$\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}W_{k}\right)\textstyle\bigsqcup_{\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}Z_{k}\right)}\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}Z_{k}\right)=\frac{\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}W_{k}\right)\oplus\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}Z_{k}\right)}{\left\\{\left(\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}\mu_{k}\right)(z),-\overline{\nu_{m}}(z)\right)\;|\;z\in\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}Z_{k}\right\\}}$

Let
$x=\left(x_{f,g}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}\in\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}W_{k}$
and
$y=\left(y_{f,g}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}\in\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}Z_{k}$.
For any $(f,g)\in\mathcal{S}_{m}^{\prime}$, define:

$x_{f,g}^{\prime}:=\begin{dcases}x_{f,g}&\textrm{if
}(f,g)\in\mathcal{S}_{m}^{\prime\prime}\\\ 0&\textrm{if
}(f,g)\in\mathcal{S}_{m}^{\prime}\backslash\mathcal{S}_{m}^{\prime\prime}\end{dcases}$

Then
$\overline{\nu_{m}}(x)=\left(x_{f,g}^{\prime}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}$.
Thus we have:

$q_{m}\left(\overline{(x,y)}\right)=\overline{\nu_{m}}(x)+\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)(y)=\left(x_{f,g}^{\prime}+\mu_{k}\left(y_{f,g}\right)\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}$

Now if $q_{m}\left(\overline{(x,y)}\right)=0$, then
$x_{f,g}^{\prime}=-\mu_{k}\left(y_{f,g}\right)$ for every
$(f,g)\in\mathcal{S}_{m}^{\prime}$. Therefore,
$\mu_{k}\left(y_{f,g}\right)=-x_{f,g}^{\prime}=-x_{f,g}$ for every
$(f,g)\in\mathcal{S}_{m}^{\prime\prime}$. Moreover,
$\mu_{k}\left(y_{f,g}\right)=-x_{f,g}^{\prime}=0$ for every
$(f,g)\in\mathcal{S}_{m}^{\prime}\backslash\mathcal{S}_{m}^{\prime\prime}$.
But $\mu$ is a cofibration, so $\mu_{k}$ is injective, implying that
$y_{f,g}=0$ for every
$(f,g)\in\mathcal{S}_{m}^{\prime}\backslash\mathcal{S}_{m}^{\prime\prime}$.
Set
$z=\left(-y_{f,g}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}\in\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}Z_{k}$.
Then
$\overline{\nu_{m}}(z)=\left(-y_{f,g}^{\prime}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}=\left(-y_{f,g}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}$
where $y_{f,g}^{\prime}$ is defined similar to $x_{f,g}^{\prime}$. Thus we
have
$x=\left(x_{f,g}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}=\left(-\mu_{k}\left(y_{f,g}\right)\right)_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}=\left(\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}\mu_{k}\right)(z)$
and
$y=\left(y_{f,g}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}=-\overline{\nu_{m}}(z)$.
This shows that $\overline{(x,y)}=0$. Therefore, $q_{m}$ is injective. On the
other hand, from the above diagram, we have
$q_{m}p_{m}=\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}$, so
$\operatorname{Im}\left(\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)\subseteq\operatorname{Im}(q_{m})$.
Therefore, we get an $R$-epimorphism
$\psi_{m}:\operatorname{Coker}\left(\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)\rightarrow\operatorname{Coker}(q_{m})$
given by
$\psi_{m}\left(x+\operatorname{Im}\left(\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)\right)=x+\operatorname{Im}(q_{m})$
for every $x\in\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}W_{k}$. We next
define an $R$-homomorphism
$\chi_{m}:\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}W_{k}\rightarrow\operatorname{Coker}\left(\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)$
as follows. Let
$w=\left(w_{f,g}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}\in\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}W_{k}$.
For any $(f,g)\in\mathcal{S}_{m}^{\prime}$, define:

$w_{f,g}^{\prime\prime}:=\begin{dcases}w_{f,g}&\textrm{if
}(f,g)\in\mathcal{S}_{m}^{\prime}\backslash\mathcal{S}_{m}^{\prime\prime}\\\
0&\textrm{if }(f,g)\in\mathcal{S}_{m}^{\prime\prime}\end{dcases}$

Set
$\chi_{m}(w):=\left(w_{f,g}^{\prime\prime}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}+\operatorname{Im}\left(\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)$.
If $w\in\operatorname{Im}(q_{m})$, then there are elements
$x=\left(x_{f,g}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}\in\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}W_{k}$
and
$y=\left(y_{f,g}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}\in\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}Z_{k}$
such that
$w=q_{m}\left(\overline{(x,y)}\right)=\left(x_{f,g}^{\prime}+\mu_{k}\left(y_{f,g}\right)\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}$.
We note that $x_{f,g}^{\prime}=0$ for every
$(f,g)\in\mathcal{S}_{m}^{\prime}\backslash\mathcal{S}_{m}^{\prime\prime}$, so
by the above construction, $\left(x_{f,g}^{\prime}\right)^{\prime\prime}=0$
for every $(f,g)\in\mathcal{S}_{m}^{\prime}$. Therefore, we get:

$\begin{split}\chi_{m}(w)&=\left(\left(x_{f,g}^{\prime}\right)^{\prime\prime}+\mu_{k}\left(y_{f,g}^{\prime\prime}\right)\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}+\operatorname{Im}\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)\\\
&=\left(\mu_{k}\left(y_{f,g}^{\prime\prime}\right)\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}+\operatorname{Im}\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)\\\
&=\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)\left(\left(y_{f,g}^{\prime\prime}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}\right)+\operatorname{Im}\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)\\\
&=0\end{split}$

As a result, $\chi_{m}\left(\operatorname{Im}(q_{m})\right)=0$, so $\chi_{m}$
induces an $R$-homomorphism
$\overline{\chi_{m}}:\operatorname{Coker}(q_{m})\rightarrow\operatorname{Coker}\left(\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)$
given by
$\overline{\chi_{m}}\left(w+\operatorname{Im}(q_{m})\right)=\left(w_{f,g}^{\prime\prime}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}+\operatorname{Im}\left(\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)$.
Now we have for every
$w=\left(w_{f,g}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}\in\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}W_{k}$:

