \(V\) which are not contained in \(U\). Hence we win by induction on this number.

Lemma

Let \(\mathcal{A}\) be an abelian category Let \(U\) be an \(m\)-truncated simplicial object of \(\mathcal{A}\). For \(n > m\) we have \(N(i_{m!}U)_n = 0\).

Proof

Write \(V = i_{m!}U\). Let \(V' \subset V\) be the simplicial subobject of \(V\) with \(V'_i = V_i\) for \(i \leq m\) and \(N(V'_i) = 0\) for \(i > m\), see Lemma 017Y. By the adjunction formula, since \(\text{sk}_m V' = U\), there is an inverse to the injection \(V' \to V\). Hence \(V' = V\).

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(U\) be a simplicial object of \(\mathcal{A}\). Let \(n \geq 0\) be an integer. The morphism \(i_{n!} \text{sk}_n U \to U\) identifies \(i_{n!} \text{sk}_n U\) with the simplicial subobject \(U' \subset U\) defined in Lemma 017Y.

Proof

By Lemma 018T we have \(N(i_{n!} \text{sk}_n U)_i = 0\) for \(i > n\). The map \(i_{n!} \text{sk}_n U \to U\) is an isomorphism in degrees \(\leq n\), see Lemma 018N. Combined we conclude that the map \(i_{n!} \text{sk}_n U \to U\) induces injective maps \(N(i_{n!} \text{sk}_n U)_i \to N(U)_i\) for all \(i\). Hence Lemma 017W applies. Thus \(i_{n!} \text{sk}_n U \to U\) is injective. The result follows easily from this.

Here is another way to think about the coskeleton functor using the material above.

Lemma

Let \(\mathcal{C}\) be a category with finite coproducts and finite limits. Let \(V\) be a simplicial object of \(\mathcal{C}\). In this case \[(\text{cosk}_n \text{sk}_n V)_{n + 1} = \Hom(i_{n !}\text{sk}_n \Delta[n + 1], V)_0.\]

Proof

By Lemma 017F the object on the left represents the functor which assigns to \(X\) the first set of the following equalities \[\begin{eqnarray*} \Mor(X \times \Delta[n + 1], \text{cosk}_n \text{sk}_n V) & = & \Mor(X \times \text{sk}_n \Delta[n + 1], \text{sk}_n V) \\ & = & \Mor(X \times i_{n !} \text{sk}_n \Delta[n + 1], V). \end{eqnarray*}\] The object on the right in the formula of the lemma is represented by the functor which assigns to \(X\) the last set in the sequence of equalities. This proves the result.

In the sequence of equalities we have used that \(\text{sk}_n (X \times \Delta[n + 1]) = X \times \text{sk}_n \Delta[n + 1]\) and that \(i_{n!}(X \times \text{sk}_n \Delta[n + 1]) = X \times i_{n !} \text{sk}_n \Delta[n + 1]\). The first equality is obvious. For any (possibly truncated) simplicial object \(W\) of \(\mathcal{C}\) and any object \(X\) of \(\mathcal{C}\) denote temporarily \(\Mor_\mathcal{C}(X, W)\) the (possibly truncated) simplicial set \([n] \mapsto \Mor_\mathcal{C}(X, W_n)\). From the definitions it follows that \(\Mor(U \times X, W) = \Mor(U, \Mor_\mathcal{C}(X, W))\) for any (possibly truncated) simplicial set \(U\). Hence \[\begin{eqnarray*} \Mor(X \times i_{n !} \text{sk}_n \Delta[n + 1], W) & = & \Mor(i_{n !} \text{sk}_n \Delta[n + 1], \Mor_\mathcal{C}(X, W)) \\ & = & \Mor(\text{sk}_n \Delta[n + 1], \text{sk}_n\Mor_\mathcal{C}(X, W)) \\ & = & \Mor(X \times \text{sk}_n \Delta[n + 1], \text{sk}_nW) \\ & = & \Mor(i_{n!}(X \times \text{sk}_n \Delta[n + 1]), W). \end{eqnarray*}\] This proves the second equality used, and ends the proof of the lemma.

Simplicial objects in abelian categories

Recall that an abelian category is defined in Homology, Section 00ZX.

Lemma

Let \(\mathcal{A}\) be an abelian category.

1. The categories \(\text{Simp}(\mathcal{A})\) and \(\text{CoSimp}(\mathcal{A})\) are abelian.

2. A morphism of (co)simplicial objects \(f : A \to B\) is injective if and only if each \(f_n : A_n \to B_n\) is injective.

3. A morphism of (co)simplicial objects \(f : A \to B\) is surjective if and only if each \(f_n : A_n \to B_n\) is surjective.

4. A sequence of (co)simplicial objects \[A \xrightarrow{f} B \xrightarrow{g} C\] is exact at \(B\) if and only if each sequence \[A_i \xrightarrow{f_i} B_i \xrightarrow{g_i} C_i\] is exact at \(B_i\).

Proof

Pre-additivity is easy. A final object is given by \(U_n = 0\) in all degrees. Existence of direct products we saw in Lemmas 016R and 0170. Kernels and cokernels are obtained by taking termwise kernels and cokernels.

For an object \(A\) of \(\mathcal{A}\) and an integer \(k\) consider the \(k\)-truncated simplicial object \(U\) with

1. \(U_i = 0\) for \(i < k\),

2. \(U_k = A\),

3. all morphisms \(U(\varphi)\) equal to zero, except \(U(\text{id}_{[k]}) = \text{id}_A\).

Since \(\mathcal{A}\) has both finite limits and finite colimits we see that both \(\text{cosk}_k U\) and \(i_{k!}U\) exist. We will describe both of these and the canonical map \(i_{k!}U \to \text{cosk}_kU\).

Lemma

With \(A\), \(k\) and \(U\) as above, so \(U_i = 0\), \(i < k\) and \(U_k = A\).

1. Given a \(k\)-truncated simplicial object \(V\) we have \[\Mor(U, V) = \{ f : A \to V_k \mid d^k_i \circ f = 0, \ i = 0, \ldots, k \}\] and \[\Mor(V, U) = \{ f : V_k \to A \mid f \circ s^{k - 1}_i = 0, \ i = 0, \ldots, k - 1 \}.\]

2. The object \(i_{k!} U\) has \(n\)th term equal to \(\bigoplus_\alpha A\) where \(\alpha\) runs over all surjective morphisms \(\alpha : [n] \to [k]\).

3. For any \(\varphi : [m] \to [n]\) the map \(i_{k!} U(\varphi)\) is described as the mapping \(\bigoplus_\alpha A \to \bigoplus_{\alpha'} A\) which maps to component corresponding to \(\alpha : [n] \to [k]\) to zero if \(\alpha \circ \varphi\) is not surjective and by the identity to the component corresponding to \(\alpha \circ \varphi\) if it is surjective.

4. The object \(\text{cosk}_k U\) has \(n\)th term equal to \(\bigoplus_\beta A\), where \(\beta\) runs over all injective morphisms \(\beta : [k] \to [n]\).

5. For any \(\varphi : [m] \to [n]\) the map \(\text{cosk}_k U(\varphi)\) is described as the mapping \(\bigoplus_\beta A \to \bigoplus_{\beta'} A\) which maps to component corresponding to \(\beta : [k] \to [n]\) to zero if \(\beta\) does not factor through \(\varphi\) and by the identity to each of the components corresponding to \(\beta'\) such that \(\beta = \varphi \circ \beta'\) if it does.

6. The canonical map \(c : i_{k !} U \to \text{cosk}_k U\) in degree \(n\) has \((\alpha, \beta)\) coefficient \(A \to A\) equal to zero if \(\alpha \circ \beta\) is not the identity and equal to \(\text{id}_A\) if it is.

7. The canonical map \(c : i_{k !} U \to \text{cosk}_k U\) is injective.

Proof

The proof of (1) is left to the reader.

Let us take the rules of (2) and (3) as the definition of a simplicial object, call it \(\tilde U\). We will show that it is an incarnation of \(i_{k!}U\). This will prove (2), (3) at the same time. We have to show that given a morphism \(f : U \to \text{sk}_kV\) there exists a unique morphism \(\tilde f : \tilde U \to V\) which recovers \(f\) upon taking the \(k\)-skeleton. From (1) we see that \(f\) corresponds with a morphism \(f_k : A \to V_k\) which maps into the kernel of \(d^k_i\) for all \(i\). For any surjective \(\alpha : [n] \to [k]\) we set \(\tilde f_\alpha : A \to V_n\) equal to the composition \(\tilde f_\alpha = V(\alpha) \circ f_k : A \to V_n\). We define \(\tilde f_n : \tilde U_n \to V_n\) as the sum of the \(\tilde f_\alpha\) over \(\alpha : [n] \to [k]\) surjective. Such a collection of \(\tilde f_\alpha\) defines a morphism of simplicial objects if and only if for any \(\varphi : [m] \to [n]\) the diagram \[\xymatrix{ \bigoplus_{\alpha : [n] \to [k]\text{ surjective}} A \ar[r]_-{\tilde f_n} \ar[d]_{(3)} & V_n \ar[d]^{V(\varphi)} \\ \bigoplus_{\alpha' : [m] \to [k]\text{ surjective}} A \ar[r]^-{\tilde f_m} & V_m }\] is commutative. Choosing \(\varphi = \alpha\) shows our choice of \(\tilde f_\alpha\) is uniquely determined by \(f_k\). The commutativity in general may be checked for each summand of the left upper corner separately. It is clear for the summands corresponding to \(\alpha\) where \(\alpha \circ \varphi\) is surjective, because those get mapped by \(\text{id}_A\) to the summand with \(\alpha' = \alpha \circ \varphi\), and we have \(\tilde f_{\alpha'} = V(\alpha') \circ f_k = V(\alpha \circ \varphi) \circ f_k = V(\varphi) \circ \tilde f_\alpha\). For those where \(\alpha \circ \varphi\) is not surjective, we have to show that \(V(\varphi) \circ \tilde f_\alpha = 0\). By definition this is equal to \(V(\varphi) \circ V(\alpha) \circ f_k = V(\alpha \circ \varphi) \circ f_k\). Since \(\alpha \circ \varphi\) is not surjective we can write it as \(\delta^k_i \circ \psi\), and we deduce that \(V(\varphi) \circ V(\alpha) \circ f_k = V(\psi) \circ d^k_i \circ f_k = 0\) see above.

Let us take the rules of (4) and (5) as the definition of a simplicial object, call it \(\tilde U\). We will show that it is an incarnation of \(\text{cosk}_k U\). This will prove (4), (5) at the same time. The argument is completely dual to the proof of (2), (3) above, but we give it anyway. We have to show that given a morphism \(f : \text{sk}_kV \to U\) there exists a unique morphism \(\tilde f : V \to \tilde U\) which recovers \(f\) upon taking the \(k\)-skeleton. From (1) we see that \(f\) corresponds with a morphism \(f_k : V_k \to A\) which is zero on the image of \(s^{k - 1}_i\) for all \(i\). For any injective \(\beta : [k] \to [n]\) we set \(\tilde f_\beta : V_n \to A\) equal to the composition \(\tilde f_\beta = f_k \circ V(\beta) : V_n \to A\). We define \(\tilde f_n : V_n \to \tilde U_n\) as the sum of the \(\tilde f_\beta\) over \(\beta : [k] \to [n]\) injective. Such a collection of \(\tilde f_\beta\) defines a morphism of simplicial objects if and only if for any \(\varphi : [m] \to [n]\) the diagram \[\xymatrix{ V_n \ar[d]_{V(\varphi)} \ar[r]_-{\tilde f_n} & \bigoplus_{\beta : [k] \to [n]\text{ injective}} A \ar[d]^{(5)} \\ V_m \ar[r]^-{\tilde f_m} & \bigoplus_{\beta' : [k] \to [m]\text{ injective}} A }\] is commutative. Choosing \(\varphi = \beta\) shows our choice of \(\tilde f_\beta\) is uniquely determined by \(f_k\). The commutativity in general may be checked for each summand of the right lower corner separately. It is clear for the summands corresponding to \(\beta'\) where \(\varphi \circ \beta'\) is injective, because these summands get mapped into by exactly the summand with \(\beta = \varphi \circ \beta'\) and we have in that case \(\tilde f_{\beta'} \circ V(\varphi) = f_k \circ V(\beta') \circ V(\varphi) = f_k \circ V(\beta) = \tilde f_\beta\). For those where \(\varphi \circ \beta'\) is not injective, we have to show that \(\tilde f_{\beta'} \circ V(\varphi) = 0\). By definition this is equal to \(f_k \circ V(\beta') \circ V(\varphi) = f_k \circ V(\varphi \circ \beta')\). Since \(\varphi \circ \beta'\) is not injective we can write it as \(\psi \circ \sigma^{k - 1}_i\), and we deduce that \(f_k \circ V(\beta') \circ V(\varphi) = f_k \circ s^{k - 1}_i \circ V(\psi) = 0\) see above.

The composition \(i_{k!}U \to \text{cosk}_kU\) is the unique map of simplicial objects which is the identity on \(A = U_k = (i_{k!}U)_k = (\text{cosk}_kU)_k\). Hence it suffices to check that the proposed rule defines a morphism of simplicial objects. To see this we have to show that for any \(\varphi : [m] \to [n]\) the diagram \[\xymatrix{ \bigoplus_{\alpha : [n] \to [k]\text{ surjective}} A \ar[d]_{(3)} \ar[r]_{(6)} & \bigoplus_{\beta : [k] \to [n]\text{ injective}} A \ar[d]^{(5)} \\ \bigoplus_{\alpha' : [m] \to [k]\text{ surjective}} A \ar[r]^{(6)} & \bigoplus_{\beta' : [k] \to [m]\text{ injective}} A }\] is commutative. Now we can think of this in terms of matrices filled with only \(0\)’s and \(1\)’s as follows: The matrix of (3) has a nonzero \((\alpha', \alpha)\) entry if and only if \(\alpha' = \alpha \circ \varphi\). Likewise the matrix of (5) has a nonzero \((\beta', \beta)\) entry if and only if \(\beta = \varphi \circ \beta'\). The upper matrix of (6) has a nonzero \((\alpha, \beta)\) entry if and only if \(\alpha \circ \beta = \text{id}_{[k]}\). Similarly for the lower matrix of (6). The commutativity of the diagram then comes down to computing the \((\alpha, \beta')\) entry for both compositions and seeing they are equal. This comes down to the following equality \[\# \left\{ \beta \mid \beta = \varphi \circ \beta' \text{ and } \alpha \circ \beta = \text{id}_{[k]} \right\} = \# \left\{ \alpha' \mid \alpha' = \alpha \circ \varphi \text{ and } \alpha' \circ \beta' = \text{id}_{[k]} \right\}\] whose proof may safely be left to the reader.

Finally, we prove (7). This follows directly from Lemmas 017W, 0185, 018N and 018T.

Definition

Let \(\mathcal{A}\) be an abelian category. Let \(A\) be an object of \(\mathcal{A}\) and let \(k\) be an integer \(\geq 0\). The Eilenberg-Maclane object \(K(A, k)\) is given by the object \(K(A, k) = i_{k!}U\) which is described in Lemma 0190 above.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(A\) be an object of \(\mathcal{A}\) and let \(k\) be an integer \(\geq 0\). Consider the simplicial object \(E\) defined by the following rules

1. \(E_n = \bigoplus_\alpha A\), where the sum is over \(\alpha : [n] \to [k + 1]\) whose image is either \([k]\) or \([k + 1]\).

2. Given \(\varphi : [m] \to [n]\) the map \(E_n \to E_m\) maps the summand corresponding to \(\alpha\) via \(\text{id}_A\) to the summand corresponding to \(\alpha \circ \varphi\), provided \(\Im(\alpha \circ \varphi)\) is equal to \([k]\) or \([k + 1]\).

Then there exists a short exact sequence \[0 \to K(A, k) \to E \to K(A, k + 1) \to 0\] which is term by term split exact.

Proof

The maps \(K(A, k)_n \to E_n\) resp. \(E_n \to K(A, k + 1)_n\) are given by the inclusion of direct sums, resp. projection of direct sums which is obvious from the inclusions of index sets. It is clear that these are maps of simplicial objects.

Lemma

Let \(\mathcal{A}\) be an abelian category. For any simplicial object \(V\) of \(\mathcal{A}\) we have \[V = \colim_n i_{n!}\text{sk}_n V\] where all the transition maps are injections.

Proof

This is true simply because each \(V_m\) is equal to \((i_{n!}\text{sk}_n V)_m\) as soon as \(n \geq m\). See also Lemma 018U for the transition maps.

Simplicial objects and chain complexes

Let \(\mathcal{A}\) be an abelian category. See Homology, Section 010V for conventions and notation regarding chain complexes. Let \(U\) be a simplicial object of \(\mathcal{A}\). The associated chain complex \(s(U)\) of \(U\), sometimes called the Moore complex, is the chain complex \[\ldots \to U_2 \to U_1 \to U_0 \to 0 \to 0 \to \ldots\] with boundary maps \(d_n : U_n \to U_{n - 1}\) given by the formula \[d_n = \sum\nolimits_{i = 0}^n (-1)^i d^n_i.\] This is a complex because, by the relations listed in Remark 016C, we have \[\begin{eqnarray*} d_n \circ d_{n + 1} & = & (\sum\nolimits_{i = 0}^n (-1)^i d^n_i) \circ (\sum\nolimits_{j = 0}^{n + 1} (-1)^j d^{n + 1}_j) \\ & = & \sum\nolimits_{0 \leq i < j \leq n + 1} (-1)^{i + j} d^n_{j - 1} \circ d^{n + 1}_i + \sum\nolimits_{n \geq i \geq j \geq 0} (-1)^{i + j} d^n_i \circ d^{n + 1}_j \\ & = & 0. \end{eqnarray*}\] The signs cancel! We denote the associated chain complex \(s(U)\). Clearly, the construction is functorial and hence defines a functor \[s : \text{Simp}(\mathcal{A}) \longrightarrow \text{Ch}_{\geq 0}(\mathcal{A}).\] Thus we have the confusing but correct formula \(s(U)_n = U_n\).

Lemma

The functor \(s\) is exact.

Proof

Clear from Lemma 018Z.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(A\) be an object of \(\mathcal{A}\) and let \(k\) be an integer. Let \(E\) be the object described in Lemma 0192. Then the complex \(s(E)\) is acyclic.

Proof

For a morphism \(\alpha : [n] \to [k + 1]\) we define \(\alpha' : [n + 1] \to [k + 1]\) to be the map such that \(\alpha'|_{[n]} = \alpha\) and \(\alpha'(n + 1) = k + 1\). Note that if the image of \(\alpha\) is \([k]\) or \([k + 1]\), then the image of \(\alpha'\) is \([k + 1]\). Consider the family of maps \(h_n : E_n \to E_{n + 1}\) which maps the summand corresponding to \(\alpha\) to the summand corresponding to \(\alpha'\) via the identity on \(A\). Let us compute \(d_{n + 1} \circ h_n - h_{n - 1} \circ d_n\). We will first do this in case the category \(\mathcal{A}\) is the category of abelian groups. Let us use the notation \(x_\alpha\) to indicate the element \(x \in A\) in the summand of \(E_n\) corresponding to the map \(\alpha\) occurring in the index set. Let us also adopt the convention that \(x_\alpha\) designates the zero element of \(E_n\) whenever \(\Im(\alpha)\) is not \([k]\) or \([k + 1]\). With these conventions we see that \[d_{n + 1}(h_n(x_\alpha)) = \sum\nolimits_{i = 0}^{n + 1} (-1)^i x_{\alpha' \circ \delta^{n + 1}_i}\] and \[h_{n - 1}(d_n(x_\alpha)) = \sum\nolimits_{i = 0}^n (-1)^i x_{(\alpha \circ \delta_i^n)'}\] It is easy to see that \(\alpha' \circ \delta^{n + 1}_i = (\alpha \circ \delta_i^n)'\) for \(i = 0, \ldots, n\). It is also easy to see that \(\alpha' \circ \delta^{n + 1}_{n + 1} = \alpha\). Thus we see that \[(d_{n + 1} \circ h_n - h_{n - 1} \circ d_n)(x_\alpha) = (-1)^{n + 1} x_\alpha\] These identities continue to hold if \(\mathcal{A}\) is any abelian category because they hold in the simplicial abelian group \([n] \mapsto \Hom(A, E_n)\); details left to the reader. We conclude that the identity map on \(E\) is homotopic to zero, with homotopy given by the system of maps \(h'_n = (-1)^{n + 1}h_n : E_n \to E_{n + 1}\). Hence we see that \(E\) is acyclic, for example by Homology, Lemma 0110.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(A\) be an object of \(\mathcal{A}\) and let \(k\) be an integer. We have \(H_i(s(K(A, k))) = A\) if \(i = k\) and \(0\) else.

Proof

First, let us prove this if \(k = 0\). In this case we have \(K(A, 0)_n = A\) for all \(n\). Furthermore, all the maps in this simplicial abelian group are \(\text{id}_A\), in other words \(K(A, 0)\) is the constant simplicial object with value \(A\). The boundary maps \(d_n = \sum_{i = 0}^n (-1)^i \text{id}_A = 0\) if \(n\) odd and \(= \text{id}_A\) if \(n\) is even. Thus \(s(K(A, 0))\) looks like this \[\ldots \to A \xrightarrow{0} A \xrightarrow{1} A \xrightarrow{0} A \to 0\] and the result is clear.

Next, we prove the result for all \(k\) by induction. Given the result for \(k\) consider the short exact sequence \[0 \to K(A, k) \to E \to K(A, k + 1) \to 0\] from Lemma 0192. By Lemma 018Z the associated sequence of chain complexes is exact. By Lemma 0196 we see that \(s(E)\) is acyclic. Hence the result for \(k + 1\) follows from the long exact sequence of homology, see Homology, Lemma 0111.

There is a second chain complex we can associate to a simplicial object of \(\mathcal{A}\). Recall that by Lemma 017V any simplicial object \(U\) of \(\mathcal{A}\) is canonically split with \(N(U_m) = \bigcap_{i = 0}^{m - 1} \Ker(d^m_i)\). We define the normalized chain complex \(N(U)\) to be the chain complex \[\ldots \to N(U_2) \to N(U_1) \to N(U_0) \to 0 \to 0 \to \ldots\] with boundary map \(d_n : N(U_n) \to N(U_{n - 1})\) given by the restriction of \((-1)^nd^n_n\) to the direct summand \(N(U_n)\) of \(U_n\). Note that Lemma 017X implies that \(d^n_n(N(U_n)) \subset N(U_{n - 1})\). It is a complex because \(d^n_n \circ d^{n + 1}_{n + 1} = d^n_n \circ d^{n + 1}_n\) and \(d^{n + 1}_n\) is zero on \(N(U_{n + 1})\) by definition. Thus we obtain a second functor \[N : \text{Simp}(\mathcal{A}) \longrightarrow \text{Ch}_{\geq 0}(\mathcal{A}).\] Here is the reason for the sign in the differential.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(U\) be a simplicial object of \(\mathcal{A}\). The canonical map \(N(U_n) \to U_n\) gives rise to a morphism of complexes \(N(U) \to s(U)\).

