injective, then there is a unique factorization
$f=\delta_{m,i_{1}}\cdots\delta_{n+1,i_{m-n}}$ where $0\leq
i_{m-n}<\cdots<i_{1}\leq m$ are the elements of $[m]$ that are not in the
image of $f$. If $f$ is surjective, then there is a unique factorization
$f=\sigma_{m,j_{1}}\cdots\sigma_{n-1,j_{n-m}}$ where $0\leq
j_{1}<\cdots<j_{n-m}<n$ are the elements of $[n]$ for which $f(j)=f(j+1)$. In
general, there is a unique factorization $f=gh$ where
$g\in\operatorname{Mor}_{\Delta}\left([k],[m]\right)$ is injective and
$h\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$ is surjective; See [We,
Lemma 8.1.2].

A trivial example of a simplicial object is the constant one in which we have
the same object in each degree and the face and the degeneracy morphisms are
all identity morphisms. We next look at some non-trivial examples.

######  Example 4.3.

Fix $n\geq 0$. We have the contravariant functor
$\Delta^{n}=\operatorname{Mor}_{\Delta}\left(-,[n]\right):\Delta\rightarrow\mathcal{S}\mathpzc{et}$,
so $\Delta^{n}$ is a simplicial set called the combinatorial $n$-simplex. If
$m\geq 0$ and $f\in\operatorname{Mor}_{\Delta}\left([n],[m]\right)$, then
$\Delta^{f}=\operatorname{Mor}_{\Delta}(-,f):\Delta^{n}=\operatorname{Mor}_{\Delta}\left(-,[n]\right)\rightarrow\operatorname{Mor}_{\Delta}\left(-,[m]\right)=\Delta^{m}$
is a natural transformation of functors, i.e. a morphism of simplicial sets.
For any $m\geq 0$, considering the images of order-preserving maps
$[m]\rightarrow[n]$, we can make the following identification:

$\Delta_{m}^{n}=\operatorname{Mor}_{\Delta}\left([m],[n]\right)\cong\left\\{(\alpha_{0},\alpha_{1},...,\alpha_{m})\in\mathbb{N}^{m+1}\;|\;0\leq\alpha_{0}\leq\alpha_{1}\leq\cdots\leq\alpha_{m}\leq
n\right\\}$

The above identification shows that $|\Delta_{m}^{n}|=\binom{n+m+1}{n}$. On
the other hand, under this identification, for any $m\geq 0$ and $0\leq i\leq
m$, the face morphism
$d_{m,i}^{\Delta^{n}}:\Delta_{m}^{n}\rightarrow\Delta_{m-1}^{n}$ is given by
$d_{m,i}^{\Delta^{n}}(\alpha_{0},\alpha_{1},...,\alpha_{m})=(\alpha_{0},...,\alpha_{i-1},\alpha_{i+1},...,\alpha_{m})$,
and the degeneracy morphism
$s_{m,i}^{\Delta^{n}}:\Delta_{m}^{n}\rightarrow\Delta_{m+1}^{n}$ is given by
$s_{m,i}^{\Delta^{n}}(\alpha_{0},\alpha_{1},...,\alpha_{m})=(\alpha_{0},...,\alpha_{i-1},\alpha_{i},\alpha_{i},\alpha_{i+1},...,\alpha_{m})$.
Given $m\geq 0$, any element of $\Delta_{m}^{n}$ is called an $m$-cell. The
$m$-cells that have repeated coordinates, i.e. the ones that belong to the
images of degeneracy morphisms, are called degenerate, and the rest are called
non-degenerate.

We further have the contravariant functor
$\partial\Delta^{n}:\Delta\rightarrow\mathcal{S}\mathpzc{et}$, given by

$\partial\Delta^{n}_{m}=\left\\{f\in\Delta_{m}^{n}=\operatorname{Mor}_{\Delta}\left([m],[n]\right)\;|\;f\textrm{
is not surjective}\right\\}$

for every $m\geq 0$, and
$(\partial\Delta^{n})(\eta):=\Delta^{n}(\eta)=\operatorname{Mor}_{\Delta}\left(\eta,[n]\right)$
for every $\eta\in\operatorname{Mor}(\Delta)$. Thus $\partial\Delta^{n}$ is a
simplicial set called the boundary of $\Delta^{n}$. If
$\iota_{m}:\partial\Delta_{m}^{n}\rightarrow\Delta_{m}^{n}$ is the inclusion
map for every $m\geq 0$, then $\iota=(\iota_{m})_{m\geq
0}:\partial\Delta^{n}\rightarrow\Delta^{n}$ is a morphism of simplicial sets.
Under the above identification, we have for every $m\geq 0$:

$\partial\Delta_{m}^{n}\cong\left\\{(\alpha_{0},\alpha_{1},...,\alpha_{m})\in\Delta_{m}^{n}\;|\;\\{\alpha_{0},\alpha_{1},...,\alpha_{m}\\}\subset[n]\right\\}$

For any $m\geq 0$, we notice that
$\partial\Delta_{m}^{1}=\\{(\underbrace{0,...,0}_{m+1\textrm{
times}}),(\underbrace{1,...,1}_{m+1\textrm{ times}})\\}$, so in light of
$\Delta_{m}^{0}=\\{(\underbrace{0,...,0}_{m+1\textrm{ times}})\\}$, we see
that there is an isomorphism
$\partial\Delta^{1}\cong\Delta^{0}\sqcup\Delta^{0}$ of simplicial sets.

We next recall some basic constructions of simplicial objects which will be
needed later.

######  Remark 4.4.

If $R$ is a ring, $M$ is a simplicial right $R$-module, and $N$ is a
simplicial left $R$-module, then the simplicial tensor product $M\otimes_{R}N$
is defined as follows. If $n\geq 0$, then
$(M\otimes_{R}N)_{n}=M_{n}\otimes_{R}N_{n}$. If
$\eta\in\operatorname{Mor}_{\Delta}\left([n],[m]\right)$, then
$(M\otimes_{R}N)(\eta)=M(\eta)\otimes_{R}N(\eta):M_{m}\otimes_{R}N_{m}\rightarrow
M_{n}\otimes_{R}N_{n}$. This construction is functorial in each argument.

If $\mathcal{C}$ is a locally small bicomplete category, $A$ is a simplicial
object over $\mathcal{C}$, and $U$ is a simplicial set, then the copower
simplicial object $A^{(U)}$ is defined as follows. If $n\geq 0$, then
$(A^{(U)})_{n}=A_{n}^{(U_{n})}$. If
$\eta\in\operatorname{Mor}_{\Delta}\left([n],[m]\right)$, then we consider the
morphism $A(\eta):A_{m}\rightarrow A_{n}$ in $\mathcal{C}$ as well as the map
$U(\eta):U_{m}\rightarrow U_{n}$, and then define
$A^{(U)}(\eta):A_{m}^{(U_{m})}\rightarrow A_{n}^{(U_{n})}$ to be the diagonal
of the following commutative diagram:

${A_{m}^{(U_{m})}}$${A_{n}^{(U_{m})}}$${A_{m}^{(U_{n})}}$${A_{n}^{(U_{n})}}$$\scriptstyle{A(\eta)^{(U_{m})}}$$\scriptstyle{\overline{U(\eta)}}$$\scriptstyle{\overline{U(\eta)}}$$\scriptstyle{A(\eta)^{(U_{n})}}$

This construction is functorial in each argument. Moreover, if $R$ is a ring,
$M$ is a simplicial left $R$-module, and $U$ is a simplicial set, then there
is a natural isomorphism $M^{(U)}\cong R^{(U)}\otimes_{R}M$ of simplicial left
$R$-modules.

If $\mathcal{C}$ is a locally small bicomplete category, and $A$ and $B$ are
two simplicial objects over $\mathcal{C}$, then the mapping simplicial set
$\operatorname{Map}_{\mathpzc{s}\mathcal{C}}(A,B)$ is defined as follows. If
$n\geq 0$, then
$\operatorname{Map}_{\mathpzc{s}\mathcal{C}}(A,B)_{n}=\operatorname{Mor}_{\mathpzc{s}\mathcal{C}}\left(A^{(\Delta^{n})},B\right)$.
If $\eta\in\operatorname{Mor}_{\Delta}\left([n],[m]\right)$ and
$f\in\operatorname{Mor}_{\mathpzc{s}\mathcal{C}}\left(A^{(\Delta^{m})},B\right)$,
then we consider the morphisms

$A^{(\Delta^{n})}\xrightarrow{A^{(\Delta^{\eta})}}A^{(\Delta^{m})}\xrightarrow{f}B$

and define the map
$\operatorname{Map}_{\mathpzc{s}\mathcal{C}}(A,B)(\eta):\operatorname{Mor}_{\mathpzc{s}\mathcal{C}}\left(A^{(\Delta^{m})},B\right)\rightarrow\operatorname{Mor}_{\mathpzc{s}\mathcal{C}}\left(A^{(\Delta^{n})},B\right)$
by setting
$\operatorname{Map}_{\mathpzc{s}\mathcal{C}}(A,B)(\eta)(f)=fA^{(\Delta^{\eta})}$.
This construction is functorial in each argument. Moreover, if $R$ is a ring,
$M$ and $N$ are two simplicial left $R$-modules, and $U$ is a simplicial set,
then there is a natural bijection of sets as follows:

$\operatorname{Mor}_{\mathpzc{s}\mathcal{M}\mathpzc{od}(R)}\left(M^{(U)},N\right)\cong\operatorname{Mor}_{\mathpzc{s}\mathcal{M}\mathpzc{od}(R)}\left(M,\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,N)\right)$

Let $U$ be a simplicial set. If $R$ is a ring and $M$ is a simplicial left
$R$-module, then we can forget the module structure of $M$, consider it as a
simplicial set, and then form
$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,M)$. Then
$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,M)$ is a simplicial
set which becomes a simplicial left $R$-module with the module structure
inherited from $M$ by pointwise operations. Similarly, if $R$ is a commutative
ring and $A$ is a simplicial commutative $R$-algebra, then we can forget the
algebra structure of $A$, consider it as a simplicial set, and then form
$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,A)$. Then
$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,A)$ is a simplicial
set which becomes a simplicial commutative $R$-algebra with the algebra
structure inherited from $A$ by pointwise operations.

We now recall the normalization functor; see [We, Definition 8.3.6].

######  Definition 4.5.

Let $R$ be a ring. The normalization functor
$\operatorname{Nor}:\mathpzc{s}\mathcal{M}\mathpzc{od}(R)\rightarrow\mathcal{C}_{\geq
0}(R)$ is defined as follows:

  1. (i)

$\operatorname{Nor}$ assigns to each simplicial left $R$-module $M$, the
normalized $R$-complex $\operatorname{Nor}(M)$, whose terms are
$\operatorname{Nor}(M)_{0}=M_{0}$ and
$\operatorname{Nor}(M)_{n}=\bigcap_{i=0}^{n-1}\operatorname{Ker}\left(d_{n,i}^{M}\right)\subseteq
M_{n}$ for every $n\geq 1$, and whose differential
$\partial_{n}^{\operatorname{Nor}(M)}:\operatorname{Nor}(M)_{n}\rightarrow\operatorname{Nor}(M)_{n-1}$
is given by $\partial_{n}^{\operatorname{Nor}(M)}(x)=(-1)^{n}d_{n,n}^{M}(x)$
for every $n\geq 1$ and $x\in\operatorname{Nor}(M)_{n}$.

  2. (ii)

$\operatorname{Nor}$ assigns to each morphism $f:M\rightarrow N$ of simplicial
left $R$-modules, the morphism
$\operatorname{Nor}(f):\operatorname{Nor}(M)\rightarrow\operatorname{Nor}(N)$
of $R$-complexes, where
$\operatorname{Nor}(f)_{n}:\operatorname{Nor}(M)_{n}\rightarrow\operatorname{Nor}(N)_{n}$
is given by $\operatorname{Nor}(f)_{n}(x)=f_{n}(x)$ for every $n\geq 0$ and
$x\in\operatorname{Nor}(M)_{n}$.

######  Remark 4.6.

Let $R$ be a ring, and $M$ a simplicial left $R$-module. The Moore complex
$C(M)$ is defined as follows. If $n\geq 0$, then $C(M)_{n}=M_{n}$. If $n\geq
1$, then
$\partial_{n}^{C(M)}=\sum_{i=0}^{n}(-1)^{i}d_{n,i}^{M}:C(M)_{n}\rightarrow
C(M)_{n-1}$. It can be verified by inspection that $C(M)$ is a connective
$R$-complex. Moreover, if we set $D(M)_{0}=0$ and
$D(M)_{n}=\sum_{i=0}^{n-1}\operatorname{Im}\left(s_{n-1,i}^{M}\right)$ for
every $n\geq 1$, and also $\partial_{n}^{D(M)}(x)=\partial_{n}^{C(M)}(x)$ for
every $n\geq 1$ and $x\in D(M)_{n}$, then $D(M)$ turns out to be a subcomplex
of $C(M)$. Moreover, we have $C(M)=\operatorname{Nor}(M)\oplus D(M)$ and
$\operatorname{Nor}(M)\cong C(M)/D(M)$ naturally in $M$. See [We, Lemma
8.3.7].

We recall a quintessential example of simplicial sets.

######  Example 4.7.

Let $\mathcal{C}$ be a small category. The nerve of $\mathcal{C}$ is a
simplicial set defined as follows. If $[n]\in\operatorname{Obj}(\Delta)$, then
viewing the poset $[n]$ as a small category, we set
$\operatorname{nrv}_{n}(\mathcal{C})=\operatorname{Fun}\left([n],\mathcal{C}\right)$,
i.e. the set of covariant functors $[n]\rightarrow\mathcal{C}$. If
$\eta\in\operatorname{Mor}_{\Delta}\left([n],[m]\right)$, then the map
$\operatorname{nrv}(\mathcal{C})(\eta):\operatorname{Fun}\left([m],\mathcal{C}\right)\rightarrow\operatorname{Fun}\left([n],\mathcal{C}\right)$
is defined by $\operatorname{nrv}(\mathcal{C})(\eta)(\theta)=\theta\eta$ for
every $\theta\in\operatorname{Fun}\left([m],\mathcal{C}\right)$. Then
$\operatorname{nrv}(\mathcal{C}):\Delta\rightarrow\mathcal{S}\mathpzc{et}$ is
a contravariant functor, so it is a simplicial set. More specifically, we have
$\operatorname{nrv}_{0}(\mathcal{C})=\operatorname{Obj}(\mathcal{C})$, and for
any $n\geq 1$, $\operatorname{nrv}_{n}(\mathcal{C})$ is the set of $n$-tuples
of composable morphisms

$A_{0}\rightarrow A_{1}\rightarrow\cdots\rightarrow A_{n-1}\rightarrow A_{n}$

in $\mathcal{C}$. Moreover, the face morphism
$d_{n,i}^{\operatorname{nrv}(\mathcal{C})}:\operatorname{nrv}_{n}(\mathcal{C})\rightarrow\operatorname{nrv}_{n-1}(\mathcal{C})$
is given by

$d_{n,i}^{\operatorname{nrv}(\mathcal{C})}\left(A_{0}\rightarrow\cdots\rightarrow
A_{i-1}\xrightarrow{f_{i-1}}A_{i}\xrightarrow{f_{i}}A_{i+1}\rightarrow\cdots\rightarrow
A_{n}\right)=\left(A_{0}\rightarrow\cdots\rightarrow
A_{i-1}\xrightarrow{f_{i}f_{i-1}}A_{i+1}\rightarrow\cdots\rightarrow
A_{n}\right)$

and the degeneracy morphism
$s_{n,i}^{\operatorname{nrv}(\mathcal{C})}:\operatorname{nrv}_{n}(\mathcal{C})\rightarrow\operatorname{nrv}_{n+1}(\mathcal{C})$
is given by

$s_{n,i}^{\operatorname{nrv}(\mathcal{C})}\left(A_{0}\rightarrow\cdots\rightarrow
A_{i-1}\rightarrow A_{i}\rightarrow A_{i+1}\rightarrow\cdots\rightarrow
A_{n}\right)=\left(A_{0}\rightarrow\cdots\rightarrow A_{i-1}\rightarrow
A_{i}\xrightarrow{1^{A_{i}}}A_{i}\rightarrow
A_{i+1}\rightarrow\cdots\rightarrow A_{n}\right)$

for every $0\leq i\leq n$.

We need the next lemma in the sequel.

######  Lemma 4.8.

Let $R$ be a ring, and $\mathcal{P}$ a poset with a least element $e$. View
$\mathcal{P}$ as a small category, and let
$\zeta:\Delta^{0}\rightarrow\operatorname{nrv}(\mathcal{P})$ be the morphism
of simplicial sets given by $\zeta_{m}(0,...,0)=(e\leq\cdots\leq e)$ for every
$m\geq 0$. Then
$\operatorname{Nor}\left(R^{(\zeta)}\right):\operatorname{Nor}\left(R^{(\Delta^{0})}\right)\rightarrow\operatorname{Nor}\left(R^{\left(\operatorname{nrv}(\mathcal{P})\right)}\right)$
is a homotopy equivalence of $R$-complexes. As a consequence, we have:

$H_{i}\left(\operatorname{Nor}\left(R^{\left(\operatorname{nrv}(\mathcal{P})\right)}\right)\right)\cong\begin{dcases}R&\textrm{if
}i=0\\\ 0&\textrm{if }i>0\end{dcases}$

###### Proof.

Set $U=\operatorname{nrv}(\mathcal{P})$. Since
$\Delta_{m}^{0}=\\{(\underbrace{0,...,0}_{m+1\textrm{ times}})\\}$ for every
$m\geq 0$, we can define a morphism $\zeta:\Delta^{0}\rightarrow U$ of
simplicial sets by setting $\zeta_{m}(0,...,0)=(e\leq\cdots\leq e)$ for every
$m\geq 0$. We show that the induced morphism

$\textstyle\overline{R^{(\zeta)}}:\frac{C\left(R^{(\Delta^{0})}\right)}{D\left(R^{(\Delta^{0})}\right)}\rightarrow\frac{C\left(R^{(U)}\right)}{D\left(R^{(U)}\right)}$

is a homotopy equivalence. Since $\Delta^{0}$ is a terminal object in
$\mathpzc{s}\mathcal{S}\mathpzc{et}$, there exists a unique morphism
$\varepsilon:U\rightarrow\Delta^{0}$ of simplicial sets. It is clear that
$\varepsilon\zeta=1^{\Delta^{0}}$, so
$R^{(\varepsilon)}R^{(\zeta)}=1^{R^{(\Delta^{0})}}$, whence

$\overline{R^{(\varepsilon)}}\overline{R^{(\zeta)}}=1^{\frac{C\left(R^{(\Delta^{0})}\right)}{D\left(R^{(\Delta^{0})}\right)}}.$

Now let $m\geq 0$. Then $R^{(U_{m})}$ is a free left $R$-module with a basis
that is in 1-1 correspondence with the set $U_{m}$. We identify this basis
with elements of $U_{m}$ in what follows. Define an $R$-homomorphism

$t_{m}:C\left(R^{(U)}\right)_{m}=R^{(U_{m})}\rightarrow
R^{(U_{m+1})}=C\left(R^{(U)}\right)_{m+1}$

by setting

$t_{m}(a_{0}\leq\cdots\leq a_{m})=(e\leq a_{0}\leq\cdots\leq a_{m})$

for every $(a_{0}\leq\cdots\leq a_{m})\in U_{m}$, and then extending linearly.
We note that $(a_{0}\leq\cdots\leq
a_{m})\in\operatorname{Im}\left(s_{m-1,i}^{U}\right)$ for some $0\leq i\leq
m-1$ if and only if $a_{i}=a_{i+1}$. This implies that if
$(a_{0}\leq\cdots\leq a_{m})\in\operatorname{Im}\left(s_{m-1,i}^{U}\right)$,
then $(e\leq a_{0}\leq\cdots\leq
a_{m})\in\operatorname{Im}\left(s_{m,i+1}^{U}\right)$. It follows that
$t_{m}\left(D\left(R^{(U)}\right)_{m}\right)\subseteq
D\left(R^{(U)}\right)_{m+1}$, so $t_{m}$ induces an $R$-homomorphism

$\textstyle\overline{t_{m}}:\frac{C\left(R^{(U)}\right)_{m}}{D\left(R^{(U)}\right)_{m}}\rightarrow\frac{C\left(R^{(U)}\right)_{m+1}}{D\left(R^{(U)}\right)_{m+1}}.$

We also set $\overline{t_{m}}=0$ for every $m<0$. We show that

$\textstyle
1^{\frac{C\left(R^{(U)}\right)_{m}}{D\left(R^{(U)}\right)_{m}}}-\overline{R^{(\zeta_{m})}}\overline{R^{(\varepsilon_{m})}}=\partial_{m+1}^{\frac{C\left(R^{(U)}\right)}{D\left(R^{(U)}\right)}}\overline{t_{m}}+\overline{t_{m-1}}\partial_{m}^{\frac{C\left(R^{(U)}\right)}{D\left(R^{(U)}\right)}}$

every $m\geq 0$. To this end, it suffices to establish the equality for the
elements of $U_{m}$. Let $(a_{0}\leq\cdots\leq a_{m})\in U_{m}$. We consider
two case:

Case I: $m=0$

In this case, we have:

$(a_{0})-\overline{R^{(\zeta_{0})}}\left(\overline{R^{(\varepsilon_{0})}}(a_{0})\right)=(a_{0})-\overline{R^{(\zeta_{0})}}(0)=(a_{0})-(e)$

On the other hand, since $\overline{t_{-1}}=0$, we have:

$\begin{split}\partial_{1}^{\frac{C\left(R^{(U)}\right)}{D\left(R^{(U)}\right)}}\left(\overline{t_{0}}(a_{0})\right)+\overline{t_{-1}}\left(\partial_{0}^{\frac{C\left(R^{(U)}\right)}{D\left(R^{(U)}\right)}}(a_{0})\right)&=\partial_{1}^{\frac{C\left(R^{(U)}\right)}{D\left(R^{(U)}\right)}}\left((e\leq
a_{0})+D\left(R^{(U)}\right)_{1}\right)=\partial_{1}^{C\left(R^{(U)}\right)}(e\leq
a_{0})\\\ &=d_{1,0}^{R^{(U)}}(e\leq a_{0})-d_{1,1}^{R^{(U)}}(e\leq
a_{0})=(a_{0})-(e)\end{split}$

Consequently, we see that the desired equality holds in this case.

