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14.33. Standard resolutions

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

Situation 14.33.1. Let $\mathcal{A}$, $\mathcal{S}$ be categories and let $i : \mathcal{A} \to \mathcal{S}$ be a functor with a left adjoint $F : \mathcal{S} \to \mathcal{A}$.

In this very general situation we will construct a simplicial object $X$ in the category of functors from $\mathcal{A}$ to $\mathcal{A}$. Please keep the following example in mind while we do this.

Example 14.33.2. As an example of the above we can take $i : \textit{Rings} \to \textit{Sets}$ to be the forgetful functor and $F : \textit{Sets} \to \textit{Rings}$ to be the functor that associates to a set $E$ the polynomial algebra $\mathbf{Z}[E]$ on $E$ over $\mathbf{Z}$. The simplicial object $X$ when evaluated on an ring $A$ will give the simplicial ring $$ \xymatrix{ \mathbf{Z}[\mathbf{Z}[\mathbf{Z}[A]]] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & \mathbf{Z}[\mathbf{Z}[A]] \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & \mathbf{Z}[A] \ar@<0ex>[l] } $$ which comes with an augmentation towards $A$. We will also show this augmentation is a homotopy equivalence.

For the general construction we will use the horizontal composition as defined in Categories, Section 4.27. The definition of the adjunction morphisms $k : F \circ i \to \text{id}_\mathcal{A}$ and $t : \text{id}_\mathcal{S} \to i \circ F$ in Categories, Section 4.24 shows that the compositions \begin{equation} \tag{14.33.2.1} i \xrightarrow{t \star 1_i} i \circ F \circ i \xrightarrow{1_i \star k} i \quad\text{and}\quad F \xrightarrow{1_F \star t} F \circ i \circ F \xrightarrow{k \star 1_F} F \end{equation} are the identity morphisms. Here to define the morphism $t \star 1$ we silently identify $i$ with $\text{id}_\mathcal{S} \circ i$ and $1$ stands for $\text{id}_i : i \to i$. We will use this notation and these relations repeatedly in what follows. For $n \geq 0$ we set $$ X_n = (F \circ i)^{\circ (n + 1)} = F \circ i \circ F \circ \ldots \circ i \circ F $$ In other words, $X_n$ is the $(n + 1)$-fold composition of $F \circ i$ with itself. We also set $X_{-1} = \text{id}_\mathcal{A}$. We have $X_{n + m + 1} = X_n \circ X_m$ for all $n, m \geq -1$. We will endow this sequence of functors with the structure of a simplicial object of $\text{Fun}(\mathcal{A}, \mathcal{A})$ by constructing the morphisms of functors $$ d^n_j : X_n \to X_{n - 1},\quad s^n_j : X_n \to X_{n + 1} $$ satisfying the relations displayed in Lemma 14.2.3. Namely, we set $$ d^n_j = 1_{X_{j - 1}} \star k \star 1_{X_{n - j - 1}} \quad\text{and}\quad s^n_j = 1_{X_{j - 1} \circ F} \star t \star 1_{i \circ X_{n - j - 1}} $$ Finally, write $\epsilon_0 = k : X_0 \to X_{-1}$.

Example 14.33.3. In Example 14.33.2 we have $X_n(A) = \mathbf{Z}[\mathbf{Z}[\ldots [A]\ldots]]$ with $n + 1$ brackets. We describe the maps constructed above using a typical element $\xi = \sum n_i[n_{ij}[a_{ij}]]$ of $X_1(A)$. The maps $d_0, d_1 : \mathbf{Z}[\mathbf{Z}[A]] \to \mathbf{Z}[A]$ are given by $$ d_0(\xi) = \sum n_in_{ij}[a_{ij}] \quad\text{and}\quad d_1(\xi) = \sum n_i[n_{ij}a_{ij}]. $$ The maps $s_0, s_1 : \mathbf{Z}[\mathbf{Z}[A]] \to \mathbf{Z}[\mathbf{Z}[\mathbf{Z}[A]]]$ are given by $$ s_0(\xi) = \sum n_i[[n_{ij}[a_{ij}]]] \quad\text{and}\quad s_1(\xi) = \sum n_i[n_{ij}[[a_{ij}]]]. $$

Lemma 14.33.4. In Situation 14.33.1 the system $X = (X_n, d^n_j, s^n_j)$ is a simplicial object of $\text{Fun}(\mathcal{A}, \mathcal{A})$ and $\epsilon_0$ defines an augmentation $\epsilon$ from $X$ to the constant simplicial object with value $X_{-1} = \text{id}_\mathcal{A}$.

