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\input{preamble}
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\begin{document}
\title{Simplicial Methods}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
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 \cite{SimpHom}.
An example of the things you can do is the paper
by Quillen on Homotopical Algebra, see \cite{Quillen}
or the paper on \'Etale Homotopy by Artin and Mazur, see \cite{ArtinMazur}.
\section{The category of finite ordered sets}
\label{section-Delta}
\noindent
The category $\Delta$ is the category with
\begin{enumerate}
\item objects $[0], [1], [2], \ldots$ with
$[n] = \{0, 1, 2, \ldots, n\}$ and
\item a morphism $[n] \to [m]$ is a nondecreasing map
$\{0, 1, 2, \ldots, n\} \to \{0, 1, 2, \ldots, m\}$
between the corresponding sets.
\end{enumerate}
Here {\it 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 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.
\begin{definition}
\label{definition-face-degeneracy}
For any integer $n\geq 1$, and any $0\leq j \leq n$ we let
{\it $\delta^n_j : [n-1] \to [n]$}
denote the injective order preserving map skipping $j$. For any
integer $n\geq 0$, and any $0\leq j \leq n$ we denote
{\it $\sigma^n_j : [n + 1] \to [n]$}
the surjective order preserving map with
$(\sigma^n_j)^{-1}(\{j\}) = \{j, j + 1\}$.
\end{definition}
\begin{lemma}
\label{lemma-face-degeneracy}
Any morphism in $\Delta$ can be written as a composition
of the morphisms $\delta^n_j$ and $\sigma^n_j$.
\end{lemma}
\begin{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.
\end{proof}
\begin{lemma}
\label{lemma-relations-face-degeneracy}
The morphisms $\delta^n_j$ and $\sigma^n_j$ satisfy the following relations.
\begin{enumerate}
\item 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.
\item 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.
\item 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.
\item 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.
\item 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.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-face-degeneracy-category}
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 \ref{lemma-relations-face-degeneracy} are satisfied,
and (c) any relation among the morphisms is a consequence of
those relations.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Simplicial objects}
\label{section-simplicial-object}
\begin{definition}
\label{definition-simplicial-object}
Let $\mathcal{C}$ be a category.
\begin{enumerate}
\item A {\it simplicial object $U$ of $\mathcal{C}$}
is a contravariant functor $U$ from $\Delta$ to
$\mathcal{C}$, in a formula:
$$
U : \Delta^{opp} \longrightarrow \mathcal{C}
$$
\item If $\mathcal{C}$ is the category of sets, then we call
$U$ a {\it simplicial set}.
\item If $\mathcal{C}$ is the category of abelian groups,
then we call $U$ a {\it simplicial abelian group}.
\item A {\it morphism of simplicial objects $U \to U'$}
is a transformation of functors.
\item The {\it category of simplicial objects of $\mathcal{C}$}
is denoted $\text{Simp}(\mathcal{C})$.
\end{enumerate}
\end{definition}
\noindent
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
$U(\varphi \circ \psi) = U(\psi) \circ U(\varphi)$.
\medskip\noindent
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 \ref{section-Delta} above.
\begin{lemma}
\label{lemma-characterize-simplicial-object}
Let $\mathcal{C}$ be a category.
\begin{enumerate}
\item 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 \ref{lemma-relations-face-degeneracy}.
\item Conversely, given a sequence of objects $U_n$ and morphisms
$d^n_j$, $s^n_j$ satisfying these relations there exists a unique
simplicial object $U$ in $\mathcal{C}$ such that $U_n = U([n])$,
$d^n_j = U(\delta^n_j)$, and $s^n_j = U(\sigma^n_j)$.
\item 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 $s^n_j$.
\end{enumerate}
\end{lemma}
\begin{proof}
This follows from Lemma \ref{lemma-face-degeneracy-category}.
\end{proof}
\begin{remark}
\label{remark-relations}
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:
\begin{enumerate}
\item If $i < j$, then $d_i \circ d_j = d_{j - 1} \circ d_i$.
\item If $i < j$, then $d_i \circ s_j = s_{j - 1} \circ d_i$.
\item We have $\text{id} = d_j \circ s_j = d_{j + 1} \circ s_j$.
\item If $i > j + 1$, then $d_i \circ s_j = s_j \circ d_{i - 1}$.
\item If $i \leq j$, then $s_i \circ s_j = s_{j + 1} \circ s_i$.
\end{enumerate}
This means that whenever the compositions on both the left and the
right are defined then the corresponding equality should hold.
\end{remark}
\noindent
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.
\medskip\noindent
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.
\begin{example}
\label{example-constant-simplicial-object}
The simplest example is the {\it constant} simplicial object with
value $X \in \Ob(\mathcal{C})$. In other words, $U_n = X$ and
all maps are $\text{id}_X$.
\end{example}
\begin{example}
\label{example-fibre-products-simplicial-object}
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.
\end{example}
\noindent
Geometrically Example \ref{example-fibre-products-simplicial-object}
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.
\begin{lemma}
\label{lemma-si-injective}
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.
\end{lemma}
\begin{proof}
This is true because $d_i^{n + 1} \circ s^n_i = \text{id}_{U_n}$.
\end{proof}
\section{Simplicial objects as presheaves}
\label{section-simplicial-presheaves}
\noindent
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 \ref{sites-definition-presheaf}.