$\psi_{m}\left(\overline{\chi_{m}}\left(w+\operatorname{Im}(q_{m})\right)\right)=\psi_{m}\left(\left(w_{f,g}^{\prime\prime}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}+\operatorname{Im}\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)\right)=\left(w_{f,g}^{\prime\prime}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}+\operatorname{Im}(q_{m})$

Let
$\widetilde{\nu_{m}}:\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}W_{k}\rightarrow\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}W_{k}$
be the induced $R$-homomorphism which projects a tuple onto its components
from $\mathcal{S}_{m}^{\prime\prime}$. Setting
$x=\widetilde{\nu_{m}}\left(\left(w_{f,g}-w_{f,g}^{\prime\prime}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}\right)\in\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}W_{k}$,
we see that:

$w-\left(w_{f,g}^{\prime\prime}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}=\left(w_{f,g}-w_{f,g}^{\prime\prime}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}=\overline{\nu_{m}}(x)=q_{m}\left(\overline{(x,0)}\right)$

Therefore,
$w-\left(w_{f,g}^{\prime\prime}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}\in\operatorname{Im}(q_{m})$,
so we get:

$\psi_{m}\left(\overline{\chi_{m}}\left(w+\operatorname{Im}(q_{m})\right)\right)=\left(w_{f,g}^{\prime\prime}\right)_{(f,g)\in\mathcal{S}_{m}^{\prime}}+\operatorname{Im}(q_{m})=w+\operatorname{Im}(q_{m})$

This shows that $\psi_{m}\overline{\chi_{m}}=1^{\operatorname{Coker}(q_{m})}$.
Thus the short exact sequence

$0\rightarrow\operatorname{Ker}(\psi_{m})\rightarrow\operatorname{Coker}\left(\textstyle\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)\xrightarrow{\psi_{m}}\operatorname{Coker}(q_{m})\rightarrow
0$

splits, so in particular, $\operatorname{Coker}(q_{m})$ is isomorphic to a
direct summand of
$\operatorname{Coker}\left(\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)$.
But
$\operatorname{Coker}\left(\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\mu_{k}\right)\cong\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime}}\operatorname{Coker}(\mu_{k})$
is projective as $\mu$ is a cofibration. As a consequence,
$\operatorname{Coker}(q_{m})$ is projective. This shows that $\zeta^{\prime}$
is a cofibration.

If $\rho:M\rightarrow N$ is a trivial fibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, then we are done. If
$R^{(\iota)}:R^{(U)}\rightarrow R^{(V)}$ is a trivial cofibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, i.e. $\mu:Z\rightarrow W$ is a
trivial cofibration in $\mathcal{C}_{\geq 0}(R)$, then we need to further show
that $\zeta^{\prime}$ is a weak equivalence, i.e. a quasi-isomorphism. Since
for any $n\geq 1$, $D(n)$ and $S(n-1)$ are connective $R$-complexes of flat
modules, and $\mu$ is a weak equivalence i.e. a quasi-isomorphism, Corollary
5.11 implies that $\mu\boxtimes_{R}S(n-1)$ and $\mu\boxtimes_{R}D(n)$ are
quasi-isomorphisms. On the other hand, since $\mu$ is a cofibration,
$\mu_{k}:Z_{k}\rightarrow W_{k}$ is injective with projective cokernel for
every $k\geq 0$, so
$\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}\mu_{k}:\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}Z_{k}\rightarrow\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}W_{k}$
is injective with projective cokernel. It follows that
$\mu\boxtimes_{R}S(n-1)$ is a cofibration. Thus $\mu\boxtimes_{R}S(n-1)$ is a
trivial cofibration, so by [Hi, Proposition 7.2.12], we deduce that $\varpi$
is a trivial cofibration, whence $\varpi$ is a quasi-isomorphism. Now
$\zeta^{\prime}\varpi=\mu\boxtimes_{R}D(n)$, so we conclude that
$\zeta^{\prime}$ is a quasi-isomorphism.

Now let $\varrho:K\rightarrow L$ be the morphism
$\operatorname{DK}(\lambda):\operatorname{DK}(0)\rightarrow\operatorname{DK}\left(S(0)\right)$.
We show that

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

is a cofibration. We have the following commutative diagram:

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

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:

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

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

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

Now the second row is exact and
$\bigoplus_{(f,g)\in\mathcal{S}_{m}^{\prime\prime}}\operatorname{Coker}(\mu_{k})$
is projective, so it follows that the first row is exact and
$\left(\operatorname{Coker}(\mu)\boxtimes_{R}S(0)\right)_{m}$ is projective.
As a result, the previous sequence is exact and
$\operatorname{Coker}(\mu)\boxtimes_{R}S(0)$ is an $R$-complex of projective
modules. This means that $\mu\boxtimes_{R}S(0)$ is a cofibration, so
$\zeta^{\prime}$ is a cofibration.