Case II: $m>0$

In this case, since $(e\leq\cdots\leq e)\in D\left(R^{(U)}\right)_{m}$, we
have:

$\begin{split}(a_{0}\leq\cdots\leq
a_{m})-\overline{R^{(\zeta_{m})}}\left(\overline{R^{(\varepsilon_{m})}}\left((a_{0}\leq\cdots\leq
a_{m})+D\left(R^{(U)}\right)_{m}\right)\right)&=(a_{0}\leq\cdots\leq
a_{m})-\overline{R^{(\zeta_{m})}}\left((0,...,0)+D\left(R^{(\Delta^{0})}\right)_{m}\right)\\\
&=(a_{0}\leq\cdots\leq a_{m})-(e\leq\cdots\leq e)+D\left(R^{(U)}\right)_{m}\\\
&=(a_{0}\leq\cdots\leq a_{m})+D\left(R^{(U)}\right)_{m}\end{split}$

On the other hand, we have:

$\begin{split}\partial_{m+1}^{\frac{C\left(R^{(U)}\right)}{D\left(R^{(U)}\right)}}\left(\overline{t_{m}}\left((a_{0}\leq\cdots\leq
a_{m})+D\left(R^{(U)}\right)_{m}\right)\right)&=\partial_{m+1}^{\frac{C\left(R^{(U)}\right)}{D\left(R^{(U)}\right)}}\left((e\leq
a_{0}\leq\cdots\leq a_{m})+D\left(R^{(U)}\right)_{m+1}\right)\\\
&=\partial_{m+1}^{C\left(R^{(U)}\right)}(e\leq a_{0}\leq\cdots\leq
a_{m})+D\left(R^{(U)}\right)_{m}\\\
&=\sum_{i=0}^{m+1}(-1)^{i}d_{m+1,i}^{R^{(U)}}(e\leq a_{0}\leq\cdots\leq
a_{m})+D\left(R^{(U)}\right)_{m}\\\ &=d_{m+1,0}^{R^{(U)}}(e\leq
a_{0}\leq\cdots\leq a_{m})+\sum_{i=1}^{m+1}(-1)^{i}d_{m+1,i}^{R^{(U)}}(e\leq
a_{0}\leq\cdots\leq a_{m})+D\left(R^{(U)}\right)_{m}\\\ &=(a_{0}\leq\cdots\leq
a_{m})+\sum_{i=1}^{m+1}(-1)^{i}(e\leq a_{0}\leq\cdots\leq a_{i-2}\leq
a_{i}\leq\cdots\leq a_{m})+D\left(R^{(U)}\right)_{m}\end{split}$

and

$\begin{split}\overline{t_{m-1}}\left(\partial_{m}^{\frac{C\left(R^{(U)}\right)}{D\left(R^{(U)}\right)}}\left((a_{0}\leq\cdots\leq
a_{m})+D\left(R^{(U)}\right)_{m}\right)\right)&=\overline{t_{m-1}}\left(\partial_{m}^{C\left(R^{(U)}\right)}(a_{0}\leq\cdots\leq
a_{m})+D\left(R^{(U)}\right)_{m-1}\right)\\\
&=\overline{t_{m-1}}\left(\sum_{i=0}^{m}(-1)^{i}d_{m,i}^{R^{(U)}}(a_{0}\leq\cdots\leq
a_{m})+D\left(R^{(U)}\right)_{m-1}\right)\\\
&=\overline{t_{m-1}}\left(\sum_{i=0}^{m}(-1)^{i}(a_{0}\leq\cdots\leq
a_{i-1}\leq a_{i+1}\leq\cdots\leq a_{m})+D\left(R^{(U)}\right)_{m-1}\right)\\\
&=\sum_{i=0}^{m}(-1)^{i}(e\leq a_{0}\leq\cdots\leq a_{i-1}\leq
a_{i+1}\leq\cdots\leq a_{m})+D\left(R^{(U)}\right)_{m}\\\
&=-\sum_{i=1}^{m+1}(-1)^{i}(e\leq a_{0}\leq\cdots\leq a_{i-2}\leq
a_{i}\leq\cdots\leq a_{m})+D\left(R^{(U)}\right)_{m}\end{split}$

Adding them up, we get:

$\partial_{m+1}^{\frac{C\left(R^{(U)}\right)}{D\left(R^{(U)}\right)}}\left(\overline{t_{m}}\left((a_{0}\leq\cdots\leq
a_{m})+D\left(R^{(U)}\right)_{m}\right)\right)+\overline{t_{m-1}}\left(\partial_{m}^{\frac{C\left(R^{(U)}\right)}{D\left(R^{(U)}\right)}}\left((a_{0}\leq\cdots\leq
a_{m})+D\left(R^{(U)}\right)_{m}\right)\right)=$ $(a_{0}\leq\cdots\leq
a_{m})+D\left(R^{(U)}\right)_{m}$

Consequently, we see that the desired equality holds in this case. This means
that $\overline{R^{(\zeta)}}\overline{R^{(\varepsilon)}}\sim
1^{\frac{C\left(R^{(U)}\right)}{D\left(R^{(U)}\right)}}$. As a result,
$\overline{R^{(\zeta)}}$ is a homotopy equivalence.

Now the commutative diagram

${\operatorname{Nor}\left(R^{(\Delta^{0})}\right)}$${\operatorname{Nor}\left(R^{(U)}\right)}$${\frac{C\left(R^{(\Delta^{0})}\right)}{D\left(R^{(\Delta^{0})}\right)}}$${\frac{C\left(R^{(U)}\right)}{D\left(R^{(U)}\right)}}$$\scriptstyle{\operatorname{Nor}\left(R^{(\zeta)}\right)}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\overline{R^{(\zeta)}}}$

implies that $\operatorname{Nor}\left(R^{(\zeta)}\right)$ is a homotopy
equivalence, hence a quasi-isomorphism.

Next let $n\geq 0$. Recall that we have an identification

$\Delta_{m}^{n}=\left\\{(\alpha_{0},\alpha_{1},...,\alpha_{m})\in\mathbb{N}^{m+1}\;|\;0\leq\alpha_{0}\leq\alpha_{1}\leq\cdots\leq\alpha_{m}\leq
n\right\\}$

for every $m\geq 0$. The simplicial left $R$-module $R^{(\Delta^{n})}$ is in
degree $m$ equal to the free left $R$-module $R^{(\Delta^{n}_{m})}$ with a
basis $e_{1},...,e_{k}$ where $k=|\Delta_{m}^{n}|=\binom{n+m+1}{n}$. If an
element of $\Delta^{n}_{m}$ is degenerate, then it belongs to the image of
some degeneracy morphism $s_{m-1,i}^{\Delta^{n}}$ where $0\leq i\leq m-1$, so
the corresponding basis element belongs to the image of the degeneracy
morphism
$R^{\left(s_{m-1,i}^{\Delta^{n}}\right)}=s_{m-1,i}^{R^{(\Delta^{n})}}$. In
light of Remark 4.6, we have the isomorphism

$\operatorname{Nor}\left(R^{(\Delta^{n})}\right)_{m}\cong\frac{R^{(\Delta_{m}^{n})}}{\sum_{i=0}^{m-1}\operatorname{Im}\left(s_{m-1,i}^{R^{(\Delta^{n})}}\right)}$

and we know that $\operatorname{Nor}\left(R^{(\Delta^{n})}\right)_{m}$ is a
direct summand of $R^{(\Delta_{m}^{n})}$, so we conclude that
$\operatorname{Nor}\left(R^{(\Delta^{n})}\right)_{m}$ is a free left
$R$-module with a basis that corresponds to the non-degenerate elements of
$\Delta^{n}_{m}$ for every $m\geq 0$. As a result, since
$\Delta_{n}^{n}=\\{(0,1,...,n)\\}$, we get
$\operatorname{Nor}\left(R^{(\Delta^{n})}\right)_{n}\cong
R^{(\\{(0,1,...,n)\\})}=R$. Moreover, if $m>n$, then every element of
$\Delta^{n}_{m}$ is degenerate, so
$\operatorname{Nor}\left(R^{(\Delta^{n})}\right)_{m}=0$. In particular, we
have
$\operatorname{Nor}\left(R^{(\Delta^{0})}\right)_{0}=R^{(\Delta_{0}^{0})}=R^{(\\{0\\})}=R$
and $\operatorname{Nor}\left(R^{(\Delta^{0})}\right)_{m}=0$ for every $m\geq
1$. As a consequence, $\operatorname{Nor}\left(R^{(\Delta^{0})}\right)$ is the
$R$-complex

$0\rightarrow R\rightarrow 0$

with $R$ in degree $0$. Therefore, we have:

$H_{i}\left(\operatorname{Nor}\left(R^{(U)}\right)\right)\cong
H_{i}\left(\operatorname{Nor}\left(R^{(\Delta^{0})}\right)\right)=\begin{dcases}R&\textrm{if
}i=0\\\ 0&\textrm{if }i>0\end{dcases}$

∎

We next recall the Dold-Kan functor. We use the functorial point of view for
simplicial modules in the next definition; see [We, 8.4.4].

######  Definition 4.9.

Let $R$ be a ring. The Dold-Kan functor $\operatorname{DK}:\mathcal{C}_{\geq
0}(R)\rightarrow\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$ is defined as follows:

  1. (i)

$\operatorname{DK}$ assigns to each connective $R$-complex $X$, the
contravariant functor
$\operatorname{DK}(X):\Delta\rightarrow\mathcal{M}\mathpzc{od}(R)$ defined as
follows. If $[n]\in\operatorname{Obj}(\Delta)$, then we set

$\operatorname{DK}(X)\left([n]\right)=\textstyle\bigoplus_{\begin{subarray}{c}{f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)}\\\
{f\textrm{ surjective}}\end{subarray}}X_{k}$

which is a finite direct sum since there is no surjective map
$[n]\rightarrow[k]$ if $k>n$. If
$\eta\in\operatorname{Mor}_{\Delta}\left([n],[m]\right)$, then the
$R$-homomorphism

$\operatorname{DK}(X)(\eta):\operatorname{DK}(X)\left([m]\right)=\textstyle\bigoplus_{\begin{subarray}{c}{g\in\operatorname{Mor}_{\Delta}\left([m],[l]\right)}\\\
{g\textrm{
surjective}}\end{subarray}}X_{l}\rightarrow\textstyle\bigoplus_{\begin{subarray}{c}{f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)}\\\
{f\textrm{
surjective}}\end{subarray}}X_{k}=\operatorname{DK}(X)\left([n]\right)$

is given by a matrix $\left(\operatorname{DK}(X)(\eta)_{f,g}:X_{l}\rightarrow
X_{k}\right)$, where we consider the unique factorization of $g\eta$ as

${{[n]}}$${{[m]}}$${{[k]}}$${{[l]}}$$\scriptstyle{\eta}$$\scriptstyle{f}$$\scriptstyle{g}$$\scriptstyle{\iota}$

in which $\iota$ is injective and $f$ is surjective (see Remark 4.2), and then
we set:

$\operatorname{DK}(X)(\eta)_{f,g}=\begin{dcases}1^{X_{l}}&\textrm{if
}k=l\textrm{ and }\iota=1^{[l]}\\\ (-1)^{l}\partial_{l}^{X}&\textrm{if
}k=l-1\textrm{ and }\iota=\delta_{l,l}\\\ 0&\textrm{otherwise}\end{dcases}$

  2. (ii)

$\operatorname{DK}$ assigns to each morphism $\phi:X\rightarrow Y$ of
connective $R$-complexes, the natural transformation
$\operatorname{DK}(\phi):\operatorname{DK}(X)\rightarrow\operatorname{DK}(Y)$,
where for any $[n]\in\operatorname{Obj}(\Delta)$, the $R$-homomorphism

$\operatorname{DK}(X)(\phi)^{[n]}:\operatorname{DK}(X)\left([n]\right)=\textstyle\bigoplus_{\begin{subarray}{c}{g\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)}\\\
{g\textrm{
surjective}}\end{subarray}}X_{l}\rightarrow\textstyle\bigoplus_{\begin{subarray}{c}{f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)}\\\
{f\textrm{
surjective}}\end{subarray}}Y_{k}=\operatorname{DK}(Y)\left([n]\right)$

is given by a matrix
$\left(\operatorname{DK}(\phi)_{f,g}^{[n]}:X_{l}\rightarrow Y_{k}\right)$
where we have:

$\operatorname{DK}(X)(\phi)_{f,g}^{[n]}=\begin{dcases}\phi_{k}&\textrm{if
}f=g\\\ 0&\textrm{otherwise}\end{dcases}$

In order to get a flavor of the Dold-Kan functor, we give the following
example.

######  Example 4.10.

Let $R$ be a ring, and $X$ a connective $R$-complex. Let
$\eta\in\operatorname{Mor}_{\Delta}\left([1],[2]\right)$ be given by
$\eta(0)=0$ and $\eta(1)=1$. Consider the following $R$-homomorphism:

$\operatorname{DK}(X)(\eta):\operatorname{DK}(X)\left([2]\right)=\textstyle\bigoplus_{\begin{subarray}{c}{g\in\operatorname{Mor}_{\Delta}\left([2],[l]\right)}\\\
{g\textrm{
surjective}}\end{subarray}}X_{l}\rightarrow\textstyle\bigoplus_{\begin{subarray}{c}{f\in\operatorname{Mor}_{\Delta}\left([1],[k]\right)}\\\
{f\textrm{
surjective}}\end{subarray}}X_{k}=\operatorname{DK}(X)\left([1]\right)$

We first determine all the surjective maps $g$ in
$\operatorname{Mor}_{\Delta}\left([2],[l]\right)$ for $0\leq l\leq 2$. We
have:

Case I: $l=0$; $g_{1}:[2]\rightarrow[0]$ given by
$g_{1}(0)=g_{1}(1)=g_{1}(2)=0$.

Case II: $l=1$; $g_{2}:[2]\rightarrow[1]$ given by $g_{2}(0)=0$, $g_{2}(1)=1$,
and $g_{2}(2)=1$; and $g_{3}:[2]\rightarrow[1]$ given by $g_{3}(0)=0$,
$g_{3}(1)=0$, and $g_{3}(2)=1$.

Case III: $l=2$; If $g_{4}:[2]\rightarrow[2]$ is surjective, then it must be
injective as well, so $g_{4}=1^{[2]}$.

We next determine all the surjective maps $f$ in
$\operatorname{Mor}_{\Delta}\left([1],[k]\right)$ for $0\leq k\leq 1$. We
have:

Case I: $k=0$; $f_{1}:[1]\rightarrow[0]$ given by $f_{1}(0)=f_{1}(1)=0$.

Case II: $k=1$; If $f_{2}:[1]\rightarrow[1]$ is surjective, then it must be
injective as well, so $f_{2}=1^{[1]}$.

By direct inspection, we observe that the only possible commutative diagrams
are as follows:

${{[1]}}$${{[2]}}$${{[0]}}$${{[0]}}$$\scriptstyle{\eta}$$\scriptstyle{f_{1}}$$\scriptstyle{g_{1}}$$\scriptstyle{1^{[0]}}$
,
${{[1]}}$${{[2]}}$${{[1]}}$${{[1]}}$$\scriptstyle{\eta}$$\scriptstyle{f_{2}}$$\scriptstyle{g_{2}}$$\scriptstyle{1^{[1]}}$
,
${{[1]}}$${{[2]}}$${{[0]}}$${{[1]}}$$\scriptstyle{\eta}$$\scriptstyle{f_{1}}$$\scriptstyle{g_{3}}$$\scriptstyle{\delta_{1,1}}$
,
${{[1]}}$${{[2]}}$${{[1]}}$${{[2]}}$$\scriptstyle{\eta}$$\scriptstyle{f_{2}}$$\scriptstyle{g_{4}}$$\scriptstyle{\delta_{2,2}}$

It follows that

$\operatorname{DK}(X)(\eta):\operatorname{DK}(X)\left([2]\right)=X_{0}\oplus
X_{1}\oplus X_{1}\oplus X_{2}\rightarrow X_{0}\oplus
X_{1}=\operatorname{DK}(X)\left([1]\right)$

can be represented by the following matrix:

$\begin{pmatrix}1^{X_{0}}&0&-\partial_{1}^{X}&0\\\
0&1^{X_{1}}&0&\partial_{2}^{X}\end{pmatrix}$

The following proposition describes how the Dold-Kan functor interacts with
surjections.

######  Proposition 4.11.

Let $R$ be a ring, and $X$ an $R$-complex. Let
$\lambda_{f}^{X}:X_{k}\rightarrow\operatorname{DK}(X)\left([n]\right)$ be the
canonical injection corresponding to the index
$f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$. If
$\eta\in\operatorname{Mor}_{\Delta}\left([n],[m]\right)$,
$g\in\operatorname{Mor}_{\Delta}\left([m],[l]\right)$ is surjective, and $x\in
X_{l}$, then we have:

$\operatorname{DK}(X)(\eta)\left(\lambda_{g}^{X}(x)\right)=\begin{dcases}\lambda_{g\eta}^{X}(x)&\textrm{if
}g\eta\textrm{ is surjective}\\\ 0&\textrm{if }g\eta\textrm{ is not surjective
but }m\in\operatorname{Im}(\eta)\end{dcases}$

###### Proof.

Let $A$ denote the matrix representing $\operatorname{DK}(X)(\eta)$ as in
Definition 4.9. First suppose that $g\eta$ is surjective. We focus on the
column $g$ of $A$. By the uniqueness of the face-degeneracy factorization in
Remark 4.2, we can only have the following commutative diagram:

${{[n]}}$${{[m]}}$${{[l]}}$${{[l]}}$$\scriptstyle{\eta}$$\scriptstyle{g\eta}$$\scriptstyle{g}$$\scriptstyle{1^{[l]}}$

This means that the column $g$ of $A$ has $1^{X_{l}}$ in the position
$(g\eta,g)$ and zero elsewhere. We note that $\lambda_{g}^{X}(x)$ has $x$ in
its coordinate corresponding to $g$ and zero elsewhere. Considering
$\lambda_{g}^{X}(x)$ as a column matrix, we see that the matrix product
$A\lambda_{g}^{X}(x)$ is equal to the column matrix with only one non-zero
coordinate $1^{X_{l}}(x)=x$ in position $g\eta$. It follows that
$\operatorname{DK}(X)(\eta)\left(\lambda_{g}^{X}(x)\right)=A\lambda_{g}^{X}(x)=\lambda_{g\eta}^{X}(x)$.