Proof

This is clear because the differential on \(s(U)_n = U_n\) is \(\sum (-1)^i d^n_i\) and the maps \(d^n_i\), \(i < n\) are zero on \(N(U_n)\), whereas the restriction of \((-1)^nd^n_n\) is the boundary map of \(N(U)\) by definition.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(A\) be an object of \(\mathcal{A}\) and let \(k\) be an integer. We have \(N(K(A, k))_i = A\) if \(i = k\) and \(0\) else.

Proof

It is clear that \(N(K(A, k))_i = 0\) when \(i < k\) because \(K(A, k)_i = 0\) in that case. It is clear that \(N(K(A, k))_k = A\) since \(K(A, k)_{k - 1} = 0\) and \(K(A, k)_k = A\). For \(i > k\) we have \(N(K(A, k))_i = 0\) by Lemma 018T and the definition of \(K(A, k)\), see Definition 0191.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(U\) be a simplicial object of \(\mathcal{A}\). The canonical morphism of chain complexes \(N(U) \to s(U)\) is split. In fact, \[s(U) = N(U) \oplus D(U)\] for some complex \(D(U)\). The construction \(U \mapsto D(U)\) is functorial.

Proof

Define \(D(U)_n\) to be the image of \[\bigoplus\nolimits_{\varphi : [n] \to [m]\text{ surjective}, \ m < n} N(U_m) \xrightarrow{\bigoplus U(\varphi)} U_n\] which is a subobject of \(U_n\) complementary to \(N(U_n)\) according to Lemma 017V and Definition 017P. We show that \(D(U)\) is a subcomplex. Pick a surjective map \(\varphi : [n] \to [m]\) with \(m < n\) and consider the composition \[N(U_m) \xrightarrow{U(\varphi)} U_n \xrightarrow{d_n} U_{n - 1}.\] This composition is the sum of the maps \[N(U_m) \xrightarrow{U(\varphi \circ \delta^n_i)} U_{n - 1}\] with sign \((-1)^i\), \(i = 0, \ldots, n\).

First we will prove by ascending induction on \(m\), \(0 \leq m < n - 1\) that all the maps \(U(\varphi \circ \delta^n_i)\) map \(N(U_m)\) into \(D(U)_{n - 1}\). (The case \(m = n - 1\) is treated below.) Whenever the map \(\varphi \circ \delta^n_i : [n - 1] \to [m]\) is surjective then the image of \(N(U_m)\) under \(U(\varphi \circ \delta^n_i)\) is contained in \(D(U)_{n - 1}\) by definition. If \(\varphi \circ \delta^n_i : [n - 1] \to [m]\) is not surjective, set \(j = \varphi(i)\) and observe that \(i\) is the unique index whose image under \(\varphi\) is \(j\). We may write \(\varphi \circ \delta^n_i = \delta^m_j \circ \psi \circ \delta^n_i\) for some \(\psi : [n - 1] \to [m - 1]\). Hence \(U(\varphi \circ \delta^n_i) = U(\psi \circ \delta^n_i) \circ d^m_j\) which is zero on \(N(U_m)\) unless \(j = m\). If \(j = m\), then \(d^m_m(N(U_m)) \subset N(U_{m - 1})\) and hence \(U(\varphi \circ \delta^n_i)(N(U_m)) \subset U(\psi \circ \delta^n_i)(N(U_{m - 1}))\) and we win by induction hypothesis.

To finish proving that \(D(U)\) is a subcomplex we still have to deal with the composition \[N(U_m) \xrightarrow{U(\varphi)} U_n \xrightarrow{d_n} U_{n - 1}.\] in case \(m = n - 1\). In this case \(\varphi = \sigma^{n - 1}_j\) for some \(0 \leq j \leq n - 1\) and \(U(\varphi) = s^{n - 1}_j\). Thus the composition is given by the sum \[\sum (-1)^i d^n_i \circ s^{n - 1}_j\] Recall from Remark 016C that \(d^n_j \circ s^{n - 1}_j = d^n_{j + 1} \circ s^{n - 1}_j = \text{id}\) and these drop out because the corresponding terms have opposite signs. The map \(d^n_n \circ s^{n - 1}_j\), if \(j < n - 1\), is equal to \(s^{n - 2}_j \circ d^{n - 1}_{n - 1}\). Since \(d^{n - 1}_{n - 1}\) maps \(N(U_{n - 1})\) into \(N(U_{n - 2})\), we see that the image \(d^n_n ( s^{n - 1}_j (N(U_{n - 1}))\) is contained in \(s^{n - 2}_j(N(U_{n - 2}))\) which is contained in \(D(U_{n - 1})\) by definition. For all other combinations of \((i, j)\) we have either \(d^n_i \circ s^{n - 1}_j = s^{n - 2}_{j - 1} \circ d^{n - 1}_i\) (if \(i < j\)), or \(d^n_i \circ s^{n - 1}_j = s^{n - 2}_j \circ d^{n - 1}_{i - 1}\) (if \(n > i > j + 1\)) and in these cases the map is zero because of the definition of \(N(U_{n - 1})\).

Remark

In the situation of Lemma 019A the subcomplex \(D(U) \subset s(U)\) can also be defined as the subcomplex with terms \[D(U)_n = \Im\left( \bigoplus\nolimits_{\varphi : [n] \to [m]\text{ surjective}, \ m < n} U_m \xrightarrow{\bigoplus U(\varphi)} U_n\right)\] Namely, since \(U_m\) is the direct sum of the subobject \(N(U_m)\) and the images of \(N(U_k)\) for surjections \([m] \to [k]\) with \(k < m\) this is clearly the same as the definition of \(D(U)_n\) given in the proof of Lemma 019A. Thus we see that if \(U\) is a simplicial abelian group, then elements of \(D(U)_n\) are exactly the sums of degenerate \(n\)-simplices.

Lemma

The functor \(N\) is exact.

Proof

By Lemma 0195 and the functorial decomposition of Lemma 019A.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(V\) be a simplicial object of \(\mathcal{A}\). The canonical morphism of chain complexes \(N(V) \to s(V)\) is a quasi-isomorphism. In other words, the complex \(D(V)\) of Lemma 019A is acyclic.

Proof

Note that the result holds for \(K(A, k)\) for any object \(A\) and any \(k \geq 0\), by Lemmas 0197 and 0199. Consider the hypothesis \(IH_{n, m}\): for all \(V\) such that \(V_j = 0\) for \(j \leq m\) and all \(i \leq n\) the map \(N(V) \to s(V)\) induces an isomorphism \(H_i(N(V)) \to H_i(s(V))\).

To start of the induction, note that \(IH_{n, n}\) is trivially true, because in that case \(N(V)_n = 0\) and \(s(V)_n = 0\).

Assume \(IH_{n, m}\), with \(m \leq n\). Pick a simplicial object \(V\) such that \(V_j = 0\) for \(j < m\). By Lemma 0190 and Definition 0191 we have \(K(V_m, m) = i_{m!} \text{sk}_mV\). By Lemma 018U the natural morphism \[K(V_m, m) = i_{m!} \text{sk}_mV \to V\] is injective. Thus we get a short exact sequence \[0 \to K(V_m, m) \to V \to W \to 0\] for some \(W\) with \(W_i = 0\) for \(i = 0, \ldots, m\). This short exact sequence induces a morphism of short exact sequence of associated complexes \[\xymatrix{ 0 \ar[r] & N(K(V_m, m)) \ar[r] \ar[d] & N(V) \ar[r] \ar[d] & N(W) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & s(K(V_m, m)) \ar[r] & s(V) \ar[r] & s(W) \ar[r] & 0 }\] see Lemmas 0195 and 019B. Hence we deduce the result for \(V\) from the result on the ends.

Dold-Kan

In this section we prove the Dold-Kan theorem relating simplicial objects in an abelian category with chain complexes.

Lemma

Let \(\mathcal{A}\) be an abelian category. The functor \(N\) is faithful, and reflects isomorphisms, injections and surjections.

Proof

The faithfulness is immediate from the canonical splitting of Lemma 017V. The statement on reflecting injections, surjections, and isomorphisms follows from Lemma 017W.

Lemma

Let \(\mathcal{A}\) and \(\mathcal{B}\) be abelian categories. Let \(N : \mathcal{A} \to \mathcal{B}\), and \(S : \mathcal{B} \to \mathcal{A}\) be functors. Suppose that

1. the functors \(S\) and \(N\) are exact,

2. there is an isomorphism \(g : N \circ S \to \text{id}_\mathcal{B}\) to the identity functor of \(\mathcal{B}\),

3. \(N\) is faithful, and

4. \(S\) is essentially surjective.

Then \(S\) and \(N\) are quasi-inverse equivalences of categories.

Proof

It suffices to construct a functorial isomorphism \(S(N(A)) \cong A\). To do this choose \(B\) and an isomorphism \(f : A \to S(B)\). Consider the map \[f^{-1} \circ g_{S(B)} \circ S(N(f)) : S(N(A)) \to S(N(S(B))) \to S(B) \to A.\] It is easy to show this does not depend on the choice of \(f, B\) and gives the desired isomorphism \(S \circ N \to \text{id}_\mathcal{A}\).

Theorem

Let \(\mathcal{A}\) be an abelian category. The functor \(N\) induces an equivalence of categories \[N : \text{Simp}(\mathcal{A}) \longrightarrow \text{Ch}_{\geq 0}(\mathcal{A})\]

Proof

We will describe a functor in the reverse direction inspired by the construction of Lemma 0192 (except that we throw in a sign to get the boundaries right). Let \(A_\bullet\) be a chain complex with boundary maps \(d_{A, n} : A_n \to A_{n - 1}\). For each \(n \geq 0\) denote \[I_n = \Big\{ \alpha : [n] \to \{0, 1, 2, \ldots\} \mid \Im(\alpha) = [k]\text{ for some }k \Big\}.\] For \(\alpha \in I_n\) we denote \(k(\alpha)\) the unique integer such that \(\Im(\alpha) = [k]\). We define a simplicial object \(S(A_\bullet)\) as follows:

1. \(S(A_\bullet)_n = \bigoplus_{\alpha \in I_n} A_{k(\alpha)}\), which we will write as \(\bigoplus_{\alpha \in I_n} A_{k(\alpha)} \cdot \alpha\) to suggest thinking of “\(\alpha\)” as a basis vector for the summand corresponding to it,

2. given \(\varphi : [m] \to [n]\) we define \(S(A_\bullet)(\varphi)\) by its restriction to the direct summand \(A_{k(\alpha)} \cdot \alpha\) of \(S(A_\bullet)_n\) as follows

   1. \(\alpha \circ \varphi \not \in I_m\) then we set it equal to zero,

   2. \(\alpha \circ \varphi \in I_m\) but \(k(\alpha \circ \varphi)\) not equal to either \(k(\alpha)\) or \(k(\alpha) - 1\) then we set it equal to zero as well,

   3. if \(\alpha \circ \varphi \in I_m\) and \(k(\alpha \circ \varphi) = k(\alpha)\) then we use the identity map to the summand \(A_{k(\alpha \circ \varphi)} \cdot (\alpha \circ \varphi)\) of \(S(A_\bullet)_m\), and

   4. if \(\alpha \circ \varphi \in I_m\) and \(k(\alpha \circ \varphi) = k(\alpha) - 1\) then we use \((-1)^{k(\alpha)} d_{A, k(\alpha)}\) to the summand \(A_{k(\alpha \circ \varphi)}\cdot (\alpha \circ \varphi)\) of \(S(A_\bullet)_m\).

Let us show that \(S(A_\bullet)\) is a simplicial object of \(\mathcal{A}\). To do this, assume we have maps \(\varphi : [m] \to [n]\) and \(\psi : [n] \to [p]\). We will show that \(S(A_\bullet)(\varphi) \circ S(A_\bullet)(\psi) = S(A_\bullet)(\psi \circ \varphi)\). Choose \(\beta \in I_p\) and set \(\alpha = \beta \circ \psi\) and \(\gamma = \alpha \circ \varphi\) viewed as maps \(\alpha : [n] \to \{0, 1, 2, \ldots\}\) and \(\gamma : [m] \to \{0, 1, 2, \ldots\}\). Picture \[\xymatrix{ [m] \ar[r]_\varphi \ar[d]_\gamma & [n] \ar[r]_\psi \ar[d]_\alpha & [p] \ar[d]^\beta \\ \Im(\gamma) \ar[r] & \Im(\alpha) \ar[r] & [k(\beta)] }\] We will show that the restriction of the maps \(S(A_\bullet)(\varphi) \circ S(A_\bullet)(\psi)\) and \(S(A_\bullet)(\psi \circ \varphi)\). to the summand \(A_{k(\beta)} \cdot \beta\) agree. There are several cases to consider

1. Say \(\alpha \not \in I_n\) so the restriction of \(S(A_\bullet)(\psi)\) to \(A_{k(\beta)} \cdot \beta\) is zero. Then either \(\gamma \not \in I_m\) or we have \([k(\gamma)] = \Im(\gamma) \subset \Im(\alpha) \subset [k(\beta)]\) and the subset \(\Im(\alpha)\) of \([k(\beta)]\) has a gap so \(k(\gamma) < k(\beta) - 1\). In both cases we see that the restriction of \(S(A_\bullet)(\psi \circ \varphi)\) to \(A_{k(\beta)} \cdot \beta\) is zero as well.

2. Say \(\alpha \in I_n\) and \(k(\alpha) < k(\beta) - 1\) so the restriction of \(S(A_\bullet)(\psi)\) to \(A_{k(\beta)} \cdot \beta\) is zero. Then either \(\gamma \not \in I_m\) or we have \([k(\gamma)] \subset [k(\alpha)] \subset [k(\beta)]\) and it follows that \(k(\gamma) < k(\beta) - 1\). In both cases we see that the restriction of \(S(A_\bullet)(\psi \circ \varphi)\) to \(A_{k(\beta)} \cdot \beta\) is zero as well.

3. Say \(\alpha \in I_n\) and \(k(\alpha) = k(\beta)\) so the restriction of \(S(A_\bullet)(\psi)\) to \(A_{k(\beta)} \cdot \beta\) is the identity map from \(A_{k(\beta)} \cdot \beta\) to \(A_{k(\alpha)} \cdot \alpha\). In this case because \(\Im(\alpha) =[k(\beta)]\) the rule describing the restriction of \(S(A_\bullet)(\psi \circ \varphi)\) to the summand \(A_{k(\beta)} \cdot \beta\) is exactly the same as the rule describing the restriction of \(S(A_\bullet)(\varphi)\) to the summand \(A_{k(\alpha)} \cdot \alpha\) and hence agreement holds.

4. Say \(\alpha \in I_n\) and \(k(\alpha) = k(\beta) - 1\) so the restriction of \(S(A_\bullet)(\psi)\) to \(A_{k(\beta)} \cdot \beta\) is given by \((-1)^{k(\beta)}d_{A, k(\beta)}\) to \(A_{k(\alpha)} \cdot \alpha\). Subcases

   1. If \(\gamma \not \in I_m\), then both the restriction of \(S(A_\bullet)(\psi \circ \varphi)\) to the summand \(A_{k(\beta)} \cdot \beta\) and the restriction of \(S(A_\bullet)(\varphi)\) to the summand \(A_{k(\alpha)} \cdot \alpha\) are zero and we get agreement.

   2. If \(\gamma \in I_m\) but \(k(\gamma) < k(\alpha) - 1\), then again both restrictions are zero and we get agreement.

   3. If \(\gamma \in I_m\) and \(k(\gamma) = k(\alpha)\) then \(\Im(\gamma) = \Im(\alpha)\). In this case the restriction of \(S(A_\bullet)(\psi \circ \varphi)\) to the summand \(A_{k(\beta)} \cdot \beta\) is given by \((-1)^{k(\beta)}d_{A, k(\beta)}\) to \(A_{k(\gamma)} \cdot \gamma\) and the restriction of \(S(A_\bullet)(\varphi)\) to the summand \(A_{k(\alpha)} \cdot \alpha\) is the identity map \(A_{k(\alpha)} \cdot \alpha \to A_{k(\gamma)} \cdot \gamma\). Hence agreement holds.

   4. Finally, if \(\gamma \in I_m\) and \(k(\gamma) = k(\alpha) - 1\) then the restriction of \(S(A_\bullet)(\varphi)\) to the summand \(A_{k(\alpha)} \cdot \alpha\) is given by \((-1)^{k(\alpha)} d_{A, k(\alpha)}\) as a map \(A_{k(\alpha)} \cdot \alpha \to A_{k(\beta)} \cdot \beta\). Since \(A_\bullet\) is a complex we see that the composition \(A_{k(\beta)} \cdot \beta \to A_{k(\alpha)} \cdot \alpha \to A_{k(\gamma)} \cdot \gamma\) is zero which matches what we get for the restriction of \(S(A_\bullet)(\psi \circ \varphi)\) to the summand \(A_{k(\beta)} \cdot \beta\) because \(k(\gamma) = k(\beta) - 2 < k(\beta) - 1\).

Thus \(S(A_\bullet)\) is a simplicial object of \(\mathcal{A}\).

Let us construct an isomorphism \(A_\bullet \to N(S(A_\bullet))\) functorial in \(A_\bullet\). Recall that \[S(A_\bullet) = N(S(A_\bullet)) \oplus D(S(A_\bullet))\] as chain complexes by Lemma 019A. On the other hand it follows from Remark 0FKI and the construction of \(S(A_\bullet)\) that \[D(S(A_\bullet))_n = \bigoplus\nolimits_{\alpha \in I_n,\ k(\alpha) < n} A_{k(\alpha)} \cdot \alpha \subset \bigoplus\nolimits_{\alpha \in I_n} A_{k(\alpha)} \cdot \alpha\] However, if \(\alpha \in I_n\) then we have \(k(\alpha) \geq n \Leftrightarrow \alpha = \text{id}_{[n]} : [n] \to [n]\). Thus the summand \(A_n \cdot \text{id}_{[n]}\) of \(S(A_\bullet)_n\) is a complement to the summand \(D(S(A_\bullet))_n\). All the maps \(d^n_i : S(A_\bullet)_n \to S(A_\bullet)_n\) restrict to zero on the summand \(A_n \cdot \text{id}_{[n]}\) except for \(d^n_n\) which produces \((-1)^n d_{A, n}\) from \(A_n \cdot \text{id}_{[n]}\) to \(A_{n - 1} \cdot \text{id}_{[n - 1]}\). We conclude that \(A_n \cdot \text{id}_{[n]}\) must be equal to the summand \(N(S(A_\bullet))_n\) and moreover the restriction of the differential \(d_n = \sum (-1)^id^n_i : S(A_\bullet)_n \to S(A_\bullet)_{n - 1}\) to the summand \(A_n \cdot \text{id}_{[n]}\) gives what we want!

Finally, we have to show that \(S \circ N\) is isomorphic to the identity functor. Let \(U\) be a simplicial object of \(\mathcal{A}\). Then we can define an obvious map \[S(N(U))_n = \bigoplus\nolimits_{\alpha \in I_n} N(U)_{k(\alpha)} \cdot \alpha \longrightarrow U_n\] by using \(U(\alpha) : N(U)_{k(\alpha)} \to U_n\) on the summand corresponding to \(\alpha\). By Definition 017P this is an isomorphism. To finish the proof we have to show that this is compatible with the maps in the simplicial objects. Thus let \(\varphi : [m] \to [n]\) and let \(\alpha \in I_n\). Set \(\beta = \alpha \circ \varphi\). Picture \[\xymatrix{ [m] \ar[r]_\varphi \ar[d]_\beta & [n] \ar[d]_\alpha \\ \Im(\beta) \ar[r] & [k(\alpha)] }\] There are several cases to consider

1. Say \(\beta \not \in I_m\). Then there exists an index \(0 \leq j < k(\alpha)\) with \(j \not \in \Im(\alpha \circ \varphi)\) and hence we can choose a factorization \(\alpha \circ \varphi = \delta^{k(\alpha)}_j \circ \psi\) for some \(\psi : [m] \to [k(\alpha) - 1]\). It follows that \(U(\varphi)\) is zero on the image of the summand \(N(U)_{k(\alpha)} \cdot \alpha\) because \(U(\varphi) \circ U(\alpha) = U(\alpha \circ \varphi) = U(\psi) \circ d^{k(\alpha)}_j\) is zero on \(N(U)_{k(\alpha)}\) by construction of \(N\). This matches our rule for \(S(N(U))\) given above.

2. Say \(\beta \in I_m\) and \(k(\beta) < k(\alpha) - 1\). Here we argue exactly as in case (1) with \(j = k(\alpha) - 1\).

3. Say \(\beta \in I_m\) and \(k(\beta) = k(\alpha)\). Here the summand \(N(U)_{k(\alpha)} \cdot \alpha\) is mapped by the identity to the summand \(N(U)_{k(\beta)} \cdot \beta\). This is the same as the effect of \(U(\varphi)\) since in this case \(U(\varphi) \circ U(\alpha) = U(\beta)\).

4. Say \(\beta \in I_m\) and \(k(\beta) = k(\alpha) - 1\). Here we use the differential \((-1)^{k(\alpha)} d_{N(U), k(\alpha)}\) to map the summand \(N(U)_{k(\alpha)} \cdot \alpha\) to the summand \(N(U)_{k(\beta)} \cdot \beta\). On the other hand, since \(\Im(\beta) = [k(\beta)]\) in this case we get \(\alpha \circ \varphi = \delta^{k(\alpha)}_{k(\alpha)} \circ \beta\). Thus we see that \(U(\varphi)\) composed with the restriction of \(U(\alpha)\) to \(N(U)_{k(\alpha)}\) is equal to \(U(\beta)\) precomposed with \(d^{k(\alpha)}_{k(\alpha)}\) restricted to \(N(U)_{k(\alpha)}\). Since \(d_{N(U), k(\alpha)} = \sum (-1)^i d^{k(\alpha)}_i\) and since \(d^{k(\alpha)}_i\) restricts to zero on \(N(U)_{k(\alpha)}\) for \(i < k(\alpha)\) we see that equality holds.

This finishes the proof of the theorem.

Dold-Kan for cosimplicial objects

Let \(\mathcal{A}\) be an abelian category. According to Homology, Lemma 010A also \(\mathcal{A}^{opp}\) is abelian. It follows formally from the definitions that \[\text{CoSimp}(\mathcal{A}) = \text{Simp}(\mathcal{A}^{opp})^{opp}.\] Thus Dold-Kan (Theorem 019G) implies that \(\text{CoSimp}(\mathcal{A})\) is equivalent to the category \(\text{Ch}_{\geq 0}(\mathcal{A}^{opp})^{opp}\). And it follows formally from the definitions that \[\text{CoCh}_{\geq 0}(\mathcal{A}) = \text{Ch}_{\geq 0}(\mathcal{A}^{opp})^{opp}.\] Putting these arrows together we obtain an equivalence \[Q : \text{CoSimp}(\mathcal{A}) \longrightarrow \text{CoCh}_{\geq 0}(\mathcal{A}).\] In this section we describe \(Q\).