If $\rho:M\rightarrow N$ is a trivial fibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, then we are done. If
$R^{(\iota)}:R^{(U)}\rightarrow R^{(V)}$ is a trivial cofibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, i.e. $\mu:Z\rightarrow W$ is a
trivial cofibration in $\mathcal{C}_{\geq 0}(R)$, then we need to further show
that $\zeta^{\prime}$ is a weak equivalence, i.e. a quasi-isomorphism. Using
the fact $S(0)=\Sigma^{0}R=R$ and Corollary 5.10, we have the following
commutative diagram for every $i\geq 0$:

${H_{i}\left(Z\boxtimes_{R}S(0)\right)}$${H_{i}\left(W\boxtimes_{R}S(0)\right)}$${H_{i}\left(Z\otimes_{R}S(0)\right)}$${H_{i}\left(W\otimes_{R}S(0)\right)}$${H_{i}\left(Z\otimes_{R}R\right)}$${H_{i}\left(W\otimes_{R}R\right)}$${H_{i}(Z)}$${H_{i}(W)}$$\scriptstyle{H_{i}\left(\mu\boxtimes_{R}S(0)\right)}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{H_{i}\left(\mu\otimes_{R}S(0)\right)}$$\scriptstyle{H_{i}\left(\mu\otimes_{R}R\right)}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{H_{i}(\mu)}$

As $\mu$ is a weak equivalence, i.e. a quasi-isomorphism,
$H_{i}(\mu):H_{i}(Z)\rightarrow H_{i}(W)$ is an isomorphism for every $i\geq
0$, so the above diagram shows that $H_{i}\left(\mu\boxtimes_{R}S(0)\right)$
is an isomorphism for every $i\geq 0$, whence $\mu\boxtimes_{R}S(0)$ is a
quasi-isomorphism. Thus $\zeta^{\prime}$ is a quasi-isomorphism. ∎

######  Corollary 6.2.

Let $R$ be a commutative ring, $M$ a simplicial $R$-module, and
$\iota:U\rightarrow V$ a morphism of simplicial sets. If
$R^{(\iota)}:R^{(U)}\rightarrow R^{(V)}$ is a (trivial) cofibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, then
$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\iota,M):\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,M)\rightarrow\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,M)$
is a (trivial) fibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$.

###### Proof.

It is clear from Theorem 4.15 that the unique morphism $f:M\rightarrow 0$ is a
fibration. Consider the commutative diagram

${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,M)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,0)=0}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,M)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,0)=0}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,f)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\iota,M)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\iota,0)=0}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,f)}$

and form the following pullback diagram:

${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,M)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,M)\mathop{\mathchoice{\rotatebox[origin={c}]{180.0}{$\displaystyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\textstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptscriptstyle\bigsqcup$}}}_{0}0}$${0}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,M)}$${0}$$\scriptstyle{\eta}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\iota,M)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,f)}$$\scriptstyle{\cong}$

If $R^{(\iota)}:R^{(U)}\rightarrow R^{(V)}$ is a (trivial) cofibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, then Lemma 6.1 implies that $\eta$ is
a (trivial) fibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, so by the
above diagram, we deduce that
$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\iota,M)$ is a
(trivial) fibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$. ∎

The following lemma is dual to Lemma 6.1:

######  Lemma 6.3.

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

${M^{(U)}}$${N^{(U)}}$${M^{(V)}}$${N^{(V)}}$$\scriptstyle{\varrho^{(U)}}$$\scriptstyle{M^{(\iota)}}$$\scriptstyle{N^{(\iota)}}$$\scriptstyle{\varrho^{(V)}}$

and form the following pushout diagram:

${M^{(U)}}$${N^{(U)}}$${M^{(V)}}$${M^{(V)}\bigsqcup_{M^{(U)}}N^{(U)}}$${N^{(V)}}$$\scriptstyle{\varrho^{(U)}}$$\scriptstyle{M^{(\iota)}}$$\scriptstyle{r}$$\scriptstyle{N^{(\iota)}}$$\scriptstyle{s}$$\scriptstyle{\varrho^{(V)}}$$\scriptstyle{\zeta}$

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

###### Proof.

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

${M^{(V)}\bigsqcup_{M^{(U)}}N^{(U)}}$${K}$${N^{(V)}}$${L}$$\scriptstyle{p}$$\scriptstyle{\zeta}$$\scriptstyle{\rho}$$\scriptstyle{q}$

Considering the commutative diagram

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

there is a unique morphism
$\eta:\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,K)\rightarrow\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,K)\textstyle\mathop{\mathchoice{\rotatebox[origin={c}]{180.0}{$\displaystyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\textstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptscriptstyle\bigsqcup$}}}_{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,L)}\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,L)$
that makes the following pullback diagram commutative:

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

On the other hand, considering the natural bijections

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

and

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

from Remark 4.4, we get the following commutative diagram:

${N}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,L)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,K)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,L)}$$\scriptstyle{\phi_{NVL}^{-1}(q)}$$\scriptstyle{\phi_{NUK}^{-1}(pr)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\iota,L)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,\rho)}$

Hence there is a unique morphism
$\theta:N\rightarrow\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,K)\mathop{\mathchoice{\rotatebox[origin={c}]{180.0}{$\displaystyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\textstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptscriptstyle\bigsqcup$}}}_{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,L)}\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,L)$
that makes the following diagram commutative:

${N}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,K)\mathop{\mathchoice{\rotatebox[origin={c}]{180.0}{$\displaystyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\textstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptstyle\bigsqcup$}}{\rotatebox[origin={c}]{180.0}{$\scriptscriptstyle\bigsqcup$}}}_{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,L)}\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,L)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,L)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,K)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,L)}$$\scriptstyle{\theta}$$\scriptstyle{\phi_{NUK}^{-1}(pr)}$$\scriptstyle{\phi_{NVL}^{-1}(q)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\iota,L)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(U,\rho)}$

Then considering the natural bijection

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

we get the following commutative diagram:

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

If
$\tau:N\rightarrow\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(V,K)$
is a morphism of simplicial $R$-modules that makes the diagram

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

commutative, then considering the natural bijection

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

we see that the following diagram is commutative:

${M^{(V)}\bigsqcup_{M^{(U)}}N^{(U)}}$${K}$${N^{(V)}}$${L}$$\scriptstyle{p}$$\scriptstyle{\zeta}$$\scriptstyle{\rho}$$\scriptstyle{q}$$\scriptstyle{\phi_{NVK}(\tau)}$