Next suppose that $g\eta$ is not surjective but $m\in\operatorname{Im}(\eta)$.
We focus on the column $g$ of $A$. If there is a commutative diagram as

${{[n]}}$${{[m]}}$${{[l]}}$${{[l]}}$$\scriptstyle{\eta}$$\scriptstyle{f}$$\scriptstyle{g}$$\scriptstyle{1^{[l]}}$

in which $f$ is surjective, then $g\eta=f$ is surjective which contradicts our
assumption. Therefore, there is no such diagram. On the other hand, if there
is a commutative diagram as

${{[n]}}$${{[m]}}$${{[l-1]}}$${{[l]}}$$\scriptstyle{\eta}$$\scriptstyle{f}$$\scriptstyle{g}$$\scriptstyle{\delta_{l,l}}$

in which $f$ is surjective, then $g\eta=\delta_{l,l}f$. Let $g(m)=p\in[l]$. If
$p\neq l$, then $p<l$. Since $g$ is surjective, there is an element $r\in[m]$
such that $g(r)=l$. But then $g(m)=p<l=g(r)$ which is a contradiction since
$r\leq m$. Therefore, $g(m)=p=l$. Since $m\in\operatorname{Im}(\eta)$, there
is an element $s\in[n]$ such that $\eta(s)=m$. But then
$\delta_{l,l}\left(f(s)\right)=g\left(\eta(s)\right)=g(m)=l$, so
$l\in\operatorname{Im}(\delta_{l,l})$ which is a contradiction. As a
consequence, there is no such diagram either. This means that the column $g$
of $A$ is entirely zero. Considering $\lambda_{g}^{X}(x)$ as a column matrix,
we see that the matrix product $A\lambda_{g}^{X}(x)$ is equal to the zero
column matrix. It follows that
$\operatorname{DK}(X)(\eta)\left(\lambda_{g}^{X}(x)\right)=A\lambda_{g}^{X}(x)=0$.
∎

######  Corollary 4.12.

Let $R$ be a ring, and $X$ an $R$-complex. Let
$\lambda_{f}^{X}:X_{k}\rightarrow\operatorname{DK}(X)\left([n]\right)$ be the
canonical injection corresponding to the index
$f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$. If
$g\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)$ is surjective and $x\in
X_{l}$, then we have

$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{g}^{X}(x)\right)=\begin{dcases}\lambda_{g\delta_{n,i}}^{X}(x)&\textrm{if
}0\leq i\leq n\textrm{ and }g\delta_{n,i}\textrm{ is surjective}\\\
0&\textrm{if }0\leq i\leq n-1\textrm{ and }g\delta_{n,i}\textrm{ is not
surjective}\\\
(-1)^{l}\lambda_{\overline{g}}^{X}\left(\partial_{l}^{X}(x)\right)&\textrm{if
}i=n\textrm{ and }g\delta_{n,n}\textrm{ is not surjective}\end{dcases}$

for every $0\leq i\leq n$, where
$\overline{g}\in\operatorname{Mor}_{\Delta}\left([n-1],[l-1]\right)$ is
induced by $g$.

###### Proof.

Let $0\leq i\leq n$. We note that
$\delta_{n,i}\in\operatorname{Mor}_{\Delta}\left([n-1],[n]\right)$ and
$d_{n,i}^{\operatorname{DK}(X)}=\operatorname{DK}(X)(\delta_{n,i})$. Let
$A_{n,i}$ denote the matrix representing $d_{n,i}^{\operatorname{DK}(X)}$ as
in Definition 4.9. If $g\delta_{n,i}$ is surjective, then by Proposition 4.11,
we have
$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{g}^{X}(x)\right)=\operatorname{DK}(X)(\delta_{n,i})\left(\lambda_{g}^{X}(x)\right)=\lambda_{g\delta_{n,i}}^{X}(x)$.
If $0\leq i\leq n-1$, then $n\in\operatorname{Im}(\delta_{n,i})$, so if in
addition, $g\delta_{n,i}$ is not surjective, then Proposition 4.11 implies
that
$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{g}^{X}(x)\right)=\operatorname{DK}(X)(\delta_{n,i})\left(\lambda_{g}^{X}(x)\right)=0$.
Now suppose that $g\delta_{n,n}$ is not surjective. We focus on the column $g$
of $A_{n,i}$. If there is a commutative diagram as

${{[n-1]}}$${{[n]}}$${{[l]}}$${{[l]}}$$\scriptstyle{\delta_{n,n}}$$\scriptstyle{f}$$\scriptstyle{g}$$\scriptstyle{1^{[l]}}$

in which $f$ is surjective, then then $g\delta_{n,n}=f$ is surjective which
contradicts our assumption. Therefore, there is no such diagram. On the other
hand, as we saw in the proof of Proposition 4.11, since $g$ is surjective, we
must have $g(n)=l$. But $g\delta_{n,n}$ is not surjective which means that if
we omit $n$ from the domain of $g$, then it will no longer remain surjective.
That is to say, $l$ is assumed by $g$ only at $n$. It follows that $g$ induces
a surjective map
$\overline{g}\in\operatorname{Mor}_{\Delta}\left([n-1],[l-1]\right)$ given by
$\overline{g}(i)=g(i)$ for every $i\in[n-1]$. Therefore, we get the following
commutative diagram:

${{[n-1]}}$${{[n]}}$${{[l-1]}}$${{[l]}}$$\scriptstyle{\delta_{n,n}}$$\scriptstyle{\overline{g}}$$\scriptstyle{g}$$\scriptstyle{\delta_{l,l}}$

This implies that the column of the matrix $A_{n,i}$ corresponding to $g$ has
all its entries zero except for the entry in the position $(\overline{g},g)$
which is equal to $(-1)^{l}\partial_{l}^{X}$. As a result, we have
$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{g}^{X}(x)\right)=A_{n,i}\lambda_{g}^{X}(x)=(-1)^{l}\lambda_{\overline{g}}^{X}\left(\partial_{l}^{X}(x)\right)$.
∎

######  Corollary 4.13.

Let $R$ be a ring, and $X$ an $R$-complex. Let
$\lambda_{f}^{X}:X_{k}\rightarrow\operatorname{DK}(X)\left([n]\right)$ be the
canonical injection corresponding to the index
$f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$. If
$g\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)$ is surjective and $x\in
X_{l}$, then we have

$s_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{g}^{X}(x)\right)=\lambda_{g\sigma_{n,i}}^{X}(x)$

for every $0\leq i\leq n$.

###### Proof.

Let $0\leq i\leq n$. We note that
$\sigma_{n,i}\in\operatorname{Mor}_{\Delta}\left([n+1],[n]\right)$ and
$s_{n,i}^{\operatorname{DK}(X)}=\operatorname{DK}(X)(\sigma_{n,i})$. Since
$\sigma_{n,i}$ is surjective, we conclude that $g\sigma_{n,i}$ is surjective,
so by Proposition 4.11, we have
$s_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{g}^{X}(x)\right)=\operatorname{DK}(X)(\sigma_{n,i})\left(\lambda_{g}^{X}(x)\right)=\lambda_{g\sigma_{n,i}}^{X}(x)$.
∎

Here is the celebrated Dold-Kan Correspondence.

######  Theorem 4.14.

Let $R$ be a ring. Then there is an adjoint equivalence of categories as
follows:

$(\operatorname{DK},\operatorname{Nor}):\mathcal{C}_{\geq
0}(R)\leftrightarrows\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$

###### Proof.

See [We, Theorem 8.4.1 and Exercise 8.4.2]. ∎

It is worth noting that under the Dold-Kan Correspondence, the connective
chain complex concentrated in degree zero corresponds to the constant
simplicial module.

The cofibrant generation property of the model structure of connective chain
complexes in conjunction with the Dold-Kan Correspondence allow us to transfer
the model structure of connective chain complexes to simplicial modules.

######  Theorem 4.15.

Let $R$ be a ring. Then $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$ is a
cofibrantly generated model category with the model structure given as
follows:

  1. (i)

Fibrations are morphisms $f:M\rightarrow N$ in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$ such that
$\operatorname{Nor}(f)_{i}:\operatorname{Nor}(M)_{i}\rightarrow\operatorname{Nor}(N)_{i}$
is surjective for every $i\geq 1$.

  2. (ii)

Cofibrations are morphisms $f:M\rightarrow N$ in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$ such that
$\operatorname{Nor}(f)_{i}:\operatorname{Nor}(M)_{i}\rightarrow\operatorname{Nor}(N)_{i}$
is injective and $\operatorname{Coker}\left(\operatorname{Nor}(f)_{i}\right)$
is a projective left $R$-module for every $i\geq 0$.

  3. (iii)

Weak equivalences are morphisms $f:M\rightarrow N$ in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$ such that
$\operatorname{Nor}(f):\operatorname{Nor}(M)\rightarrow\operatorname{Nor}(N)$
is a quasi-isomorphism in $\mathcal{C}_{\geq 0}(R)$.

###### Proof.

Follows from Corollary 2.16, Theorem 3.1, and Theorem 4.14. ∎

##  5\. Shuffle Product and a Variant of Eilenberg-Zilber Theorem

In this section, we introduce the notion of shuffle product of connective
chain complexes and prove a variant of the Eilenberg-Zilber Theorem. This will
equip us with a formidable tool to overcome the acyclicity condition in
Theorem 2.9.

We first review the concept of a shuffle. For any $p,q\geq 0$, let $S_{p+q}$
denote the symmetric group on the set $\\{1,2,...,p+q\\}$, and
$\operatorname{Sh}(p,q)$ denote the set of $(p,q)$-shuffles in $S_{p+q}$, i.e.

$\operatorname{Sh}(p,q)=\left\\{\nu\in S_{p+q}\mid\nu(1)<\cdots<\nu(p)\textrm{
and }\nu(p+1)<\cdots<\nu(p+q)\right\\}.$

We note that $|\operatorname{Sh}(p,q)|=\binom{p+q}{p}$. For any
$\nu\in\operatorname{Sh}(p,q)$, consider the set:

$A_{\nu}=\left\\{\left(\nu(\alpha),\nu(\beta)\right)\mid 1\leq\alpha\leq
p\textrm{, }p+1\leq\beta\leq p+q,\textrm{ and
}\nu(\alpha)>\nu(\beta)\right\\}$

Then $\operatorname{sgn}(\nu)=(-1)^{|A_{\nu}|}$. Indeed, we fix the elements
$\nu(p+1)<\cdots<\nu(p+q)$, and then for each $\alpha=p,p-1,...,1$, we compare
$\nu(\alpha)$ with these elements to see how many moves to the right are
needed in order to put $\nu(\alpha)$ in the right place.

We next fix some notations in the next remark.

######  Remark 5.1.

Let $n\geq 0$ and $0\leq k,l\leq n$. Moreover, let
$f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$ and
$g\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)$ be surjective. Define a
map $\phi_{f,g}:[n]\rightarrow[k]\times[l]$ by setting
$\phi_{f,g}(i)=\left(f(i),g(i)\right)$ for every $i\in[n]$. Suppose that $f$
and $g$ are not bijective. As in Remark 4.2, let $0\leq
j_{1}<\cdots<j_{n-k}<n$ be the elements of $[n]$ for which $f(j)=f(j+1)$ so
that $f=\sigma_{k,j_{1}}\cdots\sigma_{n-1,j_{n-k}}$, and $0\leq
i_{1}<\cdots<i_{n-l}<n$ the elements of $[n]$ for which $g(i)=g(i+1)$ so that
$g=\sigma_{l,i_{1}}\cdots\sigma_{n-1,i_{n-l}}$. If $\phi_{f,g}$ is not
injective, then there are integers $m,m^{\prime}\in[n]$ with $m<m^{\prime}$
such that $\phi_{f,g}(m)=\phi_{f,g}(m^{\prime})$, so we get
$f(m)=f(m^{\prime})$ and $g(m)=g(m^{\prime})$. But $f(m)\leq f(m+1)\leq
f(m^{\prime})$ and $g(m)\leq g(m+1)\leq g(m^{\prime})$, so $f(m)=f(m+1)$ and
$g(m)=g(m+1)$. This means that
$m\in\\{i_{1},...,i_{n-l}\\}\cap\\{j_{1},...,j_{n-k}\\}$. Conversely, if
$m\in\\{i_{1},...,i_{n-l}\\}\cap\\{j_{1},...,j_{n-k}\\}$, then
$\phi_{f,g}(m)=\phi_{f,g}(m+1)$, so $\phi_{f,g}$ is not injective. As a
result, $\phi_{f,g}$ is injective if and only if
$\\{i_{1},...,i_{n-l}\\}\cap\\{j_{1},...,j_{n-k}\\}=\emptyset$. If $f$ is
bijective, then $f=1^{[n]}$ , so $\\{j_{1},...,j_{n-k}\\}=\emptyset$, and
$\phi_{f,g}$ is injective. A similar result holds when $g$ is bijective.
Including these two degenerate cases as well, we can state in general that
$\phi_{f,g}$ is injective if and only if
$\\{i_{1},...,i_{n-l}\\}\cap\\{j_{1},...,j_{n-k}\\}=\emptyset$.

Next let $n\geq 0$ and $0\leq k,l\leq n$ with $k+l=n$. Set:

$\mathcal{S}_{n}^{k,l}=\left\\{(f,g)\in\operatorname{Mor}_{\Delta}([n],[k])\times\operatorname{Mor}_{\Delta}([n],[l])\;|\;\begin{tabular}[]{ccc}$f$
and $g$ are surjective, and $\phi_{f,g}:[n]\rightarrow[k]\times[l]$ given\\\
by $\phi_{f,g}(i)=(f(i),g(i))$ for every $i\in[n]$ is
injective\end{tabular}\right\\}$

Let $(f,g)\in\mathcal{S}_{n}^{k,l}$. As before, let $0\leq
j_{1}<\cdots<j_{l}<n$ be the elements of $[n]$ for which $f(j)=f(j+1)$ so that
$f=\sigma_{k,j_{1}}\cdots\sigma_{n-1,j_{l}}$, and $0\leq i_{1}<\cdots<i_{k}<n$
the elements of $[n]$ for which $g(i)=g(i+1)$ so that
$g=\sigma_{l,i_{1}}\cdots\sigma_{n-1,i_{k}}$. Define a shuffle
$\nu_{f,g}\in\operatorname{Sh}(k,l)$ by setting
$\nu_{f,g}(\alpha)=i_{\alpha}+1$ for every $1\leq\alpha\leq k$ and
$\nu_{f,g}(\alpha)=j_{\alpha-k}+1$ for every $k+1\leq\alpha\leq n$. In other
words, $\nu_{f,g}$ can be represented as follows:

$\nu_{f,g}=\begin{pmatrix}1&\cdots&k&k+1&\cdots&n\\\
i_{1}+1&\cdots&i_{k}+1&j_{1}+1&\cdots&j_{l}+1\end{pmatrix}$

In the degenerate case where $f$ is bijective, we have $k=n$, so $l=0$, whence
$g=0$, implying that $\\{i_{1},...,i_{n}\\}=\\{0,...,n-1\\}$. Therefore,
$\nu_{f,g}=1^{[n]}$, so $\operatorname{sgn}(\nu_{f,g})=1$. The degenerate case
where $g$ is bijective leads to a similar conclusion. Now define a map
$\eta:\mathcal{S}_{n}^{k,l}\rightarrow\operatorname{Sh}(k,l)$ by setting
$\eta(f,g)=\nu_{f,g}$ for every $(f,g)\in\mathcal{S}_{n}^{k,l}$. On the other
hand, for any shuffle $\nu\in\operatorname{Sh}(k,l)$, define two maps as
follows:

$f_{n,\nu}=\sigma_{k,\nu(k+1)-1}\cdots\sigma_{n-1,\nu(n)-1}:[n]\rightarrow[k]$

and

$g_{n,\nu}=\sigma_{l,\nu(1)-1}\cdots\sigma_{n-1,\nu(k)-1}:[n]\rightarrow[l]$

Then it is clear that $(f_{n,\nu},g_{n,\nu})\in\mathcal{S}_{n}^{k,l}$. Next
define a map $\theta:\operatorname{Sh}(k,l)\rightarrow\mathcal{S}_{n}^{k,l}$
by setting $\theta(\nu)=(f_{n,\nu},g_{n,\nu})$ for every
$\nu\in\operatorname{Sh}(k,l)$. It is easily seen that $\eta$ and $\theta$ are
mutually inverse maps. Hence the two sets $\mathcal{S}_{n}^{k,l}$ and
$\operatorname{Sh}(k,l)$ are in 1-1 correspondence.

We now define the shuffle product of connective chain complexes using the
Dold-Kan functor. The terminology is justified by Remark 5.1.

######  Definition 5.2.

Let $R$ be a ring, $X$ a connective $R^{\operatorname{op}}$-complex, and $Y$ a
connective $R$-complex. Then the shuffle product $X\boxtimes_{R}Y$ is defined
as follows. For any $n\geq 0$, set:

$\mathcal{S}_{n}=\left\\{(f,g)\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)\times\operatorname{Mor}_{\Delta}\left([n],[l]\right)\;|\;\begin{tabular}[]{ccc}$0\leq
k,l\leq n$, $f$ and $g$ are surjective, and
$\phi_{f,g}:[n]\rightarrow[k]\times[l]$\\\ given by
$\phi_{f,g}(i)=\left(f(i),g(i)\right)$ for every $i\in[n]$ is
injective\end{tabular}\right\\}$

Then we have

$(X\boxtimes_{R}Y)_{n}:=\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{n}}(X_{k}\otimes_{R}Y_{l})$

for every $n\geq 0$. Moreover, if
$\lambda_{f}^{X}:X_{k}\rightarrow\operatorname{DK}(X)\left([n]\right)$ is the
canonical injection corresponding to the index
$f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$,
$\lambda_{g}^{Y}:Y_{l}\rightarrow\operatorname{DK}(Y)\left([n]\right)$ is the
canonical injection corresponding to the index
$g\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)$, and
$\lambda_{f,g}^{X,Y}:X_{k}\otimes_{R}Y_{l}\rightarrow(X\boxtimes_{R}Y)_{n}$ is
the canonical injection corresponding to the indices
$f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$ and
$g\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)$, then the differential

$\partial_{n}^{X\boxtimes_{R}Y}:(X\boxtimes_{R}Y)_{n}\rightarrow(X\boxtimes_{R}Y)_{n-1}$

is given by

$\partial_{n}^{X\boxtimes_{R}Y}\left(\lambda_{f,g}^{X,Y}(x\otimes
y)\right):=\textstyle\sum_{i=0}^{n}(-1)^{i}d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)$

for every $n\geq 0$, $(f,g)\in\mathcal{S}_{n}$, $x\in X_{k}$, and $y\in
Y_{l}$.

We need to first show that the shuffle product of two complexes is indeed a
complex.

######  Proposition 5.3.

Let $R$ be a ring, $X$ a connective $R^{\operatorname{op}}$-complex, and $Y$ a
connective $R$-complex. Then $X\boxtimes_{R}Y$ is a connective
$\mathbb{Z}$-complex.

###### Proof.

Let $n\geq 0$. Then we have:

$\begin{split}C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)_{n}&=\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)_{n}=\operatorname{DK}(X)\left([n]\right)\otimes_{R}\operatorname{DK}(Y)\left([n]\right)\\\
&=\left(\textstyle\bigoplus_{\begin{subarray}{c}{f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)}\\\
{f\textrm{
surjective}}\end{subarray}}X_{k}\right)\otimes_{R}\left(\textstyle\bigoplus_{\begin{subarray}{c}{g\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)}\\\
{g\textrm{ surjective}}\end{subarray}}Y_{l}\right)\\\
&\cong\textstyle\bigoplus_{\begin{subarray}{c}{f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)}\\\
{g\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)}\\\ {f\textrm{ and
}g\textrm{ surjective}}\end{subarray}}(X_{k}\otimes_{R}Y_{l})\end{split}$

We suppress the above isomorphism and consider it as an equality in what
follows. Let
$\lambda_{f}^{X}:X_{k}\rightarrow\operatorname{DK}(X)\left([n]\right)$ be the
canonical injection corresponding to the index
$f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$,
$\lambda_{g}^{Y}:Y_{l}\rightarrow\operatorname{DK}(Y)\left([n]\right)$ the
canonical injection corresponding to the index
$g\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)$, and
$\lambda_{f,g}^{X,Y}:X_{k}\otimes_{R}Y_{l}\rightarrow
C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)_{n}$ the
canonical injection corresponding to the indices
$f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$ and
$g\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)$. Then it is clear that
$\lambda_{f}^{X}(x)\otimes\lambda_{g}^{Y}(y)=\lambda_{f,g}^{X,Y}(x\otimes y)$
for every $x\in X_{k}$ and $y\in Y_{l}$, so we have
$\lambda_{f}^{X}\otimes_{R}\lambda_{g}^{Y}=\lambda_{f,g}^{X,Y}$. As a result,
we get

$\begin{split}\partial_{n}^{C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)}\left(\lambda_{f,g}^{X,Y}(x\otimes
y)\right)&=\textstyle\sum_{i=0}^{n}(-1)^{i}d_{n,i}^{\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)}\left(\lambda_{f,g}^{X,Y}(x\otimes
y)\right)\\\
&=\textstyle\sum_{i=0}^{n}(-1)^{i}\left(d_{n,i}^{\operatorname{DK}(X)}\otimes_{R}d_{n,i}^{\operatorname{DK}(Y)}\right)\left(\lambda_{f}^{X}(x)\otimes\lambda_{g}^{Y}(y)\right)\\\
&=\textstyle\sum_{i=0}^{n}(-1)^{i}d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)\end{split}$

for every $\lambda_{f,g}^{X,Y}(x\otimes y)\in
C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)_{n}$. We
show that $X\boxtimes_{R}Y$ is a subcomplex of
$C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)$. Let
$n\geq 0$, and let $\mathcal{S}_{n}$ be as in Definition 5.2. Then we can
write:

$\begin{split}C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)_{n}&=\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{n}}(X_{k}\otimes_{R}Y_{l})\right)\oplus\left(\textstyle\bigoplus_{(f,g)\notin\mathcal{S}_{n}}(X_{k}\otimes_{R}Y_{l})\right)\\\
&=(X\boxtimes_{R}Y)_{n}\oplus\left(\textstyle\bigoplus_{(f,g)\notin\mathcal{S}_{n}}(X_{k}\otimes_{R}Y_{l})\right)\end{split}$

Next let $(f,g)\in\mathcal{S}_{n}$. We show that
$\partial_{n}^{C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)}\left(\lambda_{f,g}^{X,Y}(x\otimes
y)\right)\in(X\boxtimes_{R}Y)_{n-1}$ for every $x\in X_{k}$ and $y\in Y_{l}$.
To this end, fix $0\leq i\leq n$, and consider the following cases:

Case I: $f\delta_{n,i}$ is not surjective but $g\delta_{n,i}$ is surjective

If $0\leq i\leq n-1$, then by Corollary 4.12,
$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)=0$, so
$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)=0\in(X\boxtimes_{R}Y)_{n-1}$.
Hence we can assume that $i=n$. In this case, as in the proof of Corollary
4.12, $f$ induces a surjective map
$\overline{f}\in\operatorname{Mor}_{\Delta}\left([n-1],[k-1]\right)$ given by
$\overline{f}(i)=f(i)$ for every $i\in[n-1]$. Accordingly, using Corollary
4.12, we have:

$d_{n,n}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,n}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)=(-1)^{k}\lambda_{\overline{f}}^{X}\left(\partial_{k}^{X}(x)\right)\otimes\lambda_{g\delta_{n,n}}^{Y}(y)=(-1)^{k}\lambda_{\overline{f},g\delta_{n,n}}^{X,Y}\left(\partial_{k}^{X}(x)\otimes
y\right)$

If $\left(\overline{f},g\delta_{n,n}\right)\notin\mathcal{S}_{n-1}$, then
$\phi_{\overline{f},g\delta_{n,n}}$ is not injective, so there is an $0\leq
m\leq n-2$ such that
$\phi_{\overline{f},g\delta_{n,n}}(m)=\phi_{\overline{f},g\delta_{n,n}}(m+1)$.
But we note that:

$\phi_{\overline{f},g\delta_{n,n}}(m)=\left(\overline{f}(m),(g\delta_{n,n})(m)\right)=\left(f(m),g(m)\right)=\phi_{f,g}(m)$

and

$\phi_{\overline{f},g\delta_{n,n}}(m+1)=\left(\overline{f}(m+1),(g\delta_{n,n})(m+1)\right)=\left(f(m+1),g(m+1)\right)=\phi_{f,g}(m+1)$

It follows that $\phi_{f,g}$ is not injective, so $(f,g)\notin\mathcal{S}_{n}$
which is a contradiction. Therefore,
$\left(\overline{f},g\delta_{n,n}\right)\in\mathcal{S}_{n-1}$, so
$\lambda_{\overline{f},g\delta_{n,n}}^{X,Y}\left(\partial_{k}^{X}(x)\otimes
y\right)\in(X\boxtimes_{R}Y)_{n-1}$.