And in fact, if $U$, $U'$ are simplicial objects
of $\mathcal{C}$, then we have
\begin{equation}
\label{equation-simplicial-set-presheaf}
\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.
\section{Cosimplicial objects}
\label{section-cosimplicial-object}
\noindent
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.
\begin{definition}
\label{definition-cosimplicial-object}
Let $\mathcal{C}$ be a category.
\begin{enumerate}
\item A {\it cosimplicial object $U$ of $\mathcal{C}$}
is a covariant functor $U$ from $\Delta$ to
$\mathcal{C}$, in a formula:
$$
U : \Delta \longrightarrow \mathcal{C}
$$
\item If $\mathcal{C}$ is the category of sets, then we call
$U$ a {\it cosimplicial set}.
\item If $\mathcal{C}$ is the category of abelian groups,
then we call $U$ a {\it cosimplicial abelian group}.
\item A {\it morphism of cosimplicial objects $U \to U'$}
is a transformation of functors.
\item The {\it category of cosimplicial objects of $\mathcal{C}$}
is denoted $\text{CoSimp}(\mathcal{C})$.
\end{enumerate}
\end{definition}
\noindent
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
$U(\varphi \circ \psi) = U(\varphi) \circ U(\psi)$.
\medskip\noindent
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 \ref{section-Delta} above.
\begin{lemma}
\label{lemma-characterize-cosimplicial-object}
Let $\mathcal{C}$ be a category.
\begin{enumerate}
\item 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 \ref{lemma-relations-face-degeneracy}.
\item 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 $\sigma^n_j = U(\sigma^n_j)$.
\item 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 $\sigma^n_j$.
\end{enumerate}
\end{lemma}
\begin{proof}
This follows from Lemma \ref{lemma-face-degeneracy-category}.
\end{proof}
\begin{remark}
\label{remark-relations-cosimplicial}
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:
\begin{enumerate}
\item If $i < j$, then
$\delta_j \circ \delta_i = \delta_i \circ \delta_{j - 1}$.
\item If $i < j$, then
$\sigma_j \circ \delta_i = \delta_i \circ \sigma_{j - 1}$.
\item We have
$\text{id} = \sigma_j \circ \delta_j = \sigma_j \circ \delta_{j + 1}$.
\item If $i > j + 1$, then
$\sigma_j \circ \delta_i = \delta_{i - 1} \circ \sigma_j$.
\item If $i \leq j$, then
$\sigma_j \circ \sigma_i = \sigma_i \circ \sigma_{j + 1}$.
\end{enumerate}
This means that whenever the compositions on both the left and the
right are defined then the corresponding equality should hold.
\end{remark}
\noindent
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.
\medskip\noindent
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.
\begin{example}
\label{example-constant-cosimplicial-object}
The simplest example is the {\it constant} cosimplicial object with
value $X \in \Ob(\mathcal{C})$. In other words, $U_n = X$ and
all maps are $\text{id}_X$.
\end{example}
\begin{example}
\label{example-push-outs-simplicial-object}
Suppose that $Y\to X$ 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
natural maps.
\end{example}
\begin{example}
\label{example-simplex-cosimplicial-set}
For every $n \geq 0$ we denote $C[n]$ the cosimplicial set
$$
\Delta \longrightarrow \textit{Sets},\quad
[k] \longmapsto \Mor_{\Delta}([n], [k])
$$
This example is dual to Example \ref{example-simplex-simplicial-set}.
\end{example}
\begin{lemma}
\label{lemma-di-injective}
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.
\end{lemma}
\begin{proof}
This is true because
$\sigma_i^{n - 1} \circ \delta^n_i = \text{id}_{U_n}$
for $j < n$.
\end{proof}
\section{Products of simplicial objects}
\label{section-products}
\noindent
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.
\begin{definition}
\label{definition-product}
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 {\it product of $U$ and $V$} is the simplicial object
$U \times V$ defined as follows:
\begin{enumerate}
\item $(U \times V)_n = U_n \times V_n$,
\item $d^n_i = (d^n_i, d^n_i)$, and
\item $s^n_i = (s^n_i, s^n_i)$.
\end{enumerate}
In other words, $U \times V$ is the product of the presheaves
$U$ and $V$ on $\Delta$.
\end{definition}
\begin{lemma}
\label{lemma-product}
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 $\mathcal{C}$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Fibre products of simplicial objects}
\label{section-fibre-products}
\noindent
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.
\begin{definition}
\label{definition-fibre-product}
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 {\it fibre product of $V$ and $W$ over $U$} is the simplicial object
$V \times_U W$ defined as follows:
\begin{enumerate}
\item $(V \times_U W)_n = V_n \times_{U_n} W_n$,
\item $d^n_i = (d^n_i, d^n_i)$, and
\item $s^n_i = (s^n_i, s^n_i)$.
\end{enumerate}
In other words, $V \times_U W$ is the fibre product of the presheaves
$V$ and $W$ over the presheaf $U$ on $\Delta$.
\end{definition}
\begin{lemma}
\label{lemma-fibre-product}
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 $\mathcal{C}$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Pushouts of simplicial objects}
\label{section-push-outs}
\noindent
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.
\begin{definition}
\label{definition-push-out}
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 {\it pushout of $V$ and $W$ over $U$} is the simplicial object
$V\amalg_U W$ defined as follows:
\begin{enumerate}
\item $(V \amalg_U W)_n = V_n \amalg_{U_n} W_n$,
\item $d^n_i = (d^n_i, d^n_i)$, and
\item $s^n_i = (s^n_i, s^n_i)$.