Now if $\varrho$ and $R^{(\iota)}$ are cofibrations, and $\rho$ is a trivial
fibration, then Lemma 6.1 implies that $\eta$ is a trivial fibration, so
$\tau$ exists, whence we conclude from the above diagram that $\zeta$ is a
cofibration. If $\varrho$ is a trivial cofibration, $R^{(\iota)}$ is a
cofibration, and $\rho$ is a fibration, then Lemma 6.1 implies that $\eta$ is
a fibration, so $\tau$ exists, whence we deduce from the above diagram that
$\zeta$ is a trivial cofibration. If $\varrho$ is a cofibration, $R^{(\iota)}$
is a trivial cofibration, and $\rho$ is a fibration, then Lemma 6.1 implies
that $\eta$ is a trivial fibration, so $\tau$ exists, whence we infer from the
above diagram that $\zeta$ is a trivial cofibration. ∎

######  Corollary 6.4.

Let $R$ be a commutative ring, $M$ a cofibrant simplicial $R$-module, and
$\iota:U\rightarrow V$ a morphism of simplicial sets. If
$R^{(\iota)}:R^{(U)}\rightarrow R^{(V)}$ is a (trivial) cofibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, then $M^{(\iota)}:M^{(U)}\rightarrow
M^{(V)}$ is a (trivial) cofibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$.

###### Proof.

Since $M$ is cofibrant, the unique morphism $f:0\rightarrow M$ is a
cofibration. Consider the commutative diagram

${0=0^{(U)}}$${M^{(U)}}$${0=0^{(V)}}$${M^{(V)}}$$\scriptstyle{f^{(U)}}$$\scriptstyle{0=0^{(\iota)}}$$\scriptstyle{M^{(\iota)}}$$\scriptstyle{f^{(V)}}$

and form the following pushout diagram:

${0}$${M^{(U)}}$${0}$${0\bigsqcup_{0}M^{(U)}}$${M^{(V)}}$$\scriptstyle{\cong}$$\scriptstyle{M^{(\iota)}}$$\scriptstyle{f^{(V)}}$$\scriptstyle{\zeta}$

If $R^{(\iota)}:R^{(U)}\rightarrow R^{(V)}$ is a (trivial) cofibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, then Lemma 6.3 implies that $\zeta$
is a (trivial) cofibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, so by
the above diagram, we deduce that $M^{(\iota)}$ is a (trivial) cofibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$. ∎

Now we can construct specific cylinder and path objects in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$. These objects are introduced in [GS,
Corollary 4.14] but the proof outlined there relies heavily on the model
structure of simplicial sets. Here we provide a proof based on Lemmas 6.1 and
6.3 whose proofs are independent of the model structure of simplicial sets.

######  Proposition 6.5.

Let $R$ be a commutative ring, and $M$ a simplicial $R$-module. Then the
following assertions hold:

  1. (i)

If $M$ is cofibrant in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, then
$M^{(\Delta^{1})}$ is a cylinder object for $M$.

  2. (ii)

$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{1},M\right)$
is a path object for $M$.

###### Proof.

Let $\varepsilon:\Delta^{1}\rightarrow\Delta^{0}$ and
$\delta:\partial\Delta^{1}\rightarrow\Delta^{0}$ be the unique morphisms, and
$\iota:\partial\Delta^{1}\rightarrow\Delta^{1}$ the inclusion morphism. Then
we have the following commutative diagram of simplicial sets:

(6.5.1)
${\partial\Delta^{1}}$${\Delta^{0}}$${\Delta^{1}}$$\scriptstyle{\delta}$$\scriptstyle{\iota}$$\scriptstyle{\varepsilon}$

We note that $\Delta_{n}^{0}=\\{(\underbrace{0,...,0}_{n+1\textrm{
times}})\\}$ and
$\Delta_{n}^{1}=\\{(\alpha_{0},\alpha_{1},\ldots,\alpha_{n})\;|\;0\leq\alpha_{0}\leq\alpha_{1}\leq\ldots\leq\alpha_{n}\leq
1\\}$ for every $n\geq 0$. Define a morphism
$\zeta=:\Delta^{0}\rightarrow\Delta^{1}$ of simplicial sets by setting
$\zeta_{n}(0,...,0)=(0,...,0)$ for every $n\geq 0$. Then it is clear that
$\varepsilon\zeta=1^{\Delta^{0}}$. We show that the induced morphism
$R^{(\zeta)}:R^{(\Delta^{0})}\rightarrow R^{(\Delta^{1})}$ is a trivial
cofibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, i.e.
$\operatorname{Nor}\left(R^{(\zeta)}\right):\operatorname{Nor}\left(R^{(\Delta^{0})}\right)\rightarrow\operatorname{Nor}\left(R^{(\Delta^{1})}\right)$
is a trivial cofibration in $\mathcal{C}_{\geq 0}(R)$. To this end, we analyze
this morphism more closely. We have observed in the proof of Lemma 4.8 that
$\operatorname{Nor}\left(R^{(\Delta^{0})}\right)$ is the $R$-complex

$0\rightarrow R\rightarrow 0$

with $R$ in degree $0$. On the other hand, we have
$\operatorname{Nor}\left(R^{(\Delta^{1})}\right)_{0}=R^{(\Delta_{0}^{1})}=R^{(\\{0,1\\})}=R^{2}$
with a basis $\\{e_{1},e_{2}\\}$ corresponding to $\Delta_{0}^{1}=\\{0,1\\}$,
and $R^{(\Delta_{1}^{1})}=R^{(\\{(0,0),(0,1),(1,1)\\})}=R^{3}$ with a basis
$\\{e_{1}^{\prime},e_{2}^{\prime},e_{3}^{\prime}\\}$ corresponding to
$\Delta_{1}^{1}=\left\\{(0,0),(0,1),(1,1)\right\\}$. We describe the morphisms