First we define the cochain complex \(s(U)\) associated to a cosimplicial object \(U\). It is the cochain complex with terms zero in negative degrees, and \(s(U)^n = U_n\) for \(n \geq 0\). As differentials we use the maps \(d^n : s(U)^n \to s(U)^{n + 1}\) defined by \(d^n = \sum_{i = 0}^{n + 1} (-1)^i \delta^{n + 1}_i\). In other words the complex \(s(U)\) looks like \[\xymatrix{ 0 \ar[r] & U_0 \ar[rr]^{\delta^1_0 - \delta^1_1} & & U_1 \ar[rr]^{\delta^2_0 - \delta^2_1 + \delta^2_2} & & U_2 \ar[r] & \ldots }\] This is sometimes also called the Moore complex associated to \(U\).

On the other hand, given a cosimplicial object \(U\) of \(\mathcal{A}\) set \(Q(U)^0 = U_0\) and \[Q(U)^n = \Coker( \xymatrix{ \bigoplus_{i = 0}^{n - 1} U_{n - 1} \ar[r]^-{\delta^n_i} & U_n }).\] The differential \(d^n : Q(U)^n \to Q(U)^{n + 1}\) is induced by \((-1)^{n + 1}\delta^{n + 1}_{n + 1}\), i.e., by fitting the morphism \((-1)^{n + 1}\delta^{n + 1}_{n + 1}\) into a commutative diagram \[\xymatrix{ U_n \ar[rr]_{(-1)^{n + 1}\delta^{n + 1}_{n + 1}} \ar[d] & & U_{n + 1} \ar[d] \\ Q(U)^n \ar[rr]^{d_n} & & Q(U)^{n + 1}. }\] We leave it to the reader to show that this diagram makes sense, i.e., that the image of \(\delta^n_i\) maps into the kernel of the right vertical arrow for \(i = 0, \ldots, n - 1\). (This is dual to Lemma 017X.) Thus our cochain complex \(Q(U)\) looks like this \[0 \to Q(U)^0 \to Q(U)^1 \to Q(U)^2 \to \ldots\] This is called the normalized cochain complex associated to \(U\). The dual to the Dold-Kan Theorem 019G is the following.

Lemma

Let \(\mathcal{A}\) be an abelian category.

1. The functor \(s : \text{CoSimp}(\mathcal{A}) \to \text{CoCh}_{\geq 0}(\mathcal{A})\) is exact.

2. The maps \(s(U)^n \to Q(U)^n\) define a morphism of cochain complexes.

3. There exists a functorial direct sum decomposition \(s(U) = D(U) \oplus Q(U)\) in \(\text{CoCh}_{\geq 0}(\mathcal{A})\).

4. The functor \(Q\) is exact.

5. The morphism of complexes \(s(U) \to Q(U)\) is a quasi-isomorphism.

6. The functor \(U \mapsto Q(U)^\bullet\) defines an equivalence of categories \(\text{CoSimp}(\mathcal{A}) \to \text{CoCh}_{\geq 0}(\mathcal{A})\).

Proof

Omitted. But the results are the exact dual statements to Lemmas 0195, 0198, 019A, 019B, 019C, and Theorem 019G.

Homotopies

Consider the simplicial sets \(\Delta[0]\) and \(\Delta[1]\). Recall that there are two morphisms \[e_0, e_1 : \Delta[0] \longrightarrow \Delta[1],\] coming from the morphisms \([0] \to [1]\) mapping \(0\) to an element of \([1] = \{0, 1\}\). Recall also that each set \(\Delta[1]_k\) is finite. Hence, if the category \(\mathcal{C}\) has finite coproducts, then we can form the product \[U \times \Delta[1]\] for any simplicial object \(U\) of \(\mathcal{C}\), see Definition 017C. Note that \(\Delta[0]\) has the property that \(\Delta[0]_k = \{*\}\) is a singleton for all \(k \geq 0\). Hence \(U \times \Delta[0] = U\). Thus \(e_0, e_1\) above gives rise to morphisms \[e_0, e_1 : U \to U \times \Delta[1].\]

Definition

Let \(\mathcal{C}\) be a category having finite coproducts. Suppose that \(U\) and \(V\) are two simplicial objects of \(\mathcal{C}\). Let \(a, b : U \to V\) be two morphisms.

1. We say a morphism \[h : U \times \Delta[1] \longrightarrow V\] is a homotopy from \(a\) to \(b\) if \(a = h \circ e_0\) and \(b = h \circ e_1\).

2. We say the morphisms \(a\) and \(b\) are homotopic or are in the same homotopy class if there exists a sequence of morphisms \(a = a_0, a_1, \ldots, a_n = b\) from \(U\) to \(V\) such that for each \(i = 1, \ldots, n\) there either exists a homotopy from \(a_{i - 1}\) to \(a_i\) or there exists a homotopy from \(a_i\) to \(a_{i - 1}\).

The relation “there is a homotopy from \(a\) to \(b\)” is in general not transitive or symmetric; we will see it is reflexive in Example 07KA. Of course, “being homotopic” is an equivalence relation on the set \(\Mor(U, V)\) and it is the equivalence relation generated by the relation “there is a homotopy from \(a\) to \(b\)” . It turns out we can define homotopies between pairs of maps of simplicial objects in any category. We will do this in Remark 019M after we work out in some detail what it means to have a morphism \(h : U \times \Delta[1] \to V\).

Let \(\mathcal{C}\) be a category with finite coproducts. Let \(U\), \(V\) be simplicial objects of \(\mathcal{C}\). Let \(a, b : U \to V\) be morphisms. Further, suppose that \(h : U \times \Delta[1] \to V\) is a homotopy from \(a\) to \(b\). For every \(n \geq 0\) let us write \[\Delta[1]_n = \{\alpha^n_0, \ldots, \alpha^n_{n + 1}\}\] where \(\alpha^n_i : [n] \to [1]\) is the map such that \[\alpha^n_i(j) = \left\{ \begin{matrix} 0 & \text{if} & j < i\\ 1 & \text{if} & j \geq i \end{matrix} \right.\] Thus \[h_n : (U \times \Delta[1])_n = \coprod U_n \cdot \alpha^n_i \longrightarrow V_n\] has a component \(h_{n, i} : U_n \to V_n\) which is the restriction to the summand corresponding to \(\alpha^n_i\) for all \(i = 0, \ldots, n + 1\).

Lemma

In the situation above, we have the following relations:

1. We have \(h_{n, 0} = b_n\) and \(h_{n, n + 1} = a_n\).

2. We have \(d^n_j \circ h_{n, i} = h_{n - 1, i - 1} \circ d^n_j\) for \(i > j\).

3. We have \(d^n_j \circ h_{n, i} = h_{n - 1, i} \circ d^n_j\) for \(i \leq j\).

4. We have \(s^n_j \circ h_{n, i} = h_{n + 1, i + 1} \circ s^n_j\) for \(i > j\).

5. We have \(s^n_j \circ h_{n, i} = h_{n + 1, i} \circ s^n_j\) for \(i \leq j\).

Conversely, given a system of maps \(h_{n, i}\) satisfying the properties listed above, then these define a morphism \(h\) which is a homotopy from \(a\) to \(b\).

Proof

Omitted. You can prove the last statement using the fact, see Lemma 0168 that to give a morphism of simplicial objects is the same as giving a sequence of morphisms \(h_n\) commuting with all \(d^n_j\) and \(s^n_j\).

Example

Suppose in the situation above \(a = b\). Then there is a trivial homotopy from \(a\) to \(b\), namely the one with \(h_{n, i} = a_n = b_n\).

Remark

Let \(\mathcal{C}\) be any category (no assumptions whatsoever). Let \(U\) and \(V\) be simplicial objects of \(\mathcal{C}\). Let \(a, b : U \to V\) be morphisms of simplicial objects of \(\mathcal{C}\). A homotopy from \(a\) to \(b\) is given by morphisms¹ \(h_{n, i} : U_n \to V_n\), for \(n \geq 0\), \(i = 0, \ldots, n + 1\) satisfying the relations of Lemma 019L. As in Definition 019K we say the morphisms \(a\) and \(b\) are homotopic if there exists a sequence of morphisms \(a = a_0, a_1, \ldots, a_n = b\) from \(U\) to \(V\) such that for each \(i = 1, \ldots, n\) there either exists a homotopy from \(a_{i - 1}\) to \(a_i\) or there exists a homotopy from \(a_i\) to \(a_{i - 1}\). Clearly, if \(F : \mathcal{C} \to \mathcal{C}'\) is any functor and \(\{h_{n, i}\}\) is a homotopy from \(a\) to \(b\), then \(\{F(h_{n, i})\}\) is a homotopy from \(F(a)\) to \(F(b)\). Similarly, if \(a\) and \(b\) are homotopic, then \(F(a)\) and \(F(b)\) are homotopic. Since the lemma says that the newer notion is the same as the old one in case finite coproduct exist, we deduce in particular that functors preserve the original notion whenever both categories have finite coproducts.

Remark

Let \(\mathcal{C}\) be any category. Suppose two morphisms \(a, a' : U \to V\) of simplicial objects are homotopic. Then for any morphism \(b : V \to W\) the two maps \(b \circ a, b \circ a' : U \to W\) are homotopic. Similarly, for any morphism \(c : X \to U\) the two maps \(a \circ c, a' \circ c : X \to V\) are homotopic. In fact the maps \(b \circ a \circ c, b \circ a' \circ c : X \to W\) are homotopic. Namely, if the maps \(h_{n, i} : U_n \to V_n\) define a homotopy from \(a\) to \(a'\) then the maps \(b \circ h_{n, i} \circ c\) define a homotopy from \(b \circ a \circ c\) to \(b \circ a' \circ c\). In this way we see that we obtain a new category \(\text{hSimp}(\mathcal{C})\) with the same objects as \(\text{Simp}(\mathcal{C})\) but whose morphisms are homotopy classes of morphisms of \(\text{Simp}(\mathcal{C})\). Thus there is a canonical functor \[\text{Simp}(\mathcal{C}) \longrightarrow \text{hSimp}(\mathcal{C})\] which is essentially surjective and surjective on sets of morphisms.

Definition

Let \(U\) and \(V\) be two simplicial objects of a category \(\mathcal{C}\). We say a morphism \(a : U \to V\) is a homotopy equivalence if there exists a morphism \(b : V \to U\) such that \(a \circ b\) is homotopic to \(\text{id}_V\) and \(b \circ a\) is homotopic to \(\text{id}_U\). We say \(U\) and \(V\) are homotopy equivalent if there exists a homotopy equivalence \(a : U \to V\).

Example

The simplicial set \(\Delta[m]\) is homotopy equivalent to \(\Delta[0]\). Namely, consider the unique morphism \(f : \Delta[m] \to \Delta[0]\) and the morphism \(g : \Delta[0] \to \Delta[m]\) given by the inclusion of the last \(0\)-simplex of \(\Delta[m]\). We have \(f \circ g = \text{id}\). We will give a homotopy \(h : \Delta[m] \times \Delta[1] \to \Delta[m]\) from \(\text{id}_{\Delta[m]}\) to \(g \circ f\). Namely \(h\) is given by the maps \[\Mor_\Delta([n], [m]) \times \Mor_\Delta([n], [1]) \to \Mor_\Delta([n], [m])\] which send \((\varphi, \alpha)\) to \[k \mapsto \left\{ \begin{matrix} \varphi(k) & \text{if} & \alpha(k) = 0 \\ m & \text{if} & \alpha(k) = 1 \end{matrix} \right.\] Note that this only works because we took \(g\) to be the inclusion of the last \(0\)-simplex. If we took \(g\) to be the inclusion of the first \(0\)-simplex we could find a homotopy from \(g \circ f\) to \(\text{id}_{\Delta[m]}\). This is an illustration of the asymmetry inherent in homotopies in the category of simplicial sets.

The following lemma says that \(U \times \Delta[1]\) is homotopy equivalent to \(U\).

Lemma

Let \(\mathcal{C}\) be a category with finite coproducts. Let \(U\) be a simplicial object of \(\mathcal{C}\). Consider the maps \(e_1, e_0 : U \to U \times \Delta[1]\), and \(\pi : U \times \Delta[1] \to U\), see Lemma 017E.

1. We have \(\pi \circ e_1 = \pi \circ e_0 = \text{id}_U\), and

2. The morphisms \(\text{id}_{U \times \Delta[1]}\), and \(e_0 \circ \pi\) are homotopic.

3. The morphisms \(\text{id}_{U \times \Delta[1]}\), and \(e_1 \circ \pi\) are homotopic.

Proof

The first assertion is trivial. Consider the map of simplicial sets \(\Delta[1] \times \Delta[1] \longrightarrow \Delta[1]\) which in degree \(n\) assigns to a pair \((\beta_1, \beta_2)\), \(\beta_i : [n] \to [1]\) the morphism \(\beta : [n] \to [1]\) defined by the rule \[\beta(i) = \max\{\beta_1(i), \beta_2(i)\}.\] It is a morphism of simplicial sets, because the action \(\Delta[1](\varphi) : \Delta[1]_n \to \Delta[1]_m\) of \(\varphi : [m] \to [n]\) is by precomposing. Using the notation \(\Delta[1]_n = \{\alpha^n_0, \ldots, \alpha^n_{n + 1}\}\) introduced above, we have \(\beta = \alpha^n_0\) if \(\beta_2 = \alpha^n_0\) and \(\beta = \beta_1\) if \(\beta_2 = \alpha^n_{n + 1}\). Since \(\alpha^n_0\), resp. \(\alpha^n_{n + 1}\) is the constant map with value \(1\), resp. \(0\), this implies that the induced morphism \[U \times \Delta[1] \times \Delta[1] \longrightarrow U \times \Delta[1]\] of Lemma 017E is a homotopy from \(\text{id}_{U \times \Delta[1]}\) to \(e_1 \circ \pi\). Similarly for \(e_0 \circ \pi\) (use minimum instead of maximum).

Lemma

Let \(f : Y \to X\) be a morphism of a category \(\mathcal{C}\) with fibre products. Assume \(f\) has a section \(s\). Consider the simplicial object \(U\) constructed in Example 016E starting with \(f\). The morphism \(U \to U\) which in each degree is the self map \((s \circ f)^{n + 1}\) of \(Y \times_X \ldots \times_X Y\) given by \(s \circ f\) on each factor is homotopic to the identity on \(U\). In particular, \(U\) is homotopy equivalent to the constant simplicial object \(X\).

Proof

Set \(g^0 = \text{id}_Y\) and \(g^1 = s \circ f\). We use the morphisms \[\begin{eqnarray*} Y \times_X \ldots \times_X Y \times \Mor([n], [1]) & \to & Y \times_X \ldots \times_X Y \\ (y_0, \ldots, y_n) \times \alpha & \mapsto & (g^{\alpha(0)}(y_0), \ldots, g^{\alpha(n)}(y_n)) \end{eqnarray*}\] where we use the functor of points point of view to define the maps. Another way to say this is to say that \(h_{n, 0} = \text{id}\), \(h_{n, n + 1} = (s \circ f)^{n + 1}\) and \(h_{n, i} = \text{id}_Y^{i + 1} \times (s \circ f)^{n + 1 - i}\). We leave it to the reader to show that these satisfy the relations of Lemma 019L. Hence they define the desired homotopy. See also Remark 019M which shows that we do not need to assume anything else on the category \(\mathcal{C}\).

Lemma

Let \(\mathcal{C}\) be a category. Let \(T\) be a set. For \(t \in T\) let \(X_t\), \(Y_t\) be simplicial objects of \(\mathcal{C}\). Assume \(X = \prod_{t \in T} X_t\) and \(Y = \prod_{t \in T} Y_t\) exist.

1. If \(X_t\) and \(Y_t\) are homotopy equivalent for all \(t \in T\) and \(T\) is finite, then \(X\) and \(Y\) are homotopy equivalent.

For \(t \in T\) let \(a_t, b_t : X_t \to Y_t\) be morphisms. Set \(a = \prod a_t : X \to Y\) and \(b = \prod b_t : X \to Y\).

1. If there exists a homotopy from \(a_t\) to \(b_t\) for all \(t \in T\), then there exists a homotopy from \(a\) to \(b\).

2. If \(T\) is finite and \(a_t, b_t : X_t \to Y_t\) for \(t \in T\) are homotopic, then \(a\) and \(b\) are homotopic.

Proof

If \(h_t = (h_{t, n , i})\) is a homotopy from \(a_t\) to \(b_t\) (see Remark 019M), then \(h = (\prod_t h_{t, n, i})\) is a homotopy from \(\prod a_t\) to \(\prod b_t\). This proves (2).

Proof of (3). Choose \(t \in T\). There exists an integer \(n \geq 0\) and a chain \(a_t = a_{t, 0}, a_{t, 1}, \ldots, a_{t, n} = b_t\) such that for every \(1 \leq i \leq n\) either there is a homotopy from \(a_{t, i - 1}\) to \(a_{t, i}\) or there is a homotopy from \(a_{t, i}\) to \(a_{t, i - 1}\). If \(n = 0\), then we pick another \(t\). (We’re done if \(a_t = b_t\) for all \(t \in T\).) So assume \(n > 0\). By Example 07KA there are is a homotopy from \(b_{t'}\) to \(b_{t'}\) for all \(t' \in T \setminus \{t\}\). Thus by (2) there is a homotopy from \(a_{t, n - 1} \times \prod_{t'} b_{t'}\) to \(b\) or there is a homotopy from \(b\) to \(a_{t, n - 1} \times \prod_{t'} b_{t'}\). In this way we can decrease \(n\) by \(1\). This proves (3).

Part (1) follows from part (3) and the definitions.

Relative simplicial homotopies and base change

The following construction makes explicit the relative homotopy formalism of Illusie, Complexe cotangent et déformations I, I.1.1.6. The underlying absolute notion is that of Remark 019M; strongly cartesian morphisms and fibred categories are as in Tag 02XK and Tag 02XM.

Let \(p:\mathcal E\to\mathcal B\) be a functor. Write \(\mathrm{Simp}(\mathcal E)\) for the category of functors \(\Delta^{\mathrm{op}}\to\mathcal E\). If \(X,Y\) are simplicial objects, a simplicial homotopy is a family \[h_{n,\alpha}:X_n\longrightarrow Y_n, \qquad \alpha:[n]\longrightarrow[1],\] whose values at the two constant maps are its endpoints and such that \[Y(\theta)h_{n,\alpha} =h_{m,\alpha\theta}X(\theta) \quad\text{for every }\theta:[m]\longrightarrow[n].\] This is the component formulation of simplicial homotopy; no coproducts in \(\mathcal E\) are required. Suppose \(pX=S\), \(pY=T\), and that both endpoints lie over \(f:S\to T\). Call the homotopy relative to \(f\) if \(p(h_{n,\alpha})=f_n\) for every \(n,\alpha\). On each set of arrows over a fixed \(f\), take the equivalence relation generated by these homotopies in either direction.

Lemma

The relative homotopy relations are compatible with composition. They define a category \(\mathrm{hSimp}(\mathcal E/\mathcal B)\) over \(\mathrm{Simp}(\mathcal B)\), with the same objects as \(\mathrm{Simp}(\mathcal E)\). Its fibre over \(S\) has as morphisms the \(\mathrm{id}_S\)-relative homotopy classes of arrows over \(S\). Every functor \(F:\mathcal E\to\mathcal E'\) over \(\mathcal B\) induces a functor between these quotient categories over \(\mathrm{Simp}(\mathcal B)\).

Proof

Let \(h\) be relative to \(f:S\to T\). Precomposition with an arrow \(v\) over \(g:S'\to S\) replaces its components by \(h_{n,\alpha}v_n\); their images under \(p\) are \(f_ng_n\). Postcomposition with an arrow \(w\) over \(k:T\to T'\) gives \(w_nh_{n,\alpha}\) over \(k_nf_n\). The displayed naturality relation is preserved in each case. Thus precomposition and postcomposition preserve elementary relative homotopies, and hence their generated equivalence relations. The quotient has well-defined associative composition and identities inherited from the original category. Applying \(F\) to every component preserves naturality, endpoints and the base map, proving the last assertion.

Lemma

Assume \(p:\mathcal E\to\mathcal B\) is a fibred category. Then:

1. For every \(f:S\to T\) in \(\mathrm{Simp}(\mathcal B)\) and every \(Y\) over \(T\), there is a lift \(c:f^*Y\to Y\) over \(f\) whose components are strongly cartesian. For every \(X\) over \(S\) and pair of arrows \(a,b:X\to Y\) over \(f\), composition with \(c\) bijects homotopies over \(\mathrm{id}_S\) between the lifted arrows \(\widetilde a,\widetilde b:X\to f^*Y\) with homotopies relative to \(f\) between \(a,b\).

2. Every \(g:S'\to S\) induces a pullback functor \[g^*:\mathrm{hSimp}(\mathcal E)_S \longrightarrow\mathrm{hSimp}(\mathcal E)_{S'},\] where these are the fibres of the preceding quotient category. The canonical comparisons between successive pullbacks are well-defined on these fibres and satisfy the usual identity and associativity constraints.

3. The functor \(\mathrm{hSimp}(\mathcal E/\mathcal B)\to\mathrm{Simp}(\mathcal B)\) is fibred. The class of a componentwise strongly cartesian lift is strongly cartesian in this quotient.

Proof

Choose a strongly cartesian arrow \(c_n:(f^*Y)_n\to Y_n\) over \(f_n\) in every degree. For \(\theta:[m]\to[n]\), define \((f^*Y)(\theta)\) as the unique arrow over \(S(\theta)\) satisfying \[c_m(f^*Y)(\theta)=Y(\theta)c_n.\] It exists since \(T(\theta)f_n=f_mS(\theta)\). Uniqueness proves the identity and composition laws, so \(f^*Y\) is a simplicial object and \(c\) is a simplicial arrow. The same argument lifts every arrow to \(Y\) over \(f\circ u\) uniquely through \(c\) over \(u\), proving that \(c\) is strongly cartesian in the simplicial category.

Given an \(f\)-relative homotopy \(h\) between \(a\) and \(b\), lift each component uniquely through \(c_n\) over \(\mathrm{id}_{S_n}\): \[c_n\widetilde h_{n,\alpha}=h_{n,\alpha}.\] To check naturality, compose \((f^*Y)(\theta)\widetilde h_{n,\alpha}\) and \(\widetilde h_{m,\alpha\theta}X(\theta)\) with \(c_m\). Their composites agree by the naturality of \(h\); they lie over the same map \(S(\theta)\), so strong cartesianness of \(c_m\) gives equality. Uniqueness also identifies the endpoints with \(\widetilde a,\widetilde b\). Conversely, composition with \(c\) gives a relative homotopy. These two constructions are inverse, proving (1).