\end{enumerate}
In other words, $V\amalg_U W$ is the pushout of the presheaves
$V$ and $W$ over the presheaf $U$ on $\Delta$.
\end{definition}
\begin{lemma}
\label{lemma-push-out}
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 $\mathcal{C}$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Products of cosimplicial objects}
\label{section-products-cosimplicial}
\noindent
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.
\begin{definition}
\label{definition-product-cosimplicial-objects}
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 {\it product of $U$ and $V$} is the cosimplicial object
$U \times V$ defined as follows:
\begin{enumerate}
\item $(U \times V)_n = U_n \times V_n$,
\item 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 $U(\varphi) \times V(\varphi)$.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma-product-cosimplicial-objects}
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 $\mathcal{C}$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Fibre products of cosimplicial objects}
\label{section-fibre-products-cosimplicial}
\noindent
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.
\begin{definition}
\label{definition-fibre-product-cosimplicial-objects}
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 {\it fibre product of $V$ and $W$ over $U$} is the cosimplicial object
$V \times_U W$ defined as follows:
\begin{enumerate}
\item $(V \times_U W)_n = V_n \times_{U_n} W_n$,
\item 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 $V(\varphi) \times_{U(\varphi)} W(\varphi)$.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma-fibre-product-cosimplicial-objects}
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 $\mathcal{C}$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Simplicial sets}
\label{section-simplicial-set}
\noindent
Let $U$ be a simplicial set. It is a good idea to think of
$U_0$ as the {\it $0$-simplices}, the set $U_1$ as the
{\it $1$-simplices},
the set $U_2$ as the {\it $2$-simplices}, and so on.
\medskip\noindent
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.
\begin{definition}
\label{definition-terminology-simplicial-sets}
Let $U$ be a simplicial set.
We say $x$ is an {\it $n$-simplex of $U$} to signify that
$x$ is an element of $U_n$. We say that $y$ is the $j$the
{\it face of $x$} to signify that $d^n_jx = y$. We say that
$z$ is the $j$th {\it degeneracy of $x$} if $z = s^n_jx$.
A simplex is called {\it degenerate} if it is the degeneracy
of another simplex.
\end{definition}
\noindent
Here are a few fundamental examples.
\begin{example}
\label{example-simplex-simplicial-set}
For every $n \geq 0$ we denote $\Delta[n]$ the simplicial set
\begin{align*}
\Delta^{opp} & \longrightarrow \textit{Sets} \\
[k] & \longmapsto \Mor_{\Delta}([k], [n])
\end{align*}
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 $\text{id}_{[n]}$.
\end{example}
\begin{lemma}
\label{lemma-simplex-map}
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 $\Delta[n]$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{example}
\label{example-simplex-category}
Consider the category $\Delta/[n]$ of objects over $[n]$
in $\Delta$, see
Categories, Example \ref{categories-example-category-over-X}.
There is a functor $p : \Delta/[n] \to \Delta$.
The fibre category of $p$ over $[k]$, see
Categories, Section \ref{categories-section-fibred-groupoids},
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 \ref{categories-example-fibred-category-from-functor-of-points}.
And this presheaf of sets agrees with the simplicial set
$\Delta[n]$. In particular, from Equation
(\ref{equation-simplicial-set-presheaf}) and
Lemma \ref{lemma-simplex-map} above
we get the formula
$$
\Mor_{\textit{PSh}(\Delta)}(\Delta/[n], U) = U_n
$$
for any simplicial set $U$.
\end{example}
\begin{lemma}
\label{lemma-product-degenerate}
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 $n > a + b$.
\end{lemma}
\begin{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 $s^{n - 1}_i$.
\end{proof}
\section{Truncated simplicial objects and skeleton functors}
\label{section-skeleton}
\noindent
Let $\Delta_{\leq n}$ denote the full subcategory of
$\Delta$ with objects $[0], [1], [2], \ldots, [n]$.
Let $\mathcal{C}$ be a category.
\begin{definition}
\label{definition-truncated-simplicial-object}
An {\it $n$-truncated simplicial object of $\mathcal{C}$}
is a contravariant functor from $\Delta_{\leq n}$ to
$\mathcal{C}$. A {\it 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 $\text{Simp}_n(\mathcal{C})$.
\end{definition}
\noindent
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 {\it 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 \ref{remark-sk-literature} to avoid possible confusion
with other functors in the literature.
\section{Products with simplicial sets}
\label{section-product-with-simplicial-sets}
\noindent
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
\begin{equation}
\label{equation-functor-product-with-simplicial-set}
\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.
\begin{definition}
\label{definition-product-with-simplicial-set}
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 {\it 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 {\it product $U \times V$ exists}.
\end{definition}
\begin{lemma}
\label{lemma-check-product-with-simplicial-set}
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 (\ref{equation-functor-product-with-simplicial-set}).
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-back-to-U}
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.
\begin{enumerate}
\item The rule $(U, V) \mapsto U \times V$
defines a functor
$\textit{FSSets} \times \text{Simp}(\mathcal{C})
\to \text{Simp}(\mathcal{C})$.