Proof. Suppose that we have shown that $X$ is a simplicial object. Then to prove that $\epsilon_0 = k$ defines an augmentation we have to check that $\epsilon_0 \circ d^1_0 = \epsilon_0 \circ d^1_1$ as morphisms $X_1 \to X_{-1}$, see Lemma 14.20.2. In other words, we have to check that the diagram $$ \xymatrix{ F \circ i \circ F \circ i \ar[rr]_{1_{F \circ i} \star k} \ar[d]_{k \star 1_{F \circ i}} & & F \circ i \ar[d]^k \\ F \circ i \ar[rr]^k & & \text{id}_\mathcal{A} } $$ is commutative. More precisely we should write this as the equality $$ (k \star 1_{\text{id}_\mathcal{A}}) \circ (1_{F \circ i} \star k) = (1_{\text{id}_\mathcal{A}} \star k) \circ (k \star 1_{F \circ i}) $$ as morphisms $(F \circ i) \circ (F \circ i) \to \text{id}_\mathcal{A} \circ \text{id}_\mathcal{A}$. Applying the general property of Categories, Lemma 4.27.2 both sides expand to $k \star k$ when equality holds.

To prove that $X$ is a simplicial object we have to check (see Remark 14.3.3):

  1. If $i < j$, then $d_i \circ d_j = d_{j - 1} \circ d_i$.
  2. If $i < j$, then $d_i \circ s_j = s_{j - 1} \circ d_i$.
  3. We have $\text{id} = d_j \circ s_j = d_{j + 1} \circ s_j$.
  4. If $i > j + 1$, then $d_i \circ s_j = s_j \circ d_{i - 1}$.
  5. If $i \leq j$, then $s_i \circ s_j = s_{j + 1} \circ s_i$.

Relation (1) is proved in exactly the same manner as the proof of the equality $\epsilon_0 \circ d^1_0 = \epsilon_0 \circ d^1_1$ above.

The simplest case of equality (5) is the commutativity of the diagram $$ \xymatrix{ F \circ i \ar[rrr]_{1_F \star t \star 1_i} \ar[d]_{1_F \star t \star 1_i} & & & F \circ i \circ F \circ i \ar[d]^{1_F \star t \star 1_{i \circ F \circ i}} \\ F \circ i \circ F \circ i \ar[rrr]^{1_{F \circ i \circ F} \star t \star 1_i} & & & F \circ i \circ F \circ i \circ F \circ i } $$ which holds because both compositions expand to the morphism $1_F \star t \star t \star 1_i$ from $F \circ \text{id}_\mathcal{A} \circ \text{id}_\mathcal{A} \circ i$ to $F \circ (i \circ F) \circ (i \circ F) \circ i$. All other cases of (5) are proved in the same manner.

The simplest case of equalities (2) and (4) is the commutativity of the diagram $$ \xymatrix{ F \circ i \circ F \circ i \ar[rrr]_{1_F \star k \star 1_i} \ar[d]_{1_{F \circ i \circ F \circ i} \star t} & & & F \circ i \ar[d]_{1_{F \circ i} \star t} \\ F \circ i \circ F \circ i \circ F \circ i \ar[rrr]^{1_F \star k \star 1_{i \circ F \circ i \circ F}} & & & F \circ i \circ F \circ i } $$ which again holds because both compositions expand to give $1_F \star k \star 1_i \star t$ as maps from $F \circ (i \circ F) \circ i \circ \text{id}_\mathcal{A}$ to $F \circ \text{id}_\mathcal{A} \circ i \circ (F \circ i)$. All other cases of (2) and (4) are proved in the same manner.

The relations (3) are the only nontrivial ones and these are consequences of the fact that the compositions in (14.33.2.1) are the identity. For example, the simplest case of (3) states that the compositions $$ \xymatrix{ F \circ i \ar[rr]_{1_{F \circ i} \star t} \ar[d]_{t \star 1_{F \circ i}} & & F \circ i \circ F \circ i \ar[d]^{1_F \star k \star 1_i} \\ F \circ i \circ F \circ i \ar[rr]^{1_F \star k \star 1_i} & & F \circ i } $$ go around the diagram either way evaluate out to the identity. Going around the top the composition evaluates to $1_F \star ((k \star 1_i) \circ (1_i \star t))$ which is the identity by what was said above. The other cases of (3) are proved in the same manner. $\square$

Before reading the proof of the following lemma, we strongly urge the reader to look at the example discussed in Example 14.33.6 in order to understand the purpose of the lemma.

Lemma 14.33.5. In Situation 14.33.1 the maps $$ 1_i \star \epsilon : i \circ X \to i, \quad\text{and}\quad \epsilon \star 1_F : X \circ F \to F $$ are homotopy equivalences.