For (2), choose lifts \(c_X:g^*X\to X\) and \(c_Y:g^*Y\to Y\). An \(S\)-relative homotopy \(h\) between arrows \(X\to Y\) gives an \(S'\)-relative homotopy by the unique componentwise solutions of \[(c_Y)_n(g^*h)_{n,\alpha}=h_{n,\alpha}(c_X)_n.\] The naturality and endpoint checks are the same as in (1). Pullback therefore preserves elementary homotopies and their equivalence closure, and hence induces the displayed functor. Different choices of lifts have unique compatible isomorphisms. Those isomorphisms identify the lifted homotopies by uniqueness as well. A composite of strongly cartesian arrows is strongly cartesian; comparing a successive lift with a lift of the composite base map thus gives the canonical comparison isomorphism. Identity and associativity constraints follow because both routes are isomorphisms with the same composite to the original object over the same base map. They remain equal after passing to the quotient.

Finally, take an arrow \(Z\to Y\) over \(f\circ u\), where \(pZ=R\) and \(u:R\to S\). The strong cartesian property of \(c\) gives its unique lift \(Z\to f^*Y\) over \(u\). The componentwise argument of (1), now with \(u_n\) in place of \(\mathrm{id}_{S_n}\), gives a bijection on the corresponding sets of elementary relative homotopies. Every finite chain of such homotopies also lifts uniquely, and composition preserves every such chain. Consequently the bijection on arrows induces a bijection on relative homotopy classes over each \(u\). This is exactly the strongly cartesian universal property in the quotient, proving (3).

Remark

A functor over \(\mathcal B\) always induces the quotient functor of the first lemma. If both categories over \(\mathcal B\) are fibred and the functor preserves strongly cartesian arrows, the images of the chosen lifts provide the canonical comparison with base change, and the induced quotient functor preserves strongly cartesian arrows. Indeed, every strongly cartesian arrow in the quotient is a chosen componentwise strongly cartesian lift composed with a vertical isomorphism. The quotient functor preserves this isomorphism as well as the chosen lift. The latter claim is not asserted for an arbitrary functor over the base. For a cofibred category and \(g:S\to S'\), choose componentwise strongly cocartesian lifts \(d_X:X\to g_!X\). A relative homotopy \(h:X\to Y\) over \(S\) lifts by the uniquely determined equations \[(g_!h)_{n,\alpha}(d_X)_n=(d_Y)_nh_{n,\alpha}.\] Uniqueness gives naturality and the endpoints. The same equations with an arbitrary subsequent base map prove the quotient cocartesian universal property, just as in the preceding proof. Thus the quotient is cofibred and \(g_!\) points from the fibre over \(S\) to the fibre over \(S'\).

Cosimplicial variants follow without new lifting assumptions. A cosimplicial object of \(\mathcal E\) is a simplicial object of \(\mathcal E^{\mathrm{op}}\) after taking opposites, and this correspondence reverses morphisms. Use the dual homotopy notion of Lemma 019X. Passing to opposites changes strongly cartesian arrows into strongly cocartesian arrows and conversely. Apply the corresponding simplicial result over \(\mathcal B^{\mathrm{op}}\) and then take opposites again. This produces the relative cosimplicial quotient, its base-change functors, and their coherence, with fibred and cofibred hypotheses in their original directions.

Definition

Let \(\mathcal A\) be additive. An \(n\)-complex is a \(\mathbb Z^n\)-graded object \(L\) with maps \(d_i:L^{\mathbf p}\to L^{\mathbf p+\mathbf e_i}\) satisfying \(d_i^2=0\) and \(d_id_j=d_jd_i\). Suppose that, for every \(r\), only finitely many \(L^{\mathbf p}\) with \(|\mathbf p|=r\) are nonzero, or, more generally, that each indicated coproduct exists. Its signed total is \[\operatorname{Tot}(L)^r=\bigoplus_{|\mathbf p|=r}L^{\mathbf p}, \qquad d|_{L^{\mathbf p}}=\sum_{j=1}^n (-1)^{p_1+\cdots+p_{j-1}}d_j.\] For a chain \(n\)-complex \(L_{\mathbf p}\), the differentials lower the corresponding index. Use \(\operatorname{Tot}(L)_r=\bigoplus_{|\mathbf p|=r} L_{\mathbf p}\) with the same prefix signs. Equivalently, apply the cochain convention to \(L^{\mathbf p}=L_{-\mathbf p}\). Regrouping statements below assume the intermediate coproducts exist; this is automatic for first-quadrant chain multicomplexes.

Lemma

The signed maps in Definition simplicial-definition-illusie-I-signed-total square to zero. Totalization is functorial in maps of \(n\)-complexes and is canonically unchanged by grouping adjacent coordinates and totalizing again, whenever these totals exist.

Proof

On \(L^{\mathbf p}\), the two terms involving \(d_jd_k\) with \(j<k\) have opposite signs: after the \(j\)th differential the prefix exponent for the \(k\)th term has gained one. They therefore cancel because \(d_jd_k=d_kd_j\). The terms \(d_j^2\) vanish. The same calculation proves functoriality. For a grouping, the exponent contributed by the outer differential is the sum of the degrees preceding the grouped block, while the inner exponent is the sum inside it; their sum is exactly the displayed prefix exponent. The direct sum indexing is the same, so the two totalizations agree canonically.

Lemma

Let \(X\) be an \(n\)-simplicial object of an additive category \(\mathcal A\). Put \(\widetilde X_{\mathbf p}=X_{\mathbf p}\) for \(\mathbf p\geq0\) and zero otherwise, and let the \(i\)th differential be the alternating sum of the faces in the \(i\)th variable. Then \(\widetilde X\) is an \(n\)-complex, and every total degree has only finitely many summands. If \(\mathcal A\) is abelian, this construction is an exact additive functor in each simplicial variable and its signed total is exact.

Proof

The simplicial identities give \(d_i^2=0\) in each variable. Faces in different variables commute, so the \(d_i\) commute. In total degree \(r\) the indices are \(n\)-tuples of nonnegative integers summing to \(r\), a finite set; thus only finite biproducts are used. Exactness follows degreewise from exactness of finite biproducts in an abelian category and from the usual alternating-face complex in each variable (Stacks Project, Tags 0194 and 0195). The final assertion is the resulting finite total complex.

Definition

For a bisimplicial object \(X\) in an additive category, put \(T_n(X)=\bigoplus_{p+q=n}X_{p,q}\), with differential \(\partial_h+(-1)^p\partial_v\) on \(X_{p,q}\), and put \(C_n(X)=X_{n,n}\) with its diagonal alternating-face differential. For a lattice path \(w\) from \((0,0)\) to \((p,q)\), let \(\alpha_w(k)\) and \(\beta_w(k)\) count its horizontal and vertical steps among the first \(k\) steps. These give ordinal maps \([p+q]\to[p]\) and \([p+q]\to[q]\). Let \(I(w)\) count pairs consisting of a vertical step preceding a horizontal step. Define \[\operatorname{Sh}_{p,q}=\sum_w(-1)^{I(w)}X(\alpha_w,\beta_w): X_{p,q}\longrightarrow X_{p+q,p+q}.\] For \(p+q=n\) define \(j_p:[p]\to[n]\) by \(i\mapsto i\) and \(k_q:[q]\to[n]\) by \(i\mapsto p+i\). The \((p,q)\) component of the other map is \[\operatorname{AW}_{p,q}=X(j_p,k_q):X_{n,n}\longrightarrow X_{p,q}.\] Thus \(\operatorname{Sh}:T(X)\to C(X)\) and \(\operatorname{AW}:C(X)\to T(X)\). The ordinal-map notation fixes both the degeneracy indexing and the contravariant order of composition.

Theorem

The maps in Definition simplicial-definition-illusie-I-shuffle-aw are natural chain maps and are the identity in degree zero. Both composites are naturally chain-homotopic to the appropriate identity. The shuffle maps are associative and the Alexander–Whitney maps are coassociative for three simplicial variables in a fixed order, under the canonical identifications of adjacent chain totalizations.

Proof

We first check chain-map identities as identities between finite integral sums of simplex operators. On a pair of strings of vertices the Alexander–Whitney formula reads \[\sum_{p=0}^n[a_0,\ldots,a_p]\otimes[b_p,\ldots,b_n].\] Deleting a horizontal vertex strictly before the cut or a vertical vertex strictly after it gives exactly the terms of \(\operatorname{AW}\partial\). The remaining terms at consecutive cuts are \([a_0,\ldots,a_{p-1}]\otimes[b_p,\ldots,b_n]\) with coefficients \((-1)^p\) and \((-1)^{p-1}\), so cancel.

For shuffle, deleting an interior vertex between unlike steps yields the same shortcut from the two paths interchanging those steps. Their inversion parities differ by one, so these terms cancel. At a vertex \((a,b)\) between two horizontal steps, the deletion has boundary sign \((-1)^{a+b}\) and removes \(b\) inversions. Its total sign is therefore \((-1)^{I(w')+a}\), the horizontal source-boundary sign. Between two vertical steps it removes \(p-a\) inversions, giving \((-1)^{I(w')+p+b}\), the signed vertical source-boundary sign. Endpoint deletions obey the same formulas. These account for every boundary term, proving both chain-map identities.

We give recursive, finite formulas for the homotopies. Use the free bisimplicial abelian-group models \[M^{p,q}_{a,b}=\mathbb Z[ \operatorname{Hom}_\Delta([a],[p])\times \operatorname{Hom}_\Delta([b],[q])].\] A basis of \(C(M^{p,q})_n\) consists of weakly increasing vertex strings in \([p]\times[q]\). Prepending \((0,0)\) defines \(K^\Delta\) of degree \(1\). If \(e^\Delta\) sends every vertex to \((0,0)\) in degree zero and is zero in positive degrees, cancellation of alternating faces gives \(\partial K^\Delta+K^\Delta\partial=1-e^\Delta\).

For \(T(M^{p,q})\), write \(P(t)\) for the unnormalized chains on \(\mathbb Z[\Delta[t]]\). Then \(T(M^{p,q})=P(p)\otimes P(q)\). Prepending \(0\) gives \(h_t\) on \(P(t)\) with \(\partial h_t+h_t\partial=1-e_t\), where \(e_t\) sends all degree-zero vertices to \(0\) and vanishes in higher degrees. For \(x\) of degree \(a\) put \[K^T(x\otimes y)=h_p(x)\otimes y+(-1)^a e_p(x)\otimes h_q(y).\] The second summand is zero unless \(a=0\). The tensor differential gives \(\partial K^T+K^T\partial=1-e_p\otimes e_q\). In particular each positive-degree cycle in either model has the specified primitive \(K\) of that cycle.

Apply the following recursion separately to \[L^C=\operatorname{Sh}\operatorname{AW}-1, \qquad L^T=\operatorname{AW}\operatorname{Sh}-1.\] Set \(H_{-1}=H_0=0\); both \(L\) vanish in degree zero. Inductively, use the universal generator \(z=(1,1)\) of \(C_n(M^{n,n})\) in the first case, and that of the \((p,q)\) summand of \(T_n(M^{p,q})\), \(p+q=n\), in the second. For \(n>0\) put \[r=L_n(z)-H_{n-1}(\partial z),\qquad H_n(z)=K(r).\] Since \(L\) is a chain map and the homotopy equation holds one degree lower, \(\partial r=L_{n-1}(\partial z)-\partial H_{n-1}(\partial z)=0\). Thus \(\partial H_n(z)+H_{n-1}(\partial z)=L_n(z)\).

The universal value just constructed is a finite integral combination of pairs of ordinal maps. Evaluate those maps through \(X\) to define \(H_n(X)\). This makes \(H_n\) natural in \(X\). Equality on the free universal generators proves the homotopy equation for these formal operators, and hence in any additive category. Free abelian groups and tensor products were used only to compute the universal formulas, not as hypotheses on that category. The contractions need not be natural in the model indices.

Finally both iterated shuffles enumerate exactly the words having the prescribed numbers of steps of each of three types. In both orders their sign is the inversion sign of that word. Both iterated Alexander–Whitney maps restrict to \([0,p]\), \([p,p+q]\), and \([p+q,p+q+r]\) in the respective coordinates. This proves associativity and coassociativity. Iteration also supplies the natural total-to-diagonal homotopy equivalence for any finite number of simplicial variables.

Remark

The composites above need not equal the identity before normalization. For the constant bisimplicial abelian group \(\mathbb Z\), degree one gives \(T_1=\mathbb Z^2\), \(C_1=\mathbb Z\), \(\operatorname{Sh}_1=(1\ \ 1)\) and \(\operatorname{AW}_1=(1\ \ 1)^{\mathsf t}\). Thus \(\operatorname{AW}_1\operatorname{Sh}_1\) is the all-ones matrix. This distinction is already respected by Illusie’s statement on printed p.7, which asserts inverseness up to functorial homotopy.

Theorem

Let \(\mathcal A\) be abelian. For every \(n\geq1\), normalization in the \(n\) directions gives an equivalence from \(n\)-simplicial objects to commuting \(n\)-fold chain complexes concentrated in nonnegative degrees in each direction. Iterated Dold–Kan inverse functors give a quasi-inverse. The successive orders are canonically isomorphic after the corresponding reindexing of coordinates. For the fixed coordinate order, grouping adjacent directions of the resulting chain complex commutes with signed totalization.

In Illusie’s Moore convention this normal object has multidegree \[(NX)_{\mathbf p} =\bigcap_{j=1}^n\ \bigcap_{i=1}^{p_j}\ker(d_i^{(j)}), \qquad \partial_j=d_0^{(j)}.\] Let \(DX\) be the sum of images of all degeneracies in any direction. There is a natural splitting of multicomplexes \(\widetilde X=NX\oplus DX\). For one direction the two Moore conventions used here and earlier in this chapter are compared through the same quotient \(\widetilde X/DX\), not by a reversal morphism in \(\Delta\).

Proof

For one direction put \(D_m=\sum_{i=0}^{m-1}\operatorname{Im}s_i\) and \(N^+_m=\bigcap_{i=1}^m\ker d_i\). On \(X_m\) use the projection \[P_m=(1-s_0d_1)(1-s_1d_2)\cdots(1-s_{m-1}d_m),\] with the rightmost factor applied first and \(P_0=1\). Successive factors kill the faces in descending index order without restoring those already killed. Use the face identity \[d_i d_j=d_{j-1}d_i\quad(i<j)\] and the face-degeneracy identities. The same identities give \(P_ms_i=0\) for every \(i<m\). Moreover \(x-P_mx\) is a sum of degeneracies and \(P_m\) is the identity on \(N^+_m\). These identities, interpreted as morphism identities in \(\mathcal A\), show \(X_m=N^+_m\oplus D_m\) and identify \(N^+_m\) with \(X_m/D_m\). Both summands are subcomplexes: on \(N^+\) the alternating boundary is \(d_0\), and on degeneracies the two identity face terms cancel. Hence the quotient identification is one of chain complexes.

The earlier normal functor \(N^-_m=\bigcap_{i<m}\ker d_i\), with differential \((-1)^m d_m\), is also canonically isomorphic to this quotient by Lemma 019A. Thus the one-variable Dold–Kan equivalence of Theorem 019G supplies the equivalence for the stated convention.

Apply that equivalence first in the abelian functor category on the remaining simplicial directions, and subsequently in the abelian categories of chain complexes already obtained. This proves the iterated equivalence. The projection operators in distinct directions commute, since every face and degeneracy in one direction commutes with every operator in the others. Their product has image the displayed intersection and kernel the sum of their kernels, namely \(DX\). This proves the splitting and canonical order-independence for normalization. The inverse functors have degreewise finite sums indexed by tuples of ordinal surjections; exchanging the order of these finite sums gives their canonical comparisons and the one-variable units and counits give the iterated units and counits.

Finally the prefix sign of a grouped block plus the prefix sign within that block equals the original prefix sign. The biproduct summands agree as well, proving the assertion about adjacent chain totals by Lemma simplicial-lemma-illusie-I-signed-total-differential. This statement does not identify normalization of a simplicial diagonal with a chain total.

Localization and simplicial derived categories

We use small categories, or a fixed universe enlargement in which all localizations below exist. A right multiplicative system and its roofs are as in Tag 04VC and Tag 04VH. The following statements spell out the generality and the fixed degree bound used in Illusie I.1.4; neither requires enough injectives or projectives.

Lemma

Let \(\mathcal C\) be small and \(S\) a collection of its arrows. There is a functor \(Q:\mathcal C\to\mathcal C[S^{-1}]\), identical on objects, universal for functors sending \(S\) to isomorphisms. Natural transformations between such functors extend uniquely as well.

Suppose now that \(S\) is a right multiplicative system and \(F:\mathcal C\to\mathcal B\) is any functor. If for every \(X\) the pro-object \((F(X'))_{s:X'\to X\text{ in }S}\) is isomorphic to a constant object \(L_X\), then these objects define \(LF:\mathcal C[S^{-1}]\to\mathcal B\) and a natural map \(u:LF\circ Q\to F\) such that \[\operatorname{Nat}(G,LF)\longrightarrow \operatorname{Nat}(G\circ Q,F),\qquad a\longmapsto u\circ(aQ),\] is bijective for every \(G:\mathcal C[S^{-1}]\to\mathcal B\). The dual assertion holds for a left multiplicative system and essentially constant ind-objects: \(F\to RF\circ Q\) is then universal in the opposite direction. No additive or triangulated structure is assumed here.

Proof

Adjoin a formal reverse edge \(s^{-1}\) for each \(s\in S\) to the underlying graph of \(\mathcal C\). Take finite paths and quotient by the congruence generated by composition and identities in \(\mathcal C\), and by \(s^{-1}s=1\) and \(ss^{-1}=1\). This constructs the small category \(\mathcal C[S^{-1}]\). The image of every new edge is forced for a functor inverting \(S\), which proves existence and uniqueness of its extension. Naturality for an invertible arrow implies naturality for its inverse; hence transformations extend uniquely too.

For the second assertion write \(I_X=S/X\), with arbitrary commuting morphisms over \(X\). This category is cofiltered: \(1_X\) gives an object, right Ore squares give common predecessors, and right cancellation equalizes parallel arrows \(a,b\) because \(s''a=s''b\) with \(s''\in S\). Let \(J_X=(QX\downarrow Q)\). An object of \(J_X\) is a pair \((Y,a:QX\to QY)\). The functor \[I_X\longrightarrow J_X,\qquad (s:X'\to X)\longmapsto (X',Q(s)^{-1})\] is initial. Indeed, objects of its comma category over \((Y,a)\) are precisely roofs \(X\xleftarrow{s}X'\xrightarrow{b}Y\) representing \(a\). There is at least one such roof by right fractions. Any two roofs have a common refinement giving the same numerator and denominator: first use the Ore square for the denominators and then the equality criterion for right fractions (Tag 04VJ). Thus this comma category is nonempty and connected, proving initiality. Equivalently, a compatible cone on \(F|_{I_X}\) extends to \(F|_{J_X}\) by choosing a roof for \(a\) and composing its cone component with \(F(b)\); common refinements prove independence and compatibility.

An essentially constant pro-object has its representing object as a limit in \(\mathcal B\). Explicitly, full faithfulness of the constant embedding in the pro-category gives, for every \(B\), \[\operatorname{Hom}(B,L_X) =\varprojlim_{I_X}\operatorname{Hom}(B,F(X')),\] with compatible identifications. Initiality therefore makes \(L_X\) a limit of \(F\) over \(J_X\). Precomposition by an arrow \(QX\to QY\) induces \(J_Y\to J_X\) and hence \(L_X\to L_Y\). The uniqueness of limit maps proves functoriality. Evaluation of the universal cone at \((X,1)\) defines \(u_X\).

For \(\eta:GQ\to F\), the arrows \(G(QX)\xrightarrow{G(a)}G(QY)\xrightarrow{\eta_Y}F(Y)\) form a cone on \(J_X\). Its unique factorization through \(L_X\) defines \(a_X\); the same uniqueness proves naturality and \(u\circ(aQ)=\eta\). Conversely, this equation forces every component of the cone and hence forces \(a_X\). This proves the stated bijection. Passing to opposite categories proves the final assertion.

Lemma

Let \(\mathcal A\) be abelian. Write \(C_{\geq0}(\mathcal A)\) for chain complexes concentrated in nonnegative degrees and \(W\) for quasi-isomorphisms. In \(\mathrm{Simp}(\mathcal A)\), let \(W_s\) be the maps inducing quasi-isomorphisms of associated chain complexes. Normalization induces equivalences \[\mathrm{Simp}(\mathcal A)[W_s^{-1}] \simeq C_{\geq0}(\mathcal A)[W^{-1}] \simeq D^{\leq0}(\mathcal A),\] where \(D^{\leq0}(\mathcal A)\) is the full subcategory of \(D(\mathcal A)\) with \(H^i=0\) for \(i>0\), and chains are converted to cochains by \(C^i=C_{-i}\). Both localizations may first be factored through their respective homotopy categories. The induced equivalences commute with these quotient functors. Quasi-isomorphisms in \(K^{\leq0}(\mathcal A)\) admit both calculi of fractions.

Proof

The natural quasi-isomorphism \(N(X)\to s(X)\) of Lemma 019C shows that \(W_s\) is exactly the class detected by \(N\). By Theorem 019G, \(N\) and its inverse carry the respective classes to each other. Their units and counits are isomorphisms, and descend along localization by the preceding lemma. They remain quasi-inverse, proving the first equivalence.

For clarity the homotopy factorization does not require a calculus of fractions in the category of complexes itself. For a cochain complex \(C\) concentrated in degrees at most zero put \(E^n=C^n\oplus C^{n+1}\), with differential \((a,b)\mapsto(d a+b,-d b)\). The maps \(j:C\to E\), \(j(a)=(a,0)\), and \(t:E^n\to E^{n-1}\), \(t(a,b)=(0,a)\), satisfy \(dt+td=1\). Thus \(E\) is contractible and still concentrated in degrees at most zero. Set \(\operatorname{Cyl}(C)=C\oplus E\), with \(i_0(x)=(x,0)\), \(i_1(x)=(x,jx)\), and projection \(p\) to \(C\). Since \(p\) is a homotopy equivalence and \(pi_0=pi_1=1\), any functor inverting homotopy equivalences identifies \(i_0\) and \(i_1\). If \(f-g=dh+hd\), the chain map \(E\to D\) given by \((a,b)\mapsto(f-g)(a)+h(b)\), together with \(g:C\to D\), extends \(g,f\) along \(i_0,i_1\). Hence that functor identifies \(f\) and \(g\). Conversely a functor identifying homotopic maps inverts homotopy equivalences. This proves that passage to the homotopy category is such a localization. Dold–Kan transports the argument to simplicial objects by Lemmas 019S and 01A4. Since homotopy equivalences are quasi-isomorphisms, the asserted factorizations and commuting comparisons follow.

Let \(K^{\leq0}\) be the full homotopy subcategory on complexes concentrated in degrees at most zero. In \(K(\mathcal A)\), quasi-isomorphisms form a bilateral multiplicative system (Tag 05RT). Canonical truncation \(\tau_{\leq0}\) is a functor on homotopy categories: restrict a homotopy in degree zero to \(\ker d^0\) and leave its lower components unchanged. It fixes \(K^{\leq0}\) and preserves quasi-isomorphisms. Apply this functor to the Ore and cancellation diagrams in \(K(\mathcal A)\). All their equalities remain equalities in \(K^{\leq0}\) and all required denominators remain quasi-isomorphisms. This proves the bilateral calculus claimed above without supposing \(K^{\leq0}\) is triangulated.