\item 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
$(U \times V)_n = \coprod_u V_n$.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\noindent
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.
\begin{lemma}
\label{lemma-morphism-from-coproduct}
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
$$
\Mor_{\text{Simp}_n(\mathcal{C})}(\text{sk}_n(X \times \Delta[k]), W)
=
\Mor_\mathcal{C}(X, W_k).
$$
\end{lemma}
\begin{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
$\gamma_\alpha = V(\alpha) \circ f$.
\medskip\noindent
The truncated case is similar, and left to the reader.
\end{proof}
\noindent
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.
\section{Hom from simplicial sets into cosimplicial objects}
\label{section-hom-from-simplicial-sets-into-cosimplicial}
\noindent
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:
\begin{enumerate}
\item we set
$\Hom_\mathcal{C}(U, V)_n = \Mor_\mathcal{C}(U_n, V_n)$, and
\item for $\varphi : [m] \to [n]$ we take the map
$\Hom_\mathcal{C}(U, V)_m \to \Hom_\mathcal{C}(U, V)_n$
given by $f \mapsto V(\varphi) \circ f \circ U(\varphi)$.
\end{enumerate}
This is our motivation for the following definition.
\begin{definition}
\label{definition-hom-deltak-cosimplicial}
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 {\it $\Hom(U, V)$} to be
the cosimplicial object of $\mathcal{C}$ defined
as follows:
\begin{enumerate}
\item 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
\item 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.
\end{enumerate}
\end{definition}
\noindent
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 \ref{lemma-back-to-U} in the case of
products of simplicial sets with simplicial objects.
\section{Hom from cosimplicial sets into simplicial objects}
\label{section-hom-from-cosimplicial-sets-into-simplicial}
\noindent
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:
\begin{enumerate}
\item we set
$\Hom_\mathcal{C}(U, V)_n = \Mor_\mathcal{C}(U_n, V_n)$, and
\item for $\varphi : [m] \to [n]$ we take the map
$\Hom_\mathcal{C}(U, V)_n \to \Hom_\mathcal{C}(U, V)_m$
given by $f \mapsto V(\varphi) \circ f \circ U(\varphi)$.
\end{enumerate}
This is our motivation for the following definition.
\begin{definition}
\label{definition-hom-deltak-simplicial}
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 {\it $\Hom(U, V)$} to be
the simplicial object of $\mathcal{C}$ defined
as follows:
\begin{enumerate}
\item 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
\item 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.
\end{enumerate}
\end{definition}
\noindent
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 \ref{lemma-back-to-U} in the case of
products of simplicial sets with simplicial objects.
\medskip\noindent
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 \ref{example-simplex-cosimplicial-set}.
\begin{lemma}
\label{lemma-morphism-into-product}
With $X$, $k$ and $U$ as above.
\begin{enumerate}
\item For any simplicial object $V$ of
$\mathcal{C}$ we have the following
canonical bijection
$$
\Mor_{\text{Simp}(\mathcal{C})}(V, U)
\longrightarrow
\Mor_\mathcal{C}(V_k, X).
$$
wich maps $\gamma$ to the morphism $\gamma_k$ composed with
the projection onto the factor corresponding to $\text{id}_{[k]}$.
\item Similarly, if $W$ is an $k$-truncated simplicial object
of $\mathcal{C}$, then we have
$$
\Mor_{\text{Simp}_k(\mathcal{C})}(W, \text{sk}_k U)
=
\Mor_\mathcal{C}(W_k, X).
$$
\item The object $U$ constructed above is an
incarnation of $\Hom(C[k], X)$ where $C[k]$ is the cosimplicial set from
Example \ref{example-simplex-cosimplicial-set}.
\end{enumerate}
\end{lemma}
\begin{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
$\gamma_\alpha = f \circ V(\alpha)$.
\medskip\noindent
The truncated case is similar, and left to the reader.
\medskip\noindent
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 $U_n$.
\end{proof}
\section{Internal Hom}
\label{section-internal-hom}
\noindent
Let $\mathcal{C}$ be a category with finite nonempty
products. Let $U$, $V$ be simplicial objects $\mathcal{C}$.
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 {\it 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 \ref{lemma-morphism-from-coproduct}.
\medskip\noindent
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
``$\Hom(\Delta[n], U)$''.
\section{Hom from simplicial sets into simplicial objects}
\label{section-hom-from-simplicial-sets}
\noindent
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 $\mathcal{C}$.
\begin{definition}
\label{definition-hom-from-simplicial-set}
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 {\it $\Hom(U, V)$} 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 $\mathcal{C}$.
\end{definition}
\noindent
Of course $\Hom(U, V)$ need not exist.
Also, by the discussion in Section \ref{section-internal-hom}
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.
\begin{lemma}
\label{lemma-exists-hom-0-from-simplicial-set}
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.
\end{lemma}
\begin{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:
\begin{enumerate}
\item The set of objects of $\mathcal{U}$ is
$\coprod_{n \geq 0} U_n$,
\item a morphism from $u' \in U_m$ to $u \in U_n$
is a $\varphi : [m] \to [n]$ such that $U(\varphi)(u) = u'$
\item for $u \in U_n$ we set $\mathcal{V}(u) = V_n$, and
\item for $\varphi : [m] \to [n]$ such that $U(\varphi)(u) = u'$
we set $\mathcal{V}(\varphi) = V(\varphi) : V_n \to V_m$.