Case II: $f\delta_{n,i}$ is surjective but $g\delta_{n,i}$ is not surjective

This case is similar to Case I.

Case III: $f\delta_{n,i}$ and $g\delta_{n,i}$ are both surjective

In this case, using Corollary 4.12, we have:

$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)=\lambda_{f\delta_{n,i}}^{X}(x)\otimes\lambda_{g\delta_{n,i}}^{Y}(y)=\lambda_{f\delta_{n,i},g\delta_{n,i}}^{X,Y}\left(x\otimes
y\right)$

If $\left(f\delta_{n,i},g\delta_{n,i}\right)\notin\mathcal{S}_{n-1}$, then
$\phi_{f\delta_{n,i},g\delta_{n,i}}$ is not injective, so there is an $0\leq
m\leq n-2$ such that
$\phi_{f\delta_{n,i},g\delta_{n,i}}(m)=\phi_{f\delta_{n,i},g\delta_{n,i}}(m+1)$.
But we note that:

$\phi_{f\delta_{n,i},g\delta_{n,i}}(m)=\left((f\delta_{n,i})(m),(g\delta_{n,i})(m)\right)=\begin{dcases}\left(f(m+1),g(m+1)\right)=\phi_{f,g}(m+1)&\textrm{if
}i\leq m\\\ \left(f(m),g(m)\right)=\phi_{f,g}(m)&\textrm{if }i=m+1\\\
\left(f(m),g(m)\right)=\phi_{f,g}(m)&\textrm{if }m+1<i\end{dcases}$

and

$\phi_{f\delta_{n,i},g\delta_{n,i}}(m+1)=\left((f\delta_{n,i})(m+1),(g\delta_{n,i})(m+1)\right)=\begin{dcases}\left(f(m+2),g(m+2)\right)=\phi_{f,g}(m+2)&\textrm{if
}i\leq m\\\ \left(f(m+2),g(m+2)\right)=\phi_{f,g}(m+2)&\textrm{if }i=m+1\\\
\left(f(m+1),g(m+1)\right)=\phi_{f,g}(m+1)&\textrm{if }m+1<i\end{dcases}$

In any case, it follows that $\phi_{f,g}$ is not injective, so
$(f,g)\notin\mathcal{S}_{n}$ which is a contradiction. Therefore,
$\left(f\delta_{n,i},g\delta_{n,i}\right)\in\mathcal{S}_{n-1}$, so
$\lambda_{f\delta_{n,i},g\delta_{n,i}}^{X,Y}\left(x\otimes
y\right)\in(X\boxtimes_{R}Y)_{n-1}$.

Case IV: $f\delta_{n,i}$ and $g\delta_{n,i}$ are neither surjective

If $0\leq i\leq n-1$, then by Corollary 4.12,
$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)=0=d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)$,
so $d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)=0\in(X\boxtimes_{R}Y)_{n-1}$.
Hence we can assume that $i=n$. In this case, as in the proof of Corollary
4.12, $f$ induces a surjective map
$\overline{f}\in\operatorname{Mor}_{\Delta}\left([n-1],[k-1]\right)$ given by
$\overline{f}(i)=f(i)$ for every $i\in[n-1]$, and $g$ induces a surjective map
$\overline{g}\in\operatorname{Mor}_{\Delta}\left([n-1],[l-1]\right)$ given by
$\overline{g}(i)=g(i)$ for every $i\in[n-1]$. Accordingly, using Corollary
4.12, we have:

$d_{n,n}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,n}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)=(-1)^{k}\lambda_{\overline{f}}^{X}\left(\partial_{k}^{X}(x)\right)\otimes(-1)^{l}\lambda_{\overline{g}}^{Y}\left(\partial_{l}^{Y}(y)\right)=(-1)^{k+l}\lambda_{\overline{f},\overline{g}}^{X,Y}\left(\partial_{k}^{X}(x)\otimes\partial_{l}^{Y}(y)\right)$

If $\left(\overline{f},\overline{g}\right)\notin\mathcal{S}_{n-1}$, then
$\phi_{\overline{f},\overline{g}}$ is not injective, so there is an $0\leq
m\leq n-2$ such that
$\phi_{\overline{f},\overline{g}}(m)=\phi_{\overline{f},\overline{g}}(m+1)$.
But we note that:

$\phi_{\overline{f},\overline{g}}(m)=\left(\overline{f}(m),\overline{g}(m)\right)=\left(f(m),g(m)\right)=\phi_{f,g}(m)$

and

$\phi_{\overline{f},\overline{g}}(m+1)=\left(\overline{f}(m+1),\overline{g}(m+1)\right)=\left(f(m+1),g(m+1)\right)=\phi_{f,g}(m+1)$

It follows that $\phi_{f,g}$ is not injective, so $(f,g)\notin\mathcal{S}_{n}$
which is a contradiction. Therefore,
$\left(\overline{f},\overline{g}\right)\in\mathcal{S}_{n-1}$, so
$\lambda_{\overline{f},\overline{g}}^{X,Y}\left(\partial_{k}^{X}(x)\otimes\partial_{l}^{Y}(y)\right)\in(X\boxtimes_{R}Y)_{n-1}$.

All in all, we conclude that

$\partial_{n}^{C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)}\left(\lambda_{f,g}^{X,Y}(x\otimes
y)\right)=\textstyle\sum_{i=0}^{n}(-1)^{i}d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)\in(X\boxtimes_{R}Y)_{n-1}$

for every $x\in X_{k}$ and $y\in Y_{l}$. It now follows that $X\boxtimes_{R}Y$
is a subcomplex of
$C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)$. ∎

We next establish the functoriality of the shuffle product.

######  Proposition 5.4.

Let $R$ be a ring. Then the following assertions hold:

  1. (i)

If $X$ is a connective $R^{\operatorname{op}}$-complex, then
$X\boxtimes_{R}-:\mathcal{C}_{\geq 0}(R)\rightarrow\mathcal{C}_{\geq
0}(\mathbb{Z})$ defines a covariant functor.

  2. (ii)

If $Y$ is a connective $R$-complex, then $-\boxtimes_{R}Y:\mathcal{C}_{\geq
0}(R^{\operatorname{op}})\rightarrow\mathcal{C}_{\geq 0}(\mathbb{Z})$ defines
a covariant functor.

###### Proof.

(i): Let $n\geq 0$, and let $\mathcal{S}_{n}$ be as in Definition 5.2. Let $X$
be a connective $R^{\operatorname{op}}$-complex. If $Y$ is a connective
$R$-complex, then by Proposition 5.3, $X\boxtimes_{R}Y$ is a connective
$\mathbb{Z}$-complex. On the other hand, if $\theta:Y\rightarrow Y^{\prime}$
is a morphism of connective $R$-complexes, then we define the morphism
$X\boxtimes_{R}\theta:X\boxtimes_{R}Y\rightarrow X\boxtimes_{R}Y^{\prime}$ of
$\mathbb{Z}$-complexes as follows. Let $n\geq 0$. Then we have
$(X\boxtimes_{R}Y)_{n}=\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{n}}(X_{k}\otimes_{R}Y_{l})$
and
$(X\boxtimes_{R}Y^{\prime})_{n}=\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{n}}(X_{k}\otimes_{R}Y_{l}^{\prime})$.
We set

$(X\boxtimes_{R}\theta)_{n}\left(\lambda_{f,g}^{X,Y}(x\otimes
y)\right):=\lambda_{f,g}^{X,Y^{\prime}}\left(x\otimes\theta_{l}(y)\right)$

for every $(f,g)\in\mathcal{S}_{n}$, $x\in X_{k}$, and $y\in Y_{l}$. Extending
it linearly to $(X\boxtimes_{R}Y)_{n}$, we see that
$(X\boxtimes_{R}\theta)_{n}$ is a $\mathbb{Z}$-homomorphism. Moreover, the
following diagram is commutative:

${(X\boxtimes_{R}Y)_{n}}$${(X\boxtimes_{R}Y)_{n-1}}$${(X\boxtimes_{R}Y^{\prime})_{n}}$${(X\boxtimes_{R}Y^{\prime})_{n-1}}$$\scriptstyle{\partial_{n}^{X\boxtimes_{R}Y}}$$\scriptstyle{(X\boxtimes_{R}\theta)_{n}}$$\scriptstyle{(X\boxtimes_{R}\theta)_{n-1}}$$\scriptstyle{\partial_{n}^{X\boxtimes_{R}Y^{\prime}}}$

Indeed, we have for every $(f,g)\in\mathcal{S}_{n}$, $x\in X_{k}$, and $y\in
Y_{l}$:

$\begin{split}\partial_{n}^{X\boxtimes_{R}Y^{\prime}}\left((X\boxtimes_{R}\theta)_{n}\left(\lambda_{f,g}^{X,Y}(x\otimes
y)\right)\right)&=\partial_{n}^{X\boxtimes_{R}Y^{\prime}}\left(\lambda_{f,g}^{X,Y^{\prime}}\left(x\otimes\theta_{l}(y)\right)\right)\\\
&=\textstyle\sum_{i=0}^{n}(-1)^{i}d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y^{\prime})}\left(\lambda_{g}^{Y^{\prime}}\left(\theta_{l}(y)\right)\right)\end{split}$

and

$\begin{split}(X\boxtimes_{R}\theta)_{n-1}\left(\partial_{n}^{X\boxtimes_{R}Y}\left(\lambda_{f,g}^{X,Y}(x\otimes
y)\right)\right)&=(X\boxtimes_{R}\theta)_{n-1}\left(\textstyle\sum_{i=0}^{n}(-1)^{i}d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)\right)\\\
&=\textstyle\sum_{i=0}^{n}(-1)^{i}(X\boxtimes_{R}\theta)_{n-1}\left(d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)\right)\end{split}$

Let $0\leq i\leq n$, and consider the following cases from Proposition 5.3:

Case I: $f\delta_{n,i}$ is not surjective but $g\delta_{n,i}$ is surjective

If $0\leq i\leq n-1$, then
$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)=0$, so we have:

$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y^{\prime})}\left(\lambda_{g}^{Y^{\prime}}\left(\theta_{l}(y)\right)\right)=0=d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)$

If $i=n$, then we have:

$\begin{split}d_{n,n}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,n}^{\operatorname{DK}(Y^{\prime})}\left(\lambda_{g}^{Y^{\prime}}\left(\theta_{l}(y)\right)\right)&=(-1)^{k}\lambda_{\overline{f},g\delta_{n,n}}^{X,Y^{\prime}}\left(\partial_{k}^{X}(x)\otimes\theta_{l}(y)\right)\\\
&=(X\boxtimes_{R}\theta)_{n-1}\left((-1)^{k}\lambda_{\overline{f},g\delta_{n,n}}^{X,Y}\left(\partial_{k}^{X}(x)\otimes
y\right)\right)\\\
&=(X\boxtimes_{R}\theta)_{n-1}\left(d_{n,n}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,n}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)\right)\end{split}$

Case II: $f\delta_{n,i}$ is surjective but $g\delta_{n,i}$ is not surjective

This case is similar to Case I.

Case III: $f\delta_{n,i}$ and $g\delta_{n,i}$ are both surjective

In this case, we have:

$\begin{split}d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y^{\prime})}\left(\lambda_{g}^{Y^{\prime}}\left(\theta_{l}(y)\right)\right)&=\lambda_{f\delta_{n,i},g\delta_{n,i}}^{X,Y^{\prime}}\left(x\otimes\theta_{l}(y)\right)=(X\boxtimes_{R}\theta)_{n-1}\left(\lambda_{f\delta_{n,i},g\delta_{n,i}}^{X,Y}\left(x\otimes
y\right)\right)\\\
&=(X\boxtimes_{R}\theta)_{n-1}\left(d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)\right)\end{split}$

Case IV: $f\delta_{n,i}$ and $g\delta_{n,i}$ are neither surjective

If $0\leq i\leq n-1$, then
$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)=d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)=d_{n,i}^{\operatorname{DK}(Y^{\prime})}\left(\lambda_{g}^{Y^{\prime}}\left(\theta_{l}(y)\right)\right)=0$,
so we have:

$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y^{\prime})}\left(\lambda_{g}^{Y^{\prime}}\left(\theta_{l}(y)\right)\right)=0=d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)$

If $i=n$, then using the fact that $\theta:Y\rightarrow Y^{\prime}$ is a
morphism of $R$-complexes, we get:

$\begin{split}d_{n,n}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,n}^{\operatorname{DK}(Y^{\prime})}\left(\lambda_{g}^{Y^{\prime}}\left(\theta_{l}(y)\right)\right)&=(-1)^{k+l}\lambda_{\overline{f},\overline{g}}^{X,Y^{\prime}}\left(\partial_{k}^{X}(x)\otimes\partial_{l}^{Y^{\prime}}\left(\theta_{l}(y)\right)\right)\\\
&=(-1)^{k+l}\lambda_{\overline{f},\overline{g}}^{X,Y^{\prime}}\left(\partial_{k}^{X}(x)\otimes\theta_{l-1}\left(\partial_{l}^{Y}(y)\right)\right)\\\
&=(X\boxtimes_{R}\theta)_{n-1}\left((-1)^{k+l}\lambda_{\overline{f},\overline{g}}^{X,Y}\left(\partial_{k}^{X}(x)\otimes\partial_{l}^{Y}(y)\right)\right)\\\
&=(X\boxtimes_{R}\theta)_{n-1}\left(d_{n,n}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,n}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)\right)\end{split}$

We observe from the above analysis that in any case we have:

$\partial_{n}^{X\boxtimes_{R}Y^{\prime}}\left((X\boxtimes_{R}\theta)_{n}\left(\lambda_{f,g}^{X,Y}(x\otimes
y)\right)\right)=(X\boxtimes_{R}\theta)_{n-1}\left(\partial_{n}^{X\boxtimes_{R}Y}\left(\lambda_{f,g}^{X,Y}(x\otimes
y)\right)\right)$

Therefore, it follows that
$\partial_{n}^{X\boxtimes_{R}Y^{\prime}}(X\boxtimes_{R}\theta)_{n}=(X\boxtimes_{R}\theta)_{n-1}\partial_{n}^{X\boxtimes_{R}Y}$.
As a result, $X\boxtimes_{R}\theta:X\boxtimes_{R}Y\rightarrow
X\boxtimes_{R}Y^{\prime}$ is a morphism of $\mathbb{Z}$-complexes. Finally, we
note that $X\boxtimes_{R}-:\mathcal{C}_{\geq 0}(R)\rightarrow\mathcal{C}_{\geq
0}(\mathbb{Z})$ preserves the composition of morphisms as well as the identity
morphisms, so it defines a covariant functor.

(ii): Similar to (i). ∎

We have the following special cases of the shuffle product.

######  Proposition 5.5.

Let $R$ be a ring, $X$ a connective $R^{\operatorname{op}}$-complex, $Y$ a
connective $R$-complex, $M$ a right $R$-module, and $N$ a left $R$-module.
Then the following assertions hold:

  1. (i)

$X\boxtimes_{R}N=X\otimes_{R}N$.

  2. (ii)

$M\boxtimes_{R}Y=M\otimes_{R}Y$.

###### Proof.

(i): Let $n\geq 0$, and let $\mathcal{S}_{n}$ be as in Definition 5.2. Then we
have
$(X\boxtimes_{R}N)_{n}=\bigoplus_{(f,g)\in\mathcal{S}_{n}}(X_{k}\otimes_{R}N_{l})$.
But $N_{l}=0$ for every $l\neq 0$, so we should only consider those terms in
the latter direct sum that correspond to $l=0$. However, if
$(f,g)\in\mathcal{S}_{n}$ and $l=0$, then $g=0$. But $\phi_{f,g}$ is
injective, so $f$ must be injective. As $f$ is already surjective, we conclude
that $k=n$ and $f=1^{[n]}$. As a result, we only have one possibly non-zero
term, i.e. $(X\boxtimes_{R}N)_{n}=X_{n}\otimes_{R}N=(X\otimes_{R}N)_{n}$. On
the other hand, since $1^{[n]}\delta_{n,i}=\delta_{n,i}$ is not surjective,
Corollary 4.12 implies that
$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{1^{[n]}}^{X}(x)\right)=0$ for
every $0\leq i\leq n-1$ and $x\in X_{n}$. Furthermore, $\operatorname{DK}(N)$
is a constant simplicial module, so $d_{n,i}^{\operatorname{DK}(N)}=1^{N}$ for
every $0\leq i\leq n$. Therefore, using Corollary 4.12, we have for every
$x\in X_{n}$ and $y\in N$:

$\begin{split}\partial_{n}^{X\boxtimes_{R}N}(x\otimes
y)&=\textstyle\sum_{i=0}^{n}(-1)^{i}d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{1^{[n]}}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(N)}\left(\lambda_{0}^{N}(y)\right)=(-1)^{n}d_{n,n}^{\operatorname{DK}(X)}\left(\lambda_{1^{[n]}}^{X}(x)\right)\otimes
d_{n,n}^{\operatorname{DK}(N)}\left(\lambda_{0}^{N}(y)\right)\\\
&=(-1)^{n}(-1)^{n}\lambda_{1^{[n-1]}}^{X}\left(\partial_{n}^{X}(x)\right)\otimes_{R}\lambda_{0}^{N}(y)=\partial_{n}^{X}(x)\otimes
y=\partial_{n}^{X\otimes_{R}N}(x\otimes y)\\\ \end{split}$

It follows that $\partial_{n}^{X\boxtimes_{R}N}=\partial_{n}^{X\otimes_{R}N}$.
Thus $X\boxtimes_{R}N=X\otimes_{R}N$.

(ii): Similar to (i). ∎

The next proposition describes the shuffle product in terms of the
normalization and Dold-Kan functors. This in turn provides some useful
corollaries.