Every morphism \(C\to D\) in \(D(\mathcal A)\) between such complexes has a roof \(C\xleftarrow{w}Z\xrightarrow{f}D\). Since \(w\) is a quasi-isomorphism, \(H^{>0}(Z)=0\), and \(\tau_{\leq0}Z\to Z\) is a quasi-isomorphism. Truncating the roof represents it inside \(K^{\leq0}[W^{-1}]\), proving fullness. If two roofs there represent the same arrow in \(D(\mathcal A)\), a common refinement in \(K(\mathcal A)\) witnesses equality. Its middle complex also has vanishing positive cohomology. Truncating that refinement witnesses the same equality in \(K^{\leq0}[W^{-1}]\), proving faithfulness. Finally any \(Z\) with \(H^{>0}(Z)=0\) is isomorphic to \(\tau_{\leq0}Z\), proving that the essential image is exactly \(D^{\leq0}\). This proves the second equivalence.

Homotopies in abelian categories

Let \(\mathcal{A}\) be an additive category. Let \(U\), \(V\) be simplicial objects of \(\mathcal{A}\). Let \(a, b : U \to V\) be morphisms. Further, suppose that \(h : U \times \Delta[1] \to V\) is a homotopy from \(a\) to \(b\). Let us prove the two morphisms of chain complexes \(s(a), s(b) : s(U) \longrightarrow s(V)\) are homotopic in the sense of Homology, Section 010V. Using the notation introduced in Section 019J we define \[s(h)_n : U_n \longrightarrow V_{n + 1}\] by the formula [019R]\[\begin{equation} s(h)_n = \sum\nolimits_{i = 0}^n (-1)^{i + 1} h_{n + 1, i + 1} \circ s^n_i. \end{equation}\] Let us compute \(d_{n + 1} \circ s(h)_n + s(h)_{n - 1} \circ d_n\). We first compute \[\begin{eqnarray*} d_{n + 1} \circ s(h)_n & = & \sum\nolimits_{j = 0}^{n + 1} \sum\nolimits_{i = 0}^n (-1)^{j + i + 1} d^{n + 1}_j \circ h_{n + 1, i + 1} \circ s^n_i \\ & = & \sum\nolimits_{1 \leq i + 1 \leq j \leq n + 1} (-1)^{j + i + 1} h_{n, i + 1} \circ d^{n + 1}_j \circ s^n_i \\ & & + \sum\nolimits_{n \geq i \geq j \geq 0} (-1)^{i + j + 1} h_{n, i} \circ d^{n + 1}_j \circ s^n_i \\ & = & \sum\nolimits_{1 \leq i + 1 < j \leq n + 1} (-1)^{j + i + 1} h_{n, i + 1} \circ s^{n - 1}_i \circ d^n_{j - 1} \\ & & + \sum\nolimits_{1 \leq i + 1 = j \leq n + 1} (-1)^{j + i + 1} h_{n, i + 1} \\ & & + \sum\nolimits_{n \geq i = j \geq 0} (-1)^{i + j + 1} h_{n, i} \\ & & + \sum\nolimits_{n \geq i > j \geq 0} (-1)^{i + j + 1} h_{n, i} \circ s^{n - 1}_{i - 1} \circ d^n_j \end{eqnarray*}\] We leave it to the reader to see that the first and the last of the four sums cancel exactly against all the terms of \[s(h)_{n - 1} \circ d_n = \sum_{i = 0}^{n - 1} \sum_{j = 0}^n (-1)^{i + 1 + j} h_{n, i + 1} \circ s^{n - 1}_i \circ d^n_j.\] Hence we obtain \[\begin{eqnarray*} d_{n + 1} \circ s(h)_n + s(h)_{n - 1} \circ d_n & = & \sum_{j = 1}^{n + 1} (-1)^{2j} h_{n, j} + \sum_{i = 0}^n (-1)^{2i + 1} h_{n, i} \\ & = & h_{n, n + 1} - h_{n , 0} \\ & = & a_n - b_n \end{eqnarray*}\] as desired.

Lemma

Let \(\mathcal{A}\) be an additive category. Let \(a, b : U \to V\) be morphisms of simplicial objects of \(\mathcal{A}\). If \(a\), \(b\) are homotopic, then \(s(a), s(b) : s(U) \to s(V)\) are homotopic maps of chain complexes. If \(\mathcal{A}\) is abelian, then also \(N(a), N(b) : N(U) \to N(V)\) are homotopic maps of chain complexes.

Proof

We may choose a sequence \(a = a_0, a_1, \ldots, a_n = b\) of morphisms from \(U\) to \(V\) such that for each \(i = 1, \ldots, n\) either there is a homotopy from \(a_i\) to \(a_{i - 1}\) or there is a homotopy from \(a_{i - 1}\) to \(a_i\). The calculation above shows that in this case either \(s(a_i)\) is homotopic to \(s(a_{i - 1})\) as a map of chain complexes or \(s(a_{i - 1})\) is homotopic to \(s(a_i)\) as a map of chain complexes. Of course, these things are equivalent and moreover being homotopic is an equivalence relation on the set of maps of chain complexes, see Homology, Section 010V. This proves that \(s(a)\) and \(s(b)\) are homotopic as maps of chain complexes.

Next, we turn to \(N(a)\) and \(N(b)\). It follows from Lemma 019A that \(N(a)\), \(N(b)\) are compositions \[N(U) \to s(U) \to s(V) \to N(V)\] where we use \(s(a)\), \(s(b)\) in the middle. Hence the assertion follows from Homology, Lemma 010W.

Lemma

Let \(\mathcal{A}\) be an additive category. Let \(a : U \to V\) be a morphism of simplicial objects of \(\mathcal{A}\). If \(a\) is a homotopy equivalence, then \(s(a) : s(U) \to s(V)\) is a homotopy equivalence of chain complexes. If in addition \(\mathcal{A}\) is abelian, then also \(N(a) : N(U) \to N(V)\) is a homotopy equivalence of chain complexes.

Proof

Omitted. See Lemma 019S above.

Homotopies and cosimplicial objects

Let \(\mathcal{C}\) be a category with finite products. Let \(V\) be a cosimplicial object and consider \(\Hom(\Delta[1], V)\), see Section 07K9. The morphisms \(e_0, e_1 : \Delta[0] \to \Delta[1]\) produce two morphisms \(e_0, e_1 : \Hom(\Delta[1], V) \to V\).

Definition

Let \(\mathcal{C}\) be a category having finite products. Let \(U\) and \(V\) be two cosimplicial objects of \(\mathcal{C}\). Let \(a, b : U \to V\) be two morphisms of cosimplicial objects of \(\mathcal{C}\).

1. We say a morphism \[h : U \longrightarrow \Hom(\Delta[1], V)\] such that \(a = e_0 \circ h\) and \(b = e_1 \circ h\) is a homotopy from \(a\) to \(b\).

2. We say \(a\) and \(b\) are homotopic or are in the same homotopy class if there exists a sequence \(a = a_0, a_1, \ldots, a_n = b\) of morphisms from \(U\) to \(V\) such that for each \(i = 1, \ldots, n\) there either exists a homotopy from \(a_i\) to \(a_{i - 1}\) or there exists a homotopy from \(a_{i - 1}\) to \(a_i\).

This is dual to the notion we introduced for simplicial objects in Section 019J. To explain this, consider a homotopy \(h : U \to \Hom(\Delta[1], V)\) from \(a\) to \(b\) as in the definition. Recall that \(\Delta[1]_n\) is a finite set. The degree \(n\) component of \(h\) is a morphism \[h_n = (h_{n, \alpha}) : U \longrightarrow \Hom(\Delta[1], V)_n = \prod\nolimits_{\alpha \in \Delta[1]_n} V_n\] The morphisms \(h_{n, \alpha} : U_n \to V_n\) of \(\mathcal{C}\) have the property that for every morphism \(f : [n] \to [m]\) of \(\Delta\) we have [07KB]\[\begin{equation} h_{m, \alpha} \circ U(f) = V(f) \circ h_{n, \alpha \circ f} \end{equation}\] Moreover, the condition that \(a = e_0 \circ h\) means that \(a_n = h_{n, 0 : [n] \to [1]}\) where \(0 : [n] \to [1]\) is the constant map with value \(0\). Similarly, the condition that \(b = e_1 \circ h\) means that \(b_n = h_{n, 1 : [n] \to [1]}\) where \(1 : [n] \to [1]\) is the constant map with value \(1\). Conversely, given a family of morphisms \(\{h_{n, \alpha}\}\) such that (07KB) holds for all morphisms \(f\) of \(\Delta\) and such that \(a_n = h_{n, 0 : [n] \to [1]}\) and \(b_n = h_{n, 1 : [n] \to [1]}\) for all \(n \geq 0\), then we obtain a homotopy \(h\) from \(a\) to \(b\) by setting \(h = \prod_{\alpha \in \Delta[1]_n} h_{n, \alpha}\).

Remark

Let \(\mathcal{C}\) be any category (no assumptions whatsoever). Let \(U\) and \(V\) be cosimplicial objects of \(\mathcal{C}\). Let \(a, b : U \to V\) be morphisms of cosimplicial objects of \(\mathcal{C}\). A homotopy from \(a\) to \(b\) is given by morphisms \(h_{n, \alpha} : U_n \to V_n\), for \(n \geq 0\), \(\alpha \in \Delta[1]_n\) satisfying (07KB) for all morphisms \(f\) of \(\Delta\) and such that \(a_n = h_{n, 0 : [n] \to [1]}\) and \(b_n = h_{n, 1 : [n] \to [1]}\) for all \(n \geq 0\). As in Definition 019W we say the morphisms \(a\) and \(b\) are homotopic if there exists a sequence of morphisms \(a = a_0, a_1, \ldots, a_n = b\) from \(U\) to \(V\) such that for each \(i = 1, \ldots, n\) there either exists a homotopy from \(a_{i - 1}\) to \(a_i\) or there exists a homotopy from \(a_i\) to \(a_{i - 1}\). Clearly, if \(F : \mathcal{C} \to \mathcal{C}'\) is any functor and \(\{h_{n, i}\}\) is a homotopy from \(a\) to \(b\), then \(\{F(h_{n, i})\}\) is a homotopy from \(F(a)\) to \(F(b)\). Similarly, if \(a\) and \(b\) are homotopic, then \(F(a)\) and \(F(b)\) are homotopic. This new notion is the same as the old one in case finite products exist. We deduce in particular that functors preserve the original notion whenever both categories have finite products.

Lemma

Let \(\mathcal{C}\) be a category. Suppose that \(U\) and \(V\) are two cosimplicial objects of \(\mathcal{C}\). Let \(a, b : U \to V\) be morphisms of cosimplicial objects. Recall that \(U\), \(V\) correspond to simplicial objects \(U'\), \(V'\) of \(\mathcal{C}^{opp}\). Moreover \(a, b\) correspond to morphisms \(a', b' : V' \to U'\). The following are equivalent

1. There exists a homotopy \(h = \{h_{n, \alpha}\}\) from \(a\) to \(b\) as in Remark 0FKJ.

2. There exists a homotopy \(h = \{h_{n, i}\}\) from \(a'\) to \(b'\) as in Remark 019M.

Thus \(a\) is homotopic to \(b\) as in Remark 0FKJ if and only if \(a'\) is homotopic to \(b'\) as in Remark 019M.

Proof

In case \(\mathcal{C}\) has finite products, then \(\mathcal{C}^{opp}\) has finite coproducts and we may use Definitions 019W and 019K instead of Remarks 0FKJ and 019M. In this case \(h : U \to \Hom(\Delta[1], V)\) is the same as a morphism \(h' : \Hom(\Delta[1], V)' \to U'\). Since products and coproducts get switched too, it is immediate that \((\Hom(\Delta[1], V))' = V' \times \Delta[1]\). Moreover, the “primed” version of the morphisms \(e_0, e_1 : \Hom(\Delta[1], V) \to V\) are the morphisms \(e_0, e_1 : V' \to \Delta[1] \times V\). Thus \(e_0 \circ h = a\) translates into \(h' \circ e_0 = a'\) and similarly \(e_1 \circ h = b\) translates into \(h' \circ e_1 = b'\). This proves the lemma in this case.

In the general case, one needs to translate the relations given by (07KB) into the relations given in Lemma 019L. We omit the details.

The final assertion is formal from the equivalence of (1) and (2).

Lemma

Let \(\mathcal{C}, \mathcal{C}', \mathcal{D}, \mathcal{D}'\) be categories. With terminology as in Remarks 0FKJ and 019M.

1. Let \(a, b : U \to V\) be morphisms of simplicial objects of \(\mathcal{D}\). Let \(F : \mathcal{D} \to \mathcal{D}'\) be a covariant functor. If \(a\) and \(b\) are homotopic, then \(F(a)\), \(F(b)\) are homotopic morphisms \(F(U) \to F(V)\) of simplicial objects.

2. Let \(a, b : U \to V\) be morphisms of cosimplicial objects of \(\mathcal{C}\). Let \(F : \mathcal{C} \to \mathcal{C}'\) be a covariant functor. If \(a\) and \(b\) are homotopic, then \(F(a)\), \(F(b)\) are homotopic morphisms \(F(U) \to F(V)\) of cosimplicial objects.

3. Let \(a, b : U \to V\) be morphisms of simplicial objects of \(\mathcal{D}\). Let \(F : \mathcal{D} \to \mathcal{C}\) be a contravariant functor. If \(a\) and \(b\) are homotopic, then \(F(a)\), \(F(b)\) are homotopic morphisms \(F(V) \to F(U)\) of cosimplicial objects.

4. Let \(a, b : U \to V\) be morphisms of cosimplicial objects of \(\mathcal{C}\). Let \(F : \mathcal{C} \to \mathcal{D}\) be a contravariant functor. If \(a\) and \(b\) are homotopic, then \(F(a)\), \(F(b)\) are homotopic morphisms \(F(V) \to F(U)\) of simplicial objects.

Proof

By Lemma 019X above, we can turn \(F\) into a covariant functor between a pair of categories, and we have to show that the functor preserves homotopic pairs of maps. This is explained in Remark 019M.

Lemma

Let \(f : X \to Y\) be a morphism of a category \(\mathcal{C}\) with pushouts. Assume there is a morphism \(s : Y \to X\) with \(s \circ f = \text{id}_X\). Consider the cosimplicial object \(U\) constructed in Example 016N starting with \(f\). The morphism \(U \to U\) which in each degree is the self map of \(Y \amalg_X \ldots \amalg_X Y\) given by \(f \circ s\) on each factor is homotopic to the identity on \(U\). In particular, \(U\) is homotopy equivalent to the constant cosimplicial object \(X\).

Proof

This lemma is dual to Lemma 019P. Hence this lemma follows on applying Lemma 019X.

Lemma

Let \(\mathcal{A}\) be an additive category. Let \(a, b : U \to V\) be morphisms of cosimplicial objects of \(\mathcal{A}\). If \(a\), \(b\) are homotopic, then \(s(a), s(b) : s(U) \to s(V)\) are homotopic maps of cochain complexes. If in addition \(\mathcal{A}\) is abelian, then \(Q(a), Q(b) : Q(U) \to Q(V)\) are homotopic maps of cochain complexes.

Proof

Let \((-)' : \mathcal{A} \to \mathcal{A}^{opp}\) be the contravariant functor \(A \mapsto A\). By Lemma 019Z the maps \(a'\) and \(b'\) are homotopic. By Lemma 019S we see that \(s(a')\) and \(s(b')\) are homotopic maps of chain complexes. Since \(s(a') = (s(a))'\) and \(s(b') = (s(b))'\) we conclude that also \(s(a)\) and \(s(b)\) are homotopic by applying the additive contravariant functor \((-)'' : \mathcal{A}^{opp} \to \mathcal{A}\). The result for the \(Q\)-complexes follows in the same manner using that \(Q(U)' = N(U')\).

Lemma

Let \(\mathcal{A}\) be an additive category. Let \(a : U \to V\) be a morphism of cosimplicial objects of \(\mathcal{A}\). If \(a\) is a homotopy equivalence, then \(s(a) : s(U) \to s(V)\) is a homotopy equivalence of chain complexes. If in addition \(\mathcal{A}\) is abelian, then also \(Q(a) : Q(U) \to Q(V)\) is a homotopy equivalence of chain complexes.

Proof

Omitted. See Lemma 01A0 above.

More homotopies in abelian categories

Let \(\mathcal{A}\) be an abelian category. In this section we show that a homotopy between morphisms in \(\text{Ch}_{\geq 0}(\mathcal{A})\) always comes from a morphism \(U \times \Delta[1] \to V\) in the category of simplicial objects. In some sense this will provide a converse to Lemma 019S. We first develop some material on homotopies between morphisms of chain complexes.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(A\) be a chain complex. Consider the covariant functor \[B \longmapsto \{ (a, b, h) \mid a, b : A \to B\text{ and }h\text{ a homotopy between }a, b \}\] There exists a chain complex \(\diamond A\) such that \(\Mor_{\text{Ch}(\mathcal{A})}(\diamond A, -)\) is isomorphic to the displayed functor. The construction \(A \mapsto \diamond A\) is functorial.

Proof

We set \(\diamond A_n = A_n \oplus A_n \oplus A_{n - 1}\), and we define \(d_{\diamond A, n}\) by the matrix \[d_{\diamond A, n} = \left( \begin{matrix} d_{A, n} & 0 & \text{id}_{A_{n - 1}} \\ 0 & d_{A, n} & -\text{id}_{A_{n - 1}} \\ 0 & 0 & -d_{A, n - 1} \end{matrix} \right) : A_n \oplus A_n \oplus A_{n - 1} \to A_{n - 1} \oplus A_{n - 1} \oplus A_{n - 2}\] If \(\mathcal{A}\) is the category of abelian groups, and \((x, y, z) \in A_n \oplus A_n \oplus A_{n - 1}\) then \(d_{\diamond A, n}(x, y, z) = (d_n(x) + z, d_n(y) - z, -d_{n - 1}(z))\). It is easy to verify that \(d^2 = 0\). Clearly, there are two maps \(\diamond a, \diamond b : A \to \diamond A\) (first summand and second summand), and a map \(\diamond A \to A[-1]\) which give a short exact sequence \[0 \to A \oplus A \to \diamond A \to A[-1] \to 0\] which is termwise split. Moreover, there is a sequence of maps \(\diamond h_n : A_n \to \diamond A_{n + 1}\), namely the identity from \(A_n\) to the summand \(A_n\) of \(\diamond A_{n + 1}\), such that \(\diamond h\) is a homotopy between \(\diamond a\) and \(\diamond b\).

We conclude that any morphism \(f : \diamond A \to B\) gives rise to a triple \((a, b, h)\) by setting \(a = f \circ \diamond a\), \(b = f \circ \diamond b\) and \(h_n = f_{n + 1} \circ \diamond h_n\). Conversely, given a triple \((a, b, h)\) we get a morphism \(f : \diamond A \to B\) by taking \[f_n = (a_n, b_n, h_{n - 1}).\] To see that this is a morphism of chain complexes you have to do a calculation. We only do this in case \(\mathcal{A}\) is the category of abelian groups: Say \((x, y, z) \in \diamond A_n = A_n \oplus A_n \oplus A_{n - 1}\). Then \[\begin{eqnarray*} f_{n - 1}(d_n(x, y, z)) & = & f_{n - 1}(d_n(x) + z, d_n(y) - z, -d_{n - 1}(z)) \\ & = & a_n(d_n(x)) + a_n(z) + b_n(d_n(y)) - b_n(z) - h_{n - 2}(d_{n - 1}(z)) \end{eqnarray*}\] and \[\begin{eqnarray*} d_n(f_n(x, y, z) & = & d_n(a_n(x) + b_n(y) + h_{n - 1}(z)) \\ & = & d_n(a_n(x)) + d_n(b_n(y)) + d_n(h_{n - 1}(z)) \end{eqnarray*}\] which are the same by definition of a homotopy.

Note that the extension \[0 \to A \oplus A \to \diamond A \to A[-1] \to 0\] comes with sections of the morphisms \(\diamond A_n \to A[-1]_n\) with the property that the associated morphism \(\delta : A[-1] \to (A \oplus A)[-1]\), see Homology, Lemma 011D equals the morphism \((1, -1) : A[-1] \to A[-1] \oplus A[-1]\).

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \[0 \to A \oplus A \to B \to C \to 0\] be a short exact sequence of chain complexes of \(\mathcal{A}\). Suppose given in addition morphisms \(s_n : C_n \to B_n\) splitting the associated short exact sequence in degree \(n\). Let \(\delta(s) : C \to (A \oplus A)[-1] = A[-1] \oplus A[-1]\) be the associated morphism of complexes, see Homology, Lemma 011D. If \(\delta(s)\) factors through the morphism \((1, -1) : A[-1] \to A[-1] \oplus A[-1]\), then there is a unique morphism \(B \to \diamond A\) fitting into a commutative diagram \[\xymatrix{ 0 \ar[r] & A \oplus A \ar[d] \ar[r] & B \ar[r] \ar[d] & C \ar[d] \ar[r] & 0 \\ 0 \ar[r] & A \oplus A \ar[r] & \diamond A \ar[r] & A[-1] \ar[r] & 0 }\] where the vertical maps are compatible with the splittings \(s_n\) and the splittings of \(\diamond A_n \to A[-1]_n\) as well.

Proof

Denote \((p_n, q_n) : B_n \to A_n \oplus A_n\) the morphism \(\pi_n\) of Homology, Lemma 011D. Also write \((a, b) : A \oplus A \to B\), and \(r : B \to C\) for the maps in the short exact sequence. Write the factorization of \(\delta(s)\) as \(\delta(s) = (1, -1) \circ f\). This means that \(p_{n - 1} \circ d_{B, n} \circ s_n = f_n\), and \(q_{n - 1} \circ d_{B, n} \circ s_n = - f_n\), and Set \(B_n \to \diamond A_n = A_n \oplus A_n \oplus A_{n - 1}\) equal to \((p_n, q_n, f_n \circ r_n)\).

Now we have to check that this actually defines a morphism of complexes. We will only do this in the case of abelian groups. Pick \(x \in B_n\). Then \(x = a_n(x_1) + b_n(x_2) + s_n(x_3)\) and it suffices to show that our definition commutes with differential for each term separately. For the term \(a_n(x_1)\) we have \((p_n, q_n, f_n \circ r_n)(a_n(x_1)) = (x_1, 0, 0)\) and the result is obvious. Similarly for the term \(b_n(x_2)\). For the term \(s_n(x_3)\) we have \[\begin{eqnarray*} (p_n, q_n, f_n \circ r_n)(d_n(s_n(x_3))) & = & (p_n, q_n, f_n \circ r_n)( \\ & & \ \ \ \ \ a_n(f_n(x_3)) - b_n(f_n(x_3)) + s_n(d_n(x_3))) \\ & = & (f_n(x_3), -f_n(x_3), f_n(d_n(x_3))) \end{eqnarray*}\] by definition of \(f_n\). And \[\begin{eqnarray*} d_n(p_n, q_n, f_n \circ r_n)(s_n(x_3)) & = & d_n(0, 0, f_n(x_3)) \\ & = & (f_n(x_3), - f_n(x_3), d_{A[-1], n}(f_n(x_3))) \end{eqnarray*}\] The result follows as \(f\) is a morphism of complexes.