$R^{3}=R^{(\Delta_{1}^{1})}\xrightarrow{d_{1,0}^{R^{(\Delta^{1})}}=R^{\left(d_{1,0}^{\Delta^{1}}\right)}}R^{(\Delta_{0}^{1})}=R^{2}$

and

$R^{3}=R^{(\Delta_{1}^{1})}\xrightarrow{d_{1,1}^{R^{(\Delta^{1})}}=R^{\left(d_{1,1}^{\Delta^{1}}\right)}}R^{(\Delta_{0}^{1})}=R^{2}$

as follows. We note that $e_{1}^{\prime}$ corresponds to $(0,0)$, so since
$d_{1,0}^{\Delta^{1}}(0,0)=0$ and $0$ corresponds to $e_{1}$, we get
$d_{1,0}^{R^{(\Delta^{1})}}(e_{1}^{\prime})=e_{1}$. Similarly, we have:

$d_{1,0}^{\Delta^{1}}:\begin{dcases}(0,0)\mapsto 0\\\ (0,1)\mapsto 1\\\
(1,1)\mapsto 1\end{dcases}\quad\text{and}\quad
d_{1,1}^{\Delta^{1}}:\begin{dcases}(0,0)\mapsto 0\\\ (0,1)\mapsto 0\\\
(1,1)\mapsto 1\end{dcases}$

Hence we get:

$d_{1,0}^{R^{(\Delta^{1})}}:\begin{dcases}e_{1}^{\prime}\mapsto e_{1}\\\
e_{2}^{\prime}\mapsto e_{2}\\\ e_{3}^{\prime}\mapsto
e_{2}\end{dcases}\quad\text{and}\quad
d_{1,1}^{R^{(\Delta^{1})}}:\begin{dcases}e_{1}^{\prime}\mapsto e_{1}\\\
e_{2}^{\prime}\mapsto e_{1}\\\ e_{3}^{\prime}\mapsto e_{2}\end{dcases}$

It follows that $d_{1,0}^{R^{(\Delta^{1})}}(a,b,c)=(a,b+c)$ and
$d_{1,1}^{R^{(\Delta^{1})}}(a,b,c)=(a+b,c)$ for every $(a,b,c)\in R^{3}$. Thus
we have:

$\operatorname{Nor}\left(R^{(\Delta^{1})}\right)_{1}=\operatorname{Ker}\left(d_{1,0}^{R^{(\Delta^{1})}}\right)=\left\\{(a,b,c)\in
R^{3}\;|\;a=0\textrm{ and }b=-c\right\\}=\left\\{(0,-a,a)\;|\;a\in
R\right\\}\cong R$

Moreover, in light of the above isomorphism,
$\partial_{1}=\partial_{1}^{\operatorname{Nor}\left(R^{(\Delta^{1})}\right)}:R\rightarrow
R^{2}$ is given by
$\partial_{1}(a)=-d_{1,1}^{R^{(\Delta^{1})}}(0,-a,a)=(a,-a)$ for every $a\in
R$. As we saw in the proof of Lemma 4.8, if $n\geq 2$, then
$\operatorname{Nor}\left(R^{(\Delta^{1})}\right)_{n}=0$. As a consequence,
$\operatorname{Nor}\left(R^{(\Delta^{1})}\right)$ is the $R$-complex

$0\rightarrow R\xrightarrow{\partial_{1}}R^{2}\rightarrow 0$

with $R$ in degree $1$ and $R^{2}$ in degree $0$. Now consider
$\operatorname{Nor}\left(R^{(\zeta)}\right):\operatorname{Nor}\left(R^{(\Delta^{0})}\right)\rightarrow\operatorname{Nor}\left(R^{(\Delta^{1})}\right)$.
We note that $\zeta_{0}:\Delta_{0}^{0}\rightarrow\Delta_{0}^{1}$ is given by
$\zeta_{0}(0)=(0)$, so
$\operatorname{Nor}\left(R^{(\zeta)}\right)_{0}=R^{(\zeta_{0})}:R\rightarrow
R^{2}$ is given by $R^{(\zeta_{0})}(a)=(a,0)$ for every $a\in R$. Indeed,
$\operatorname{Nor}\left(R^{(\zeta)}\right)$ is as follows:

${0}$${0}$${R}$${0}$${0}$${R}$${R^{2}}$${0}$$\scriptstyle{R^{(\zeta_{0})}}$$\scriptstyle{\partial_{1}}$

It is now clear that $\operatorname{Nor}\left(R^{(\zeta)}\right)$ is injective
whose cokernel is an $R$-complex of projective modules, so
$\operatorname{Nor}\left(R^{(\zeta)}\right)$ is a cofibration. Furthermore, by
Lemma 4.8, $\operatorname{Nor}\left(R^{(\zeta)}\right)$ is a homotopy
equivalence, hence a quasi-isomorphism, i.e. a weak equivalence. Therefore,
$\operatorname{Nor}\left(R^{(\zeta)}\right)$ is a trivial cofibration in
$\mathcal{C}_{\geq 0}(R)$.

We next show that $R^{(\iota)}:R^{(\partial\Delta^{1})}\rightarrow
R^{(\Delta^{1})}$ is a cofibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$,
i.e.
$\operatorname{Nor}\left(R^{(\iota)}\right):\operatorname{Nor}\left(R^{(\partial\Delta^{1})}\right)\rightarrow\operatorname{Nor}\left(R^{(\Delta^{1})}\right)$
is a cofibration in $\mathcal{C}_{\geq 0}(R)$. To this end, we analyze this
morphism more closely. We have
$\operatorname{Nor}\left(R^{(\partial\Delta^{1})}\right)_{0}=R^{(\partial\Delta_{0}^{1})}=R^{(\\{0,1\\})}=R^{2}$.
If $n\geq 1$, then every element of $\partial\Delta_{n}^{1}$ is degenerate, so
$\operatorname{Nor}\left(R^{(\partial\Delta^{1})}\right)_{n}=0$. As a
consequence, $\operatorname{Nor}\left(R^{(\partial\Delta^{1})}\right)$ is the
$R$-complex

