Section 14.5 (016I): Cosimplicial objects

14.5 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 14.5.1. Let $\mathcal{C}$ be a category.

A cosimplicial object $U$ of $\mathcal{C}$ is a covariant functor $U$ from $\Delta$ to $\mathcal{C}$ , in a formula:

$U : \Delta \longrightarrow \mathcal{C}$
If $\mathcal{C}$ is the category of sets, then we call $U$ a cosimplicial set.
If $\mathcal{C}$ is the category of abelian groups, then we call $U$ a cosimplicial abelian group.
A morphism of cosimplicial objects $U \to U'$ is a transformation of functors.
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 14.2 above.

Lemma 14.5.2. Let $\mathcal{C}$ be a category.

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 14.2.3.
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 $\sigma ^ n_ j = U(\sigma ^ n_ j)$ .
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 14.2.4. $\square$

Remark 14.5.3. 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:

If $i < j$ , then $\delta _ j \circ \delta _ i = \delta _ i \circ \delta _{j - 1}$ .
If $i < j$ , then $\sigma _ j \circ \delta _ i = \delta _ i \circ \sigma _{j - 1}$ .
We have $\text{id} = \sigma _ j \circ \delta _ j = \sigma _ j \circ \delta _{j + 1}$ .
If $i > j + 1$ , then $\sigma _ j \circ \delta _ i = \delta _{i - 1} \circ \sigma _ j$ .
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 $U_2 \to U_3$ . 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 14.5.4. The simplest example is the constant cosimplicial object with value $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ . In other words, $U_ n = X$ and all maps are $\text{id}_ X$ .

Example 14.5.5. Suppose that $X\to Y$ is a morphism of $C$ 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 14.5.6. For every $n \geq 0$ we denote $C[n]$ the cosimplicial set

$\Delta \longrightarrow \textit{Sets},\quad [k] \longmapsto \mathop{\mathrm{Mor}}\nolimits _{\Delta }([n], [k])$

This example is dual to Example 14.11.2.

Lemma 14.5.7. 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 $\sigma _ i^{n - 1} \circ \delta ^ n_ i = \text{id}_{U_ n}$ for $j < n$ . $\square$

Comments (2)

Comment #255 by Keenan Kidwell on July 27, 2013 at 19:25

In part 3 of Lemma 016K, "simplicial" should be "cosimplicial."

Comment #259 by Johan on July 31, 2013 at 01:27

Fixed here. Thanks!

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14.5 Cosimplicial objects

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