Proof. Denote $\epsilon_n : X_n \to X_{-1}$ the components of the augmentation morphism. We observe that $\epsilon_n = k^{\star (n + 1)}$, the $(n + 1)$-fold $\star$-composition of $k$. Recall that $t : \text{id}_\mathcal{S} \to i \circ F$ is the adjunction map. We have the morphisms $$ t^{\star (n + 1)} \star 1_i : i \longrightarrow i \circ (F \circ i)^{\circ (n + 1)} = i \circ X_n $$ which are right inverse to $1_i \star \epsilon_n$ and the morphisms $$ 1_F \star t^{\star (n + 1)} : F \longrightarrow (F \circ i)^{\circ (n + 1)} \circ F = X_n \circ F $$ which are right inverse to $\epsilon_n \star 1_F$. These morphisms determine morphisms of simplicial objects $b : i \to i \circ X$ and $c : F \to X \circ F$ (proof omitted). To finish it suffices to construct a homotopy between the morphisms $1, b \circ (1_i \star \epsilon) : i \circ X \to i \circ X$ and between the two morphisms $1, c \circ (\epsilon \star 1_F) : X \circ F \to X \circ F$.

To show the morphisms $b \circ (1_i \star \epsilon), 1 : i \circ X \to i \circ X$ are homotopic we have to construct morphisms $$ h_{n, j} : i \circ X_n \to i \circ X_n $$ for $n \geq 0$ and $0 \leq j \leq n + 1$ satisfying the relations described in Lemma 14.26.2. See also Remark 14.26.4. We are forced to set $h_{n, 0} = 1$ and $$ h_{n , n + 1} = b_n \circ (1_i \star \epsilon_n) = (t^{\star (n + 1)} \star 1_i) \circ (1_i \star k^{\star (n + 1)}) $$ Thus a logical choice is $$ h_{n , j} = (t^{\star (j)} \star 1) \circ (1_i \star k^{\star (j)} \star 1) $$ Here and in the rest of the proof we drop the subscript from $1$ if it is clear by knowing the source and the target of the morphism what this subscript should be. Writing $$ i \circ X_n = i \circ F \circ i \circ \ldots \circ F \circ i $$ we can think of the morphism $h_{n, j}$ as collapsing the first $j$ pairs $(F \circ i)$ to $\text{id}_\mathcal{S}$ using $k^{\star (j)}$, then adding a $\text{id}_\mathcal{S}$ in front and expanding this to $j$ pairs $(i \circ F)$ using $t^{\star (j)}$. We have to prove

  1. We have $d^n_m \circ h_{n, j} = h_{n - 1, j - 1} \circ d^n_m$ for $j > m$.
  2. We have $d^n_m \circ h_{n, j} = h_{n - 1, j} \circ d^n_m$ for $j \leq m$.
  3. We have $s^n_m \circ h_{n, j} = h_{n + 1, j + 1} \circ s^n_m$ for $j > m$.
  4. We have $s^n_m \circ h_{n, j} = h_{n + 1, j} \circ s^n_m$ for $j \leq m$.

Recall that $d^n_m$ is given by applying $k$ to the $(m + 1)$st pair $(F \circ i)$ in the functor $X_n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (2) holds (because $k$ does $\star$-commute with $k$, but not with $t$). Similarly, $s^n_m$ is given by applying $1_F \star t \star i_i$ to the $(m + 1)$st pair $(F \circ i)$ in $X_n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (4) holds. In the two remaining cases one uses the fact that the compositions in (14.33.2.1) are the identity causes the drop in the index $j$. Some details omitted.

To show the morphisms $1, c \circ (\epsilon \star 1_F) : X \circ F \to X \circ F$ are homotopic we have to construct morphisms $$ h_{n, j} : X_n \circ F \longrightarrow X_n \circ F $$ for $n \geq 0$ and $0 \leq j \leq n + 1$ satisfying the relations described in Lemma 14.26.2. See also Remark 14.26.4. We are forced to set $h_{n, 0} = 1$ and $$ h_{n , n + 1} = c_n \circ (\epsilon_n \star 1_F) = (1_F \star t^{\star (n + 1)}) \circ (k^{\star (n + 1)} \star 1_F) $$ Thus a logical choice is $$ h_{n , j} = (1_F \star t^{\star (j)} \star 1) \circ (k^{\star (j)} \star 1) $$ Here and in the rest of the proof we drop the subscript from $1$ if it is clear by knowing the source and the target of the morphism what this subscript should be. Writing $$ X_n \circ F = F \circ i \circ F \circ \ldots \circ i \circ F $$ we can think of the morphism $h_{n, j}$ as collapsing the first $j$ pairs $(F \circ i)$ to $\text{id}_\mathcal{S}$ using $k^{\star (j)}$, then inserting a $\text{id}_\mathcal{S}$ just after the first $F$ and expanding this to $j$ pairs $(i \circ F)$ using $t^{\star (j)}$. We have to prove

  1. We have $d^n_m \circ h_{n, j} = h_{n - 1, j - 1} \circ d^n_m$ for $j > m$.
  2. We have $d^n_m \circ h_{n, j} = h_{n - 1, j} \circ d^n_m$ for $j \leq m$.
  3. We have $s^n_m \circ h_{n, j} = h_{n + 1, j + 1} \circ s^n_m$ for $j > m$.
  4. We have $s^n_m \circ h_{n, j} = h_{n + 1, j} \circ s^n_m$ for $j \leq m$.