$0\rightarrow R^{2}\rightarrow 0$

with $R^{2}$ in degree $0$. Now consider
$\operatorname{Nor}\left(R^{(\iota)}\right):\operatorname{Nor}\left(R^{(\partial\Delta^{1})}\right)\rightarrow\operatorname{Nor}\left(R^{(\Delta^{1})}\right)$.
We note that $\partial\Delta_{0}^{1}=\Delta_{0}^{1}=\\{0,1\\}$, so
$\iota_{0}:\partial\Delta_{0}^{1}\rightarrow\Delta_{0}^{1}$ is the identity
map, thereby $R^{(\iota_{0})}=1^{R^{2}}:R^{2}\rightarrow R^{2}$. Indeed,
$\operatorname{Nor}\left(R^{(\iota)}\right)$ is as follows:

${0}$${0}$${R^{2}}$${0}$${0}$${R}$${R^{2}}$${0}$$\scriptstyle{R^{(\iota_{0})}}$$\scriptstyle{\partial_{1}}$

It is now clear that $\operatorname{Nor}\left(R^{(\iota)}\right)$ is injective
whose cokernel is an $R$-complex of projective modules, so it is a
cofibration. Now we have:

(i): The diagram (6.5.1) yields the following commutative diagram of
simplicial $R$-modules:

${M^{(\partial\Delta^{1})}}$${M^{(\Delta^{0})}}$${M^{(\Delta^{1})}}$$\scriptstyle{M^{(\delta)}}$$\scriptstyle{M^{(\iota)}}$$\scriptstyle{M^{(\varepsilon)}}$

From the relation $\varepsilon\zeta=1^{\Delta^{0}}$, we get
$M^{(\varepsilon)}M^{(\zeta)}=1^{M^{(\Delta^{0})}}$. Since
$R^{(\zeta)}:R^{(\Delta^{0})}\rightarrow R^{(\Delta^{1})}$ is a trivial
cofibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, Corollary 6.4 shows
that $M^{(\zeta)}:M^{(\Delta^{0})}\rightarrow M^{(\Delta^{1})}$ is a trivial
cofibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, hence in particular a
weak equivalence. Thus $M^{(\varepsilon)}$ is a weak equivalence in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$. On the other hand,
$R^{(\iota)}:R^{(\partial\Delta^{1})}\rightarrow R^{(\Delta^{1})}$ is a
cofibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, so Corollary 6.4
implies that $M^{(\iota)}:M^{(\partial\Delta^{1})}\rightarrow
M^{(\Delta^{1})}$ is a cofibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$.
We finally note that $M^{(\Delta^{0})}\cong M$, and also we have
$M^{(\partial\Delta^{1})}\cong M^{(\Delta^{0}\sqcup\Delta^{0})}\cong
M^{(\Delta^{0})}\oplus M^{(\Delta^{0})}\cong M\oplus M$ by Example 4.3. Taking
these isomorphisms into account, the above diagram yields a commutative
diagram

${M\oplus
M}$${M}$${M^{\left(\Delta^{1}\right)}}$$\scriptstyle{\nabla^{M}}$$\scriptstyle{\kappa^{M}}$$\scriptstyle{\xi^{M}}$

in which $\kappa^{M}$ is a cofibration and $\xi^{M}$ is a weak equivalence.
This means that $M^{(\Delta^{1})}$ is a cylinder object for $M$.

(ii): The diagram (6.5.1) yields the following commutative diagram of
simplicial $R$-modules:

${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{0},M\right)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\partial\Delta^{1},M\right)}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{1},M\right)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\delta,M)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\varepsilon,M)}$$\scriptstyle{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\iota,M)}$

From the relation $\varepsilon\zeta=1^{\Delta^{0}}$, we get
$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\zeta,M)\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\varepsilon,M)=1^{\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{0},M\right)}$.
Since $R^{(\zeta)}:R^{(\Delta^{0})}\rightarrow R^{(\Delta^{1})}$ is a trivial
cofibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, the Corollary 6.2
shows that
$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\zeta,M):\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{1},M\right)\rightarrow\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{0},M\right)$
is a trivial fibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, hence in
particular a weak equivalence. Thus
$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\varepsilon,M)$ is a
weak equivalence in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$. On the other
hand, $R^{(\iota)}:R^{(\partial\Delta^{1})}\rightarrow R^{(\Delta^{1})}$ is a
cofibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, so Corollary 6.2
implies that
$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}(\iota,M):\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{1},M\right)\rightarrow\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\partial\Delta^{1},M\right)$
is a fibration in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$. We finally note
that
$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{0},M\right)\cong
M$, and also we have
$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\partial\Delta^{1},M\right)\cong\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{0}\sqcup\Delta^{0},M\right)\cong\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{0},M\right)\times\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{0},M\right)\cong
M\times M$. Taking these isomorphisms into account, the above diagram yields a
commutative diagram

${M}$${M\times
M}$${\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{1},M\right)}$$\scriptstyle{\Delta^{M}}$$\scriptstyle{\zeta^{M}}$$\scriptstyle{\nu^{M}}$

in which $\nu^{M}$ is a fibration and $\zeta^{M}$ is a weak equivalence. This
means that
$\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{1},M\right)$
is a path object for $M$. ∎

We are now ready to prove the main theorem.

######  Theorem 6.6.

Let $R$ be a commutative ring, and
$\sharp:\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)\rightarrow\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$
the forgetful functor. Then
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$ is a
cofibrantly generated model category with the model structure given as
follows:

  1. (i)

Fibrations are morphisms $f:A\rightarrow B$ in
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$ such that
$\operatorname{Nor}\left(f^{\sharp}\right)_{i}:\operatorname{Nor}\left(A^{\sharp}\right)_{i}\rightarrow\operatorname{Nor}\left(B^{\sharp}\right)_{i}$
is surjective for every $i\geq 1$.

  2. (ii)

Weak equivalences are morphisms $f:A\rightarrow B$ in
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$ such that
$\operatorname{Nor}\left(f^{\sharp}\right):\operatorname{Nor}\left(A^{\sharp}\right)\rightarrow\operatorname{Nor}\left(B^{\sharp}\right)$
is a quasi-isomorphism in $\mathcal{C}_{\geq 0}(R)$.

