Section 14.3 (0169): Simplicial objects

14.3 Simplicial objects

Definition 14.3.1. Let $\mathcal{C}$ be a category.

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

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

Lemma 14.3.2. Let $\mathcal{C}$ be a category.

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 14.2.3, 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 $s^ n_ i \circ s^{n - 1}_ j = s^ n_{j + 1} \circ s^{n - 1}_ i$ .
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 $s^ n_ j = U(\sigma ^ n_ j)$ .
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 14.2.4. $\square$

Remark 14.3.3. 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:

If $i < j$ , then $d_ i \circ d_ j = d_{j - 1} \circ d_ i$ .
If $i < j$ , then $d_ i \circ s_ j = s_{j - 1} \circ d_ i$ .
We have $\text{id} = d_ j \circ s_ j = d_{j + 1} \circ s_ j$ .
If $i > j + 1$ , then $d_ i \circ s_ j = s_ j \circ d_{i - 1}$ .
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 $U_3 \to U_2$ . 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 14.3.4. The simplest example is the constant simplicial 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 14.3.5. 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 14.3.5 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 14.3.6. 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 $d_ i^{n + 1} \circ s^ n_ i = \text{id}_{U_ n}$ . $\square$

Comments (0)

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$ ). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

The Stacks project

14.3 Simplicial objects

Comments (0)

Post a comment