Recall that $d^n_m$ is given by applying $k$ to the $(m + 1)$st pair $(F \circ i)$ in the functor $X_n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (2) holds (because $k$ does $\star$-commute with $k$, but not with $t$). Similarly, $s^n_m$ is given by applying $1_F \star t \star i_i$ to the $(m + 1)$st pair $(F \circ i)$ in $X_n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (4) holds. In the two remaining cases one uses the fact that the compositions in (14.33.2.1) are the identity causes the drop in the index $j$. Some details omitted. $\square$

Example 14.33.6. Going back to the example discussed in Example 14.33.2 our Lemma 14.33.5 signifies that for any ring $A$ the map of simplicial rings $$ \xymatrix{ \mathbf{Z}[\mathbf{Z}[\mathbf{Z}[A]]] \ar[d] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & \mathbf{Z}[\mathbf{Z}[A]] \ar[d] \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & \mathbf{Z}[A] \ar[d] \ar@<0ex>[l] \\ A \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & A \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & A \ar@<0ex>[l] } $$ is a homotopy equivalence on underlying simplicial sets. Moreover, the inverse map constructed in Lemma 14.33.5 is in degree $n$ given by $$ a \longmapsto [\ldots[a]\ldots] $$ with obvious notation. In the other direction the lemma tells us that for every set $E$ there is a homotopy equivalence $$ \xymatrix{ \mathbf{Z}[\mathbf{Z}[\mathbf{Z}[\mathbf{Z}[E]]]] \ar[d] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & \mathbf{Z}[\mathbf{Z}[\mathbf{Z}[E]]] \ar[d] \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & \mathbf{Z}[\mathbf{Z}[E]] \ar[d] \ar@<0ex>[l] \\ \mathbf{Z}[E] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & \mathbf{Z}[E] \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & \mathbf{Z}[E] \ar@<0ex>[l] } $$ of rings. The inverse map constructed in the lemma is in degree $n$ given by the ring map $$ \sum m_{e_1,\ldots,e_p}[e_1][e_2] \ldots [e_p] \longmapsto \sum m_{e_1,\ldots,e_p}[\ldots[e_1]\ldots][\ldots[e_2]\ldots] \ldots [\ldots[e_p]\ldots] $$ (with obvious notation).

    The code snippet corresponding to this tag is a part of the file simplicial.tex and is located in lines 6065–6522 (see updates for more information).

    \section{Standard resolutions}
    \label{section-standard}
    
    \noindent
    Some of the material in this section can be found in
    \cite[Appendix 1]{Godement} and \cite[I 1.5]{cotangent}.
    
    \begin{situation}
    \label{situation-adjoint-functors}
    Let $\mathcal{A}$, $\mathcal{S}$ be categories and let
    $i : \mathcal{A} \to \mathcal{S}$ be a functor with a left adjoint
    $F : \mathcal{S} \to \mathcal{A}$.
    \end{situation}
    
    \noindent
    In this very general situation we will construct a simplicial
    object $X$ in the category of functors from $\mathcal{A}$ to $\mathcal{A}$.
    Please keep the following example in mind while we do this.
    
    \begin{example}
    \label{example-polynomial-algebra}
    As an example of the above we can take
    $i : \textit{Rings} \to \textit{Sets}$ to be the forgetful functor
    and $F : \textit{Sets} \to \textit{Rings}$ to be the functor that associates
    to a set $E$ the polynomial algebra $\mathbf{Z}[E]$ on $E$ over $\mathbf{Z}$.
    The simplicial object $X$ when evaluated on an ring $A$ will give
    the simplicial ring
    $$
    \xymatrix{
    \mathbf{Z}[\mathbf{Z}[\mathbf{Z}[A]]]
    \ar@<2ex>[r]
    \ar@<0ex>[r]
    \ar@<-2ex>[r]
    &
    \mathbf{Z}[\mathbf{Z}[A]]
    \ar@<1ex>[r]
    \ar@<-1ex>[r]
    \ar@<1ex>[l]
    \ar@<-1ex>[l]
    &
    \mathbf{Z}[A]
    \ar@<0ex>[l]
    }
    $$
    which comes with an augmentation towards $A$. We will also show this
    augmentation is a homotopy equivalence.
    \end{example}
    