\end{enumerate}
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.
\end{proof}
\begin{lemma}
\label{lemma-exists-hom-0-from-simplicial-set-finite}
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.
\end{lemma}
\begin{proof}
We have to show that the category $\mathcal{U}$ described
in the proof of Lemma \ref{lemma-exists-hom-0-from-simplicial-set}
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 \ref{categories-lemma-initial}.
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 \ref{categories-definition-initial}.
\medskip\noindent
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 \ref{lemma-si-injective}, 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 $a \in \mathcal{U}_{\leq 2m_0}$.
\medskip\noindent
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 \ref{categories-definition-initial}.
\end{proof}
\begin{lemma}
\label{lemma-exists-hom-from-simplicial-set-finite}
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
$$
\Hom(U, V)_n = \Hom(U \times \Delta[n], V)_0.
$$
\end{lemma}
\begin{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 \ref{lemma-exists-hom-0-from-simplicial-set-finite}
because $U \times \Delta[n]$ is a simplicial set with finite
sets of simplices and no nondegenerate simplices in high enough degree,
see Lemma \ref{lemma-product-degenerate}.
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 $\mathcal{C}$.
\medskip\noindent
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.
\medskip\noindent
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 \ref{lemma-morphism-from-coproduct} 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.
\medskip\noindent
We also leave it to the reader to see that
$f \mapsto f'$ and $f' \mapsto f$ are mutually inverse
operations.
\end{proof}
\begin{lemma}
\label{lemma-hom-from-coprod}
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.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-exists-hom-from-simplicial-set-finite}
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
$$
(X \times \Delta[n] \times V)
\times_{X \times \Delta[n] \times U}
(X \times \Delta[n] \times W)
=
X \times \Delta[n] \times (V \amalg_U W)
$$
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 $\Hom(V \amalg_U W, T)_n$.
\end{proof}
\section{Splitting simplicial objects}
\label{section-splitting}
\noindent
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.)
\begin{definition}
\label{definition-split}
Let $\mathcal{C}$ be a category which admits finite nonempty coproducts.
We say a simplicial object $U$ of $\mathcal{C}$ is {\it split}
if there exist subobjects $N(U_m)$ of $U_m$, $m \geq 0$
with the property that
\begin{equation}
\label{equation-splitting}
\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 {\it split}
if there exist subobjects $N(U_m)$ of $U_m$, $r \geq m \geq 0$
with the property that (\ref{equation-splitting})
is an isomorphism for $r \geq n \geq 0$.
\end{definition}
\noindent
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.
\begin{lemma}
\label{lemma-splitting-simplicial-sets}
Let $U$ be a simplicial set. Then $U$ has a unique splitting
with $N(U_m)$ equal to the set of nondegenerate $m$-simplices.
\end{lemma}
\begin{proof}
From the definition it follows immediately, that if there is a
splitting then $N(U_m)$ has the 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 $\psi = \varphi$.
\end{proof}
\noindent
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 \ref{lemma-splitting-simplicial-sets}
is not functorial. Here is a case where it is functorial.
\begin{lemma}
\label{lemma-injective-map-simplicial-sets}
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)
no two nondegenerate simplices of $U$ are mapped
to the same simplex of $V$.
Then $f_n$ is injective for all $n$.
Same holds with ``injective'' replaced by
``surjective'' or ``bijective''.
\end{lemma}
\begin{proof}
Under hypothesis (a) we see that the map $f$ preserves
the disjoint union decompositions of the splitting
of Lemma \ref{lemma-splitting-simplicial-sets}, 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).
\end{proof}
\begin{lemma}
\label{lemma-simplicial-set-n-skel-sub}
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 $m > n$.
\end{lemma}
\begin{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.
\end{proof}
\begin{lemma}
\label{lemma-splitting-simplicial-groups}
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.
\end{lemma}
\begin{proof}
By induction on $n$ we will show that the choice of $N(U_m)$
in the lemma guarantees that (\ref{equation-splitting}) 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 \ref{remark-relations}.
\medskip\noindent
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.
\medskip\noindent
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
$d_n(x_1) = 0$. 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.
\medskip\noindent
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 \ref{remark-relations}.
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.
\medskip\noindent
The claim implies we can uniquely write
$$
x = s_0(z_0) + s_1(z_1) + \ldots + s_n(z_n) + x_0
$$
with $x_0 \in N(U_{n + 1})$ 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
$$
\Ker(d_0) \cap \ldots \cap \Ker(d_j)
=
N(U_{n + 1}) \oplus
\bigoplus\nolimits_{i = j + 1}^{i = n}
s_i\Big(\Ker(d_n) \cap \ldots \cap \Ker(d_{i - 1})\Big)
$$
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
$$
x = x_0 + s_0(z_{0, 0}) + s_1(z_{1, 0}) + \ldots + s_n(z_{n, 0}) +
\sum\nolimits_{0 \leq i \leq j \leq n - 1} s_i(s_j(z'_{i, j})).
$$
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 $0 \leq i_1 \leq i_2 \leq \ldots \leq i_k \leq n - k + 1$.
\end{proof}
\begin{lemma}
\label{lemma-splitting-abelian-category}
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 $\mathcal{A}$.