######  Proposition 5.6.

Let $R$ be a ring, $X$ a connective $R^{\operatorname{op}}$-complex, and $Y$ a
connective $R$-complex. Then the following natural isomorphism of
$\mathbb{Z}$-complexes holds:

$X\boxtimes_{R}Y\cong\operatorname{Nor}\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)$

###### Proof.

Let $n\geq 0$, and let $\mathcal{S}_{n}$ be as in Definition 5.2. As in the
proof of Proposition 5.3, we can write:

(5.6.1)
$C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)_{n}=(X\boxtimes_{R}Y)_{n}\oplus\left(\textstyle\bigoplus_{(f,g)\notin\mathcal{S}_{n}}(X_{k}\otimes_{R}Y_{l})\right)$

Let $f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$ and
$g\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)$ be surjective. Let
$0\leq j_{1}<\cdots<j_{n-k}<n$ be the elements of $[n]$ for which
$f(j)=f(j+1)$ so that $f=\sigma_{k,j_{1}}\cdots\sigma_{n-1,j_{n-k}}$, and
$0\leq i_{1}<\cdots<i_{n-l}<n$ the elements of $[n]$ for which $g(i)=g(i+1)$
so that $g=\sigma_{l,i_{1}}\cdots\sigma_{n-1,i_{n-l}}$. If
$(f,g)\notin\mathcal{S}_{n}$, then $\phi_{f,g}$ is not injective, so there is
an element $m\in\\{i_{1},...,i_{n-l}\\}\cap\\{j_{1},...,j_{n-k}\\}$. This
implies that $f=\sigma_{k,j_{1}}\cdots\sigma_{p,m}\cdots\sigma_{n-1,j_{n-k}}$
and $g=\sigma_{l,i_{1}}\cdots\sigma_{q,m}\cdots\sigma_{n-1,i_{n-l}}$ where
$k\leq p\leq n-1$ and $l\leq q\leq n-1$. Using the face-degeneracy relations
of Remark 4.2, we can successively swap the face map $\sigma_{p,m}$ with the
terms appearing on its right to get
$f=\sigma_{k,j_{1}}\cdots\sigma_{n-2,j_{n-k}-1}\sigma_{n-1,m}$. Similarly, we
get $g=\sigma_{l,i_{1}}\cdots\sigma_{n-2,i_{n-l}-1}\sigma_{n-1,m}$. Applying
the contravariant functor $\operatorname{DK}(X)$, we get
$\operatorname{DK}(X)(f)=s_{n-1,m}^{\operatorname{DK}(X)}s_{n-2,j_{n-k}-1}^{\operatorname{DK}(X)}\cdots
s_{k,j_{1}}^{\operatorname{DK}(X)}$ and
$\operatorname{DK}(Y)(g)=s_{n-1,m}^{\operatorname{DK}(Y)}s_{n-2,i_{n-l}-1}^{\operatorname{DK}(Y)}\cdots
s_{l,i_{1}}^{\operatorname{DK}(Y)}$. Moreover, since $f$ and $g$ are
surjective, we can invoke Proposition 4.11 to write:

$\lambda_{f}^{X}=\operatorname{DK}(X)(f)\lambda_{1^{[k]}}^{X}=s_{n-1,m}^{\operatorname{DK}(X)}s_{n-2,j_{n-k}-1}^{\operatorname{DK}(X)}\cdots
s_{k,j_{1}}^{\operatorname{DK}(X)}\lambda_{1^{[k]}}^{X}$

and

$\lambda_{g}^{Y}=\operatorname{DK}(Y)(g)\lambda_{1^{[l]}}^{Y}=s_{n-1,m}^{\operatorname{DK}(Y)}s_{n-2,i_{n-l}-1}^{\operatorname{DK}(Y)}\cdots
s_{l,i_{1}}^{\operatorname{DK}(Y)}\lambda_{1^{[l]}}^{Y}$

Therefore, we obtain:

$\begin{split}\lambda_{f,g}^{X,Y}&=\lambda_{f}^{X}\otimes_{R}\lambda_{g}^{Y}\\\
&=s_{n-1,m}^{\operatorname{DK}(X)}s_{n-2,j_{n-k}-1}^{\operatorname{DK}(X)}\cdots
s_{k,j_{1}}^{\operatorname{DK}(X)}\lambda_{1^{[k]}}^{X}\otimes_{R}s_{n-1,m}^{\operatorname{DK}(Y)}s_{n-2,i_{n-l}-1}^{\operatorname{DK}(Y)}\cdots
s_{l,i_{1}}^{\operatorname{DK}(Y)}\lambda_{1^{[l]}}^{Y}\\\
&=\left(s_{n-1,m}^{\operatorname{DK}(X)}\otimes_{R}s_{n-1,m}^{\operatorname{DK}(Y)}\right)\left(s_{n-2,j_{n-k}-1}^{\operatorname{DK}(X)}\cdots
s_{k,j_{1}}^{\operatorname{DK}(X)}\lambda_{1^{[k]}}^{X}\otimes_{R}s_{n-2,i_{n-l}-1}^{\operatorname{DK}(Y)}\cdots
s_{l,i_{1}}^{\operatorname{DK}(Y)}\lambda_{1^{[l]}}^{Y}\right)\\\
&=\left(s_{n-1,m}^{\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)}\right)\left(s_{n-2,j_{n-k}-1}^{\operatorname{DK}(X)}\cdots
s_{k,j_{1}}^{\operatorname{DK}(X)}\lambda_{1^{[k]}}^{X}\otimes_{R}s_{n-2,i_{n-l}-1}^{\operatorname{DK}(Y)}\cdots
s_{l,i_{1}}^{\operatorname{DK}(Y)}\lambda_{1^{[l]}}^{Y}\right)\end{split}$

This shows that:

$\operatorname{Im}\left(\lambda_{f,g}^{X,Y}\right)\subseteq\operatorname{Im}\left(s_{n-1,m}^{\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)}\right)\subseteq\textstyle\sum_{i=0}^{n-1}\operatorname{Im}\left(s_{n-1,i}^{\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)}\right)$

Thus $\lambda_{f,g}^{X,Y}(x\otimes
y)\in\sum_{i=0}^{n-1}\operatorname{Im}\left(s_{n-1,i}^{\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)}\right)$
for every $x\in X_{k}$ and $y\in Y_{l}$. It follows that:

$\textstyle\bigoplus_{(f,g)\notin\mathcal{S}_{n}}(X_{k}\otimes_{R}Y_{l})\subseteq\textstyle\sum_{i=0}^{n-1}\operatorname{Im}\left(s_{n-1,i}^{\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)}\right)$

On the other hand, let
$f^{\prime}\in\operatorname{Mor}_{\Delta}\left([n-1],[k]\right)$ and
$g^{\prime}\in\operatorname{Mor}_{\Delta}\left([n-1],[l]\right)$ be
surjective. Using Corollary 4.13, we have for every $0\leq i\leq n-1$, $x\in
X_{k}$, and $y\in Y_{l}$:

$\begin{split}s_{n-1,i}^{\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)}\left(\lambda_{f^{\prime},g^{\prime}}^{X,Y}(x\otimes
y)\right)&=\left(s_{n-1,i}^{\operatorname{DK}(X)}\otimes_{R}s_{n-1,i}^{\operatorname{DK}(Y)}\right)\left(\left(\lambda_{f^{\prime}}^{X}\otimes_{R}\lambda_{g^{\prime}}^{Y}\right)(x\otimes
y)\right)\\\
&=\left(s_{n-1,i}^{\operatorname{DK}(X)}\lambda_{f^{\prime}}^{X}\otimes_{R}s_{n-1,i}^{\operatorname{DK}(Y)}\lambda_{g^{\prime}}^{Y}\right)\left(x\otimes
y\right)\\\
&=\left(\lambda_{f^{\prime}\sigma_{n-1,i}}^{X}\otimes_{R}\lambda_{g^{\prime}\sigma_{n-1,i}}^{Y}\right)\left(x\otimes
y\right)\\\
&=\lambda_{f^{\prime}\sigma_{n-1,i},g^{\prime}\sigma_{n-1,i}}^{X,Y}\left(x\otimes
y\right)\end{split}$

Set
$f=f^{\prime}\sigma_{n-1,i}\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$
and
$g=g^{\prime}\sigma_{n-1,i}\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)$.
As $\sigma_{n-1,i}(i)=i=\sigma_{n-1,i}(i+1)$, we see that
$\phi_{f,g}(i)=\left(f(i),g(i)\right)=\left(f(i+1),g(i+1)\right)=\phi_{f,g}(i+1)$.
Thus $\phi_{f,g}$ is not injective, so $(f,g)\notin\mathcal{S}_{n}$. It
follows that:

$\textstyle\sum_{i=0}^{n-1}\operatorname{Im}\left(s_{n-1,i}^{\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)}\right)\subseteq\textstyle\bigoplus_{(f,g)\notin\mathcal{S}_{n}}(X_{k}\otimes_{R}Y_{l})$

Therefore, we have:

(5.6.2)
$\textstyle\bigoplus_{(f,g)\notin\mathcal{S}_{n}}(X_{k}\otimes_{R}Y_{l})=\textstyle\sum_{i=0}^{n-1}\operatorname{Im}\left(s_{n-1,i}^{\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)}\right)=D\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)_{n}$

Hence we obtain from (5.6.1) and (5.6.2) that:

$C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)_{n}=(X\boxtimes_{R}Y)_{n}\oplus
D\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)_{n}$

In light of the proof of Proposition 5.3 and Remark 4.6, we observe that
$X\boxtimes_{R}Y$ and
$D\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)$ are
subcomplexes of
$C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)$, so we
deduce that:

(5.6.3)
$C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)=(X\boxtimes_{R}Y)\oplus
D\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)$

On the other hand, we know from Remark 4.6 that:

(5.6.4)
$C\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)=\operatorname{Nor}\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)\oplus
D\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)$

Therefore, it follows from (5.6.3) and (5.6.4) that
$X\boxtimes_{R}Y\cong\operatorname{Nor}\left(\operatorname{DK}(X)\otimes_{R}\operatorname{DK}(Y)\right)$.
∎

######  Corollary 5.7.

Let $R$ be a ring, $M$ a simplicial right $R$-module, and $N$ a simplicial
left $R$-module. Then the following natural isomorphism of
$\mathbb{Z}$-complexes holds:

$\operatorname{Nor}(M\otimes_{R}N)\cong\operatorname{Nor}(M)\boxtimes_{R}\operatorname{Nor}(N)$

###### Proof.

In view of Theorem 4.14 and Proposition 5.6, we have the following natural
isomorphisms:

$\operatorname{Nor}(M\otimes_{R}N)\cong\operatorname{Nor}\left(\operatorname{DK}\left(\operatorname{Nor}(M)\right)\otimes_{R}\operatorname{DK}\left(\operatorname{Nor}(N)\right)\right)\cong\operatorname{Nor}(M)\boxtimes_{R}\operatorname{Nor}(N)$

∎

######  Corollary 5.8.

Let $R$ be a ring, $M$ a simplicial right $R$-module, $N$ a simplicial left
$R$-module, $K$ a right $R$-module, and $L$ a left $R$-module. Then the
following natural isomorphisms of $\mathbb{Z}$-complexes hold:

  1. (i)

$\operatorname{Nor}(M\otimes_{R}L)\cong\operatorname{Nor}(M)\otimes_{R}L$.

  2. (ii)

$\operatorname{Nor}(K\otimes_{R}N)\cong K\otimes_{R}\operatorname{Nor}(N)$.

###### Proof.

(i): If we consider $L$ as a constant simplicial left $R$-module, then
$\operatorname{Nor}(L)=L$ is the $R$-complex with only $L$ concentrated in
degree zero. Therefore, in light of Corollary 5.7 and Proposition 5.5, we get
the following natural isomorphism:

$\operatorname{Nor}(M\otimes_{R}L)\cong\operatorname{Nor}(M)\boxtimes_{R}\operatorname{Nor}(L)=\operatorname{Nor}(M)\boxtimes_{R}L=\operatorname{Nor}(M)\otimes_{R}L$

(ii): Similar to (i). ∎

Now we are ready to establish a variant of the Eilenberg-Zilber Theorem; see
[We, Chapter 8, Section 8.5] for the original version.

######  Theorem 5.9.

Let $R$ be a ring, $X$ a connective $R^{\operatorname{op}}$-complex, and $Y$ a
connective $R$-complex. Then there exists a natural quasi-isomorphism
$\nabla^{X,Y}:X\otimes_{R}Y\rightarrow X\boxtimes_{R}Y$ of
$\mathbb{Z}$-complexes.

###### Proof.

Let $n\geq 0$. Then
$(X\otimes_{R}Y)_{n}=\bigoplus_{k+l=n}(X_{k}\otimes_{R}Y_{l})$. For any $0\leq
k,l\leq n$, set:

$\mathcal{S}_{n}^{k,l}=\left\\{(f,g)\in\operatorname{Mor}_{\Delta}([n],[k])\times\operatorname{Mor}_{\Delta}([n],[l])\;|\;\begin{tabular}[]{ccc}$f$
and $g$ are surjective, and $\phi_{f,g}:[n]\rightarrow[k]\times[l]$ given\\\
by $\phi_{f,g}(i)=(f(i),g(i))$ for every $i\in[n]$ is
injective\end{tabular}\right\\}$

Moreover, set $\mathcal{S}_{n}=\bigcup_{0\leq k,l\leq
n}\mathcal{S}_{n}^{k,l}$. Then
$(X\boxtimes_{R}Y)_{n}=\bigoplus_{(f,g)\in\mathcal{S}_{n}}(X_{k}\otimes_{R}Y_{l})$.
Now let $k,l\geq 0$ be such that $k+l=n$. Let
$\iota_{k,l}^{X,Y}:X_{k}\otimes_{R}Y_{l}\rightarrow(X\otimes_{R}Y)_{n}$ be the
canonical injection. Also, let
$\lambda_{f}^{X}:X_{k}\rightarrow\operatorname{DK}(X)([n])$ be the canonical
injection corresponding to the index
$f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$,
$\lambda_{g}^{Y}:Y_{l}\rightarrow\operatorname{DK}(Y)([n])$ the canonical
injection corresponding to the index
$g\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)$, and
$\lambda_{f,g}^{X,Y}:X_{k}\otimes_{R}Y_{l}\rightarrow(X\boxtimes_{R}Y)_{n}$
the canonical injection corresponding to the indices
$f\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$ and
$g\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)$. Moreover, let
$\nu_{f,g}\in\operatorname{Sh}(k,l)$ denote the shuffle determined by a pair
$(f,g)\in\mathcal{S}_{n}^{k,l}$ as in Remark 5.1. Define a map
$\psi_{k,l}:X_{k}\times Y_{l}\rightarrow(X\boxtimes_{R}Y)_{n}$ by setting

$\psi_{k,l}(x,y)=\textstyle\sum_{(f,g)\in\mathcal{S}_{n}^{k,l}}\operatorname{sgn}(\nu_{f,g})\lambda_{f,g}^{X,Y}(x\otimes
y)$

for every $(x,y)\in X_{k}\times Y_{l}$. It is clear that $\psi_{k,l}$ is an
$R$-balanced map. Therefore, the universal property of tensor product yields a
unique $\mathbb{Z}$-homomorphism
$\overline{\psi_{k,l}}:X_{k}\otimes_{R}Y_{l}\rightarrow(X\boxtimes_{R}Y)_{n}$
that makes the following diagram commutative:

${X_{k}\times
Y_{l}}$${(X\boxtimes_{R}Y)_{n}}$${X_{k}\otimes_{R}Y_{l}}$$\scriptstyle{\psi_{k,l}}$$\scriptstyle{\omega_{X_{k},Y_{l}}}$$\scriptstyle{\overline{\psi_{k,l}}}$

Then we have

$\overline{\psi_{k,l}}(x\otimes
y)=\textstyle\sum_{(f,g)\in\mathcal{S}_{n}^{k,l}}\operatorname{sgn}(\nu_{f,g})\lambda_{f,g}^{X,Y}(x\otimes
y)$

for every $x\in X_{k}$ and $y\in Y_{l}$. By the universal property of direct
sum, there is a unique $\mathbb{Z}$-homomorphism
$\nabla_{n}^{X,Y}:(X\otimes_{R}Y)_{n}\rightarrow(X\boxtimes_{R}Y)_{n}$ that
makes the following diagram commutative for every $0\leq k,l\leq n$ with
$k+l=n$:

${X_{k}\otimes_{R}Y_{l}}$${(X\otimes_{R}Y)_{n}}$${(X\boxtimes_{R}Y)_{n}}$$\scriptstyle{\iota_{k,l}^{X,Y}}$$\scriptstyle{\overline{\psi_{k,l}}}$$\scriptstyle{\nabla_{n}^{X,Y}}$

Then we have

$\nabla_{n}^{X,Y}\left(\textstyle\sum_{k+l=n}\iota_{k,l}^{X,Y}(x_{k}\otimes
y_{l})\right)=\textstyle\sum_{k+l=n}\textstyle\sum_{(f,g)\in\mathcal{S}_{n}^{k,l}}\operatorname{sgn}(\nu_{f,g})\lambda_{f,g}^{X,Y}(x_{k}\otimes
y_{l})$

for every $\sum_{k+l=n}\iota_{k,l}^{X,Y}(x_{k}\otimes
y_{l})\in(X\otimes_{R}Y)_{n}$.

Now set $\nabla^{X,Y}=\left(\nabla_{n}^{X,Y}\right)_{n\geq
0}:X\otimes_{R}Y\rightarrow X\boxtimes_{R}Y$. We show that $\nabla^{X,Y}$ is a
morphism of $\mathbb{Z}$-complexes. Let $n\geq 0$. Then the following diagram
is commutative:

${(X\otimes_{R}Y)_{n}}$${(X\otimes_{R}Y)_{n-1}}$${(X\boxtimes_{R}Y)_{n}}$${(X\boxtimes_{R}Y)_{n-1}}$$\scriptstyle{\partial_{n}^{X\otimes_{R}Y}}$$\scriptstyle{\nabla_{n}^{X,Y}}$$\scriptstyle{\nabla_{n-1}^{X,Y}}$$\scriptstyle{\partial_{n}^{X\boxtimes_{R}Y}}$

Indeed, we have for every $x\in X_{k}$ and $y\in Y_{l}$ with $k+l=n$:

$\begin{split}\partial_{n}^{X\boxtimes_{R}Y}\left(\nabla_{n}^{X,Y}\left(\iota_{k,l}^{X,Y}(x\otimes
y)\right)\right)&=\partial_{n}^{X\boxtimes_{R}Y}\left(\textstyle\sum_{(f,g)\in\mathcal{S}_{n}^{k,l}}\operatorname{sgn}(\nu_{f,g})\lambda_{f,g}^{X,Y}(x\otimes
y)\right)\\\
&=\textstyle\sum_{(f,g)\in\mathcal{S}_{n}^{k,l}}\operatorname{sgn}(\nu_{f,g})\partial_{n}^{X\boxtimes_{R}Y}\left(\lambda_{f,g}^{X,Y}(x\otimes
y)\right)\\\
&=\textstyle\sum_{i=0}^{n}(-1)^{i}\textstyle\sum_{(f,g)\in\mathcal{S}_{n}^{k,l}}\operatorname{sgn}(\nu_{f,g})d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)\\\ \end{split}$

Let $0\leq i\leq n$ and $(f,g)\in\mathcal{S}_{n}^{k,l}$, and consider the
following cases:

Case I: $f\delta_{n,i}$ is not surjective but $g\delta_{n,i}$ is surjective

If $0\leq i\leq n-1$, then by Corollary 4.12,
$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)=0$, so
$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)=0$. Hence we can
assume that $i=n$. In this case, by Corollary 4.12, $f$ induces a surjective
map $\overline{f}\in\operatorname{Mor}_{\Delta}\left([n-1],[k-1]\right)$ given
by $\overline{f}(i)=f(i)$ for every $i\in[n-1]$. Moreover, we note that
$g\delta_{n,n}=g\mid_{[n-1]}:[n-1]\rightarrow[l]$. If $0\leq
j_{1}<\cdots<j_{l}<n$ are the elements of $[n]$ for which $f(j)=f(j+1)$, and
$0\leq i_{1}<\cdots<i_{k}<n$ are the elements of $[n]$ for which
$g(i)=g(i+1)$, then we have:

$\nu_{f,g}=\begin{pmatrix}1&\cdots&k&k+1&\cdots&n\\\
i_{1}+1&\cdots&i_{k}+1&j_{1}+1&\cdots&j_{l}+1\end{pmatrix}$

Set:

$A=\left\\{(i_{\alpha}+1,j_{\beta}+1)\mid 1\leq\alpha\leq k\textrm{,
}1\leq\beta\leq l\textrm{, and }i_{\alpha}>j_{\beta}\right\\}$

If $|A|=r$, then $\operatorname{sgn}(\nu_{f,g})=(-1)^{r}$. As $f$ is
surjective but $f\delta_{n,n}$ is not surjective, we see that $j_{l}<n-1$
since otherwise, $f(n-1)=f(n)=k$ which would imply that
$f\delta_{n,n}=f\mid_{[n-1]}:[n-1]\rightarrow[k]$ is surjective. Also, since
$g$ and $g\delta_{n,n}$ are both surjective, we notice that $i_{k}=n-1$ as
$g(n-1)=l=g(n)$. As a result, $0\leq j_{1}<\cdots<j_{l}<n-1$ are the elements
of $[n-1]$ for which $\overline{f}(j)=\overline{f}(j+1)$, and $0\leq
i_{1}<\cdots<i_{k-1}<n-1$ are the elements of $[n-1]$ for which
$(g\delta_{n,n})(i)=(g\delta_{n,n})(i+1)$, so we have:

$\nu_{\overline{f},g\delta_{n,n}}=\begin{pmatrix}1&\cdots&k-1&k&\cdots&n-1\\\
i_{1}+1&\cdots&i_{k-1}+1&j_{1}+1&\cdots&j_{l}+1\end{pmatrix}$

Set:

$B=\left\\{(i_{\alpha}+1,j_{\beta}+1)\mid 1\leq\alpha\leq k-1\textrm{,
}1\leq\beta\leq l\textrm{, and }i_{\alpha}>j_{\beta}\right\\}$

If $|B|=s$, then
$\operatorname{sgn}\left(\nu_{\overline{f},g\delta_{n,n}}\right)=(-1)^{s}$.
Since $i_{k}=n-1>j_{l}>\cdots>j_{1}$, we conclude that

$A=B\mathbin{\mathchoice{\leavevmode\vtop{\halign{\hfil$\m@th\displaystyle#$\hfil\cr\cup\cr\cdot\crcr}}}{\leavevmode\vtop{\halign{\hfil$\m@th\textstyle#$\hfil\cr\cup\cr\cdot\crcr}}}{\leavevmode\vtop{\halign{\hfil$\m@th\scriptstyle#$\hfil\cr\cup\cr\cdot\crcr}}}{\leavevmode\vtop{\halign{\hfil$\m@th\scriptscriptstyle#$\hfil\cr\cup\cr\cdot\crcr}}}}\\{(i_{k}+1,j_{\beta}+1)\mid
1\leq\beta\leq l\\},$

which in turn implies that $r=s+l$. Therefore, we have
$\operatorname{sgn}(\nu_{f,g})=(-1)^{r}=(-1)^{l}(-1)^{s}=(-1)^{l}\operatorname{sgn}\left(\nu_{\overline{f},g\delta_{n,n}}\right)$.
Thus using Corollary 4.12, we get:

$\begin{split}(-1)^{n}\operatorname{sgn}(\nu_{f,g})d_{n,n}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,n}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)&=(-1)^{n}(-1)^{l}\operatorname{sgn}\left(\nu_{\overline{f},g\delta_{n,n}}\right)(-1)^{k}\lambda_{\overline{f}}^{X}\left(\partial_{k}^{X}(x)\right)\otimes\lambda_{g\delta_{n,n}}^{Y}(y)\\\
&=\operatorname{sgn}\left(\nu_{\overline{f},g\delta_{n,n}}\right)\lambda_{\overline{f},g\delta_{n,n}}^{X,Y}\left(\partial_{k}^{X}(x)\otimes
y\right)\end{split}$

Case II: $f\delta_{n,i}$ is surjective but $g\delta_{n,i}$ is not surjective

If $0\leq i\leq n-1$, then by Corollary 4.12,
$d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)=0$, so
$d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)=0$. Hence we can
assume that $i=n$. In this case, we have
$f\delta_{n,n}=f\mid_{[n-1]}:[n-1]\rightarrow[k]$. Moreover, by Corollary
4.12, $g$ induces a surjective map
$\overline{g}\in\operatorname{Mor}_{\Delta}\left([n-1],[l-1]\right)$ given by
$\overline{g}(i)=g(i)$ for every $i\in[n-1]$. If $0\leq j_{1}<\cdots<j_{l}<n$
are the elements of $[n]$ for which $f(j)=f(j+1)$, and $0\leq
i_{1}<\cdots<i_{k}<n$ are the elements of $[n]$ for which $g(i)=g(i+1)$, then
we have:

$\nu_{f,g}=\begin{pmatrix}1&\cdots&k&k+1&\cdots&n\\\
i_{1}+1&\cdots&i_{k}+1&j_{1}+1&\cdots&j_{l}+1\end{pmatrix}$

Set:

$A=\left\\{(i_{\alpha}+1,j_{\beta}+1)\mid 1\leq\alpha\leq k\textrm{,
}1\leq\beta\leq l\textrm{, and }i_{\alpha}>j_{\beta}\right\\}$

If $|A|=r$, then $\operatorname{sgn}(\nu_{f,g})=(-1)^{r}$. As $f$ and
$f\delta_{n,n}$ are both surjective, we notice that $j_{l}=n-1$ as
$f(n-1)=k=f(n)$. Also, since $g$ is surjective but $g\delta_{n,n}$ is not
surjective, we see that $i_{k}<n-1$ since otherwise, $g(n-1)=g(n)=l$ which
would imply that $g\delta_{n,n}=g\mid_{[n-1]}:[n-1]\rightarrow[l]$ is
surjective. As a result, $0\leq j_{1}<\cdots<j_{l-1}<n-1$ are the elements of
$[n-1]$ for which $(f\delta_{n,n})(j)=(f\delta_{n,n})(j+1)$, and $0\leq
i_{1}<\cdots<i_{k}<n-1$ are the elements of $[n-1]$ for which
$\overline{g}(i)=\overline{g}(i+1)$, so we have:

$\nu_{f\delta_{n,n},\overline{g}}=\begin{pmatrix}1&\cdots&k&k+1&\cdots&n-1\\\
i_{1}+1&\cdots&i_{k}+1&j_{1}+1&\cdots&j_{l-1}+1\end{pmatrix}$

Set:

$B=\left\\{(i_{\alpha}+1,j_{\beta}+1)\mid 1\leq\alpha\leq k\textrm{,
}1\leq\beta\leq l-1\textrm{, and }i_{\alpha}>j_{\beta}\right\\}$

If $|B|=s$, then
$\operatorname{sgn}\left(\nu_{f\delta_{n,n},\overline{g}}\right)=(-1)^{s}$.
Since $j_{l}=n-1>i_{k}>\cdots>i_{1}$, we deduce that $A=B$, which in turn
implies that $r=s$. Therefore, we have
$\operatorname{sgn}(\nu_{f,g})=(-1)^{r}=(-1)^{s}=\operatorname{sgn}\left(\nu_{f\delta_{n,n},\overline{g}}\right)$.
Thus using Corollary 4.12, we get:

$\begin{split}(-1)^{n}\operatorname{sgn}(\nu_{f,g})d_{n,n}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,n}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)&=(-1)^{n}\operatorname{sgn}\left(\nu_{f\delta_{n,n},\overline{g}}\right)\lambda_{f\delta_{n,n}}^{X}(x)\otimes(-1)^{l}\lambda_{\overline{g}}^{Y}\left(\partial_{l}^{Y}(y)\right)\\\
&=(-1)^{k}\operatorname{sgn}\left(\nu_{f\delta_{n,n},\overline{g}}\right)\lambda_{f\delta_{n,n},\overline{g}}^{X,Y}\left(x\otimes\partial_{l}^{Y}(y)\right)\end{split}$

Case III: $f\delta_{n,i}$ and $g\delta_{n,i}$ are both surjective

Let $0\leq j_{1}<\cdots<j_{l}<n$ be the elements of $[n]$ for which
$f(j)=f(j+1)$, and $0\leq i_{1}<\cdots<i_{k}<n$ the elements of $[n]$ for
which $g(i)=g(i+1)$. If $i=0$, then $f\delta_{n,0}$ is surjective which means
that if $0$ is omitted from the domain of $f$, then it will remain surjective.
This implies that $f(0)=0=f(1)$, so $j_{1}=0$. Similarly, we see that
$i_{1}=0$. This is a contradiction as
$\\{i_{1},...,i_{k}\\}\cap\\{j_{1},...,j_{l}\\}=\emptyset$. If $i=n$, then
$f\delta_{n,n}$ is surjective which means that if $n$ is omitted from the
domain of $f$, then it will remain surjective. This yields that
$f(n-1)=k=f(n)$, so $j_{l}=n-1$. Similarly, we see that $i_{k}=n-1$. This is a
contradiction as $\\{i_{1},...,i_{k}\\}\cap\\{j_{1},...,j_{l}\\}=\emptyset$.
Hence we can assume that $1\leq i\leq n-1$. By Corollary 4.12, we have:

$\operatorname{sgn}(\nu_{f,g})d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)=\operatorname{sgn}(\nu_{f,g})\lambda_{f\delta_{n,i}}^{X}(x)\otimes\lambda_{g\delta_{n,i}}^{Y}(y)=\operatorname{sgn}(\nu_{f,g})\lambda_{f\delta_{n,i},g\delta_{n,i}}^{X,Y}(x\otimes
y)$

Since $f\delta_{n,i}$ is surjective, we observe that if $i$ is omitted from
the domain of $f$, then it will remain surjective. This means that
$f(i-1)=f(i)$ or $f(i)=f(i+1)$, so $i-1\in\\{j_{1},...,j_{l}\\}$ or
$i\in\\{j_{1},...,j_{l}\\}$. Similarly, we have $i-1\in\\{i_{1},...,i_{k}\\}$
or $i\in\\{i_{1},...,i_{k}\\}$. Since
$\\{i_{1},...,i_{k}\\}\cap\\{j_{1},...,j_{l}\\}=\emptyset$, we should have
either $i-1\in\\{j_{1},...,j_{l}\\}$ and $i\in\\{i_{1},...,i_{k}\\}$, or
$i\in\\{j_{1},...,j_{l}\\}$ and $i-1\in\\{i_{1},...,i_{k}\\}$. Without loss of
generality, suppose that $i-1\in\\{j_{1},...,j_{l}\\}$ and
$i\in\\{i_{1},...,i_{k}\\}$. Define a map $f^{\prime}:[n]\rightarrow[k]$ by
setting $f^{\prime}(t)=f(t)$ for every $t\neq i$ and $f^{\prime}(i)=f(i+1)$.
Since $f^{\prime}(i-1)=f(i-1)=f(i)<f(i+1)=f^{\prime}(i)=f^{\prime}(i+1)$, it
follows that $f^{\prime}$ is order-preserving. That is,
$f^{\prime}\in\operatorname{Mor}_{\Delta}\left([n],[k]\right)$. Moreover,
$f^{\prime}(i-1)=f(i-1)=f(i)$, so $f(i)\in\operatorname{Im}(f^{\prime})$,
whence we see that $f^{\prime}$ is surjective. Also, since $f$ and
$f^{\prime}$ have the same values except at $i$, we have
$f\delta_{n,i}=f^{\prime}\delta_{n,i}$. Similarly, define a map
$g^{\prime}:[n]\rightarrow[l]$ by setting $g^{\prime}(t)=g(t)$ for every
$t\neq i$ and $g^{\prime}(i)=g(i-1)$. Since
$g^{\prime}(i-1)=g^{\prime}(i)=g(i-1)<g(i)=g(i+1)=g^{\prime}(i+1)$, it follows
that $g^{\prime}$ is order-preserving. That is,
$g^{\prime}\in\operatorname{Mor}_{\Delta}\left([n],[l]\right)$. Moreover,
$g^{\prime}(i+1)=g(i+1)=g(i)$, so $g(i)\in\operatorname{Im}(g^{\prime})$,
thereby we notice that $g^{\prime}$ is surjective. Also, since $g$ and
$g^{\prime}$ have the same values except at $i$, we have
$g\delta_{n,i}=g^{\prime}\delta_{n,i}$. Now if
$(f^{\prime},g^{\prime})\notin\mathcal{S}_{n}^{k,l}$, then
$\phi_{f^{\prime},g^{\prime}}$ is not injective, so there is an $0\leq m\leq
n-1$ such that
$\phi_{f^{\prime},g^{\prime}}(m)=\phi_{f^{\prime},g^{\prime}}(m+1)$. But we
note that:

$\phi_{f^{\prime},g^{\prime}}(m)=\left(f^{\prime}(m),g^{\prime}(m)\right)=\begin{dcases}\left(f(m),g(m)\right)=\phi_{f,g}(m)&\textrm{if
}m<i-1\\\ \left(f(m),g(m)\right)&\textrm{if }m=i-1\\\
\left(f(m+1),g(m-1)\right)&\textrm{if }m=i\\\
\left(f(m),g(m)\right)=\phi_{f,g}(m)&\textrm{if }m>i\end{dcases}$

and

$\phi_{f^{\prime},g^{\prime}}(m+1)=\left(f^{\prime}(m+1),g^{\prime}(m+1)\right)=\begin{dcases}\left(f(m+1),g(m+1)\right)=\phi_{f,g}(m+1)&\textrm{if
}m<i-1\\\ \left(f(m+2),g(m)\right)&\textrm{if }m=i-1\\\
\left(f(m+1),g(m+1)\right)&\textrm{if }m=i\\\
\left(f(m+1),g(m+1)\right)=\phi_{f,g}(m+1)&\textrm{if }m>i\end{dcases}$

Taking into account that $\phi_{f,g}$ is injective, we notice that in any
case, $\phi_{f^{\prime},g^{\prime}}(m)\neq\phi_{f^{\prime},g^{\prime}}(m+1)$
which is a contradiction. Hence
$(f^{\prime},g^{\prime})\in\mathcal{S}_{n}^{k,l}$. We finally have:

$\nu_{f,g}=\begin{pmatrix}1&\cdots&p&\cdots&k&k+1&\cdots&q&\cdots&n\\\
i_{1}+1&\cdots&i&\cdots&i_{k}+1&j_{1}+1&\cdots&i-1&\cdots&j_{l}+1\end{pmatrix}$

Furthermore, we have:

$\nu_{f^{\prime},g^{\prime}}=\begin{pmatrix}1&\cdots&p&\cdots&k&k+1&\cdots&q&\cdots&n\\\
i_{1}+1&\cdots&i-1&\cdots&i_{k}+1&j_{1}+1&\cdots&i&\cdots&j_{l}+1\end{pmatrix}$

It is clear from the above displays that
$\operatorname{sgn}(\nu_{f^{\prime},g^{\prime}})=-\operatorname{sgn}(\nu_{f,g})$.
As a result, we have:

$\operatorname{sgn}(\nu_{f^{\prime},g^{\prime}})d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f^{\prime}}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g^{\prime}}^{Y}(y)\right)=\operatorname{sgn}(\nu_{f^{\prime},g^{\prime}})\lambda_{f^{\prime}\delta_{n,i}}^{X}(x)\otimes\lambda_{g^{\prime}\delta_{n,i}}^{Y}(y)=\operatorname{sgn}(\nu_{f^{\prime},g^{\prime}})\lambda_{f^{\prime}\delta_{n,i},g^{\prime}\delta_{n,i}}^{X,Y}(x\otimes
y)=$ $-sgn(\nu_{f,g})\lambda_{f\delta_{n,i},g\delta_{n,i}}^{X,Y}(x\otimes y)$

Consequently, fixing $1\leq i\leq n-1$, for any pair
$(f,g)\in\mathcal{S}_{n}^{k,l}$, there is a dual pair
$(f^{\prime},g^{\prime})\in\mathcal{S}_{n}^{k,l}$ whose corresponding term
cancels that of $(f,g)$. Therefore, the sum of the terms arising in this way
vanishes.

Case IV: $f\delta_{n,i}$ and $g\delta_{n,i}$ are neither surjective

Let $0\leq j_{1}<\cdots<j_{l}<n$ be the elements of $[n]$ for which
$f(j)=f(j+1)$, and $0\leq i_{1}<\cdots<i_{k}<n$ the elements of $[n]$ for
which $g(i)=g(i+1)$. Since $f\delta_{n,i}$ is not surjective, we conclude that
if $i$ is omitted from the domain of $f$, then it will no longer remain
surjective. This implies that $f(i)$ is not assumed by any element of $[n]$
other than $i$. If $i=0$, then $f(0)\neq f(1)$, so
$0\notin\\{j_{1},...,j_{l}\\}$. Similarly, $0\notin\\{i_{1},...,i_{k}\\}$. But
then
$\\{i_{1},...,i_{k},j_{1},...,j_{l}\\}\subseteq\\{0,1,...,n-1\\}\setminus\\{0\\}$
and $\\{i_{1},...,i_{k}\\}\cap\\{j_{1},...,j_{l}\\}=\emptyset$, so $n=k+l\leq
n-1$ which is a contradiction. If $1\leq i\leq n-1$, then $f(i-1)\neq f(i)\neq
f(i+1)$, so $i-1,i\notin\\{j_{1},...,j_{l}\\}$. Similarly,
$i-1,i\notin\\{i_{1},...,i_{k}\\}$. But then
$\\{i_{1},...,i_{k},j_{1},...,j_{l}\\}\subseteq\\{0,1,...,n-1\\}\setminus\\{i-1,i\\}$,
so $n=k+l\leq n-2$ which is a contradiction. If $i=n$, then $f(n-1)\neq f(n)$,
so $n-1\notin\\{j_{1},...,j_{l}\\}$. Similarly,
$n-1\notin\\{i_{1},...,i_{k}\\}$. But then
$\\{i_{1},...,i_{k},j_{1},...,j_{l}\\}\subseteq\\{0,1,...,n-1\\}\setminus\\{n-1\\}$,
so $n=k+l\leq n-1$ which is a contradiction. Therefore, this case never
happens.

Now in light of the above analysis, we have:

$\textstyle\sum_{i=0}^{n}(-1)^{i}\textstyle\sum_{(f,g)\in\mathcal{S}_{n}^{k,l}}\operatorname{sgn}(\nu_{f,g})d_{n,i}^{\operatorname{DK}(X)}\left(\lambda_{f}^{X}(x)\right)\otimes
d_{n,i}^{\operatorname{DK}(Y)}\left(\lambda_{g}^{Y}(y)\right)=$
$\textstyle\sum_{\left(\overline{f},g\delta_{n,n}\right)\in\mathcal{S}_{n-1}^{k-1,l}}\operatorname{sgn}\left(\nu_{\overline{f},g\delta_{n,n}}\right)\lambda_{\overline{f},g\delta_{n,n}}^{X,Y}\left(\partial_{k}^{X}(x)\otimes
y\right)+(-1)^{k}\textstyle\sum_{\left(f\delta_{n,n},\overline{g}\right)\in\mathcal{S}_{n-1}^{k,l-1}}\operatorname{sgn}\left(\nu_{f\delta_{n,n},\overline{g}}\right)\lambda_{f\delta_{n,n},\overline{g}}^{X,Y}\left(x\otimes\partial_{l}^{Y}(y)\right)$

On the other hand, we have:

$\begin{split}\nabla_{n-1}^{X,Y}\left(\partial_{n}^{X\otimes_{R}Y}\left(\iota_{k,l}^{X,Y}(x\otimes
y)\right)\right)&=\nabla_{n-1}^{X,Y}\left(\iota_{k-1,l}^{X,Y}\left(\partial_{k}^{X}(x)\otimes
y\right)\right)+(-1)^{k}\nabla_{n-1}^{X,Y}\left(\iota_{k,l-1}^{X,Y}\left(x\otimes\partial_{l}^{Y}(y)\right)\right)\\\
&=\textstyle\sum_{(f,g)\in\mathcal{S}_{n-1}^{k-1,l}}\operatorname{sgn}(\nu_{f,g})\lambda_{f,g}^{X,Y}\left(\partial_{k}^{X}(x)\otimes
y\right)+(-1)^{k}\textstyle\sum_{(f,g)\in\mathcal{S}_{n-1}^{k,l-1}}\operatorname{sgn}(\nu_{f,g})\lambda_{f,g}^{X,Y}\left(x\otimes\partial_{l}^{Y}(y)\right)\end{split}$

After reindexing properly, we observe that
$\partial_{n}^{X\boxtimes_{R}Y}\nabla_{n}^{X,Y}\iota_{k,l}^{X,Y}=\nabla_{n-1}^{X,Y}\partial_{n}^{X\otimes_{R}Y}\iota_{k,l}^{X,Y}$
for every $k,l\geq 0$ with $k+l=n$, so we infer that
$\partial_{n}^{X\boxtimes_{R}Y}\nabla_{n}^{X,Y}=\nabla_{n-1}^{X,Y}\partial_{n}^{X\otimes_{R}Y}$.
As a consequence, $\nabla^{X,Y}:X\otimes_{R}Y\rightarrow X\boxtimes_{R}Y$ is a
morphism of $\mathbb{Z}$-complexes.