Lemma

Let \(\mathcal{A}\) be an abelian category. Let \(U\), \(V\) be simplicial objects of \(\mathcal{A}\). Let \(a, b : U \to V\) be a pair of morphisms. Assume the corresponding maps of chain complexes \(N(a), N(b) : N(U) \to N(V)\) are homotopic by a homotopy \(\{N_n : N(U)_n \to N(V)_{n + 1}\}\). Then there exists a homotopy from \(a\) to \(b\) as in Definition 019K. Moreover, one can choose the homotopy \(h : U \times \Delta[1] \to V\) such that \(N_n = N(h)_n\) where \(N(h)\) is the homotopy coming from \(h\) as in Section 019Q.

Proof

Let \((\diamond N(U), \diamond a, \diamond b, \diamond h)\) be as in Lemma 01A2 and its proof. By that lemma there exists a morphism \(\diamond N(U) \to N(V)\) representing the triple \((N(a), N(b), \{N_n\})\). We will show there exists a morphism \(\psi : N(U \times \Delta[1]) \to \diamond{N(U)}\) such that \(\diamond a = \psi \circ N(e_0)\), and \(\diamond b = \psi \circ N(e_1)\). Moreover, we will show that the homotopy between \(N(e_0), N(e_1) : N(U) \to N(U \times \Delta[1])\) coming from (019R) and Lemma 019S with \(h = \text{id}_{U \times \Delta[1]}\) is mapped via \(\psi\) to the canonical homotopy \(\diamond h\) between the two maps \(\diamond a, \diamond b : N(U) \to \diamond{N(U)}\). Certainly this will imply the lemma.

Note that \(N : \text{Simp}(\mathcal{A}) \to \text{Ch}_{\geq 0}(\mathcal{A})\) as a functor is a direct summand of the functor \(s : \text{Simp}(\mathcal{A}) \to \text{Ch}_{\geq 0}(\mathcal{A})\). Also, the functor \(\diamond\) is compatible with direct sums. Thus it suffices instead to construct a morphism \(\Psi : s(U \times \Delta[1]) \to \diamond{s(U)}\) with the corresponding properties. This is what we do below.

By Definition 019K the morphisms \(e_0 : U \to U \times \Delta[1]\) and \(e_1 : U \to U \times \Delta[1]\) are homotopic with homotopy \(\text{id}_{U \times \Delta[1]}\). By Lemma 019S we get an explicit homotopy \(\{h_n : s(U)_n \to s(U \times \Delta[1])_{n + 1}\}\) between the morphisms of chain complexes \(s(e_0) : s(U) \to s(U \times \Delta[1])\) and \(s(e_1) : s(U) \to s(U \times \Delta[1])\). By Lemma 01A3 above we get a corresponding morphism \[\Phi : \diamond{s(U)} \to s(U \times \Delta[1])\] According to the construction, \(\Phi_n\) restricted to the summand \(s(U)[-1]_n = s(U)_{n - 1}\) of \(\diamond{s(U)}_n\) is equal to \(h_{n - 1}\). And \[h_{n - 1} = \sum\nolimits_{i = 0}^{n - 1} (-1)^{i + 1} s^n_i \cdot \alpha^n_{i + 1} : U_{n - 1} \to \bigoplus\nolimits_j U_n \cdot \alpha^n_j.\] with obvious notation.

On the other hand, the morphisms \(e_i : U \to U \times \Delta[1]\) induce a morphism \((e_0, e_1) : U \oplus U \to U \times \Delta[1]\). Denote \(W\) the cokernel. Note that, if we write \((U \times \Delta[1])_n = \bigoplus_{\alpha : [n] \to [1]} U_n \cdot \alpha\), then we may identify \(W_n = \bigoplus_{i = 1}^n U_n \cdot \alpha^n_i\) with \(\alpha^n_i\) as in Section 019J. We have a commutative diagram \[\xymatrix{ 0 \ar[r] & U \oplus U \ar[rd]_{(1, 1)} \ar[r] & U \times \Delta[1] \ar[d]^\pi \ar[r] & W \ar[r] & 0 \\ & & U & & }\] This implies we have a similar commutative diagram after applying the functor \(s\). Next, we choose the splittings \(\sigma_n : s(W)_n \to s(U \times \Delta[1])_n\) by mapping the summand \(U_n \cdot \alpha^n_i \subset W_n\) via \((-1, 1)\) to the summands \(U_n \cdot \alpha^n_0 \oplus U_n \cdot \alpha^n_i \subset (U \times \Delta[1])_n\). Note that \(s(\pi)_n \circ \sigma_n = 0\). It follows that \((1, 1) \circ \delta(\sigma)_n = 0\). Hence \(\delta(\sigma)\) factors as in Lemma 01A3. By that lemma we obtain a canonical morphism \(\Psi : s(U \times \Delta[1]) \to \diamond{s(U)}\).

To compute \(\Psi\) we first compute the morphism \(\delta(\sigma) : s(W) \to s(U)[-1] \oplus s(U)[-1]\). According to Homology, Lemma 011D and its proof, to do this we have compute \[d_{s(U \times \delta[1]), n} \circ \sigma_n - \sigma_{n - 1} \circ d_{s(W), n}\] and write it as a morphism into \(U_{n - 1} \cdot \alpha^{n - 1}_0 \oplus U_{n - 1} \cdot \alpha^{n - 1}_n\). We only do this in case \(\mathcal{A}\) is the category of abelian groups. We use the short hand notation \(x_{\alpha}\) for \(x \in U_n\) to denote the element \(x\) in the summand \(U_n \cdot \alpha\) of \((U \times \Delta[1])_n\). Recall that \[d_{s(U \times \delta[1]), n} = \sum\nolimits_{i = 0}^n (-1)^i d^n_i\] where \(d^n_i\) maps the summand \(U_n \cdot \alpha\) to the summand \(U_{n - 1} \cdot (\alpha \circ \delta^n_i)\) via the morphism \(d^n_i\) of the simplicial object \(U\). In terms of the notation above this means \[d_{s(U \times \delta[1]), n}(x_\alpha) = \sum\nolimits_{i = 0}^n (-1)^i (d^n_i(x))_{\alpha \circ \delta^n_i}\] Starting with \(x_\alpha \in W_n\), in other words \(\alpha = \alpha^n_j\) for some \(j \in \{1, \ldots, n\}\), we see that \(\sigma_n(x_\alpha) = x_\alpha - x_{\alpha^n_0}\) and hence \[(d_{s(U \times \delta[1]), n} \circ \sigma_n)(x_\alpha) = \sum\nolimits_{i = 0}^n (-1)^i (d^n_i(x))_{\alpha \circ \delta^n_i} - \sum\nolimits_{i = 0}^n (-1)^i (d^n_i(x))_{\alpha^n_0 \circ \delta^n_i}\] To compute \(d_{s(W), n}(x_\alpha)\), we have to omit all terms where \(\alpha \circ \delta^n_i = \alpha^{n - 1}_0, \alpha^{n - 1}_n\). Hence we get \[\begin{matrix} (\sigma_{n - 1} \circ d_{s(W), n})(x_\alpha) = \\ \sum\nolimits_{i = 0, \ldots, n\text{ and } \alpha \circ \delta^n_i \not = \alpha^{n - 1}_0\text{ or }\alpha^{n - 1}_n} \Big((-1)^i (d^n_i(x))_{\alpha \circ \delta^n_i} - (-1)^i (d^n_i(x))_{\alpha^{n - 1}_0} \Big) \end{matrix}\] Clearly the difference of the two terms is the sum \[\sum\nolimits_{i = 0, \ldots, n\text{ and } \alpha \circ \delta^n_i = \alpha^{n - 1}_0\text{ or }\alpha^{n - 1}_n} \Big((-1)^i (d^n_i(x))_{\alpha \circ \delta^n_i} - (-1)^i (d^n_i(x))_{\alpha^{n - 1}_0} \Big)\] Of course, if \(\alpha \circ \delta^n_i = \alpha^{n - 1}_0\) then the term drops out. Recall that \(\alpha = \alpha^n_j\) for some \(j \in \{1, \ldots, n\}\). The only way \(\alpha^n_j \circ \delta^n_i = \alpha^{n - 1}_n\) is if \(j = n\) and \(i = n\). Thus we actually get \(0\) unless \(j = n\) and in that case we get \((-1)^n(d^n_n(x))_{\alpha^{n - 1}_n} - (-1)^n(d^n_n(x))_{\alpha^{n - 1}_0}\). In other words, we conclude the morphism \[\delta(\sigma)_n : W_n \to (s(U)[-1] \oplus s(U)[-1])_n = U_{n - 1} \oplus U_{n - 1}\] is zero on all summands except \(U_n \cdot \alpha^n_n\) and on that summand it is equal to \(((-1)^nd^n_n, -(-1)^nd^n_n)\). (Namely, the first summand of the two corresponds to the factor with \(\alpha^{n - 1}_n\) because that is the map \([n - 1] \to [1]\) which maps everybody to \(0\), and hence corresponds to \(e_0\).)

We obtain a canonical diagram \[\xymatrix{ 0 \ar[r] & s(U) \oplus s(U) \ar[r] \ar[d] & \diamond{s(U)} \ar[r] \ar[d]^{\Phi}& s(U)[-1] \ar[r] \ar[d] & 0 \\ 0 \ar[r] & s(U) \oplus s(U) \ar[r] \ar[d] & s(U \times \Delta[1]) \ar[r] \ar[d]^\Psi & s(W) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & s(U) \oplus s(U) \ar[r] & \diamond{s(U)} \ar[r] & s(U)[-1] \ar[r] & 0 }\] We claim that \(\Phi \circ \Psi\) is the identity. To see this it is enough to prove that the composition of \(\Phi\) and \(\delta(\sigma)\) as a map \(s(U)[-1] \to s(W) \to s(U)[-1] \oplus s(U)[-1]\) is the identity in the first factor and minus identity in the second. By the computations above it is \(((-1)^nd^n_0, -(-1)^nd^n_0) \circ (-1)^n s^n_n = (1, -1)\) as desired.

Trivial Kan fibrations

Recall that for \(n \geq 0\) the simplicial set \(\Delta[n]\) is given by the rule \([k] \mapsto \Mor_\Delta([k], [n])\), see Example 0176. Recall that \(\Delta[n]\) has a unique nondegenerate \(n\)-simplex and all nondegenerate simplices are faces of this \(n\)-simplex. In fact, the nondegenerate simplices of \(\Delta[n]\) correspond exactly to injective morphisms \([k] \to [n]\), which we may identify with subsets of \([n]\). Moreover, recall that \(\Mor(\Delta[n], X) = X_n\) for any simplicial set \(X\) (Lemma 0177). We set \[\partial \Delta[n] = i_{(n - 1)!}\text{sk}_{n - 1}\Delta[n]\] and we call it the boundary of \(\Delta[n]\). From Lemma 018P we see that \(\partial \Delta[n] \subset \Delta[n]\) is the simplicial subset having the same nondegenerate simplices in degrees \(\leq n - 1\) but not containing the nondegenerate \(n\)-simplex.

Definition

A map \(X \to Y\) of simplicial sets is called a trivial Kan fibration if \(X_0 \to Y_0\) is surjective and for all \(n \geq 1\) and any commutative solid diagram \[\xymatrix{ \partial \Delta[n] \ar[r] \ar[d] & X \ar[d] \\ \Delta[n] \ar[r] \ar@{-->}[ru] & Y }\] a dotted arrow exists making the diagram commute.

A trivial Kan fibration satisfies a very general lifting property.

Lemma

Let \(f : X \to Y\) be a trivial Kan fibration of simplicial sets. For any solid commutative diagram \[\xymatrix{ Z \ar[r]_b \ar[d] & X \ar[d] \\ W \ar[r]^a \ar@{-->}[ru] & Y }\] of simplicial sets with \(Z \to W\) (termwise) injective a dotted arrow exists making the diagram commute.

Proof

Suppose that \(Z \not = W\). Let \(n\) be the smallest integer such that \(Z_n \not = W_n\). Let \(x \in W_n\), \(x \not \in Z_n\). Denote \(Z' \subset W\) the simplicial subset containing \(Z\), \(x\), and all degeneracies of \(x\). Let \(\varphi : \Delta[n] \to Z'\) be the morphism corresponding to \(x\) (Lemma 0177). Then \(\varphi|_{\partial \Delta[n]}\) maps into \(Z\) as all the nondegenerate simplices of \(\partial \Delta[n]\) end up in \(Z\). By assumption we can extend \(b \circ \varphi|_{\partial \Delta[n]}\) to \(\beta : \Delta[n] \to X\). By Lemma 018R the simplicial set \(Z'\) is the pushout of \(\Delta[n]\) and \(Z\) along \(\partial \Delta[n]\). Hence \(b\) and \(\beta\) define a morphism \(b' : Z' \to X\). In other words, we have extended the morphism \(b\) to a bigger simplicial subset of \(Z\).

The proof is finished by an application of Zorn’s lemma (omitted).

Lemma

Let \(f : X \to Y\) be a trivial Kan fibration of simplicial sets. Let \(Y' \to Y\) be a morphism of simplicial sets. Then \(X \times_Y Y' \to Y'\) is a trivial Kan fibration.

Proof

This follows immediately from the functorial properties of the fibre product (Lemma 016U) and the definitions.

Lemma

The composition of two trivial Kan fibrations is a trivial Kan fibration.

Proof

Omitted.

Lemma

Let \(\ldots \to U^2 \to U^1 \to U^0\) be a sequence of trivial Kan fibrations. Let \(U = \lim U^t\) defined by taking \(U_n = \lim U_n^t\). Then \(U \to U^0\) is a trivial Kan fibration.

Proof

Omitted. Hint: use that for a countable sequence of surjections of sets the inverse limit is nonempty.

Lemma

Let \(X_i \to Y_i\) be a set of trivial Kan fibrations. Then \(\prod X_i \to \prod Y_i\) is a trivial Kan fibration.

Proof

Omitted.

Lemma

A filtered colimit of trivial Kan fibrations is a trivial Kan fibration.

Proof

Omitted. Hint: See description of filtered colimits of sets in Categories, Section 04AX.

Lemma

Let \(f : X \to Y\) be a trivial Kan fibration of simplicial sets. Then \(f\) is a homotopy equivalence.

Proof

By Lemma 08NM we can choose an right inverse \(g : Y \to X\) to \(f\). Consider the diagram \[\xymatrix{ \partial \Delta[1] \times X \ar[d] \ar[r] & X \ar[d] \\ \Delta[1] \times X \ar[r] \ar@{-->}[ru] & Y }\] Here the top horizontal arrow is given by \(\text{id}_X\) and \(g \circ f\) where we use that \((\partial \Delta[1] \times X)_n = X_n \amalg X_n\) for all \(n \geq 0\). The bottom horizontal arrow is given by the map \(\Delta[1] \to \Delta[0]\) and \(f : X \to Y\). The diagram commutes as \(f \circ g \circ f = f\). By Lemma 08NM we can fill in the dotted arrow and we win.

Kan fibrations

Let \(n\), \(k\) be integers with \(0 \leq k \leq n\) and \(1 \leq n\). Let \(\sigma_0, \ldots, \sigma_n\) be the \(n + 1\) faces of the unique nondegenerate \(n\)-simplex \(\sigma\) of \(\Delta[n]\), i.e., \(\sigma_i = d_i\sigma\). We let \[\Lambda_k[n] \subset \Delta[n]\] be the \(k\)th horn of the \(n\)-simplex \(\Delta[n]\). It is the simplicial subset of \(\Delta[n]\) generated by \(\sigma_0, \ldots, \hat \sigma_k, \ldots, \sigma_n\). In other words, the image of the displayed inclusion contains all the nondegenerate simplices of \(\Delta[n]\) except for \(\sigma\) and \(\sigma_k\).

Definition

A map \(X \to Y\) of simplicial sets is called a Kan fibration if for all \(k, n\) with \(1 \leq n\), \(0 \leq k \leq n\) and any commutative solid diagram \[\xymatrix{ \Lambda_k[n] \ar[r] \ar[d] & X \ar[d] \\ \Delta[n] \ar[r] \ar@{-->}[ru] & Y }\] a dotted arrow exists making the diagram commute. A Kan complex is a simplicial set \(X\) such that \(X \to *\) is a Kan fibration, where \(*\) is the constant simplicial set on a singleton.

Lemma

Let \(\mathcal{C}\) be a category. The following are equivalent.

1. The category \(\mathcal{C}\) is a groupoid.

2. The nerve \(N(\mathcal{C})\) is a Kan complex.

3. Every map \(\Lambda_k[n] \to N(\mathcal{C})\) with \(n \geq 2\) has a unique extension to \(\Delta[n]\).

Proof

Suppose \(\mathcal{C}\) is a groupoid. An inner horn in its nerve has a unique filler because composition in \(\mathcal{C}\) is defined and associative. An outer horn has a unique filler because every arrow is invertible. This proves (3). Horns in degree \(1\) can be filled using identity arrows, so (3) implies (2).

Assume (2), and let \(a : x \to y\) be an arrow of \(\mathcal{C}\). Filling the \(2\)-horn whose two given edges are \(a : x \to y\) and \(\text{id}_y : y \to y\) produces an arrow \(b : y \to x\) such that \(a \circ b = \text{id}_y\). Filling the other outer \(2\)-horn produces an arrow \(c : y \to x\) such that \(c \circ a = \text{id}_x\). Then \[b = \text{id}_x \circ b = (c \circ a) \circ b = c \circ (a \circ b) = c,\] and \(a\) is invertible. Thus \(\mathcal{C}\) is a groupoid.

Remark

The argument is internal. In the situation of Remark simplicial-remark-internal-category-nerve, the category object is a groupoid object if and only if the two outer horn morphisms in degree \(2\) are isomorphisms. The inverse morphism is recovered by filling either outer \(2\)-horn.

Note that \(\Lambda_k[n]\) is always nonempty. Thus a morphism from the empty simplicial set to any simplicial set is always a Kan fibration. It follows from Lemma 08NM that a trivial Kan fibration is a Kan fibration.

Lemma

Let \(f : X \to Y\) be a Kan fibration of simplicial sets. Let \(Y' \to Y\) be a morphism of simplicial sets. Then \(X \times_Y Y' \to Y'\) is a Kan fibration.

Proof

This follows immediately from the functorial properties of the fibre product (Lemma 016U) and the definitions.

Lemma

The composition of two Kan fibrations is a Kan fibration.

Proof

Omitted.

Lemma

Let \(\ldots \to U^2 \to U^1 \to U^0\) be a sequence of Kan fibrations. Let \(U = \lim U^t\) defined by taking \(U_n = \lim U_n^t\). Then \(U \to U^0\) is a Kan fibration.

Proof

Omitted. Hint: use that for a countable sequence of surjections of sets the inverse limit is nonempty.

Lemma

Let \(X_i \to Y_i\) be a set of Kan fibrations. Then \(\prod X_i \to \prod Y_i\) is a Kan fibration.

Proof

Omitted.

The following lemma is due to J.C. Moore, see [Moore-Cartan].

Lemma

Let \(X\) be a simplicial group. Then \(X\) is a Kan complex.

Proof

The following proof is basically just a translation into English of the proof in the reference mentioned above. Using the terminology as explained in the introduction to this section, suppose \(f : \Lambda_k[n] \to X\) is a morphism from a horn. Set \(x_i = f(\sigma_i) \in X_{n - 1}\) for \(i = 0, \ldots, \hat k, \ldots, n\). This means that for \(i < j\) we have \(d_i x_j = d_{j - 1} x_i\) whenever \(i, j \not = k\). We have to find an \(x \in X_n\) such that \(x_i = d_ix\) for \(i = 0, \ldots, \hat k, \ldots, n\).

We first prove there exists a \(u \in X_n\) such that \(d_i u = x_i\) for \(i < k\). This is trivial for \(k = 0\). If \(k > 0\), one defines by induction an element \(u^r \in X_n\) such that \(d_i u^r = x_i\) for \(0 \leq i \leq r\). Start with \(u^0 = s_0x_0\). If \(r < k - 1\), we set \[y^r = s_{r + 1}((d_{r + 1}u^r)^{-1}x_{r + 1}),\quad u^{r + 1} = u^r y^r.\] An easy calculation shows that \(d_iy^r = 1\) (unit element of the group \(X_{n - 1}\)) for \(i \leq r\) and \(d_{r + 1}y^r = (d_{r + 1}u^r)^{-1}x_{r + 1}\). It follows that \(d_iu^{r + 1} = x_i\) for \(i \leq r + 1\). Finally, take \(u = u^{k - 1}\) to get \(u\) as promised.

Next we prove, by induction on the integer \(r\), \(0 \leq r \leq n - k\), there exists a \(x^r \in X_n\) such that \[d_i x^r = x_i\quad\text{for }i < k\text{ and }i > n - r.\] Start with \(x^0 = u\) for \(r = 0\). Having defined \(x^r\) for \(r \leq n - k - 1\) we set \[z^r = s_{n - r - 1}((d_{n - r}x^r)^{-1}x_{n - r}),\quad x^{r + 1} = x^rz^r\] A simple calculation, using the given relations, shows that \(d_iz^r = 1\) for \(i < k\) and \(i > n - r\) and that \(d_{n - r}(z^r) = (d_{n - r}x^r)^{-1}x_{n - r}\). It follows that \(d_ix^{r + 1} = x_i\) for \(i < k\) and \(i > n - r - 1\). Finally, we take \(x = x^{n - k}\) which finishes the proof.

Lemma

Let \(f : X \to Y\) be a homomorphism of simplicial abelian groups which is termwise surjective. Then \(f\) is a Kan fibration of simplicial sets.

Proof

Consider a commutative solid diagram \[\xymatrix{ \Lambda_k[n] \ar[r]_a \ar[d] & X \ar[d] \\ \Delta[n] \ar[r]^b \ar@{-->}[ru] & Y }\] as in Definition 08NU. The map \(a\) corresponds to \(x_0, \ldots, \hat x_k, \ldots, x_n \in X_{n - 1}\) satisfying \(d_i x_j = d_{j - 1} x_i\) for \(i < j\), \(i, j \not = k\). The map \(b\) corresponds to an element \(y \in Y_n\) such that \(d_iy = f(x_i)\) for \(i \not = k\). Our task is to produce an \(x \in X_n\) such that \(d_ix = x_i\) for \(i \not = k\) and \(f(x) = y\).

Since \(f\) is termwise surjective we can find \(x \in X_n\) with \(f(x) = y\). Replace \(y\) by \(0 = y - f(x)\) and \(x_i\) by \(x_i - d_ix\) for \(i \not = k\). Then we see that we may assume \(y = 0\). In particular \(f(x_i) = 0\). In other words, we can replace \(X\) by \(\Ker(f) \subset X\) and \(Y\) by \(0\). In this case the statement become Lemma 08NZ.