\end{lemma}
\begin{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
\ref{lemma-splitting-simplicial-groups}. 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 (\ref{equation-splitting})
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.
\end{proof}
\begin{lemma}
\label{lemma-injective-map-simplicial-abelian}
\begin{slogan}
The Dold-Kan normalization functor reflects
injectivity, surjectivity, and isomorphy.
\end{slogan}
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''.
\end{lemma}
\begin{proof}
This is clear from Lemma \ref{lemma-splitting-abelian-category}
and the definition of a splitting.
\end{proof}
\begin{lemma}
\label{lemma-N-d-in-N}
Let $\mathcal{A}$ be an abelian category.
Let $U$ be a simplicial object in $\mathcal{A}$.
Let $N(U_m)$ as in Lemma \ref{lemma-splitting-abelian-category} above.
Then $d^m_m(N(U_m)) \subset N(U_{m - 1})$.
\end{lemma}
\begin{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 \ref{remark-relations}.
The result follows.
\end{proof}
\begin{lemma}
\label{lemma-simplicial-abelian-n-skel-sub}
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 $m > n$.
\end{lemma}
\begin{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 \ref{lemma-splitting-abelian-category},
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.
\end{proof}
\section{Coskeleton functors}
\label{section-coskeleton}
\noindent
Let $\mathcal{C}$ be a category.
The {\it 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
\begin{equation}
\label{equation-cosk}
\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 {\it $\text{cosk}_nV$ 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 \ref{categories-section-opposite}.
\begin{example}
\label{example-cosk0}
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 \ref{example-fibre-products-simplicial-object}.
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
(\ref{equation-cosk}) holds.
\end{example}
\noindent
Recall the category $\Delta/[n]$, see Example \ref{example-simplex-category}.
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 $U(n)$.
\begin{lemma}
\label{lemma-existence-cosk}
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 \ref{categories-lemma-functorial-limit}.
\end{lemma}
\begin{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.
\medskip\noindent
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 $n = k$, $\varphi = \alpha'$, 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 $\text{sk}_m V \to U$.
\medskip\noindent
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 $V \to \text{cosk}_m U$.
\end{proof}
\begin{lemma}
\label{lemma-trivial-cosk}
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 $U_n$.
\end{lemma}
\begin{proof}
This is true because the category $(\Delta/[n])_{\leq m}$
has an final object in this case, namely the identity
map $[n] \to [n]$.
\end{proof}
\begin{lemma}
\label{lemma-recover-cosk}
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.
\end{lemma}
\begin{proof}
Combine Lemmas \ref{lemma-existence-cosk} and \ref{lemma-trivial-cosk}.
\end{proof}
\noindent
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 \ref{lemma-relations-face-degeneracy}. This motivates
the following lemma.
\begin{lemma}
\label{lemma-formula-limit}
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
$$
U_{n + 1} = \lim_{(\Delta/[n + 1])_{\leq n}^{opp}} U(n)
$$
\end{lemma}
\begin{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 $T$.
\end{proof}
\begin{lemma}
\label{lemma-work-out}
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
$$
\Mor_{\text{Simp}_{n + 1}(\mathcal{C})}(V, \tilde U)
=
\Mor_{\text{Simp}_n(\mathcal{C})}(\text{sk}_nV, U)
$$
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-trivial-cosk} there are identifications
$$
U_i = \lim_{(\Delta/[i])_{\leq n}^{opp}} U(i)
$$
for $0 \leq i \leq n$. By Lemma \ref{lemma-formula-limit}
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 \ref{lemma-existence-cosk}. This defines
an $(n + 1)$-truncated simplicial object $\tilde U$
with $\text{sk}_n \tilde U = U$.
\medskip\noindent
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.
\end{proof}
\begin{remark}
\label{remark-explicit-face-degeneracy}
Let $U$, and $U_{n + 1}$ be as in Lemma \ref{lemma-work-out}.
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 \ref{lemma-work-out}.
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.
\end{remark}
\begin{remark}
\label{remark-cosk-simplicial-sets}
The construction of Lemma \ref{lemma-work-out}
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.
\medskip\noindent
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 \ref{lemma-work-out}.
\end{remark}
\begin{remark}
\label{remark-inductive-coskeleton}
{\it 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:
\begin{enumerate}
\item To start, set $U^m = U$.
\item 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
\ref{lemma-work-out}.
\end{enumerate}
Since the construction of Lemma \ref{lemma-work-out} 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 \ref{lemma-work-out}
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.
\end{remark}
\begin{lemma}
\label{lemma-cosk-up}
Let $\mathcal{C}$ be a category which has finite limits.
\begin{enumerate}
\item For every $n$ the functor $\text{sk}_n : \text{Simp}(\mathcal{C})
\to \text{Simp}_n(\mathcal{C})$ has a right adjoint $\text{cosk}_n$.
\item 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 $\text{sk}_{n'}\text{cosk}_n$.
\item For every $m \geq n \geq 0$ and every $n$-truncated simplicial
object $U$ of $\mathcal{C}$ we have
$\text{cosk}_m \text{sk}_m \text{cosk}_n U = \text{cosk}_n U$.
\item 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 $m \geq n$.
\end{enumerate}
\end{lemma}
\begin{proof}
The existence in (1) follows from Lemma \ref{lemma-existence-cosk} above.
Parts (2) and (3) follow from the discussion
in Remark \ref{remark-inductive-coskeleton}. After this (4) is obvious.