    \noindent
    For the general construction we will use the horizontal composition as
    defined in Categories, Section \ref{categories-section-formal-cat-cat}.
    The definition of the adjunction morphisms
    $k : F \circ i \to \text{id}_\mathcal{A}$ and
    $t : \text{id}_\mathcal{S} \to i \circ F$
    in Categories, Section \ref{categories-section-adjoint}
    shows that the compositions
    \begin{equation}
    \label{equation-composition}
    i \xrightarrow{t \star 1_i} i \circ F \circ i \xrightarrow{1_i \star k} i
    \quad\text{and}\quad
    F \xrightarrow{1_F \star t} F \circ i \circ F \xrightarrow{k \star 1_F} F
    \end{equation}
    are the identity morphisms. Here to define the morphism $t \star 1$ we
    silently identify $i$ with $\text{id}_\mathcal{S} \circ i$ and
    $1$ stands for $\text{id}_i : i \to i$. We will use this notation and
    these relations repeatedly in what follows.
    For $n \geq 0$ we set
    $$
    X_n = (F \circ i)^{\circ (n + 1)} =
    F \circ i \circ F \circ \ldots \circ i \circ F
    $$
    In other words, $X_n$ is the $(n + 1)$-fold composition of $F \circ i$
    with itself. We also set $X_{-1} = \text{id}_\mathcal{A}$.
    We have $X_{n + m + 1} = X_n \circ X_m$ for all $n, m \geq -1$.
    We will endow this sequence of functors with the structure of a
    simplicial object of $\text{Fun}(\mathcal{A}, \mathcal{A})$ by constructing
    the morphisms of functors
    $$
    d^n_j : X_n \to X_{n - 1},\quad s^n_j : X_n \to X_{n + 1}
    $$
    satisfying the relations displayed in
    Lemma \ref{lemma-relations-face-degeneracy}.
    Namely, we set
    $$
    d^n_j = 1_{X_{j - 1}} \star k \star 1_{X_{n - j - 1}}
    \quad\text{and}\quad
    s^n_j = 1_{X_{j - 1} \circ F} \star t \star 1_{i \circ X_{n - j - 1}}
    $$
    Finally, write $\epsilon_0 = k : X_0 \to X_{-1}$.
    
    \begin{example}
    \label{example-polynomial-algebra-maps}
    In Example \ref{example-polynomial-algebra} we have
    $X_n(A) = \mathbf{Z}[\mathbf{Z}[\ldots [A]\ldots]]$ with
    $n + 1$ brackets. We describe the maps constructed above
    using a typical element $\xi = \sum n_i[n_{ij}[a_{ij}]]$
    of $X_1(A)$. The maps
    $d_0, d_1 : \mathbf{Z}[\mathbf{Z}[A]] \to \mathbf{Z}[A]$
    are given by
    $$
    d_0(\xi) = \sum n_in_{ij}[a_{ij}]
    \quad\text{and}\quad
    d_1(\xi) = \sum n_i[n_{ij}a_{ij}].
    $$
    The maps
    $s_0, s_1 : \mathbf{Z}[\mathbf{Z}[A]] \to
    \mathbf{Z}[\mathbf{Z}[\mathbf{Z}[A]]]$
    are given by
    $$
    s_0(\xi) = \sum n_i[[n_{ij}[a_{ij}]]]
    \quad\text{and}\quad
    s_1(\xi) = \sum n_i[n_{ij}[[a_{ij}]]].
    $$
    \end{example}
    
    \begin{lemma}
    \label{lemma-standard-simplicial}
    In Situation \ref{situation-adjoint-functors}
    the system $X = (X_n, d^n_j, s^n_j)$
    is a simplicial object of $\text{Fun}(\mathcal{A}, \mathcal{A})$
    and $\epsilon_0$ defines an augmentation $\epsilon$ from $X$
    to the constant simplicial object with value $X_{-1} = \text{id}_\mathcal{A}$.
    \end{lemma}
    
    \begin{proof}
    Suppose that we have shown that $X$ is a simplicial object.
    Then to prove that $\epsilon_0 = k$ defines an augmentation
    we have to check that $\epsilon_0 \circ d^1_0 = \epsilon_0 \circ d^1_1$
    as morphisms $X_1 \to X_{-1}$, see Lemma \ref{lemma-augmentation-howto}.
    In other words, we have to check that the diagram
    $$
    \xymatrix{
    F \circ i \circ F \circ i
    \ar[rr]_{1_{F \circ i} \star k}
    \ar[d]_{k \star 1_{F \circ i}} & & F \circ i \ar[d]^k \\
    F \circ i \ar[rr]^k & & \text{id}_\mathcal{A}
    }
    $$
    is commutative. More precisely we should write this as the equality
    $$
    (k \star 1_{\text{id}_\mathcal{A}}) \circ (1_{F \circ i} \star k) =
    (1_{\text{id}_\mathcal{A}} \star k) \circ (k \star 1_{F \circ i})
    $$
    as morphisms
    $(F \circ i) \circ (F \circ i) \to
    \text{id}_\mathcal{A} \circ \text{id}_\mathcal{A}$.
    Applying the general property of
    Categories, Lemma \ref{categories-lemma-properties-2-cat-cats}
    both sides expand to $k \star k$ when equality holds.
    