Lemma

Let \(f : X \to Y\) be a homomorphism of simplicial abelian groups which is termwise surjective and induces a quasi-isomorphism on associated chain complexes. Then \(f\) is a trivial Kan fibration of simplicial sets.

Proof

Consider a commutative solid diagram \[\xymatrix{ \partial \Delta[n] \ar[r]_a \ar[d] & X \ar[d] \\ \Delta[n] \ar[r]^b \ar@{-->}[ru] & Y }\] as in Definition 08NL. The map \(a\) corresponds to \(x_0, \ldots, x_n \in X_{n - 1}\) satisfying \(d_i x_j = d_{j - 1} x_i\) for \(i < j\). The map \(b\) corresponds to an element \(y \in Y_n\) such that \(d_iy = f(x_i)\). Our task is to produce an \(x \in X_n\) such that \(d_ix = x_i\) and \(f(x) = y\).

Since \(f\) is termwise surjective we can find \(x \in X_n\) with \(f(x) = y\). Replace \(y\) by \(0 = y - f(x)\) and \(x_i\) by \(x_i - d_ix\). Then we see that we may assume \(y = 0\). In particular \(f(x_i) = 0\). In other words, we can replace \(X\) by \(\Ker(f) \subset X\) and \(Y\) by \(0\). This works, because by Homology, Lemma 0111 the homology of the chain complex associated to \(\Ker(f)\) is zero and hence \(\Ker(f) \to 0\) induces a quasi-isomorphism on associated chain complexes.

Since \(X\) is a Kan complex (Lemma 08NZ) we can find \(x \in X_n\) with \(d_i x = x_i\) for \(i = 0, \ldots, n - 1\). After replacing \(x_i\) by \(x_i - d_ix\) for \(i = 0, \ldots, n\) we may assume that \(x_0 = x_1 = \ldots = x_{n - 1} = 0\). In this case we see that \(d_i x_n = 0\) for \(i = 0, \ldots, n - 1\). Thus \(x_n \in N(X)_{n - 1}\) and lies in the kernel of the differential \(N(X)_{n - 1} \to N(X)_{n - 2}\). Here \(N(X)\) is the normalized chain complex associated to \(X\), see Section 0194. Since \(N(X)\) is quasi-isomorphic to \(s(X)\) (Lemma 019C) and thus acyclic we find \(x \in N(X_n)\) whose differential is \(x_n\). This \(x\) answers the question posed by the lemma and we are done.

Lemma

Let \(f : X \to Y\) be a map of simplicial abelian groups. If \(f\) is a homotopy equivalence of simplicial sets, then \(f\) induces a quasi-isomorphism of associated chain complexes.

Proof

In this proof we will write \(H_n(Z) = H_n(s(Z)) = H_n(N(Z))\) when \(Z\) is a simplicial abelian group, with \(s\) and \(N\) as in Section 0194. Let \(\mathbf{Z}[X]\) denote the free abelian group on \(X\) viewed as a simplicial set and similarly for \(\mathbf{Z}[Y]\). Consider the commutative diagram \[\xymatrix{ \mathbf{Z}[X] \ar[r]_g \ar[d] & \mathbf{Z}[Y] \ar[d] \\ X \ar[r]^f & Y }\] of simplicial abelian groups. Since taking the free abelian group on a set is a functor, we see that the horizontal arrow is a homotopy equivalence of simplicial abelian groups, see Lemma 019Y. By Lemma 019T we see that \(H_n(g) : H_n(\mathbf{Z}[X]) \to H_n(\mathbf{Z}[Y])\) is bijective for all \(n \geq 0\).

Let \(\xi \in H_n(Y)\). By definition of \(N(Y)\) we can represent \(\xi\) by an element \(y \in N(Y_n)\) whose boundary is zero. This means \(y \in Y_n\) with \(d^n_0(y) = \ldots = d^n_{n - 1}(y) = 0\) because \(y \in N(Y_n)\) and \(d^n_n(y) = 0\) because the boundary of \(y\) is zero. Denote \(0_n \in Y_n\) the zero element. Then we see that \[\tilde y = [y] - [0_n] \in (\mathbf{Z}[Y])_n\] is an element with \(d^n_0(\tilde y) = \ldots = d^n_{n - 1}(\tilde y) = 0\) and \(d^n_n(\tilde y) = 0\). Thus \(\tilde y\) is in \(N(\mathbf{Z}[Y])_n\) has boundary \(0\), i.e., \(\tilde y\) determines a class \(\tilde \xi \in H_n(\mathbf{Z}[Y])\) mapping to \(\xi\). Because \(H_n(\mathbf{Z}[X]) \to H_n(\mathbf{Z}[Y])\) is bijective we can lift \(\tilde \xi\) to a class in \(H_n(\mathbf{Z}[X])\). Looking at the commutative diagram above we see that \(\xi\) is in the image of \(H_n(X) \to H_n(Y)\).

Let \(\xi \in H_n(X)\) be an element mapping to zero in \(H_n(Y)\). Exactly as in the previous parapgraph we can represent \(\xi\) by an element \(x \in N(X_n)\) whose boundary is zero, i.e., \(d^n_0(x) = \ldots = d^n_{n - 1}(x) = d^n_n(x) = 0\). In particular, we see that \([x] - [0_n]\) is an element of \(N(\mathbf{Z}[X])_n\) whose boundary is zero, whence defines a lift \(\tilde \xi \in H_n(\mathbf{Z}[x])\) of \(\xi\). The fact that \(\xi\) maps to zero in \(H_n(Y)\) means there exists a \(y \in N(Y_{n + 1})\) whose boundary is \(f_n(x)\). This means \(d^{n + 1}_0(y) = \ldots = d^{n + 1}_n(y) = 0\) and \(d^{n + 1}_{n + 1}(y) = f(x)\). However, this means exactly that \(z = [y] - [0_{n + 1}]\) is in \(N(\mathbf{Z}[y])_{n + 1}\) and \[g([x] - [0_n]) = [f(x)] - [0_n] = \text{boundary of }z\] This proves that \(\tilde \xi\) maps to zero in \(H_n(\mathbf{Z}[y])\). As \(H_n(\mathbf{Z}[X]) \to H_n(\mathbf{Z}[Y])\) is bijective we conclude \(\tilde \xi = 0\) and hence \(\xi = 0\).

A homotopy equivalence

Suppose that \(A\), \(B\) are sets, and that \(f : A \to B\) is a map. Consider the associated map of simplicial sets \[\xymatrix{ \text{cosk}_0(A) \ar@{=}[r] & \Big(\ldots A \times A \times A \ar[d] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & A \times A \ar[d] \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & A \Big) \ar[d] \ar@<0ex>[l] \\ \text{cosk}_0(B) \ar@{=}[r] & \Big( \ldots B \times B \times B \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & B \times B \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & B \Big) \ar@<0ex>[l] }\] See Example 0182. The case \(n = 0\) of the following lemma says that this map of simplicial sets is a trivial Kan fibration if \(f\) is surjective.

Lemma

Let \(f : V \to U\) be a morphism of simplicial sets. Let \(n \geq 0\) be an integer. Assume

1. The map \(f_i : V_i \to U_i\) is a bijection for \(i < n\).

2. The map \(f_n : V_n \to U_n\) is a surjection.

3. The canonical morphism \(U \to \text{cosk}_n \text{sk}_n U\) is an isomorphism.

4. The canonical morphism \(V \to \text{cosk}_n \text{sk}_n V\) is an isomorphism.

Then \(f\) is a trivial Kan fibration.

Proof

Consider a solid diagram \[\xymatrix{ \partial \Delta[k] \ar[r] \ar[d] & V \ar[d] \\ \Delta[k] \ar[r] \ar@{-->}[ru] & U }\] as in Definition 08NL. Let \(x \in U_k\) be the \(k\)-simplex corresponding to the lower horizontal arrow. If \(k \leq n\) then the dotted arrow is the one corresponding to a lift \(y \in V_k\) of \(x\); the diagram will commute as the other nondegenerate simplices of \(\Delta[k]\) are in degrees \(< k\) where \(f\) is an isomorphism. If \(k > n\), then by conditions (3) and (4) we have (using adjointness of skeleton and coskeleton functors) \[\Mor(\Delta[k], U) = \Mor(\text{sk}_n\Delta[k], \text{sk}_nU) = \Mor(\text{sk}_n\partial\Delta[k], \text{sk}_nU) = \Mor(\partial \Delta[k], U)\] and similarly for \(V\) because \(\text{sk}_n\Delta[k] = \text{sk}_n\partial\Delta[k]\) for \(k > n\). Thus we obtain a unique dotted arrow fitting into the diagram in this case also.

Let \(A, B\) be sets. Let \(f^0, f^1 : A \to B\) be maps of sets. Consider the induced maps \(f^0, f^1 : \text{cosk}_0(A) \to \text{cosk}_0(B)\) abusively denoted by the same symbols. The following lemma for \(n = 0\) says that \(f^0\) is homotopic to \(f^1\). In fact, there is a homotopy \(h : \text{cosk}_0(A) \times \Delta[1] \to \text{cosk}_0(A)\) from \(f^0\) to \(f^1\) with components \[\begin{eqnarray*} h_m : A \times \ldots \times A \times \Mor_{\Delta}([m], [1]) & \longrightarrow & B \times \ldots \times B, \\ (a_0, \ldots, a_m, \alpha) & \longmapsto & (f^{\alpha(0)}(a_0), \ldots, f^{\alpha(m)}(a_m)) \end{eqnarray*}\] To check that this works, note that for a map \(\varphi : [k] \to [m]\) the induced maps are \((a_0, \ldots, a_m) \mapsto (a_{\varphi(0)}, \ldots, a_{\varphi(k)})\) and \(\alpha \mapsto \alpha \circ \varphi\). Thus \(h = (h_m)_{m \geq 0}\) is clearly a map of simplicial sets as desired.

Lemma

Let \(f^0, f^1 : V \to U\) be maps of simplicial sets. Let \(n \geq 0\) be an integer. Assume

1. The maps \(f^j_i : V_i \to U_i\), \(j = 0, 1\) are equal for \(i < n\).

2. The canonical morphism \(U \to \text{cosk}_n \text{sk}_n U\) is an isomorphism.

3. The canonical morphism \(V \to \text{cosk}_n \text{sk}_n V\) is an isomorphism.

Then \(f^0\) is homotopic to \(f^1\).

Proof

Let \(W\) be the \(n\)-truncated simplicial set with \(W_i = U_i\) for \(i < n\) and \(W_n = U_n / \sim\) where \(\sim\) is the equivalence relation generated by \(f^0(y) \sim f^1(y)\) for \(y \in V_n\). This makes sense as the morphisms \(U(\varphi) : U_n \to U_i\) corresponding to \(\varphi : [i] \to [n]\) for \(i < n\) factor through the quotient map \(U_n \to W_n\) because \(f^0\) and \(f^1\) are morphisms of simplicial sets and equal in degrees \(< n\). Next, we upgrade \(W\) to a simplicial set by taking \(\text{cosk}_n W\). By Lemma 01A6 the morphism \(g : U \to W\) is a trivial Kan fibration. Observe that \(g \circ f^0 = g \circ f^1\) by construction and denote this morphism \(f : V \to W\). Consider the diagram \[\xymatrix{ \partial \Delta[1] \times V \ar[rr]_{f^0, f^1} \ar[d] & & U \ar[d] \\ \Delta[1] \times V \ar[rr]^f \ar@{-->}[rru] & & W }\] By Lemma 08NM the dotted arrow exists and the proof is done.

Proof

We have to construct a morphism of simplicial sets \(h : V \times \Delta[1] \to U\) which recovers \(f^i\) on composing with \(e_i\). The case \(n = 0\) was dealt with above the lemma. Thus we may assume that \(n \geq 1\). The map \(\Delta[1] \to \text{cosk}_1 \text{sk}_1 \Delta[1]\) is an isomorphism, see Lemma 018E. Thus we see that \(\Delta[1] \to \text{cosk}_n \text{sk}_n \Delta[1]\) is an isomorphism as \(n \geq 1\), see Lemma 018B. And hence \(V \times \Delta[1] \to \text{cosk}_n \text{sk}_n (V \times \Delta[1])\) is an isomorphism too, see Lemma 018C. In other words, in order to construct the homotopy it suffices to construct a suitable morphism of \(n\)-truncated simplicial sets \(h : \text{sk}_n V \times \text{sk}_n \Delta[1] \to \text{sk}_n U\).

For \(k = 0, \ldots, n - 1\) we define \(h_k\) by the formula \(h_k(v, \alpha) = f^0(v) = f^1(v)\). The map \(h_n : V_n \times \Mor_{\Delta}([n], [1]) \to U_n\) is defined as follows. Pick \(v \in V_n\) and \(\alpha : [n] \to [1]\):

1. If \(\Im(\alpha) = \{0\}\), then we set \(h_n(v, \alpha) = f^0(v)\).

2. If \(\Im(\alpha) = \{0, 1\}\), then we set \(h_n(v, \alpha) = f^0(v)\).

3. If \(\Im(\alpha) = \{1\}\), then we set \(h_n(v, \alpha) = f^1(v)\).

Let \(\varphi : [k] \to [l]\) be a morphism of \(\Delta_{\leq n}\). We will show that the diagram \[\xymatrix{ V_{l} \times \Mor([l], [1]) \ar[r] \ar[d] & U_{l} \ar[d] \\ V_{k} \times \Mor([k], [1]) \ar[r] & U_{k} }\] commutes. Pick \(v \in V_{l}\) and \(\alpha : [l] \to [1]\). The commutativity means that \[h_k(V(\varphi)(v), \alpha \circ \varphi) = U(\varphi)(h_l(v, \alpha)).\] In almost every case this holds because \(h_k(V(\varphi)(v), \alpha \circ \varphi) = f^0(V(\varphi)(v))\) and \(U(\varphi)(h_l(v, \alpha)) = U(\varphi)(f^0(v))\), combined with the fact that \(f^0\) is a morphism of simplicial sets. The only cases where this does not hold is when either (A) \(\Im(\alpha) = \{1\}\) and \(l = n\) or (B) \(\Im(\alpha \circ \varphi) = \{1\}\) and \(k = n\). Observe moreover that necessarily \(f^0(v) = f^1(v)\) for any degenerate \(n\)-simplex of \(V\). Thus we can narrow the cases above down even further to the cases (A) \(\Im(\alpha) = \{1\}\), \(l = n\) and \(v\) nondegenerate, and (B) \(\Im(\alpha \circ \varphi) = \{1\}\), \(k = n\) and \(V(\varphi)(v)\) nondegenerate.

In case (A), we see that also \(\Im(\alpha \circ \varphi) = \{1\}\). Hence we see that not only \(h_l(v, \alpha) = f^1(v)\) but also \(h_k(V(\varphi)(v), \alpha \circ \varphi) = f^1(V(\varphi)(v))\). Thus we see that the relation holds because \(f^1\) is a morphism of simplicial sets.

In case (B) we conclude that \(l = k = n\) and \(\varphi\) is bijective, since otherwise \(V(\varphi)(v)\) is degenerate. Thus \(\varphi = \text{id}_{[n]}\), which is a trivial case.

Lemma

Let \(A\), \(B\) be sets, and that \(f : A \to B\) is a map. Consider the simplicial set \(U\) with \(n\)-simplices \[A \times_B A \times_B \ldots \times_B A\ (n + 1 \text{ factors)}.\] see Example 016E. If \(f\) is surjective, the morphism \(U \to B\) where \(B\) indicates the constant simplicial set with value \(B\) is a trivial Kan fibration.

Proof

Observe that \(U\) fits into a cartesian square \[\xymatrix{ U \ar[d] \ar[r] & \text{cosk}_0(A) \ar[d] \\ B \ar[r] & \text{cosk}_0(B) }\] Since the right vertical arrow is a trivial Kan fibration by Lemma 01A6, so is the left by Lemma 08NN.

Preparation for standard resolutions

The material in this section can be found in [Godement, Appendix 1]

Example

Let \(Y : \mathcal{C} \to \mathcal{C}\) be a functor from a category to itself and suppose given transformations of functors \[d : Y \longrightarrow \text{id}_\mathcal{C} \quad\text{and}\quad s : Y \longrightarrow Y \circ Y\] Using these transformations we can construct something that looks like a simplicial object. Namely, for \(n \geq 0\) we define \[X_n = Y \circ \ldots \circ Y \quad (n + 1\text{ compositions})\] Observe that \(X_{n + m + 1} = X_n \circ X_m\) for \(n, m \geq 0\). Next, for \(n \geq 0\) and \(0 \leq j \leq n\) we define using notation as in Categories, Section 003D \[d^n_j = 1_{X_{j - 1}} \star d \star 1_{X_{n - j - 1}} : X_n \to X_{n - 1} \quad\text{and}\quad s^n_j = 1_{X_{j - 1}} \star s \star 1_{X_{n - j - 1}} : X_n \to X_{n + 1}\] So \(d^n_j\), resp. \(s^n_j\) is the natural transformation using \(d\), resp. \(s\) on the \(j\)th \(Y\) (counted from the left) in the composition defining \(X_n\).

Lemma

In Example 0G5M if \[1_Y = (d \star 1_Y) \circ s = (1_Y \star d) \circ s \quad\text{and}\quad (s \star 1) \circ s = (1 \star s) \circ s\] then \(X = (X_n, d^n_j, s^n_j)\) is a simplicial object in the category of endofunctors of \(\mathcal{C}\) and \(d : X_0 = Y \to \text{id}_\mathcal{C}\) defines an augmentation.

Proof

To see that we obtain a simplicial object we have to check that the relations (1)(a) – (e) of Lemma 016B are satisfied. We will use the short hand notation \[1_a = 1_{X_{a - 1}} = 1_Y \star \ldots \star 1_Y \quad (a\text{ factors})\] for \(a \geq 0\). With this notation we have \[d^n_j = 1_j \star d \star 1_{n - j} \quad\text{and}\quad s^n_j = 1_j \star s \star 1_{n - j}\] We are repeatedly going to use the rule that for transformations of funtors \(a, a', b, b'\) we have \((a' \circ a) \star (b' \circ b) = (a' \star b') \circ (a \star b)\) provided that the \(\star\) and \(\circ\) compositions in this formula make sense, see Categories, Lemma 003F.

Condition (1)(a) always holds (no conditions needed on \(d\) and \(s\)). Namely, let \(0 \leq i < j \leq n + 1\). We have to show that \(d^n_i \circ d^{n + 1}_j = d^n_{j - 1} \circ d^{n + 1}_i\), i.e., \[(1_i \star d \star 1_{n - i}) \circ (1_j \star d \star 1_{n + 1 - j}) = (1_{j - 1} \star d \star 1_{n + 1 - j}) \circ (1_i \star d \star 1_{n + 1 - i})\] We can rewrite the left hand side as \[\begin{align*} & (1_i \star d \star 1_{j - i - 1} \star 1_{n + 1 - j}) \circ (1_i \star 1_1 \star 1_{j - i - 1} \star d \star 1_{n + 1 - j}) \\ & = 1_i \star \left((d \star 1_{j - i - 1}) \circ (1_1 \star 1_{j - i - 1} \star d)\right) \star 1_{n + 1 - j} \\ & = 1_i \star d \star 1_{j - i - 1} \star d \star 1_{n + 1 - j} \end{align*}\] The second equality is true because \(d \circ 1_1 = d\) and \(1_{j - i} \circ (1_{j - i - 1} \star d) = 1_{j - i - 1} \star d\). A similar computation gives the same result for the right hand side.

We check condition (1)(b). Let \(0 \leq i < j \leq n - 1\). We have to show that \(d^n_i \circ s^{n - 1}_j = s^{n - 2}_{j - 1} \circ d^{n - 1}_i\), i.e., \[(1_i \star d \star 1_{n - i}) \circ (1_j \star s \star 1_{n - 1 - j}) = (1_{j - 1} \star s \star 1_{n - 1 - j}) \circ (1_i \star d \star 1_{n - 1 - i})\] By the same kind of calculus as in case (1)(a) both sides simplify to \(1_i \star d \star 1_{j - i - 1} \star s \star 1_{n - j - 1}\).

We check condition (1)(c). Let \(0 \leq j \leq n - 1\). We have to show \(\text{id} = d^n_j \circ s^{n - 1}_j = d^n_{j + 1} \circ s^{n - 1}_j\), i.e., \[1_n = (1_j \star d \star 1_{n - j}) \circ (1_j \star s \star 1_{n - 1 - j}) = (1_{j + 1} \star d \star 1_{n - j - 1}) \circ (1_j \star s \star 1_{n - 1 - j})\] This is easily seen to be implied by the first assumption of the lemma.

We check condition (1)(d). Let \(0 < j + 1 < i \leq n\). We have to show \(d^n_i \circ s^{n - 1}_j = s^{n - 2}_j \circ d^{n - 1}_{i - 1}\), i.e., \[(1_i \star d \star 1_{n - i}) \circ (1_j \star s \star 1_{n - 1 - j}) = (1_j \star s \star 1_{n - 2 - j}) \circ (1_{i - 1} \star d \star 1_{n - i})\] By the same kind of calculus as in case (1)(a) both sides simplify to \(1_j \star s \star 1_{i - j - 2} \star d \star 1_{n - i}\).

We check condition (1)(e). Let \(0 \leq i \leq j \leq n - 1\). We have to show that \(s^n_i \circ s^{n - 1}_j = s^n_{j + 1} \circ s^{n - 1}_i\), i.e., \[(1_i \star s \star 1_{n - i}) \circ (1_j \star s \star 1_{n - 1 - j}) = (1_{j + 1} \star s \star 1_{n - 1 - j}) \circ (1_i \star s \star 1_{n - 1 - i})\] By the same kind of calculus as in case (1)(a) this reduces to \[(s \star 1_{j - i + 1}) \circ (1_{j - i} \star s) = (1_{j - i + 1} \star s) \circ (s \star 1_{j - i})\] If \(j = i\) this is exactly one of the two assumptions of the lemma. For \(j > i\) left and right hand side both reduce to the equality \(s \star 1_{j - i - 1} \star s\) by calculations similar to those we did in case (1)(a).

Finally, in order to show that \(d\) defines an augmentation we have to show that \(d \circ (1_1 \star d) = d \circ (d \star 1_1)\) which is true because both sides are equal to \(d \star d\).

Example

Let \(\mathcal{C}\), \(Y\), \(d\), \(s\) be as in Example 0G5M satisfying the equations of Lemma 0G5N. Given functors \(F : \mathcal{A} \to \mathcal{C}\) and \(G : \mathcal{C} \to \mathcal{B}\) we obtain a simplicial object \(G \circ X \circ F\) in the category of functors from \(\mathcal{A}\) to \(\mathcal{B}\) which comes with an augmentation to \(G \circ F\).

Lemma

Let \(\mathcal{A}\), \(\mathcal{B}\), \(\mathcal{C}\), \(Y\), \(d\), \(s\), \(F\), \(G\) be as in Example 0G5P. Given a transformation of functors \(h_0 : G \circ F \to G \circ Y \circ F\) such that \[1_{G \circ F} = (1_G \star d \star 1_F) \circ h_0\] Then there is a morphism \(h : G \circ F \to G \circ X \circ F\) of simplicial objects such that \(\epsilon \circ h = \text{id}\) where \(\epsilon : G \circ X \circ F \to G \circ F\) is the augmentation.