  3. (iii)

Cofibration are morphisms in
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$ that have the
left lifting property against trivial fibrations defined in (i) and (ii).

###### Proof.

By Theorems 4.14 and 4.15,
$(\operatorname{DK},\operatorname{Nor}):\mathcal{C}_{\geq
0}(R)\leftrightarrows\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$ is an adjoint
equivalence of categories through which
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$ becomes a cofibrantly generated model
category. If $\mathcal{X}$ and $\mathcal{Y}$ are the sets of morphisms in
$\mathcal{C}_{\geq 0}(R)$ corresponding to its cofibrant generation, then we
have observed that the domains of morphisms in $\mathcal{X}$ and $\mathcal{Y}$
are small with respect to $\operatorname{Mor}\left(\mathcal{C}_{\geq
0}(R)\right)$. It follows that the domains of morphisms in
$\operatorname{DK}(\mathcal{X})$ and $\operatorname{DK}(\mathcal{Y})$ are
small with respect to
$\operatorname{Mor}\left(\mathpzc{s}\mathcal{M}\mathpzc{od}(R)\right)$. On the
other hand, we have the adjoint pair of functors
$\left(\operatorname{Sym}_{R}(-),\sharp\right):\mathcal{M}\mathpzc{od}(R)\leftrightarrows\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$
that extends degreewise to an adjoint pair of functors
$\left(\operatorname{Sym}_{R}(-),\sharp\right):\mathpzc{s}\mathcal{M}\mathpzc{od}(R)\leftrightarrows\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$.
Also, $\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$ is locally small and
bicomplete, so $\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$
is locally small and bicomplete. By the construction of filtered direct
limits, we notice that the forgetful functor
$\sharp:\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)\rightarrow\mathcal{M}\mathpzc{od}(R)$
preserves filtered direct limits, so its extension
$\sharp:\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)\rightarrow\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$
also preserves filtered direct limits, hence sequential direct limits in
particular. Therefore, Corollary 2.11 implies that the domains of morphisms in
$\operatorname{Sym}_{R}\left(\operatorname{DK}(\mathcal{X})\right)$ and
$\operatorname{Sym}_{R}\left(\operatorname{DK}(\mathcal{Y})\right)$ are small
with respect to
$\operatorname{Mor}\left(\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)\right)$.
In particular,
$\operatorname{Sym}_{R}\left(\operatorname{DK}(\mathcal{X})\right)$ and
$\operatorname{Sym}_{R}\left(\operatorname{DK}(\mathcal{Y})\right)$ permit the
small object argument. Now we let a morphism $f:A\rightarrow B$ in
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$ be a fibration
or a weak equivalence if $f^{\sharp}:A^{\sharp}\rightarrow B^{\sharp}$ is a
fibration or a weak equivalence in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$,
respectively. As a consequence, $f:A\rightarrow B$ is a fibration in
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$ if and only if
$\operatorname{Nor}\left(f^{\sharp}\right)_{i}:\operatorname{Nor}\left(A^{\sharp}\right)_{i}\rightarrow\operatorname{Nor}\left(B^{\sharp}\right)_{i}$
is surjective for every $i\geq 1$. Also, $f:A\rightarrow B$ is a weak
equivalence in $\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$
if and only if
$\operatorname{Nor}\left(f^{\sharp}\right):\operatorname{Nor}\left(A^{\sharp}\right)\rightarrow\operatorname{Nor}\left(B^{\sharp}\right)$
is a quasi-isomorphism. Finally, we let a morphism $f:A\rightarrow B$ in
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$ be a
cofibration if it has the left lifting property against trivial fibrations.

In order to apply Theorem 2.9, we need to establish the acyclicity condition,
i.e. we need to show that any cofibration in
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$ that has the
left lifting property against fibrations is also a weak equivalence. Let
$g:A\rightarrow B$ be a cofibration in
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$ that has the
left lifting property against fibrations. If $f:A\rightarrow 0$ is the unique
morphism in $\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$,
then $f^{\sharp}:A^{\sharp}\rightarrow 0^{\sharp}$ is a fibration in
$\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$, so $f$ is a fibration in
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$. Therefore, the
commutative diagram on the left can be completed to the commutative diagram on
the right:

$\leavevmode\hbox to57.44pt{\vbox
to46.41pt{\pgfpicture\makeatletter\hbox{\hskip 28.40446pt\lower-23.2526pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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In particular, $hg=1^{A}$. On the other hand, by the universal property of
product, there is a unique morphism $l:B\rightarrow B\sqcap B$ that makes the
following diagrams commutative:

${B}$${B}$${B\sqcap
B}$${B}$$\scriptstyle{1^{B}}$$\scriptstyle{gh}$$\scriptstyle{l}$$\scriptstyle{\pi_{1}^{B}}$$\scriptstyle{\pi_{2}^{B}}$

Consider the path object
$\operatorname{Path}\left(B^{\sharp}\right)=\operatorname{Map}_{\mathpzc{s}\mathcal{S}\mathpzc{et}}\left(\Delta^{1},B^{\sharp}\right)$
in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$. Then there is a commutative
diagram

${B^{\sharp}}$${B^{\sharp}\sqcap
B^{\sharp}}$${\operatorname{Path}\left(B^{\sharp}\right)}$$\scriptstyle{\Delta_{B^{\sharp}}}$$\scriptstyle{\zeta_{B^{\sharp}}}$$\scriptstyle{\nu_{B^{\sharp}}}$

in which $\nu_{B^{\sharp}}$ is a fibration and $\zeta_{B^{\sharp}}$ is a weak
equivalence in $\mathpzc{s}\mathcal{M}\mathpzc{od}(R)$. But we observed in the
proof of Proposition 6.5 that this diagram is isomorphic to a diagram in which
morphisms could be either morphisms of simplicial modules or simplicial
commutative algebras depending on what structure we consider; See also Remark
4.4. Therefore, we can also have the commutative diagram