\end{proof}
\begin{remark}
\label{remark-existence-cosk}
We do not need all finite limits in order to be able to define
the coskeleton functors. Here are some remarks
\begin{enumerate}
\item We have seen in Examples \ref{example-cosk0} that if $\mathcal{C}$
has products of pairs of objects then $\text{cosk}_0$ exists.
\item For $k > 0$ the functor $\text{cosk}_k$ exists if
$\mathcal{C}$ has finite connected limits.
\end{enumerate}
This is clear from the inductive procedure of constructing coskeleta
(Remarks \ref{remark-cosk-simplicial-sets} and
\ref{remark-inductive-coskeleton}) 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 \ref{lemma-existence-cosk}
are connected. Observe that we do not need the categories
for $n \leq k$ by Lemma \ref{lemma-trivial-cosk} or
Lemma \ref{lemma-recover-cosk}. (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.)
\end{remark}
\begin{lemma}
\label{lemma-cosk-product}
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.
\end{lemma}
\begin{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.
\end{proof}
\begin{lemma}
\label{lemma-cosk-fibre-product}
Assume $\mathcal{C}$ has fibre products.
Let $U, V, W$ be $n$-truncated simplicial objects of the
category $\mathcal{C}$. Then
$$
\text{cosk}_n (V \times_U W)
=
\text{cosk}_nU \times_{\text{cosk}_n U} \text{cosk}_nV
$$
whenever the left and right hand side exist.
\end{lemma}
\begin{proof}
Omitted, but very similar to the proof of
Lemma \ref{lemma-cosk-product} above.
\end{proof}
\begin{lemma}
\label{lemma-cosk-above-object}
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 $k \geq 1$.
\end{lemma}
\begin{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 \ref{categories-lemma-connected-limit-over-X}
because the categories $(\Delta/[n])_{\leq k}$ for $k \geq 1$ and
$n \geq k + 1$ used in Lemma \ref{lemma-existence-cosk}
are connected. Observe that we do not need the categories
for $n \leq k$ by Lemma \ref{lemma-trivial-cosk} or
Lemma \ref{lemma-recover-cosk}.
\end{proof}
\begin{lemma}
\label{lemma-simplex-cosk}
The canonical map
$\Delta[n] \to \text{cosk}_1 \text{sk}_1 \Delta[n]$
is an isomorphism.
\end{lemma}
\begin{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 $u \in X_1$. In other words, we have
$$
\Mor(U, \Delta[n])
=
\Mor(\text{sk}_1 U, \text{sk}_1 \Delta[n])
$$
as desired.
\end{proof}
\section{Augmentations}
\label{section-augmentation}
\begin{definition}
\label{definition-augmentation}
Let $\mathcal{C}$ be a category.
Let $U$ be a simplicial object of $\mathcal{C}$.
An {\it augmentation $\epsilon : U \to X$ of
$U$ towards an object $X$ of $\mathcal{C}$}
is a morphism from $U$ into the constant simplicial
object $X$.
\end{definition}
\begin{lemma}
\label{lemma-augmentation-howto}
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 $\epsilon_0 \circ d^1_0 = \epsilon_0 \circ d^1_1$.
\end{lemma}
\begin{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.
\end{proof}
\begin{lemma}
\label{lemma-cosk-minus-one}
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 \ref{example-fibre-products-simplicial-object}.
For any simplicial object $V$ of $\mathcal{C}$ we have
\begin{eqnarray*}
\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{eqnarray*}
In particular we have $U = \text{cosk}_1 \text{sk}_1 U$.
\end{lemma}
\begin{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.
\end{proof}
\begin{remark}
\label{remark-augmentation}
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 \ref{lemma-augmentation-howto} and \ref{lemma-cosk-minus-one}
we obtain a canonical morphism
$V \to U$.
\end{remark}
\section{Left adjoints to the skeleton functors}
\label{section-adjoint-left}
\noindent
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.
\medskip\noindent
We use a similar construction as in Section \ref{section-skeleton}.
Recall the category $[n]/\Delta$ of objects
under $[n]$, see
Categories, Example \ref{categories-example-category-under-X}.
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 $U(n')$.
\begin{lemma}
\label{lemma-left-adjoint-exists}
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 \ref{categories-lemma-functorial-colimit}.
\end{lemma}
\begin{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.
\medskip\noindent
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 $\gamma' : i_{m!}U \to V$.
\medskip\noindent
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.
\end{proof}
\begin{lemma}
\label{lemma-recovering-U}
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 $U_n$.
\end{lemma}
\begin{proof}
This is so because the category $([n]/\Delta)_{\leq m}$
has an initial object, namely $\text{id} : [n] \to [n]$.
\end{proof}
\begin{lemma}
\label{lemma-recovering-U-for-real}
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.
\end{lemma}
\begin{proof}
Combine Lemmas \ref{lemma-left-adjoint-exists} and \ref{lemma-recovering-U}.
\end{proof}
\begin{lemma}
\label{lemma-imshriek-sets}
If $U$ is an $m$-truncated simplicial set and $n > m$
then all $n$-simplices of $i_{m!}U$ are degenerate.
\end{lemma}
\begin{proof}
This can be seen from the construction of
$i_{m!}U$ in Lemma \ref{lemma-left-adjoint-exists},
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 \ref{lemma-simplicial-set-n-skel-sub}.