Proof

Denote \(u_n : Y = X_0 \to X_n\) the map of the simplicial object \(X\) corresponding to the unique morphism \([n] \to [0]\) in \(\Delta\). Set \(h_n : G \circ F \to G \circ X_n \circ F\) equal to \((1_G \star u_n \star 1_F) \circ h_0\).

For any simplicial object \(X = (X_n)\) in any category \(u =(u_n) : X_0 \to X\) is a morphism from the constant simplicial object on \(X_0\) to \(X\). Hence \(h\) is a morphism of simplicial objects because it is the composition of \(1_G \star u \star 1_F\) and \(h_0\).

Let us check that \(\epsilon \circ h = \text{id}\). We compute \[\epsilon_n \circ (1_G \star u_n \star 1_F) \circ h_0 = \epsilon_0 \circ h_0 = \text{id}\] The first equality because \(\epsilon\) is a morphism of simplicial objects and the second equality because \(\epsilon_0 = (1_G \star d \star 1_F)\) and we can apply the assumption in the statement of the lemma.

Lemma

Let \(\mathcal{A}\), \(\mathcal{B}\), \(\mathcal{C}\), \(Y\), \(d\), \(s\), \(F\), \(G\) be as in Example 0G5P. Let \(F' : \mathcal{A} \to \mathcal{C}\) and \(G' : \mathcal{C} \to \mathcal{B}\) be two functors. Let \((a_n) : G \circ X \to G' \circ X\) be a morphism of simplicial objects compatible via augmentations with \(a : G \to G'\). Let \((b_n) : X \circ F \to X \circ F'\) be a morphism of simplicial objects compatible via augmentations with \(b : F \to F'\). Then the two maps \[a \star (b_n), (a_n) \star b : G \circ X \circ F \to G' \circ X \circ F'\] are homotopic.

Proof

To show the morphisms are homotopic we construct morphisms \[h_{n, i} : G \circ X_n \circ F \to G' \circ X_n \circ F'\] for \(n \geq 0\) and \(0 \leq i \leq n + 1\) satisfying the relations described in Lemma 019L. See also Remark 019M. To satisfy condition (1) of Lemma 019L we are forced to set \(h_{n, 0} = a \star b_n\) and \(h_{n , n + 1} = a_n \star b\). Thus a logical choice is \[h_{n , i} = a_{i - 1} \star b_{n - i}\] for \(1 \leq i \leq n\). Setting \(a = a_{-1}\) and \(b = b_{-1}\) we see the displayed formular holds for \(0 \leq i \leq n + 1\).

Recall that \[d^n_j = 1_G \star 1_j \star d \star 1_{n - j} \star 1_F\] on \(G \circ X \circ F\) where we use the notation \(1_a = 1_{Y \circ \ldots \circ Y}\) introduced in the proof of Lemma 0G5N. We are going to use below that we can rewrite this as \[\begin{align*} d^n_j & = d^j_j \star 1_{n - j} = d^{j + 1}_j \star 1_{n - j} = \ldots = d^{n - 1}_j \star 1_1 \\ & = 1_j \star d^{n - j}_0 = 1_{j - 1} \star d^{n - j + 1}_1 = \ldots = 1_1 \star d^{n - 1}_{j - 1} \end{align*}\] Of course we have the analogous formulae for \(d^n_j\) on \(G' \circ X \circ F'\).

We check condition (2) of Lemma 019L. Let \(i > j\). We have to show \[d^n_j \circ (a_{i - 1} \star b_{n - i}) = (a_{i - 2} \star b_{n - i}) \circ d^n_j\] Since \(i - 1 \geq j\) we can use one of the possible descriptions of \(d^n_j\) to rewrite the left hand side as \[(d^{i - 1}_j \star 1_{n - i + 1}) \circ (a_{i - 1} \star b_{n - i}) = (d^{i - 1}_j \circ a_{i - 1}) \star b_{n - i} = (a_{i - 2} \circ d^{i - 1}_j) \star b_{n - i}\] Similarly the right hand side becomes \[(a_{i - 2} \star b_{n - i}) \circ (d^{i - 1}_j \star 1_{n - i + 1}) = (a_{i - 2} \circ d^{i - 1}_j) \star b_{n - i}\] Thus we obtain the same result and (2) is checked.

We check condition (3) of Lemma 019L. Let \(i \leq j\). We have to show \[d^n_j \circ (a_{i - 1} \star b_{n - i}) = (a_{i - 1} \star b_{n - 1 - i}) \circ d^n_j\] Since \(j \geq i\) we may rewrite the left hand side as \[(1_i \star d^{n - i}_{j - i}) \circ (a_{i - 1} \star b_{n - i}) = a_{i - 1} \star (b_{n - 1 - i} \circ d^{n - i}_{j - i})\] A similar manipulation shows this agrees with the right hand side.

Recall that \[s^n_j = 1_G \star 1_j \star s \star 1_{n - j} \star 1_F\] on \(G \circ X \circ F\). We are going to use below that we can rewrite this as \[\begin{align*} s^n_j & = s^j_j \star 1_{n - j} = s^{j + 1}_j \star 1_{n - j - 1} = \ldots = s^{n - 1}_j \star 1_1 \\ & = 1_j \star s^{n - j}_0 = 1_{j - 1} \star s^{n - j + 1}_1 = \ldots = 1_1 \star s^{n - 1}_{j - 1} \end{align*}\] Of course we have the analogous formulae for \(s^n_j\) on \(G' \circ X \circ F'\).

We check condition (4) of Lemma 019L. Let \(i > j\). We have to show \[s^n_j \circ (a_{i - 1} \star b_{n - i}) = (a_i \star b_{n - i}) \circ s^n_j\] Since \(i - 1 \geq j\) we can rewrite the left hand side as \[(s^{i - 1}_j \star 1_{n - i + 1}) \circ (a_{i - 1} \star b_{n - i}) = (s^{i - 1}_j \circ a_{i - 1}) \star b_{n - i} = (a_i \circ s^{i - 1}_j) \star b_{n - i}\] Similarly the right hand side becomes \[(a_i \star b_{n - i}) \circ (s^{i - 1}_j \star 1_{n - i + 1}) = (a_i \circ s^{i - 1}_j) \star b_{n - i}\] as desired.

We check condition (5) of Lemma 019L. Let \(i \leq j\). We have to show \[s^n_j \circ (a_{i - 1} \star b_{n - i}) = (a_{i - 1} \star b_{n + 1 - i}) \circ s^n_j\] This equality holds because both sides evaluate to \(a_{i - 1} \star (s^{n - i}_{j - i} \circ b_{n - i}) = a_{i - 1} \star (b_{n + 1 - i} \circ s^{n - i}_{j - i})\) by exactly the same arguments as above.

Lemma

Let \(\mathcal{C}\), \(Y\), \(d\), \(s\) be as in Example 0G5M satisfying the equations of Lemma 0G5N. Let \(f : \text{id}_\mathcal{C} \to \text{id}_\mathcal{C}\) be an endomorphism of the identity functor. Then \(f \star 1_X, 1_X \star f : X \to X\) are maps of simplicial objects compatible with \(f\) via the augmentation \(\epsilon : X \to \text{id}_\mathcal{C}\). Moreover, \(f \star 1_X\) and \(1_X \star f\) are homotopic.

Proof

The map \(f \star 1_X\) is the map with components \[X_n = \text{id}_\mathcal{C} \circ X_n \xrightarrow{f \star 1_{X_n}} \text{id}_\mathcal{C} \circ X_n = X_n\] For a transformation \(a : F \to G\) of endofunctors of \(\mathcal{C}\) we have \(a \circ (f \star 1_F) = f \star a = (f \star 1_G) \circ a\). Thus \(f \star 1_X\) is indeed a morphism of simplicial objects. Similarly for \(1_X \star f\).

To show the morphisms are homotopic we construct morphisms \(h_{n, i} : X_n \to X_n\) for \(n \geq 0\) and \(0 \leq i \leq n + 1\) satisfying the relations described in Lemma 019L. See also Remark 019M. It turns out we can take \[h_{n, i} = 1_i \star f \star 1_{n + 1 - i}\] where \(1_i\) is the identity transformation on \(Y \circ \ldots \circ Y\) as in the proof of Lemma 0G5N. We have \(h_{n, 0} = f \star 1_{X_n}\) and \(h_{n, n + 1} = 1_{X_n} \star f\) which checks the first condition. In checking the other conditions we use the comments made in the proof of Lemma 0G5R about the maps \(d^n_j\) and \(s^n_j\).

We check condition (2) of Lemma 019L. Let \(i > j\). We have to show \[d^n_j \circ (1_i \star f \star 1_{n + 1 - i}) = (1_{i - 1} \star f \star 1_{n + 1 - i}) \circ d^n_j\] Since \(i - 1 \geq j\) we can use one of the possible descriptions of \(d^n_j\) to rewrite the left hand side as \[(d^{i - 1}_j \star 1_{n - i + 1}) \circ (1_i \star f \star 1_{n + 1 - i}) = d^{i - 1}_j \star f \star 1_{n + 1 - i}\] Similarly the right hand side becomes \[(1_{i - 1} \star f \star 1_{n + 1 - i}) \circ (d^{i - 1}_j \star 1_{n - i + 1}) = d^{i - 1}_j \star f \star 1_{n + 1 - i}\] Thus we obtain the same result and (2) is checked.

The conditions (3), (4), and (5) of Lemma 019L are checked in exactly the same manner using the strategy of the proof of Lemma 0G5R. We omit the details².

Standard resolutions

Some of the material in this section can be found in [Godement, Appendix 1] and [cotangent, I 1.5].

Situation

Let \(\mathcal{A}\), \(\mathcal{S}\) be categories and let \(V : \mathcal{A} \to \mathcal{S}\) be a functor with a left adjoint \(U : \mathcal{S} \to \mathcal{A}\).

In this very general situation we will construct a simplicial object \(X\) in the category of functors from \(\mathcal{A}\) to \(\mathcal{A}\). We suggest looking at the examples presented later on before reading the text of this section.

For the construction we will use the horizontal composition as defined in Categories, Section 003D. The definition of the adjunction morphisms³ \[d : U \circ V \to \text{id}_\mathcal{A} \quad (\text{counit}) \quad\text{and}\quad \eta : \text{id}_\mathcal{S} \to V \circ U \quad (\text{unit})\] in Categories, Section 0036 shows that the compositions [08NB]\[\begin{equation} V \xrightarrow{\eta \star 1_V} V \circ U \circ V \xrightarrow{1_V \star d} V \quad\text{and}\quad U \xrightarrow{1_U \star \eta} U \circ V \circ U \xrightarrow{d \star 1_U} U \end{equation}\] are the identity morphisms. Here to define the morphism \(\eta \star 1_V\) we silently identify \(V\) with \(\text{id}_\mathcal{S} \circ V\) and \(1_V\) stands for \(\text{id}_V : V \to V\). We will use this notation and these relations repeatedly in what follows. For \(n \geq 0\) we set \[X_n = (U \circ V)^{\circ (n + 1)} = U \circ V \circ U \circ \ldots \circ U \circ V\] In other words, \(X_n\) is the \((n + 1)\)-fold composition of \(U \circ V\) with itself. We also set \(X_{-1} = \text{id}_\mathcal{A}\). We have \(X_{n + m + 1} = X_n \circ X_m\) for all \(n, m \geq -1\). We will endow this sequence of functors with the structure of a simplicial object of \(\text{Fun}(\mathcal{A}, \mathcal{A})\) by constructing the morphisms of functors \[d^n_j : X_n \to X_{n - 1},\quad s^n_j : X_n \to X_{n + 1}\] satisfying the relations displayed in Lemma 0167. Namely, we set \[d^n_j = 1_{X_{j - 1}} \star d \star 1_{X_{n - j - 1}} \quad\text{and}\quad s^n_j = 1_{X_{j - 1} \circ U} \star \eta \star 1_{V \circ X_{n - j - 1}}\] Finally, write \(\epsilon_0 = d : X_0 \to X_{-1}\).

Lemma

In Situation 08N9 the system \(X = (X_n, d^n_j, s^n_j)\) is a simplicial object of \(\text{Fun}(\mathcal{A}, \mathcal{A})\) and \(\epsilon_0\) defines an augmentation \(\epsilon\) from \(X\) to the constant simplicial object with value \(X_{-1} = \text{id}_\mathcal{A}\).

Proof

Consider \(Y = U \circ V : \mathcal{A} \to \mathcal{A}\). We already have the transformation \(d : Y = U \circ V \to \text{id}_\mathcal{A}\). Let us denote \[s = 1_U \star \eta \star 1_V : Y = U \circ \text{id}_\mathcal{S} \circ V \longrightarrow U \circ V \circ U \circ V = Y \circ Y\] This places us in the situation of Example 0G5M. It is immediate from the formulas that the \(X, d^n_i, s^n_i\) constructed above and the \(X, s^n_i, s^n_i\) constructed from \(Y, d, s\) in Example 0G5M agree. Thus, according to Lemma 0G5N it suffices to prove that \[1_Y = (d \star 1_Y) \circ s = (1_Y \star d) \circ s \quad\text{and}\quad (s \star 1) \circ s = (1 \star s) \circ s\] The first equal sign translates into the equality \[1_U \star 1_V = (d \star 1_U \star 1_V) \circ (1_U \star \eta \star 1_V)\] which holds if we have \(1_U = (d \star 1_U) \circ (1_U \star \eta)\) which in turn holds by (08NB). Similarly for the second equal sign. For the last equation we need to prove \[(1_U \star \eta \star 1_V \star 1_U \star 1_V) \circ (1_U \star \eta \star 1_V) = (1_U \star 1_V \star 1_U \star \eta \star 1_V) \circ (1_U \star \eta \star 1_V)\] For this it suffices to prove \((\eta \star 1_V \star 1_U) \circ \eta = (1_V \star 1_U \star \eta) \circ \eta\) which is true because both sides are the same as \(\eta \star \eta\).

Before reading the proof of the following lemma, we advise the reader to look at the example discussed in Example 08NE in order to understand the purpose of the lemma.

Lemma

In Situation 08N9 the maps \[1_V \star \epsilon : V \circ X \to V, \quad\text{and}\quad \epsilon \star 1_U : X \circ U \to U\] are homotopy equivalences.

Proof

As in the proof of Lemma 08NC we set \(Y = U \circ V\) so that we are in the situation of Example 0G5M.

Proof of the first homotopy equivalence. By Lemma 0G5Q to construct a map \(h : V \to V \circ X\) right inverse to \(1_V \star \epsilon\) it suffices to construct a map \(h_0 : V \to V \circ Y = V \circ U \circ V\) such that \(1_V = (1_V \star d) \circ h_0\). Of course we take \(h_0 = \eta \star 1_V\) and the equality holds by (08NB). To finish the proof we need to show the two maps \[(1_V \star \epsilon) \circ h, 1_V \star \text{id}_X : V \circ X \longrightarrow V \circ X\] are homotopic. This follows immediately from Lemma 0G5R (with \(G = G' = V\) and \(F = F' = \text{id}_\mathcal{S}\)).

The proof of the second homotopy equivalence. By Lemma 0G5Q to construct a map \(h : U \to X \circ U\) right inverse to \(\epsilon \star 1_U\) it suffices to construct a map \(h_0 : U \to Y \circ U = U \circ V \circ U\) such that \(1_U = (d \star 1_U) \circ h_0\). Of course we take \(h_0 = 1_U \star \eta\) and the equality holds by (08NB). To finish the proof we need to show the two maps \[(\epsilon \star 1_U) \circ h, \text{id}_X \star 1_U : X \circ U \longrightarrow X \circ U\] are homotopic. This follows immediately from Lemma 0G5R (with \(G = G' = \text{id}_\mathcal{A}\) and \(F = F' = U\)).

Example

Let \(R\) be a ring. As an example of the above we can take \(i : \text{Mod}_R \to \textit{Sets}\) to be the forgetful functor and \(F : \textit{Sets} \to \text{Mod}_R\) to be the functor that associates to a set \(E\) the free \(R\)-module \(R[E]\) on \(E\). For an \(R\)-module \(M\) the simplicial \(R\)-module \(X(M)\) will have the following shape \[X(M) = \left( \ldots \xymatrix{ R[R[R[M]]] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & R[R[M]] \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & R[M] \ar@<0ex>[l] } \right)\] which comes with an augmentation towards \(M\). We will also show this augmentation is a homotopy equivalence of sets. By Lemmas 08NS, 08P2, and 08P1 this is equivalent to asking \(M\) to be the only nonzero cohomology group of the chain complex associated to the simplicial module \(X(M)\).

Example

Let \(A\) be a ring. Let \(\textit{Alg}_A\) be the category of commutative \(A\)-algebras. As an example of the above we can take \(i : \textit{Alg}_A \to \textit{Sets}\) to be the forgetful functor and \(F : \textit{Sets} \to \textit{Alg}_A\) to be the functor that associates to a set \(E\) the polynomial algebra \(A[E]\) on \(E\) over \(A\). (We apologize for the overlap in notation between this example and Example 0G5T.) For an \(A\)-algebra \(B\) the simplicial \(A\)-algebra \(X(B)\) will have the following shape \[X(B) = \left( \ldots \xymatrix{ A[A[A[B]]] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & A[A[B]] \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & A[B] \ar@<0ex>[l] } \right)\] which comes with an augmentation towards \(B\). We will also show this augmentation is a homotopy equivalence of sets. By Lemmas 08NS, 08P2, and 08P1 this is equivalent to asking \(B\) to be the only nonzero cohomology group of the chain complex of \(A\)-modules associated to \(X(B)\) viewed as a simplicial \(A\)-module.

Example

In Example 0G5T we have \(X_n(M) = R[R[\ldots [M]\ldots]]\) with \(n + 1\) brackets. We describe the maps constructed above using a typical element \[\xi = \sum\nolimits_i r_i\left[\sum\nolimits_j r_{ij}[m_{ij}]\right]\] of \(X_1(M)\). The maps \(d_0, d_1 : R[R[M]] \to R[M]\) are given by \[d_0(\xi) = \sum\nolimits_{i, j} r_ir_{ij}[m_{ij}] \quad\text{and}\quad d_1(\xi) = \sum\nolimits_i r_i\left[\sum\nolimits_j r_{ij}m_{ij}\right].\] The maps \(s_0, s_1 : R[R[M]] \to R[R[R[M]]]\) are given by \[s_0(\xi) = \sum\nolimits_i r_i\left[\left[\sum\nolimits_j r_{ij}[m_{ij}]\right]\right] \quad\text{and}\quad s_1(\xi) = \sum\nolimits_i r_i\left[\sum\nolimits_j r_{ij}[[m_{ij}]]\right].\]

Example

In Example 08NA we have \(X_n(B) = A[A[\ldots [B]\ldots]]\) with \(n + 1\) brackets. We describe the maps constructed above using a typical element \[\xi = \sum\nolimits_i a_i [x_{i, 1}] \ldots [x_{i, m_i}] \in A[A[B]] = X_1(B)\] where for each \(i, j\) we can write \[x_{i, j} = \sum a_{i, j, k} [b_{i, j, k, 1}] \ldots [b_{i, j, k, n_{i, j, k}}] \in A[B]\] Obviously this is horrendous! To ease the notation, to see what the \(A\)-algebra maps \(d_0, d_1 : A[A[B]] \to A[B]\) are doing it suffices to see what happens to the variables \([x]\) where \[x = \sum a_k [b_{k, 1}] \ldots [b_{k, n_k}] \in A[B]\] is a general element. For these we get \[d_0([x]) = x = \sum a_k [b_{k, 1}] \ldots [b_{k, n_k}] \quad\text{and}\quad d_1([x]) = \left[\sum a_k b_{k, 1} \ldots b_{k, n_k}\right]\] The maps \(s_0, s_1 : A[A[B]] \to A[A[A[B]]]\) are given by \[s_0([x]) = \left[\left[\sum a_k [b_{k, 1}] \ldots [b_{k, n_k}]\right]\right] \quad\text{and}\quad s_1([x]) = \left[\sum a_k [[b_{k, 1}]] \ldots [[b_{k, n_k}]]\right]\]

Example

Going back to the example discussed in Example 08NA our Lemma 08ND signifies that for any ring map \(A \to B\) the map of simplicial rings \[\xymatrix{ A[A[A[B]]] \ar[d] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & A[A[B]] \ar[d] \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & A[B] \ar[d] \ar@<0ex>[l] \\ B \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & B \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & B \ar@<0ex>[l] }\] is a homotopy equivalence on underlying simplicial sets. Moreover, the inverse map constructed in Lemma 08ND is in degree \(n\) given by \[b \longmapsto [\ldots[b]\ldots]\] with obvious notation. In the other direction the lemma tells us that for every set \(E\) there is a homotopy equivalence \[\xymatrix{ A[A[A[A[E]]]] \ar[d] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & A[A[A[E]]] \ar[d] \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & A[A[E]] \ar[d] \ar@<0ex>[l] \\ A[E] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & A[E] \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & A[E] \ar@<0ex>[l] }\] of rings. The inverse map constructed in the lemma is in degree \(n\) given by the ring map \[\sum a_{e_1,\ldots,e_p}[e_1][e_2] \ldots [e_p] \longmapsto \sum a_{e_1,\ldots,e_p}[\ldots[e_1]\ldots][\ldots[e_2]\ldots] \ldots [\ldots[e_p]\ldots]\] (with obvious notation).

Example

Let \(K \to R\) be a ring map. In Situation 08N9 take \[\mathcal{A} = \text{Mod}_R,\qquad \mathcal{S} = \text{Mod}_K,\] let \(V : \mathcal{A} \to \mathcal{S}\) be restriction of scalars, and let \(U : \mathcal{S} \to \mathcal{A}\) be given by \(U(N) = R \otimes_K N\). For an \(R\)-module \(M\) the resulting augmented simplicial \(R\)-module, denoted \(\operatorname{Bar}(R/K, M)\), has terms \[\operatorname{Bar}_n(R/K, M) = R \otimes_K \ldots \otimes_K R \otimes_K M\] with \(n + 1\) factors \(R\). On a pure tensor \(r_0 \otimes \ldots \otimes r_n \otimes m\), the map \(d_i\) multiplies \(r_i\) and \(r_{i + 1}\) if \(0 \leq i < n\), the map \(d_n\) applies the action of \(r_n\) to \(m\), and the map \(s_i\) inserts \(1\) immediately after \(r_i\). The augmentation in degree zero is \[R \otimes_K M \longrightarrow M,\qquad r \otimes m \longmapsto rm.\] By Lemma 08ND the augmentation is a homotopy equivalence after restriction to simplicial \(K\)-modules. Hence Lemma 08P2 shows that its associated augmented chain complex is acyclic. The same holds for its normalized augmented chain complex by Lemma 019C. Thus both are resolutions of \(M\) by \(R\)-modules.