(6.6.1) ${B}$${B\sqcap
B}$${\operatorname{Path}(B)}$$\scriptstyle{\Delta_{B}}$$\scriptstyle{\zeta_{B}}$$\scriptstyle{\nu_{B}}$

in which $\nu_{B}$ is a fibration and $\zeta_{B}$ is a weak equivalence in
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$. Then the
following diagram is commutative:

${A}$${\operatorname{Path}(B)}$${B}$${B\sqcap
B}$$\scriptstyle{\zeta_{B}g}$$\scriptstyle{g}$$\scriptstyle{\nu_{B}}$$\scriptstyle{l}$

Indeed, we have

$\pi_{1}^{B}lg=1^{B}g=\pi_{1}^{B}\Delta_{B}g=\pi_{1}^{B}\nu_{B}\zeta_{B}g$

and

$\pi_{2}^{B}lg=ghg=g1^{A}=g=1^{B}g=\pi_{2}^{B}\Delta_{B}g=\pi_{2}^{B}\nu_{B}\zeta_{B}g,$

so it follows that $lg=\nu_{B}\zeta_{B}g$. As $\nu_{B}$ is a fibration, the
above diagram can be completed to the following commutative diagram:

${A}$${\operatorname{Path}(B)}$${B}$${B\sqcap
B}$$\scriptstyle{\zeta_{B}g}$$\scriptstyle{g}$$\scriptstyle{\nu_{B}}$$\scriptstyle{l}$$\scriptstyle{\psi}$

Now the following diagram is commutative:

${B}$${B}$${B}$${\operatorname{Path}(B)}$$\scriptstyle{1^{B}}$$\scriptstyle{gh}$$\scriptstyle{\psi}$$\scriptstyle{\pi_{1}^{B}\nu_{B}}$$\scriptstyle{\pi_{2}^{B}\nu_{B}}$

Indeed, we have $\pi_{1}^{B}\nu_{B}\psi=\pi_{1}^{B}l=1^{B}$ and
$\pi_{2}^{B}\nu_{B}\psi=\pi_{2}^{B}l=gh$. Let
$\mathcal{G}=\operatorname{Nor}\left(-^{\sharp}\right):\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)\rightarrow\mathcal{C}_{\geq
0}(R)$. Applying the functor $\mathcal{G}$ to the diagram (6.6.1) yields the
commutative diagram

${\mathcal{G}(B)}$${\mathcal{G}(B)\sqcap\mathcal{G}(B)}$${\mathcal{G}\left(\operatorname{Path}(B)\right)}$$\scriptstyle{\eta\mathcal{G}(\Delta_{B})=\Delta_{\mathcal{G}(B)}}$$\scriptstyle{\mathcal{G}(\zeta_{B})}$$\scriptstyle{\eta\mathcal{G}(\nu_{B})}$

in $\mathcal{C}_{\geq 0}(R)$ in which $\eta:\mathcal{G}(B\sqcap
B)\rightarrow\mathcal{G}(B)\sqcap\mathcal{G}(B)$ is the natural isomorphism.
Since $\nu_{B}$ is a fibration and $\zeta_{B}$ is a weak equivalence in
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$, we see by
definition that $\eta\mathcal{G}(\Delta_{B})$ is a fibration and
$\mathcal{G}(\zeta_{B})$ is a weak equivalence in $\mathcal{C}_{\geq 0}(R)$.
This means that $\mathcal{G}\left(\operatorname{Path}(B)\right)$ is a path
object for $\mathcal{G}(B)$. Moreover, the following diagram is commutative:

${\mathcal{G}(B)}$${\mathcal{G}(B)}$${\mathcal{G}(B)}$${\mathcal{G}\left(\operatorname{Path}(B)\right)}$$\scriptstyle{1^{\mathcal{G}(B)}}$$\scriptstyle{\mathcal{G}(g)\mathcal{G}(h)}$$\scriptstyle{\mathcal{G}(\psi)}$$\scriptstyle{\pi_{1}^{\mathcal{G}(B)}\eta\mathcal{G}(\nu_{B})}$$\scriptstyle{\pi_{2}^{\mathcal{G}(B)}\eta\mathcal{G}(\nu_{B})}$

Since $\mathcal{C}_{\geq 0}(R)$ is a model category by Theorem 3.1, the above
diagram shows that $\mathcal{G}(g)\mathcal{G}(h)\sim_{r}1^{\mathcal{G}(B)}$.
Applying the localization functor
$\mathcal{L}_{R}:=\mathcal{L}_{\mathcal{C}_{\geq 0}(R)}:\mathcal{C}_{\geq
0}(R)\rightarrow\operatorname{Ho}\left(\mathcal{C}_{\geq 0}(R)\right)$, noting
that $\mathcal{L}_{R}$ maps weak equivalences to isomorphisms, and invoking
[Hi, Lemma 8.3.4], we get
$\mathcal{L}_{R}\left(\mathcal{G}(g)\right)\mathcal{L}_{R}\left(\mathcal{G}(h)\right)=1^{\mathcal{G}(B)}$.
On the other hand, from the relation $hg=1^{A}$, we get
$\mathcal{L}_{R}\left(\mathcal{G}(h)\right)\mathcal{L}_{R}\left(\mathcal{G}(g)\right)=1^{\mathcal{G}(A)}$.
Therefore, $\mathcal{L}_{R}\left(\mathcal{G}(g)\right)$ is an isomorphism, so
by Proposition 2.7, $\mathcal{G}(g)$ is a weak equivalence in
$\mathcal{C}_{\geq 0}(R)$, so by definition, $g$ is a weak equivalence in
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$.

It now follows from Theorem 2.9 that
$\mathpzc{s}\mathcal{C}\mathpzc{om}\mathcal{A}\mathpzc{lg}(R)$ is a
cofibrantly generated model category. ∎

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