By the adjunction formula,
since $\text{sk}_m V' = U$, there is an inverse to the
injection $V' \to V$. Hence $V' = V$.
\end{proof}
\begin{lemma}
\label{lemma-n-skeleton-sets}
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 \ref{lemma-simplicial-set-n-skel-sub}.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-imshriek-sets} 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 \ref{lemma-injective-map-simplicial-sets} applies.
Thus $i_{n!} \text{sk}_n U \to U$
is injective. The result follows easily from this.
\end{proof}
\begin{remark}
\label{remark-sk-literature}
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 \ref{lemma-n-skeleton-sets} 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 $\text{cosk}_m$.
\end{remark}
\begin{lemma}
\label{lemma-glue-simplex}
Let $U \subset V$ be simplicial sets.
Suppose $n \geq 0$ and $x \in V_n$, $x \not \in U_n$ are such that
\begin{enumerate}
\item $V_i = U_i$ for $i < n$,
\item $V_n = U_n \cup \{x\}$,
\item any $z \in V_j$, $z \not \in U_j$ for $j > n$
is degenerate.
\end{enumerate}
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.
\end{lemma}
\begin{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 \ref{lemma-injective-map-simplicial-sets}.
\end{proof}
\begin{lemma}
\label{lemma-add-simplices}
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 \ref{lemma-glue-simplex} applies to each of
the inclusions $W^i \subset W^{i + 1}$.
\end{lemma}
\begin{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
\ref{lemma-glue-simplex}. 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.
\end{proof}
\begin{lemma}
\label{lemma-imshriek-abelian}
Let $\mathcal{A}$ be an abelian category
Let $U$ be an $m$-truncated simplicial object of
$\mathcal{A}$. For $n > m$ we have $N(i_{m!}U)_n = 0$.
\end{lemma}
\begin{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 \ref{lemma-simplicial-abelian-n-skel-sub}.
By the adjunction formula,
since $\text{sk}_m V' = U$, there is an inverse to the
injection $V' \to V$. Hence $V' = V$.
\end{proof}
\begin{lemma}
\label{lemma-n-skeleton-abelian}
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 \ref{lemma-simplicial-abelian-n-skel-sub}.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-imshriek-abelian}
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 \ref{lemma-recovering-U-for-real}.
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 \ref{lemma-injective-map-simplicial-abelian}
applies. Thus $i_{n!} \text{sk}_n U \to U$
is injective. The result follows easily from this.
\end{proof}
\noindent
Here is another way to think about the coskeleton functor
using the material above.
\begin{lemma}
\label{lemma-cosk-shriek}
Let $\mathcal{C}$ be a category with finite coproducts
and finite limits. Let $V$ be a simplicial object of $\mathcal{C}$.
In this case
$$
(\text{cosk}_n \text{sk}_n V)_{n + 1}
=
\Hom(i_{n !}\text{sk}_n \Delta[n + 1], V)_0.
$$
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-morphism-from-coproduct}
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.
\medskip\noindent
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.
\end{proof}
\section{Simplicial objects in abelian categories}
\label{section-abelian}
\noindent
Recall that an abelian category is defined
in Homology, Section \ref{homology-section-abelian-categories}.
\begin{lemma}
\label{lemma-abelian}
Let $\mathcal{A}$ be an abelian category.
\begin{enumerate}
\item The categories $\text{Simp}(\mathcal{A})$ and
$\text{CoSimp}(\mathcal{A})$ are abelian.
\item 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.
\item 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.
\item 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 $B_i$.
\end{enumerate}
\end{lemma}
\begin{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 \ref{lemma-product} and
\ref{lemma-product-cosimplicial-objects}.
Kernels and cokernels are obtained by taking
termwise kernels and cokernels.
\end{proof}
\noindent
For an object $A$ of $\mathcal{A}$ and an integer
$k$ consider the $k$-truncated simplicial object
$U$ with
\begin{enumerate}
\item $U_i = 0$ for $i < k$,
\item $U_k = A$,
\item all morphisms $U(\varphi)$ equal to zero,
except $U(\text{id}_{[k]}) = \text{id}_A$.
\end{enumerate}
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 $i_{k!}U \to \text{cosk}_kU$.
\begin{lemma}
\label{lemma-eilenberg-maclane-object}
With $A$, $k$ and $U$ as above, so $U_i = 0$, $i < k$ and $U_k = A$.
\begin{enumerate}
\item 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
$$
\Mor(V, U)
=
\{ f : V_k \to A \mid f \circ s^{k - 1}_i = 0, \ i = 0, \ldots, k - 1 \}.
$$
\item The object $i_{k!} U$ has $n$th term equal to
$\bigoplus_\alpha A$ where $\alpha$ runs over all
surjective morphisms $\alpha : [n] \to [k]$.
\item 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.
\item The object $\text{cosk}_k U$ has $n$th term equal to
$\bigoplus_\beta A$, where $\beta$ runs over all
injective morphisms $\beta : [k] \to [n]$.
\item 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.
\item 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.
\item The canonical map
$
c : i_{k !} U \to \text{cosk}_k U
$
is injective.
\end{enumerate}
\end{lemma}
\begin{proof}
The proof of (1) is left to the reader.
\medskip\noindent
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.
\medskip\noindent
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{