    \medskip\noindent
    To prove that $X$ is a simplicial object we have to check
    (see Remark \ref{remark-relations}):
    \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}
    Relation (1) is proved in exactly the same manner as the proof
    of the equality $\epsilon_0 \circ d^1_0 = \epsilon_0 \circ d^1_1$ above.
    
    \medskip\noindent
    The simplest case of equality (5) is the commutativity of the diagram
    $$
    \xymatrix{
    F \circ i \ar[rrr]_{1_F \star t \star 1_i}
    \ar[d]_{1_F \star t \star 1_i} & & &
    F \circ i \circ F \circ i \ar[d]^{1_F \star t \star 1_{i \circ F \circ i}} \\
    F \circ i \circ F \circ i
    \ar[rrr]^{1_{F \circ i \circ F} \star t \star 1_i} & & &
    F \circ i \circ F \circ i \circ F \circ i
    }
    $$
    which holds because both compositions expand to the morphism
    $1_F \star t \star t \star 1_i$ from
    $F \circ \text{id}_\mathcal{A} \circ \text{id}_\mathcal{A} \circ i$
    to $F \circ (i \circ F) \circ (i \circ F) \circ i$.
    All other cases of (5) are proved in the same manner.
    
    \medskip\noindent
    The simplest case of equalities (2) and (4) is the commutativity of
    the diagram
    $$
    \xymatrix{
    F \circ i \circ F \circ i \ar[rrr]_{1_F \star k \star 1_i}
    \ar[d]_{1_{F \circ i \circ F \circ i} \star t} & & &
    F \circ i \ar[d]_{1_{F \circ i} \star t} \\
    F \circ i \circ F \circ i \circ F \circ i
    \ar[rrr]^{1_F \star k \star 1_{i \circ F \circ i \circ F}} & & &
    F \circ i \circ F \circ i
    }
    $$
    which again holds because both compositions expand to give
    $1_F \star k \star 1_i \star t$ as maps from
    $F \circ (i \circ F) \circ i \circ \text{id}_\mathcal{A}$
    to $F \circ \text{id}_\mathcal{A} \circ i \circ (F \circ i)$.
    All other cases of (2) and (4) are proved in the same manner.
    
    \medskip\noindent
    The relations (3) are the only nontrivial ones and these are
    consequences of the fact that the compositions in
    (\ref{equation-composition}) are the identity. For example, the simplest
    case of (3) states that the compositions
    $$
    \xymatrix{
    F \circ i \ar[rr]_{1_{F \circ i} \star t}
    \ar[d]_{t \star 1_{F \circ i}} & &
    F \circ i \circ F \circ i \ar[d]^{1_F \star k \star 1_i} \\
    F \circ i \circ F \circ i \ar[rr]^{1_F \star k \star 1_i} & &
    F \circ i
    }
    $$
    go around the diagram either way evaluate out to the identity. Going around
    the top the composition evaluates to
    $1_F \star ((k \star 1_i) \circ (1_i \star t))$
    which is the identity by what was said above. The other cases of (3) are
    proved in the same manner.
    \end{proof}
    
    \noindent
    Before reading the proof of the following lemma, we strongly urge the
    reader to look at the example discussed in
    Example \ref{example-polynomial-algebra-homotopy}
    in order to understand the purpose of the lemma.
    
    \begin{lemma}
    \label{lemma-standard-simplicial-homotopy}
    In Situation \ref{situation-adjoint-functors}
    the maps
    $$
    1_i \star \epsilon : i \circ X \to i,
    \quad\text{and}\quad
    \epsilon \star 1_F : X \circ F \to F
    $$
    are homotopy equivalences.
    \end{lemma}
    
    \begin{proof}
    Denote $\epsilon_n : X_n \to X_{-1}$ the components of the augmentation
    morphism. We observe that $\epsilon_n = k^{\star (n + 1)}$, the $(n + 1)$-fold
    $\star$-composition of $k$.
    Recall that $t : \text{id}_\mathcal{S} \to i \circ F$ is the
    adjunction map. We have the morphisms
    $$
    t^{\star (n + 1)} \star 1_i :
    i \longrightarrow i \circ (F \circ i)^{\circ (n + 1)} = i \circ X_n
    $$
    which are right inverse to $1_i \star \epsilon_n$ and the morphisms
    $$
    1_F \star t^{\star (n + 1)} :
    F \longrightarrow (F \circ i)^{\circ (n + 1)} \circ F = X_n \circ F
    $$
    which are right inverse to $\epsilon_n \star 1_F$. These morphisms
    determine morphisms of simplicial objects $b : i \to i \circ X$ and
    $c : F \to X \circ F$ (proof omitted).
    To finish it suffices to construct a homotopy between the morphisms
    $1, b \circ (1_i \star \epsilon) : i \circ X \to i \circ X$
    and between the two morphisms
    $1, c \circ (\epsilon \star 1_F) : X \circ F \to X \circ F$.
    
    \medskip\noindent
    To show the morphisms
    $b \circ (1_i \star \epsilon), 1 : i \circ X \to i \circ X$
    are homotopic we have to construct morphisms
    $$
    h_{n, j} : i \circ X_n \to i \circ X_n
    $$
    for $n \geq 0$ and $0 \leq j \leq n + 1$
    satisfying the relations described in Lemma \ref{lemma-relations-homotopy}.
    See also Remark \ref{remark-homotopy-better}.
    We are forced to set $h_{n, 0} = 1$ and
    $$
    h_{n , n + 1} = b_n \circ (1_i \star \epsilon_n) =
    (t^{\star (n + 1)} \star 1_i) \circ (1_i \star k^{\star (n + 1)})
    $$
    Thus a logical choice is
    $$
    h_{n , j} =
    (t^{\star (j)} \star 1) \circ (1_i \star k^{\star (j)} \star 1)
    $$
    Here and in the rest of the proof we drop the subscript from $1$
    if it is clear by knowing the source and the target of the morphism
    what this subscript should be. Writing
    $$
    i \circ X_n = i \circ F \circ i \circ \ldots \circ F \circ i
    $$
    we can think of the morphism $h_{n, j}$ as collapsing the first
    $j$ pairs $(F \circ i)$ to $\text{id}_\mathcal{S}$ using $k^{\star (j)}$,
    then adding a $\text{id}_\mathcal{S}$ in front and expanding this
    to $j$ pairs $(i \circ F)$ using $t^{\star (j)}$. We have to prove
    \begin{enumerate}
    \item We have $d^n_m \circ h_{n, j} = h_{n - 1, j - 1} \circ d^n_m$
    for $j > m$.
    \item We have $d^n_m \circ h_{n, j} = h_{n - 1, j} \circ d^n_m$
    for $j \leq m$.
    \item We have $s^n_m \circ h_{n, j} = h_{n + 1, j + 1} \circ s^n_m$
    for $j > m$.
    \item We have $s^n_m \circ h_{n, j} = h_{n + 1, j} \circ s^n_m$
    for $j \leq m$.
    \end{enumerate}
    Recall that $d^n_m$ is given by applying $k$ to the $(m + 1)$st pair
    $(F \circ i)$ in the functor $X_n = (F \circ i)^{\circ (n + 1)}$. Thus
    it is clear that (2) holds (because $k$ does $\star$-commute with $k$, but
    not with $t$). Similarly, $s^n_m$ is given by applying
    $1_F \star t \star i_i$ to the $(m + 1)$st pair $(F \circ i)$ in
    $X_n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (4) holds.
    In the two remaining cases one uses the fact that the compositions
    in (\ref{equation-composition}) are the identity causes the drop
    in the index $j$. Some details omitted.
    
    \medskip\noindent
    To show the morphisms
    $1, c \circ (\epsilon \star 1_F) : X \circ F \to X \circ F$
    are homotopic we have to construct morphisms
    $$
    h_{n, j} : X_n \circ F \longrightarrow X_n \circ F
    $$
    for $n \geq 0$ and $0 \leq j \leq n + 1$
    satisfying the relations described in Lemma \ref{lemma-relations-homotopy}.
    See also Remark \ref{remark-homotopy-better}.
    We are forced to set $h_{n, 0} = 1$ and
    $$
    h_{n , n + 1} = c_n \circ (\epsilon_n \star 1_F) =
    (1_F \star t^{\star (n + 1)}) \circ (k^{\star (n + 1)} \star 1_F)
    $$
    Thus a logical choice is
    $$
    h_{n , j} =
    (1_F \star t^{\star (j)} \star 1) \circ (k^{\star (j)} \star 1)
    $$
    Here and in the rest of the proof we drop the subscript from $1$
    if it is clear by knowing the source and the target of the morphism
    what this subscript should be. Writing
    $$
    X_n \circ F = F \circ i \circ F \circ \ldots \circ i \circ F
    $$
    we can think of the morphism $h_{n, j}$ as collapsing the first
    $j$ pairs $(F \circ i)$ to $\text{id}_\mathcal{S}$ using $k^{\star (j)}$,
    then inserting a $\text{id}_\mathcal{S}$ just after the first $F$ and
    expanding this to $j$ pairs $(i \circ F)$ using $t^{\star (j)}$.
    We have to prove
    \begin{enumerate}
    \item We have $d^n_m \circ h_{n, j} = h_{n - 1, j - 1} \circ d^n_m$
    for $j > m$.
    \item We have $d^n_m \circ h_{n, j} = h_{n - 1, j} \circ d^n_m$
    for $j \leq m$.
    \item We have $s^n_m \circ h_{n, j} = h_{n + 1, j + 1} \circ s^n_m$
    for $j > m$.
    \item We have $s^n_m \circ h_{n, j} = h_{n + 1, j} \circ s^n_m$
    for $j \leq m$.
    \end{enumerate}
    Recall that $d^n_m$ is given by applying $k$ to the $(m + 1)$st pair
    $(F \circ i)$ in the functor $X_n = (F \circ i)^{\circ (n + 1)}$. Thus
    it is clear that (2) holds (because $k$ does $\star$-commute with $k$, but
    not with $t$). Similarly, $s^n_m$ is given by applying
    $1_F \star t \star i_i$ to the $(m + 1)$st pair $(F \circ i)$ in
    $X_n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (4) holds.
    In the two remaining cases one uses the fact that the compositions
    in (\ref{equation-composition}) are the identity causes the drop
    in the index $j$. Some details omitted.
    \end{proof}
    
    \begin{example}
    \label{example-polynomial-algebra-homotopy}
    Going back to the example discussed in
    Example \ref{example-polynomial-algebra}
    our Lemma \ref{lemma-standard-simplicial-homotopy}
    signifies that for any ring $A$ the map of simplicial rings
    $$
    \xymatrix{
    \mathbf{Z}[\mathbf{Z}[\mathbf{Z}[A]]] \ar[d]
    \ar@<2ex>[r]
    \ar@<0ex>[r]
    \ar@<-2ex>[r]
    &
    \mathbf{Z}[\mathbf{Z}[A]] \ar[d]
    \ar@<1ex>[r]
    \ar@<-1ex>[r]
    \ar@<1ex>[l]
    \ar@<-1ex>[l]
    &
    \mathbf{Z}[A] \ar[d]
    \ar@<0ex>[l] \\
    A
    \ar@<2ex>[r]
    \ar@<0ex>[r]
    \ar@<-2ex>[r]
    &
    A
    \ar@<1ex>[r]
    \ar@<-1ex>[r]
    \ar@<1ex>[l]
    \ar@<-1ex>[l]
    &
    A
    \ar@<0ex>[l]
    }
    $$
    is a homotopy equivalence on underlying simplicial sets. Moreover, the inverse
    map constructed in Lemma \ref{lemma-standard-simplicial-homotopy}
    is in degree $n$ given by
    $$
    a \longmapsto [\ldots[a]\ldots]
    $$
    with obvious notation. In the other direction the lemma tells us that
    for every set $E$ there is a homotopy equivalence
    $$
    \xymatrix{
    \mathbf{Z}[\mathbf{Z}[\mathbf{Z}[\mathbf{Z}[E]]]] \ar[d]
    \ar@<2ex>[r]
    \ar@<0ex>[r]
    \ar@<-2ex>[r]
    &
    \mathbf{Z}[\mathbf{Z}[\mathbf{Z}[E]]] \ar[d]
    \ar@<1ex>[r]
    \ar@<-1ex>[r]
    \ar@<1ex>[l]
    \ar@<-1ex>[l]
    &
    \mathbf{Z}[\mathbf{Z}[E]] \ar[d]
    \ar@<0ex>[l] \\
    \mathbf{Z}[E]
    \ar@<2ex>[r]
    \ar@<0ex>[r]
    \ar@<-2ex>[r]
    &
    \mathbf{Z}[E]
    \ar@<1ex>[r]
    \ar@<-1ex>[r]
    \ar@<1ex>[l]
    \ar@<-1ex>[l]
    &
    \mathbf{Z}[E]
    \ar@<0ex>[l]
    }
    $$
    of rings. The inverse map constructed in the lemma is in degree $n$
    given by the ring map
    $$
    \sum m_{e_1,\ldots,e_p}[e_1][e_2] \ldots [e_p]
    \longmapsto
    \sum m_{e_1,\ldots,e_p}[\ldots[e_1]\ldots][\ldots[e_2]\ldots]
    \ldots [\ldots[e_p]\ldots]
    $$
    (with obvious notation).
    \end{example}
    
    
    
    
    
    
    
    
    
    
    \input{chapters}

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