We next establish the naturality of $\nabla^{X,Y}$. To this end, let
$\eta:X\rightarrow X^{\prime}$ be a morphism of
$R^{\operatorname{op}}$-complexes. Then the following diagram is commutative:

${X\otimes_{R}Y}$${X^{\prime}\otimes_{R}Y}$${X\boxtimes_{R}Y}$${X^{\prime}\boxtimes_{R}Y}$$\scriptstyle{\eta\otimes_{R}Y}$$\scriptstyle{\nabla^{X,Y}}$$\scriptstyle{\nabla^{X^{\prime},Y}}$$\scriptstyle{\eta\boxtimes_{R}Y}$

Indeed, we have for every $x\in X_{k}$ and $y\in Y_{l}$ with $k+l=n$:

$\begin{split}(\eta\boxtimes_{R}Y)_{n}\left(\nabla_{n}^{X,Y}\left(\iota_{k,l}^{X,Y}(x\otimes
y)\right)\right)&=(\eta\boxtimes_{R}Y)_{n}\left(\textstyle\sum_{(f,g)\in\mathcal{S}_{n}^{k,l}}\operatorname{sgn}(\nu_{f,g})\lambda_{f,g}^{X,Y}(x\otimes
y)\right)\\\
&=\textstyle\sum_{(f,g)\in\mathcal{S}_{n}^{k,l}}\operatorname{sgn}(\nu_{f,g})(\eta\boxtimes_{R}Y)_{n}\left(\lambda_{f,g}^{X,Y}(x\otimes
y)\right)\\\
&=\textstyle\sum_{(f,g)\in\mathcal{S}_{n}^{k,l}}\operatorname{sgn}(\nu_{f,g})\lambda_{f,g}^{X^{\prime},Y}\left(\eta_{k}(x)\otimes
y\right)\\\
&=\nabla_{n}^{X^{\prime},Y}\left(\iota_{k,l}^{X^{\prime},Y}\left(\eta_{k}(x)\otimes
y\right)\right)\\\
&=\nabla_{n}^{X^{\prime},Y}\left((\eta\otimes_{R}Y)_{n}\left(\iota_{k,l}^{X,Y}(x\otimes
y)\right)\right)\end{split}$

As a result,
$(\eta\boxtimes_{R}Y)_{n}\nabla_{n}^{X,Y}\iota_{k,l}^{X,Y}=\nabla_{n}^{X^{\prime},Y}(\eta\otimes_{R}Y)_{n}\iota_{k,l}^{X,Y}$
for every $k,l\geq 0$ with $k+l=n$, so we infer that
$(\eta\boxtimes_{R}Y)_{n}\nabla_{n}^{X,Y}=\nabla_{n}^{X^{\prime},Y}(\eta\otimes_{R}Y)_{n}$
for every $n\geq 0$, whence
$(\eta\boxtimes_{R}Y)\nabla^{X,Y}=\nabla^{X^{\prime},Y}(\eta\otimes_{R}Y)$. If
$\theta:Y\rightarrow Y^{\prime}$ is a morphism of $R$-complexes, then a
similar argument shows that the following diagram is commutative:

${X\otimes_{R}Y}$${X\otimes_{R}Y^{\prime}}$${X\boxtimes_{R}Y}$${X\boxtimes_{R}Y^{\prime}}$$\scriptstyle{X\otimes_{R}\theta}$$\scriptstyle{\nabla^{X,Y}}$$\scriptstyle{\nabla^{X,Y^{\prime}}}$$\scriptstyle{X\boxtimes_{R}\theta}$

As a consequence, $\nabla^{X,Y}$ is natural in $X$ and $Y$.

We finally show that $\nabla^{X,Y}$ is a quasi-isomorphism. The author has
learned the ideas of the following delicate argument from [Lu, proof of
Theorem 2.5.7.14]. We first consider the special case in which $X$ and $Y$ are
left-bounded. Thus we assume that $X_{i}=0$ for every $i>n$, and $Y_{i}=0$ for
every $i>m$. We argue by induction on $n$ to show that $\nabla^{X,Y}$ is a
quasi-isomorphism. If $n=0$, then $X$ has only one non-zero term in degree
$0$, so we can identify $X$ with the left $R$-module $X_{0}$. Consequently,
$X\otimes_{R}Y=X\boxtimes_{R}Y$ by Proposition 5.5, and
$\nabla_{i}^{X,Y}:(X\otimes_{R}Y)_{i}\rightarrow(X\boxtimes_{R}Y)_{i}$ is the
identity map for every $i\geq 0$. Therefore, $\nabla^{X,Y}$ is a quasi-
isomorphism. Now let $n\geq 1$, and assume the result for $n-1$. Consider the
degreewise split short exact sequence

$0\rightarrow X_{n-1\sqsubset}\rightarrow
X\rightarrow\Sigma^{n}X_{n}\rightarrow 0$

of $R$-complexes in which $\sqsubset$ designates the hard truncation from
left; see [Ch, A.1.14]. Then we obtain the following commutative diagram with
exact rows:

${0}$${X_{n-1\sqsubset}\otimes_{R}Y}$${X\otimes_{R}Y}$${\Sigma^{n}X_{n}\otimes_{R}Y}$${0}$${0}$${X_{n-1\sqsubset}\boxtimes_{R}Y}$${X\boxtimes_{R}Y}$${\Sigma^{n}X_{n}\boxtimes_{R}Y}$${0}$$\scriptstyle{\nabla^{X_{n-1\sqsubset},Y}}$$\scriptstyle{\nabla^{X,Y}}$$\scriptstyle{\nabla^{\Sigma^{n}X_{n},Y}}$

By the induction hypothesis, $\nabla^{X_{n-1\sqsubset},Y}$ is a quasi-
isomorphism. Therefore, the above diagram implies that $\nabla^{X,Y}$ is a
quasi-isomorphism if and only if $\nabla^{\Sigma^{n}X_{n},Y}$ is a quasi-
isomorphism. We note that
$X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right)$ is a left-
bounded $R$-complex with

$\left(X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\right)_{n}=X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right)_{n}=X_{n}\otimes_{R}R\cong
X_{n},$

so by the above argument,
$\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),Y}$
is a quasi-isomorphism if and only if $\nabla^{\Sigma^{n}X_{n},Y}$ is a quasi-
isomorphism. It follows that $\nabla^{X,Y}$ is a quasi-isomorphism if and only
if
$\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),Y}$
is a quasi-isomorphism. Now we argue by induction on $m$ to show that
$\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),Y}$
is a quasi-isomorphism. If $m=0$, then $Y$ has only one non-zero term in
degree $0$, so we can identify $Y$ with the left $R$-module $Y_{0}$, and thus
a similar argument as above shows that
$\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),Y}$
is a quasi-isomorphism. Let $m\geq 1$, and assume the result for $m-1$.
Consider the degreewise split short exact sequence

$0\rightarrow Y_{m-1\sqsubset}\rightarrow
Y\rightarrow\Sigma^{m}Y_{m}\rightarrow 0$

of $R$-complexes. Then we achieve the following commutative diagram with exact
rows:

${0}$${\left(X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\right)\otimes_{R}Y_{m-1\sqsubset}}$${\left(X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\right)\otimes_{R}Y}$${\left(X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\right)\otimes_{R}\Sigma^{m}Y_{m}}$${0}$${0}$${\left(X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\right)\boxtimes_{R}Y_{m-1\sqsubset}}$${\left(X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\right)\boxtimes_{R}Y}$${\left(X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\right)\boxtimes_{R}\Sigma^{m}Y_{m}}$${0}$$\scriptstyle{\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),Y_{m-1\sqsubset}}}$$\scriptstyle{\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),Y}}$$\scriptstyle{\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),\Sigma^{m}Y_{m}}}$

By the induction hypothesis,
$\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),Y_{m-1\sqsubset}}$
is a quasi-isomorphism. Therefore, the above diagram implies that
$\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),Y}$
is a quasi-isomorphism if and only if
$\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),\Sigma^{m}Y_{m}}$
is a quasi-isomorphism. We note that
$\operatorname{Nor}\left(R^{(\Delta^{m})}\right)\otimes_{R}Y_{m}$ is a left-
bounded $R$-complex with

$\left(\operatorname{Nor}\left(R^{(\Delta^{m})}\right)\otimes_{R}Y_{m}\right)_{m}=\operatorname{Nor}\left(R^{(\Delta^{m})}\right)_{m}\otimes_{R}Y_{m}=R\otimes_{R}Y_{m}\cong
Y_{m},$

so by the above argument,
$\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),\operatorname{Nor}\left(R^{(\Delta^{m})}\right)\otimes_{R}Y_{m}}$
is a quasi-isomorphism if and only if
$\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),\Sigma^{m}Y_{m}}$
is a quasi-isomorphism. It follows that
$\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),Y}$
is a quasi-isomorphism if and only if
$\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),\operatorname{Nor}\left(R^{(\Delta^{m})}\right)\otimes_{R}Y_{m}}$
is a quasi-isomorphism. Altogether, we infer that $\nabla^{X,Y}$ is a quasi-
isomorphism if and only if
$\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),\operatorname{Nor}\left(R^{(\Delta^{m})}\right)\otimes_{R}Y_{m}}$
is a quasi-isomorphism.

Now let $n,m\geq 0$. Viewing $[n]$ as a small category, we have
$\operatorname{nrv}([n])=\Delta^{n}$. Since $[n]$ has a least element $0$, we
can define a morphism $\zeta^{\Delta^{n}}:\Delta^{0}\rightarrow\Delta^{n}$ of
simplicial sets by setting $\zeta_{k}^{\Delta^{n}}(0,...,0)=(0,...,0)$ for
every $k\geq 0$. Then by Lemma 4.8,
$\operatorname{Nor}\left(R^{\left(\zeta^{\Delta^{n}}\right)}\right):\operatorname{Nor}\left(R^{(\Delta^{0})}\right)\rightarrow\operatorname{Nor}\left(R^{(\Delta^{n})}\right)$
is a homotopy equivalence. Similarly, $\operatorname{nrv}([m])=\Delta^{m}$,
$\zeta^{\Delta^{m}}:\Delta^{0}\rightarrow\Delta^{m}$ is given by
$\zeta_{k}^{\Delta^{m}}(0,...,0)=(0,...,0)$ for every $k\geq 0$, and
$\operatorname{Nor}\left(R^{\left(\zeta^{\Delta^{m}}\right)}\right):\operatorname{Nor}\left(R^{(\Delta^{0})}\right)\rightarrow\operatorname{Nor}\left(R^{(\Delta^{m})}\right)$
is a homotopy equivalence. Also, $[n]\times[m]$ is a poset with componentwise
order, so $\operatorname{nrv}([n]\times[m])=\Delta^{n}\times\Delta^{m}$,
$\zeta^{\Delta^{n}\times\Delta^{m}}:\Delta^{0}\rightarrow\Delta^{n}\times\Delta^{m}$
is given by
$\zeta_{k}^{\Delta^{n}\times\Delta^{m}}(0,...,0)=\left((0,...,0),(0,...,0)\right)$
for every $k\geq 0$, and
$\operatorname{Nor}\left(R^{\left(\zeta^{\Delta^{n}\times\Delta^{m}}\right)}\right):\operatorname{Nor}\left(R^{(\Delta^{0})}\right)\rightarrow\operatorname{Nor}\left(R^{(\Delta^{n}\times\Delta^{m})}\right)$
is a homotopy equivalence. Let

$\eta:\operatorname{Nor}\left(R^{(\Delta^{0})}\right)\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{0})}\right)\rightarrow\operatorname{Nor}\left(R^{(\Delta^{0})}\right)$

be the isomorphism given by $\eta_{0}(a\otimes b)=ab$ for every $a,b\in R$,
and $\eta_{i}=0$ for every $i\neq 0$. Furthermore, let $\chi$ be the
composition of the following morphisms of $\mathbb{Z}$-complexes:

$\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{m})}\right)\xrightarrow{\nabla^{\operatorname{Nor}\left(R^{(\Delta^{n})}\right),\operatorname{Nor}\left(R^{(\Delta^{m})}\right)}}\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\boxtimes_{R}\operatorname{Nor}\left(R^{(\Delta^{m})}\right)$
$\xrightarrow{\cong}\operatorname{Nor}\left(R^{(\Delta^{n})}\otimes_{R}R^{(\Delta^{m})}\right)\xrightarrow{\cong}\operatorname{Nor}\left(R^{(\Delta^{n}\times\Delta^{m})}\right)$

Then the following diagram is commutative:

${\operatorname{Nor}\left(R^{(\Delta^{0})}\right)\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{0})}\right)}$${\operatorname{Nor}\left(R^{(\Delta^{0})}\right)}$${\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{m})}\right)}$${\operatorname{Nor}\left(R^{(\Delta^{n}\times\Delta^{m})}\right)}$$\scriptstyle{\eta}$$\scriptstyle{\operatorname{Nor}\left(R^{\left(\zeta^{\Delta^{n}}\right)}\right)\otimes_{R}\operatorname{Nor}\left(R^{\left(\zeta^{\Delta^{m}}\right)}\right)}$$\scriptstyle{\operatorname{Nor}\left(R^{\left(\zeta^{\Delta^{n}\times\Delta^{m}}\right)}\right)}$$\scriptstyle{\chi}$

To see this, we note that $\operatorname{Nor}\left(R^{(\Delta^{0})}\right)$
has only one non-zero term in degree $0$, so we must have
$\operatorname{Nor}\left(R^{\left(\zeta^{\Delta^{n}}\right)}\right)_{i}=\operatorname{Nor}\left(R^{\left(\zeta^{\Delta^{m}}\right)}\right)_{i}=\operatorname{Nor}\left(R^{\left(\zeta^{\Delta^{n}\times\Delta^{m}}\right)}\right)_{i}=0$
for every $i\neq 0$. Therefore, it suffices to check the commutativity of the
above diagram for $i=0$. If $i=0$, then we have the following commutative
diagram:

${R\otimes_{R}R}$${R}$${R^{n}\otimes_{R}R^{m}}$${R^{nm}}$$\scriptstyle{\eta_{0}}$$\scriptstyle{R^{\left(\zeta_{0}^{\Delta^{n}}\right)}\otimes_{R}R^{\left(\zeta_{0}^{\Delta^{m}}\right)}}$$\scriptstyle{R^{\left(\zeta_{0}^{\Delta^{n}\times\Delta^{m}}\right)}}$$\scriptstyle{\chi_{0}}$

Indeed, we have for every $a,b\in R$:

$\begin{split}\chi_{0}\left(\left(R^{\left(\zeta_{0}^{\Delta^{n}}\right)}\otimes_{R}R^{\left(\zeta_{0}^{\Delta^{m}}\right)}\right)(a\otimes
b)\right)&=\chi_{0}\left(R^{\left(\zeta_{0}^{\Delta^{n}}\right)}(a)\otimes
R^{\left(\zeta_{0}^{\Delta^{m}}\right)}(b)\right)=\chi_{0}\left((a,0,...,0)\otimes(b,0,...,0)\right)\\\
&=(ab,0,...,0)=R^{\left(\zeta_{0}^{\Delta^{n}\times\Delta^{m}}\right)}(ab)=R^{\left(\zeta_{0}^{\Delta^{n}\times\Delta^{m}}\right)}\left(\eta_{0}(a\otimes
b)\right)\end{split}$

Now $\eta$ is an isomorphism, and
$\operatorname{Nor}\left(R^{\left(\zeta^{\Delta^{n}}\right)}\right)\otimes_{R}\operatorname{Nor}\left(R^{\left(\zeta^{\Delta^{m}}\right)}\right)$
and
$\operatorname{Nor}\left(R^{\left(\zeta^{\Delta^{n}\times\Delta^{m}}\right)}\right)$
are homotopy equivalences, so the above diagram implies that $\chi$ is a
homotopy equivalence. By the definition of $\chi$, we conclude that
$\nabla^{\operatorname{Nor}\left(R^{(\Delta^{n})}\right),\operatorname{Nor}\left(R^{(\Delta^{m})}\right)}$
is a homotopy equivalence. Consider the following commutative diagram:

${X_{n}\otimes_{R}\left(\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{m})}\right)\right)\otimes_{R}Y_{m}}$${\left(X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\right)\otimes_{R}\left(\operatorname{Nor}\left(R^{(\Delta^{m})}\right)\otimes_{R}Y_{m}\right)}$${X_{n}\otimes_{R}\left(\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\boxtimes_{R}\operatorname{Nor}\left(R^{(\Delta^{m})}\right)\right)\otimes_{R}Y_{m}}$${\left(X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right)\right)\boxtimes_{R}\left(\operatorname{Nor}\left(R^{(\Delta^{m})}\right)\otimes_{R}Y_{m}\right)}$$\scriptstyle{X_{n}\otimes_{R}\nabla^{\operatorname{Nor}\left(R^{(\Delta^{n})}\right),\operatorname{Nor}\left(R^{(\Delta^{m})}\right)}\otimes_{R}Y_{m}}$$\scriptstyle{\cong}$$\scriptstyle{\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),\operatorname{Nor}\left(R^{(\Delta^{m})}\right)\otimes_{R}Y_{m}}}$$\scriptstyle{\cong}$

Since tensor product functor preserves homotopy equivalences,
$X_{n}\otimes_{R}\nabla^{\operatorname{Nor}\left(R^{(\Delta^{n})}\right),\operatorname{Nor}\left(R^{(\Delta^{m})}\right)}\otimes_{R}Y_{m}$
is a homotopy equivalence, so we deduce from the above diagram that
$\nabla^{X_{n}\otimes_{R}\operatorname{Nor}\left(R^{(\Delta^{n})}\right),\operatorname{Nor}\left(R^{(\Delta^{m})}\right)\otimes_{R}Y_{m}}$
is a homotopy equivalence, so in particular, it is a quasi-isomorphism.
Therefore, the above discussion shows that $\nabla^{X,Y}$ is a quasi-
isomorphism.

Next suppose that $X$ and $Y$ are arbitrary. Then we have the following
commutative diagram:

${\underset{(n,m)\in\mathbb{Z}\times\mathbb{Z}}{\varinjlim}\left(X_{n\sqsubset}\otimes_{R}Y_{m\sqsubset}\right)}$${\underset{(n,m)\in\mathbb{Z}\times\mathbb{Z}}{\varinjlim}\left(X_{n\sqsubset}\boxtimes_{R}Y_{m\sqsubset}\right)}$${\left(\underset{n\in\mathbb{Z}}{\varinjlim}X_{n\sqsubset}\right)\otimes_{R}\left(\underset{m\in\mathbb{Z}}{\varinjlim}Y_{m\sqsubset}\right)}$${\left(\underset{n\in\mathbb{Z}}{\varinjlim}X_{n\sqsubset}\right)\boxtimes_{R}\left(\underset{m\in\mathbb{Z}}{\varinjlim}Y_{m\sqsubset}\right)}$${X\otimes_{R}Y}$${X\boxtimes_{R}Y}$$\scriptstyle{\underset{(n,m)\in\mathbb{Z}\times\mathbb{Z}}{\varinjlim}\nabla^{X_{n\sqsubset}\otimes_{R}Y_{m\sqsubset}}}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\nabla^{\underset{n\in\mathbb{Z}}{\varinjlim}X_{n\sqsubset},\underset{m\in\mathbb{Z}}{\varinjlim}Y_{m\sqsubset}}}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\nabla^{X,Y}}$

By the special case, $\nabla^{X_{n\sqsubset}\otimes_{R}Y_{m\sqsubset}}$ is a
quasi-isomorphism for every $n,m\geq 0$, so
$\underset{(n,m)\in\mathbb{Z}\times\mathbb{Z}}{\varinjlim}\nabla^{X_{n\sqsubset}\otimes_{R}Y_{m\sqsubset}}$
is a quasi-isomorphism. Then the above diagram shows that $\nabla^{X,Y}$ is a
quasi-isomorphism. ∎

######  Corollary 5.10.

Let $R$ be a ring, $X$ a connective $R^{\operatorname{op}}$-complex, and $Y$ a
connective $R$-complex. Then the following natural $\mathbb{Z}$-isomorphism
holds for every $i\geq 0$:

$H_{i}(X\boxtimes_{R}Y)\cong H_{i}(X\otimes_{R}Y)$

###### Proof.

Follows from Theorem 5.9. ∎

######  Corollary 5.11.

Let $R$ be a ring. Then the following assertions hold:

  1. (i)

If $X$ is a connective $R^{\operatorname{op}}$-complex of flat modules, then
$X\boxtimes_{R}-:\mathcal{C}_{\geq 0}(R)\rightarrow\mathcal{C}_{\geq
0}(\mathbb{Z})$ preserves quasi-isomorphisms.

  2. (ii)

If $Y$ is a connective $R$-complex of flat modules, then
$-\boxtimes_{R}Y:\mathcal{C}_{\geq
0}(R^{\operatorname{op}})\rightarrow\mathcal{C}_{\geq 0}(\mathbb{Z})$
preserves quasi-isomorphisms.

###### Proof.

(i): Since $X$ is a right-bounded $R^{\operatorname{op}}$-complex of flat
modules, [CT, Corollary A.2] or [CFH, Example 5.4.8] implies that
$X\otimes_{R}-:\mathcal{C}_{\geq 0}(R)\rightarrow\mathcal{C}_{\geq
0}(\mathbb{Z})$ preserves acyclicity, so by [AF, Examples 1.1.F, (3)], it also
preserves quasi-isomorphisms. On the other hand, by Theorem 5.9, there is a
natural quasi-isomorphism $\nabla^{X,Y}:X\otimes_{R}Y\rightarrow
X\boxtimes_{R}Y$ of $\mathbb{Z}$-complexes. Let $g:Y\rightarrow Y^{\prime}$ be
a quasi-isomorphism of $R$-complexes. Consider the following commutative
diagram:

${X\otimes_{R}Y}$${X\otimes_{R}Y^{\prime}}$${X\boxtimes_{R}Y}$${X\boxtimes_{R}Y^{\prime}}$$\scriptstyle{X\otimes_{R}g}$$\scriptstyle{\nabla^{X,Y}}$$\scriptstyle{\nabla^{X,Y^{\prime}}}$$\scriptstyle{X\boxtimes_{R}g}$

As $X\otimes_{R}g$, $\nabla^{X,Y}$, and $\nabla^{X,Y^{\prime}}$ are quasi-
isomorphisms, the above diagram implies that $X\boxtimes_{R}g$ is a quasi-
isomorphism.

(ii): Similar to (i). ∎

##  6\. Model Structure on Simplicial Commutative Algebras

In this section, we deploy Theorem 2.9 to transfer the model structure on
connective chain complexes to simplicial commutative algebras. In order to
overcome the acyclicity condition, we need to construct certain cylinder and
path objects for simplicial commutative algebras. To this end, we need Lemmas
6.1 and 6.3 below. Variants of these lemmas are incorporated as Axiom SM7 into
Quillen’s theory of model categories to define a simplicially enriched model
category; see [Qu2, Definition 2 on page 2.2] or [GJ, Proposition 11.5 and
Axiom 3.1]. He proves that the category of simplicial sets satisfy these
lemmas. However, the proofs that are available in the literature are based on
the model structure of simplicial sets which we have totally avoided in this
article. Instead, we leverage the shuffle product of connective chain
complexes developed in the previous section to prove these lemmas for
simplicial modules.

######  Lemma 6.1.

Let $R$ be a commutative ring, $\rho:M\rightarrow N$ a fibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, and $\iota:U\rightarrow V$ a morphism
of simplicial sets. Consider the commutative diagram

${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,M)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,N)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,M)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,N)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,\rho)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\iota,M)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\iota,N)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,\rho)}$

and form the following pullback diagram:

${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,M)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,M)\mathop{\mathchoice{\rotatebox[origin={c}]{180.0}{$\displaystyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\textstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptscriptstyle\bigsqcup$}}}_{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,N)}\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,N)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,N)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,M)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,N)}$$\scriptstyle{\eta}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\iota,M)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,\rho)}$$\scriptstyle{r}$$\scriptstyle{s}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\iota,N)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,\rho)}$

If $R^{(\iota)}:R^{(U)}\rightarrow R^{(V)}$ is a cofibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, then $\eta$ is a fibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$ which is trivial if either $\rho$ or
$R^{(\iota)}$ is.

###### Proof.

Let $\varrho:K\rightarrow L$ be a morphism of simplicial $R$-modules. Consider
a commutative diagram as follows:

${K}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,M)}$${L}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,M)\mathop{\mathchoice{\rotatebox[origin={c}]{180.0}{$\displaystyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\textstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptscriptstyle\bigsqcup$}}}_{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,N)}\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,N)}$$\scriptstyle{p}$$\scriptstyle{\varrho}$$\scriptstyle{\eta}$$\scriptstyle{q}$

Considering the commutative diagram

${K^{(U)}}$${L^{(U)}}$${K^{(V)}}$${L^{(V)}}$$\scriptstyle{\varrho^{(U)}}$$\scriptstyle{K^{(\iota)}}$$\scriptstyle{L^{(\iota)}}$$\scriptstyle{\varrho^{(V)}}$

there is a unique morphism $\zeta:K^{(V)}\bigsqcup_{K^{(U)}}L^{(U)}\rightarrow
L^{(V)}$ that makes the following pushout diagram commutative:

${K^{(U)}}$${L^{(U)}}$${K^{(V)}}$${K^{(V)}\bigsqcup_{K^{(U)}}L^{(U)}}$${L^{(V)}}$$\scriptstyle{\varrho^{(U)}}$$\scriptstyle{K^{(\iota)}}$$\scriptstyle{L^{(\iota)}}$$\scriptstyle{\varrho^{(V)}}$$\scriptstyle{\zeta}$

On the other hand, considering the natural bijections

$\phi_{KVM}:\operatorname{Mor}_{\mathpzc{s}\mathcal{M}\mathpzc{od}(R)}\left(K,\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,M)\right)\rightarrow\operatorname{Mor}_{\mathpzc{s}\mathcal{M}\mathpzc{od}(R)}\left(K^{(V)},M\right)$

and

$\phi_{LUM}:\operatorname{Mor}_{\mathpzc{s}\mathcal{M}\mathpzc{od}(R)}\left(L,\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,M)\right)\rightarrow\operatorname{Mor}_{\mathpzc{s}\mathcal{M}\mathpzc{od}(R)}\left(L^{(U)},M\right)$

from Remark 4.4, we get the following commutative diagram:

${K^{(U)}}$${L^{(U)}}$${K^{(V)}}$${M}$$\scriptstyle{\varrho^{(U)}}$$\scriptstyle{K^{(\iota)}}$$\scriptstyle{\phi_{LUM}(sq)}$$\scriptstyle{\phi_{KVM}(p)}$

Hence there is a unique morphism
$\xi:K^{(V)}\bigsqcup_{K^{(U)}}L^{(U)}\rightarrow M$ that makes the following
pushout diagram commutative:

${K^{(U)}}$${L^{(U)}}$${K^{(V)}}$${K^{(V)}\bigsqcup_{K^{(U)}}L^{(U)}}$${M}$$\scriptstyle{\varrho^{(U)}}$$\scriptstyle{K^{(\iota)}}$$\scriptstyle{\phi_{LUM}(sq)}$$\scriptstyle{\phi_{KVM}(p)}$$\scriptstyle{\xi}$

Then considering the natural bijection

$\phi_{LVN}:\operatorname{Mor}_{\mathpzc{s}\mathcal{M}\mathpzc{od}(R)}\left(L,\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,N)\right)\rightarrow\operatorname{Mor}_{\mathpzc{s}\mathcal{M}\mathpzc{od}(R)}\left(L^{(V)},N\right)$

we get the following commutative diagram:

${K^{(V)}\bigsqcup_{K^{(U)}}L^{(U)}}$${M}$${L^{(V)}}$${N}$$\scriptstyle{\xi}$$\scriptstyle{\zeta}$$\scriptstyle{\rho}$$\scriptstyle{\phi_{LVN}(rq)}$

If $\tau:L^{(V)}\rightarrow M$ is a morphism of simplicial $R$-modules that
makes the diagram

${K^{(V)}\bigsqcup_{K^{(U)}}L^{(U)}}$${M}$${L^{(V)}}$${N}$$\scriptstyle{\xi}$$\scriptstyle{\zeta}$$\scriptstyle{\rho}$$\scriptstyle{\phi_{LVN}(rq)}$$\scriptstyle{\tau}$

commutative, then considering the natural bijection

$\phi_{LVM}:\operatorname{Mor}_{\mathpzc{s}\mathcal{M}\mathpzc{od}(R)}\left(L,\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,M)\right)\rightarrow\operatorname{Mor}_{\mathpzc{s}\mathcal{M}\mathpzc{od}(R)}\left(L^{(V)},M\right)$

we see that the following diagram is commutative:

${K}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,M)}$${L}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,M)\mathop{\mathchoice{\rotatebox[origin={c}]{180.0}{$\displaystyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\textstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptscriptstyle\bigsqcup$}}}_{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,N)}\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,N)}$$\scriptstyle{p}$$\scriptstyle{\varrho}$$\scriptstyle{\eta}$$\scriptstyle{q}$$\scriptstyle{\phi_{LVM}^{-1}(\tau)}$

Therefore, in order to show that $\eta$ has the right lifting property against
$\varrho$, it suffices to show that $\rho$ has the right lifting property
against $\zeta$. Indeed, if $\rho$ is a fibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, then it suffices to show that $\zeta$
is a trivial cofibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, and if
$\rho$ is a trivial fibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, then
it suffices to show that $\zeta$ is a cofibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$.

Denote the morphism
$\operatorname{Nor}\left(R^{(\iota)}\right):\operatorname{Nor}\left(R^{(U)}\right)\rightarrow\operatorname{Nor}\left(R^{(V)}\right)$
in $\mathcal{C}_{\geq 0}(R)$ by $\mu:Z\rightarrow W$ for convenience. If $X$
is an $R$-complex and $T$ is a simplicial set, then in view of Remark 4.4,
Corollary 5.7, and Theorem 4.14, we have the following natural isomorphisms:

$\operatorname{Nor}\left(\operatorname{DK}(X)^{(T)}\right)\cong\operatorname{Nor}\left(R^{(T)}\otimes_{R}\operatorname{DK}(X)\right)\cong\operatorname{Nor}\left(R^{(T)}\right)\boxtimes_{R}\operatorname{Nor}\left(\operatorname{DK}(X)\right)\cong\operatorname{Nor}\left(R^{(T)}\right)\boxtimes_{R}X$

As a consequence, assuming that $\varrho:K\rightarrow L$ is of the form
$\operatorname{DK}(h):\operatorname{DK}(X)\rightarrow\operatorname{DK}(Y)$ for
some morphism $h:X\rightarrow Y$ of connective $R$-complexes, then we get a
commutative diagram as follows:

${\operatorname{Nor}\left(K^{(V)}\bigsqcup_{K^{(U)}}L^{(U)}\right)}$${\operatorname{Nor}\left(L^{(V)}\right)}$${\operatorname{Nor}\left(K^{(V)}\right)\bigsqcup_{\operatorname{Nor}\left(K^{(U)}\right)}\operatorname{Nor}\left(L^{(U)}\right)}$${\operatorname{Nor}\left(L^{(V)}\right)}$${\operatorname{Nor}\left(\operatorname{DK}(X)^{(V)}\right)\bigsqcup_{\operatorname{Nor}\left(\operatorname{DK}(X)^{(U)}\right)}\operatorname{Nor}\left(\operatorname{DK}(Y)^{(U)}\right)}$${\operatorname{Nor}\left(\operatorname{DK}(Y)^{(V)}\right)}$${\left(\operatorname{Nor}\left(R^{(V)}\right)\boxtimes_{R}X\right)\bigsqcup_{\left(\operatorname{Nor}\left(R^{(U)}\right)\boxtimes_{R}X\right)}\left(\operatorname{Nor}\left(R^{(U)}\right)\boxtimes_{R}Y\right)}$${\operatorname{Nor}\left(R^{(V)}\right)\boxtimes_{R}Y}$${\left(W\boxtimes_{R}X\right)\bigsqcup_{\left(Z\boxtimes_{R}X\right)}\left(Z\boxtimes_{R}Y\right)}$${W\boxtimes_{R}Y}$$\scriptstyle{\operatorname{Nor}(\zeta)}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\zeta^{\prime}}$

Thus in order to show that $\zeta$ is a (trivial) cofibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, i.e. $\operatorname{Nor}(\zeta)$ is a
(trivial) cofibration in $\mathcal{C}_{\geq 0}(R)$, it suffices to show that
$\zeta^{\prime}$ is a (trivial) cofibration in $\mathcal{C}_{\geq 0}(R)$.

By Theorem 3.1, $\mathcal{C}_{\geq 0}(R)$ is a cofibrantly generated model
category with the corresponding sets

$\mathcal{X}=\\{\iota^{n}:S(n-1)\rightarrow D(n)\;|\;n\geq
1\\}\cup\\{\lambda:0\rightarrow S(0)\\}$

where $\iota^{n}$ is the inclusion morphism for every $n\geq 1$, and

$\mathcal{Y}=\\{\kappa^{n}:0\rightarrow D(n)\;|\;n\geq 1\\}.$

In light of the proof of Corollary 2.16 and Theorem 4.15, we see that
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$ is a cofibrantly generated model
category with the corresponding sets $\operatorname{DK}(\mathcal{X})$ and
$\operatorname{DK}(\mathcal{Y})$. In particular,
$\operatorname{RLP}\left(\operatorname{DK}(\mathcal{X})\right)$ is the class
of trivial fibrations and
$\operatorname{RLP}\left(\operatorname{DK}(\mathcal{Y})\right)$ is the class
of fibrations in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$. As a result, in
order to show that $\eta$ is a fibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, it suffices to show that $\eta$ has
the right lifting property against morphisms in
$\operatorname{DK}(\mathcal{Y})$, i.e. to choose $\varrho:K\rightarrow L$ from
$\operatorname{DK}(\mathcal{Y})$ and solve the corresponding lifting problem.
Similarly, to show that $\eta$ is a trivial fibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, it suffices to show that $\eta$ has
the right lifting property against morphisms in
$\operatorname{DK}(\mathcal{X})$, i.e. to choose $\varrho:K\rightarrow L$ from
$\operatorname{DK}(\mathcal{X})$ and solve the corresponding lifting problem.

Let $X$ be a connective $R$-complex. We analyze the shuffle products
$X\boxtimes_{R}D(n)$ and $X\boxtimes_{R}S(n-1)$ for every $n\geq 1$ more
closely. For any $m\geq 0$, set:

$\mathcal{S}_{m}=\left\\{(f,g)\in\operatorname{Mor}_{\Delta}\left([m],[k]\right)\times\operatorname{Mor}_{\Delta}\left([m],[l]\right)\;|\;\begin{tabular}[]{ccc}$0\leq
k,l\leq m$, $f$ and $g$ are surjective, and
$\phi_{f,g}:[m]\rightarrow[k]\times[l]$\\\ given by
$\phi_{f,g}(i)=\left(f(i),g(i)\right)$ for every $i\in[m]$ is
injective\end{tabular}\right\\}$

Let $n\geq 1$ and $m\geq 0$. If $l\neq n,n-1$, then $D(n)_{l}=0$. Moreover,
$D(n)_{n}=D(n)_{n-1}=R$. Let $\mathcal{S}_{m}^{\prime}$ be the subset of
$\mathcal{S}_{m}$ in which we assume $l=n$ or $n-1$. Then we have the
following natural isomorphisms:

$\left(X\boxtimes_{R}D(n)\right)_{m}=\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}}\left(X_{k}\otimes_{R}D(n)_{l}\right)\cong\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\left(X_{k}\otimes_{R}R\right)\cong\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}X_{k}$

If $l\neq n-1$, then $S(n-1)_{l}=0$. Moreover, $S(n-1)_{n-1}=R$. Let
$\mathcal{S}_{m}^{\prime\prime}$ be the subset of $\mathcal{S}_{m}$ in which
we assume $l=n-1$. Then we have the following natural isomorphisms:

$\left(X\boxtimes_{R}S(n)\right)_{m}=\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}}\left(X_{k}\otimes_{R}S(n)_{l}\right)\cong\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}\left(X_{k}\otimes_{R}R\right)\cong\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}X_{k}$

It is clear that
$\mathcal{S}_{m}^{\prime\prime}\subseteq\mathcal{S}_{m}^{\prime}$.
Accordingly, we let
$\nu_{m}:\mathcal{S}_{m}^{\prime\prime}\rightarrow\mathcal{S}_{m}^{\prime}$ be
the inclusion map.

Now suppose that $R^{(\iota)}:R^{(U)}\rightarrow R^{(V)}$ is a cofibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, i.e. $\mu:Z\rightarrow W$ is a
cofibration in $\mathcal{C}_{\geq 0}(R)$. We will show that $\eta$ is a
fibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$. To this end, let $n\geq
1$, and let $\varrho:K\rightarrow L$ be the morphism
$\operatorname{DK}(\kappa^{n}):\operatorname{DK}(0)\rightarrow\operatorname{DK}\left(D(n)\right)$.
As we observed before, it suffices to show that

$\zeta^{\prime}:\left(W\boxtimes_{R}0\right)\textstyle\bigsqcup_{\left(Z\boxtimes_{R}0\right)}\left(Z\boxtimes_{R}D(n)\right)\rightarrow
W\boxtimes_{R}D(n)$

is a trivial cofibration. We have the following commutative diagram:

${\left(W\boxtimes_{R}0\right)\bigsqcup_{\left(Z\boxtimes_{R}0\right)}\left(Z\boxtimes_{R}D(n)\right)}$${W\boxtimes_{R}D(n)}$${0\bigsqcup_{0}\left(Z\boxtimes_{R}D(n)\right)}$${W\boxtimes_{R}D(n)}$${Z\boxtimes_{R}D(n)}$${W\boxtimes_{R}D(n)}$$\scriptstyle{\zeta^{\prime}}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\mu\boxtimes_{R}D(n)}$

As $\mu$ is a cofibration, we have the short exact sequence $0\rightarrow
Z\xrightarrow{\mu}W\rightarrow\operatorname{Coker}(\mu)\rightarrow 0$ of
$R$-complexes in which $\operatorname{Coker}(\mu)$ is an $R$-complex of
projective modules. In particular, for any $k\geq 0$, we have the short exact
sequence $0\rightarrow
Z_{k}\xrightarrow{\mu_{k}}W_{k}\rightarrow\operatorname{Coker}(\mu_{k})\rightarrow
0$ of $R$-modules in which $\operatorname{Coker}(\mu_{k})$ is projective. Now
consider the following sequence:

$0\rightarrow
Z\boxtimes_{R}D(n)\xrightarrow{\mu\boxtimes_{R}D(n)}W\boxtimes_{R}D(n)\rightarrow\operatorname{Coker}(\mu)\boxtimes_{R}D(n)\rightarrow
0$

For each $m\geq 0$, we have the following commutative diagram:

${0}$${\left(Z\boxtimes_{R}D(n)\right)_{m}}$${\left(W\boxtimes_{R}D(n)\right)_{m}}$${\left(\operatorname{Coker}(\mu)\boxtimes_{R}D(n)\right)_{m}}$${0}$${0}$${\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}Z_{k}}$${\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}W_{k}}$${\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\operatorname{Coker}(\mu_{k})}$${0}$$\scriptstyle{\left(\mu\boxtimes_{R}D(n)\right)_{m}}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}}$

Now the second row is exact and
$\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\operatorname{Coker}(\mu_{k})$ is
projective, so it follows that the first row is exact and
$\left(\operatorname{Coker}(\mu)\boxtimes_{R}D(n)\right)_{m}$ is projective.
As a result, the previous sequence is exact and
$\operatorname{Coker}(\mu)\boxtimes_{R}D(n)$ is an $R$-complex of projective
modules. This means that $\mu\boxtimes_{R}D(n)$ is a cofibration, so
$\zeta^{\prime}$ is a cofibration. On the other hand, we have the following
commutative diagram for every $i\geq 0$:

${H_{i}\left(Z\boxtimes_{R}D(n)\right)}$${H_{i}\left(W\boxtimes_{R}D(n)\right)}$${H_{i}\left(Z\otimes_{R}D(n)\right)}$${H_{i}\left(W\otimes_{R}D(n)\right)}$$\scriptstyle{H_{i}\left(\mu\boxtimes_{R}D(n)\right)}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{H_{i}\left(\mu\otimes_{R}D(n)\right)}$

We note that $D(n)$ is an exact $R$-complex whose terms and boundaries are all
flat, so by the Künneth isomorphisms in [Ro, Corollary 10.84],
$Z\otimes_{R}D(n)$ and $W\otimes_{R}D(n)$ are exact. As a result, for any
$i\geq 0$, we notice that

$0=H_{i}\left(Z\otimes_{R}D(n)\right)\xrightarrow{H_{i}\left(\mu\otimes_{R}D(n)\right)}H_{i}\left(W\otimes_{R}D(n)\right)=0$

is an isomorphism, so the above diagram implies that
$H_{i}\left(\mu\boxtimes_{R}D(n)\right)$ is an isomorphism. Thus
$\mu\boxtimes_{R}D(n)$ is a quasi-isomorphism, so $\zeta^{\prime}$ is a quasi-
isomorphism. Altogether, $\zeta^{\prime}$ is a trivial cofibration.

Next suppose that either $R^{(\iota)}:R^{(U)}\rightarrow R^{(V)}$ is a trivial
cofibration or $\rho:M\rightarrow N$ is a trivial fibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$. We will show that $\eta$ is a trivial
fibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$. To this end, let $n\geq
1$, and let $\varrho:K\rightarrow L$ be the morphism
$\operatorname{DK}(\iota^{n}):\operatorname{DK}\left(S(n-1)\right)\rightarrow\operatorname{DK}\left(D(n)\right)$.
We show that

$\zeta^{\prime}:\left(W\boxtimes_{R}S(n-1)\right)\textstyle\bigsqcup_{\left(Z\boxtimes_{R}S(n-1)\right)}\left(Z\boxtimes_{R}D(n)\right)\rightarrow
W\boxtimes_{R}D(n)$

is a cofibration. Consider the following pushout diagram:

${Z\boxtimes_{R}S(n-1)}$${Z\boxtimes_{R}D(n)}$${W\boxtimes_{R}S(n-1)}$${\left(W\boxtimes_{R}S(n-1)\right)\bigsqcup_{\left(Z\boxtimes_{R}S(n-1)\right)}\left(Z\boxtimes_{R}D(n)\right)}$${W\boxtimes_{R}D(n)}$$\scriptstyle{Z\boxtimes_{R}\iota^{n}}$$\scriptstyle{\mu\boxtimes_{R}S(n-1)}$$\scriptstyle{\varpi}$$\scriptstyle{\mu\boxtimes_{R}D(n)}$$\scriptstyle{W\boxtimes_{R}\iota^{n}}$$\scriptstyle{\zeta^{\prime}}$

In each degree $m\geq 0$, the above diagram is naturally isomorphic to the
following diagram:

${\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}Z_{k}}$${\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}Z_{k}}$${\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}W_{k}}$${\left(\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}W_{k}\right)\bigsqcup_{\left(\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}Z_{k}\right)}\left(\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}Z_{k}\right)}$${\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}W_{k}}$$\scriptstyle{\overline{\nu_{m}}}$$\scriptstyle{\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}\mu_{k}}$$\scriptstyle{p_{m}}$$\scriptstyle{\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}}$$\scriptstyle{\overline{\nu_{m}}}$$\scriptstyle{q_{m}}$

Let $m\geq 0$. By definition, we have: