2.1.1 Auslander–Reiten class and co-Auslander–Reiten class

Let $T\in \text{mod}R$ be a Wakamatsu-tilting module with $S=\text{End}(T_{R})^{op}$. There are the following two interesting classes associated with Wakamatsu-tilting modules.

The Auslander–Reiten class in $\text{mod}R$ with respect to the Wakamatsu-tilting module $T_{R}$, denoted by ${\mathcal{X}}_{T}$, is defined as follows [Reference Auslander and Reiten3].

${\mathcal{X}}_{T}:=\{M\in \text{mod}R|$ there is an infinite exact sequence $0\rightarrow M\stackrel{f_{0}}{\longrightarrow }T_{0}\stackrel{f_{1}}{\longrightarrow }T_{1}\stackrel{f_{2}}{\longrightarrow }\cdots \,$ such that $\text{Im}f_{i}\in \text{KerExt}_{R}^{{\geqslant}1}(-,T)$ for each $i\geqslant 0$, where $T_{i}\in \text{add}_{\text{mod}R}T$ for all $i\}$.

Obviously, it holds that ${\mathcal{X}}_{T}\subseteq \text{KerExt}_{R}^{{\geqslant}1}(-,T)$. Moreover, these two classes coincide with each other provided that $T$ is a cotilting $R$-module.

Dually, the co-Auslander–Reiten class in $\text{mod}R$ with respect to the Wakamatsu-tilting $R$-module $T$, denoted by $_{T}{\mathcal{X}}$, is defined as follows.

$_{T}{\mathcal{X}}:=\{M\in \text{mod}R|$ there is an infinite exact sequence $\cdots \stackrel{f_{2}}{\longrightarrow }T_{1}\stackrel{f_{1}}{\longrightarrow }T_{0}\stackrel{f_{0}}{\longrightarrow }M\rightarrow 0$ such that $\text{Im}f_{i}\in \text{KerExt}_{R}^{{\geqslant}1}(T,-)$ for each $i\geqslant 0$, where $T_{i}\in \text{add}_{\text{mod}R}T$ for all $i\}$.

Similarly, we have that $_{T}{\mathcal{X}}\subseteq \text{KerExt}_{R}^{{\geqslant}1}(T,-)$ and they coincide with each other provided that $T$ is a tilting $R$-module.

The following result gives some properties about the Auslander–Reiten class and the co-Auslander–Reiten class for a Wakamatsu-tilting module [Reference Auslander and Reiten3, Reference Mantese and Reiten22, Reference Wakamatsu26, Reference Wakamatsu27].

We remark that in case $T=R$, the class ${\mathcal{X}}_{T}={\mathcal{X}}_{R}$ is just the class of all Gorenstein projective $R$-modules. Dually, in case $T=DR$, the class $_{T}{\mathcal{X}}=_{DR}{\mathcal{X}}$ is just the class of all Gorenstein injective modules. We refer to [Reference Enochs and Jenda15] for more on Gorenstein projective and Gorenstein injective modules.

2.1.2

The following is a characterization of the Auslander–Reiten class and the co-Auslander–Reiten class, by [Reference Wakamatsu26, Section 2].

2.1.3 Useful isomorphisms

Let $T$ be a Wakamatsu-tilting $S$-$R$-bimodule. Then we have the following isomorphisms of bimodules:

$$\begin{eqnarray}_{S}DS_{S}\simeq _{S}T\otimes _{R}DT_{S}\qquad \text{and}\qquad _{R}DR_{R}\simeq _{R}DT\otimes _{S}T_{R}.\end{eqnarray}$$

Given an adjoint pair $(\text{F},\text{G})$ of functors, we denote by $\boldsymbol{\unicode[STIX]{x1D6E4}}$ the natural adjoint isomorphism

$$\begin{eqnarray}\boldsymbol{\unicode[STIX]{x1D6E4}}:\text{Hom}(\text{F}(-),-)\simeq \text{Hom}(-,\text{G}(-)).\end{eqnarray}$$

Moreover, for a homomorphism $f:\text{F}(X)\rightarrow Y$, we denote by $\boldsymbol{\unicode[STIX]{x1D6E4}}(f):X\rightarrow \text{G}(Y)$ the image of $f$ under the isomorphism $\boldsymbol{\unicode[STIX]{x1D6E4}}$. We denote by $\unicode[STIX]{x1D702}$ and $\unicode[STIX]{x1D716}$ the unit and counit of this adjoint pair, respectively, that is,

$$\begin{eqnarray}\displaystyle & \unicode[STIX]{x1D702}_{X}=\boldsymbol{\unicode[STIX]{x1D6E4}}^{T}(1_{F(X)}):X\rightarrow GF(X)\qquad \text{and} & \displaystyle \nonumber\\ \displaystyle & \unicode[STIX]{x1D716}_{Y}=(\boldsymbol{\unicode[STIX]{x1D6E4}}^{T})^{-1}(1_{G(Y)}):FG(Y)\rightarrow Y. & \displaystyle \nonumber\end{eqnarray}$$

In particular, associated with an $S$-$R$-bimodule $T$, we have the following adjoint isomorphism:

$$\begin{eqnarray}\boldsymbol{\unicode[STIX]{x1D6E4}}^{T}:\text{Hom}_{R}(-\!\otimes _{S}T,-)\simeq \text{Hom}_{S}(-,\text{Hom}_{R}(T,-)).\end{eqnarray}$$

We denote by $\unicode[STIX]{x1D702}^{T}$ and $\unicode[STIX]{x1D716}^{T}$ the unit and counit of this adjoint pair, respectively, that is, for $X\in \text{mod}S$ and $Y\in \text{mod}R$, respectively,

$$\begin{eqnarray}\displaystyle & \unicode[STIX]{x1D702}_{X}^{T}=\boldsymbol{\unicode[STIX]{x1D6E4}}^{T}(1_{X\otimes _{S}T}):X\rightarrow \text{Hom}_{R}(T,X\otimes _{S}T)\qquad \text{and} & \displaystyle \nonumber\\ \displaystyle & \unicode[STIX]{x1D716}_{Y}^{T}=(\boldsymbol{\unicode[STIX]{x1D6E4}}^{T})^{-1}(1_{\text{Hom}_{R}(T,Y)}):\text{Hom}_{R}(T,Y)\otimes _{S}T\rightarrow Y. & \displaystyle \nonumber\end{eqnarray}$$

By the naturality of the isomorphism $\boldsymbol{\unicode[STIX]{x1D6E4}}$, for all homomorphisms $f:X_{1}\rightarrow X_{2}$, $g:\text{F}(X_{2})\rightarrow Y_{1}$ and $h:Y_{1}\rightarrow Y_{2}$, it holds that $\boldsymbol{\unicode[STIX]{x1D6E4}}(\text{F}(f)\cdot g\cdot h)=f\cdot \boldsymbol{\unicode[STIX]{x1D6E4}}(g)\cdot \text{G}(h)$.

In particular, for a morphism $g:F(X)\rightarrow Y$, by applying $\boldsymbol{\unicode[STIX]{x1D6E4}}$ to the composition $F(X)\stackrel{1_{F(X)}}{\rightarrow }F(X)\stackrel{g}{\rightarrow }Y$, we have that $\boldsymbol{\unicode[STIX]{x1D6E4}}(g)=\boldsymbol{\unicode[STIX]{x1D6E4}}(1_{F(X)})\cdot \text{G}(g)=\unicode[STIX]{x1D702}_{X}\cdot \text{G}(g)$. Dually, for a morphism $f:X\rightarrow G(Y)$, we have that $\boldsymbol{\unicode[STIX]{x1D6E4}}^{-1}(f)=F(f)\unicode[STIX]{x1D716}_{Y}$.

2.2.1

We recall some basic facts on repetitive algebras mainly from [Reference Happel18].

Let $R$ be an Artin algebra. The repetitive algebra $\widehat{R}$ of $R$ was first introduced in [Reference Hughes and Waschbüsch20] and is defined to be the direct sum $\widehat{R}=\bigoplus _{n\in \mathbb{Z}}R\oplus \bigoplus _{n\in \mathbb{Z}}DR$ with the multiplication given by

$$\begin{eqnarray}(a_{n},\unicode[STIX]{x1D711}_{n})(b_{n},\unicode[STIX]{x1D713}_{n})_{n}=(a_{n}b_{n},a_{n+1}\unicode[STIX]{x1D713}_{n}+\unicode[STIX]{x1D711}_{n}b_{n})_{n}.\end{eqnarray}$$

The repetitive algebra $\widehat{R}$ can be interpreted as the following infinite matrix algebra (without the identity):

$$\begin{eqnarray}\left(\begin{array}{@{}ccccc@{}}\ddots & & & & \\ \ddots & R & & & \\ & DR & R & & \\ & & DR & R & \\ & & & \ddots & \ddots \end{array}\right).\end{eqnarray}$$

2.2.2

Consider the following two categories:

1. (1) ${\mathcal{R}}{\mathcal{C}}^{\otimes }(R):=\{X=\{X_{i},\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\}_{i\in \mathbb{Z}}\mid X_{i}\in \text{mod}R$ such that almost all $X_{i}$ are 0 and $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X):X_{i}\otimes _{R}DR\rightarrow X_{i-1}$ satisfying $(\unicode[STIX]{x1D6FF}_{i+1}^{\otimes }(X)\otimes _{R}DR)\cdot \unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)=0$, for each $i\}$, where a morphism between two objects $X$ and $Y$ is given by $f=\{f_{i}:X_{i}\rightarrow Y_{i}\}$ such that $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\cdot f_{i-1}=f_{i}\otimes _{R}DR\cdot \unicode[STIX]{x1D6FF}_{i}^{\otimes }(Y)$ for all $i$.

2. (2) ${\mathcal{R}}{\mathcal{C}}^{\text{H}}(R):=\{X=\{X_{i},\unicode[STIX]{x1D6FF}_{i}^{\text{H}}(X)\}_{i\in \mathbb{Z}}\mid X_{i}\in \text{mod}R$ such that almost all $X_{i}$ are 0 and $\unicode[STIX]{x1D6FF}_{i}^{\text{H}}(X):X_{i}\rightarrow \text{Hom}_{R}(DR,X_{i-1})$ satisfying $\unicode[STIX]{x1D6FF}_{i+1}^{\text{H}}(X)\cdot \text{Hom}_{R}(DR,\unicode[STIX]{x1D6FF}_{i}^{\text{H}}(X))=0$, for each $i\}$, where a morphism between two objects $X$ and $Y$ is given by $f=\{f_{i}:X_{i}\rightarrow Y_{i}\}$ such that $\unicode[STIX]{x1D6FF}_{i}^{\text{H}}(X)\cdot \text{Hom}_{R}(DR,f_{i-1})=f_{i}\cdot \unicode[STIX]{x1D6FF}_{i}^{\text{H}}(Y)$ for all $i$.

One can check that these two categories ${\mathcal{R}}{\mathcal{C}}^{\otimes }(R)$ and ${\mathcal{R}}{\mathcal{C}}^{\text{H}}(R)$ are both abelian categories. Moreover, they are indeed equivalent to each other as abelian categories, via the adjoint pair $(-\!\otimes _{R}DR,\text{Hom}_{R}(DR,-))$. Indeed, an object $X=\{X_{i},\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\}\in {\mathcal{R}}{\mathcal{C}}^{\otimes }(R)$ is equivalent to an object $X=\{X_{i},\boldsymbol{\unicode[STIX]{x1D6E4}}^{DR}(\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X))\}\in {\mathcal{R}}{\mathcal{C}}^{\text{H}}(R)$. We will freely use this equivalence. In particular, we often view objects $X$ in these two categories being of the following form with almost all terms $X_{i}=0$

$$\begin{eqnarray}\cdots \stackrel{\unicode[STIX]{x1D6FF}_{i+1}}{{\rightsquigarrow}}X_{i}\stackrel{\unicode[STIX]{x1D6FF}_{i}}{{\rightsquigarrow}}X_{i-1}\stackrel{\unicode[STIX]{x1D6FF}_{i-1}}{{\rightsquigarrow}}\cdots \,,\end{eqnarray}$$

where $\unicode[STIX]{x1D6FF}_{i}$ means $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)$ (resp., $\unicode[STIX]{x1D6FF}_{i}^{\text{H}}(X)$) if $X\in {\mathcal{R}}{\mathcal{C}}^{\otimes }(R)$ (resp., $X\in {\mathcal{R}}{\mathcal{C}}^{\text{H}}(R)$). We call it a (bounded chain) repe-complex with the repe-difference $\unicode[STIX]{x1D6FF}$ and denote by ${\mathcal{R}}{\mathcal{C}}(R)$ the category of all such repe-complexes and call it the repetitive category over $R$. Note that there is an obvious automorphism $[1]:{\mathcal{R}}{\mathcal{C}}(R)\rightarrow {\mathcal{R}}{\mathcal{C}}(R)$ defined by $(X[1])_{i}=X_{i-1}$ for each $i$.

Note that if $X=\{X_{i}\}$ is a repe-complex, then $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\cdot \unicode[STIX]{x1D6FF}_{i-1}^{\text{H}}(X)=0$ since $\boldsymbol{\unicode[STIX]{x1D6E4}}^{DR}(\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\cdot \unicode[STIX]{x1D6FF}_{i-1}^{\text{H}}(X))=\unicode[STIX]{x1D6FF}_{i}^{\text{H}}(X)\cdot \text{Hom}_{R}(DR,\unicode[STIX]{x1D6FF}_{i-1}^{\text{H}}(X))=0$.

We say that a repe-complex $X=\{X_{i},\unicode[STIX]{x1D6FF}_{i}\}\in {\mathcal{R}}{\mathcal{C}}(R)$ is trivial if each $\unicode[STIX]{x1D6FF}_{i}=0$. The full subcategory of all trivial repe-complexes is denoted by ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)$. Note that there is a natural forgetful functor from ${\mathcal{R}}{\mathcal{C}}(R)$ to ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)$ by forgetting the repe-difference.

Let ${\mathcal{C}}$ be a class of $R$-modules, we denote by ${\mathcal{R}}{\mathcal{C}}({\mathcal{C}})$ the class of repe-complexes with terms in ${\mathcal{C}}$. The notation ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{C}})$ is defined similarly.

The connection between the repetitive category and the algebra $\widehat{R}$ is that the repetitive category is equivalent to $\text{mod}\widehat{R}$, that is, the module category of the algebra $\widehat{R}$, as abelian categories; see [Reference Happel18] for details. Thus, we may identify ${\mathcal{R}}{\mathcal{C}}(R)=\text{mod}\widehat{R}$.

2.2.3

As shown in [Reference Happel18], $\widehat{R}$ is a self-injective algebra and then the category ${\mathcal{R}}{\mathcal{C}}(R)(=\text{mod}\widehat{R})$ is a Frobenius category, where the projective (and also injective) objects are of the form

$$\begin{eqnarray}\cdots \stackrel{\unicode[STIX]{x1D6FF}_{i+1}}{{\rightsquigarrow}}P_{i}\oplus I_{i}\stackrel{\unicode[STIX]{x1D6FF}_{i}}{{\rightsquigarrow}}P_{i-1}\oplus I_{i-1}\stackrel{\unicode[STIX]{x1D6FF}_{i-1}}{{\rightsquigarrow}}\cdots \,,\end{eqnarray}$$

where $P_{i}\in \text{proj}R$, $I_{i}\in \text{inj}R$ and $\unicode[STIX]{x1D6FF}_{i}=\big(\begin{smallmatrix}0 & \unicode[STIX]{x1D6FF}_{i}^{\prime }\\ 0 & 0\end{smallmatrix}\big)$ such that $\unicode[STIX]{x1D6FF}_{i}^{\prime }:P_{i}\otimes _{R}DR\rightarrow I_{i-1}$ is an isomorphism (considered in ${\mathcal{R}}{\mathcal{C}}^{\otimes }(R)$) or, equivalently, $\unicode[STIX]{x1D6FF}_{i}^{\prime }:P_{i}\rightarrow \text{Hom}_{R}(DR,I_{i-1})$ is an isomorphism (considered in ${\mathcal{R}}{\mathcal{C}}^{\text{H}}(R)$). Thus, its stable category $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)$ is a triangulated category. We will call it the stable repetitive category of $\text{mod}R$ (or simply, of $R$).

It was shown in [Reference Happel18] that there is a fully faithful triangle embedding from the derived category ${\mathcal{D}}^{b}(\text{mod}R)$ to the stable repetitive category $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)$. Moreover, there is a triangle equivalence between ${\mathcal{D}}^{b}(\text{mod}R)$ and $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)$ if and only if $R$ has finite global dimension. We note that this result was generalized in [Reference Yamaura28] and also a simple proof of this result was presented there.

For basic knowledge on triangulated categories, derived categories and the tilting theory, we refer to [Reference Happel18].

3.2.1

In general, the two cotorsion pairs in Proposition 3.1(1) are not complete. For instance, consider the case $T=R$. Then ${\mathcal{X}}_{R}$ is the class of all Gorenstein projective modules (note that we only consider finitely generated modules). It is well known that this class is not a precovering class in general; see, for instance, [Reference Yoshino29]. Thus, the cotorsion pair $({\mathcal{X}}_{R},\text{KerExt}_{R}^{1}({\mathcal{X}}_{R},-))$ cannot be complete. Dually, the cotorsion pair $(\text{KerExt}_{R}^{1}(-,_{DR}{\mathcal{X}}),_{DR}{\mathcal{X}})$ in case $T=DR$ is not complete in general.

However, the other cotorsion pair of the two cotorsion pairs in Proposition 3.1(1), that is, the cotorsion pair

$$\begin{eqnarray}(\text{KerExt}_{R}^{1}(-,_{R}{\mathcal{X}}),_{R}{\mathcal{X}})(=(\text{proj}R,\text{mod}R))\end{eqnarray}$$

for $T=R$ and the cotorsion pair

$$\begin{eqnarray}({\mathcal{X}}_{DR},\text{KerExt}_{R}^{1}({\mathcal{X}}_{DR},-))(=(\text{mod}R,\text{inj}R))\end{eqnarray}$$

for $T=DR$, respectively, is clearly complete. This leads to the following general definition.

For example, $R$ and $DR$ are good Wakamatsu-tilting modules. In general, if $_{S}T_{R}$ is a good Wakamatsu-tilting bimodule, then $_{R}DT_{S}$ is also a good Wakamatsu-tilting bimodule by the definition and the fact that $DDT=T$.

In general, we do not know if the following question has an affirmative answer.

However, we will see in Section 5 that the answer to the above question is ‘yes’ for algebras of finite representation type.

3.2.2

By the definition, we have the following property of Wakamatsu-tilting bimodules.

3.2.3

Recall that a subcategory ${\mathcal{A}}\subseteq \text{mod}R$ is covariantly finite (or a preenveloping class) if for any $X\in \text{mod}R$, there is an object $A_{X}\in {\mathcal{A}}$ and a homomorphism $u_{_{X}}:X\rightarrow A_{X}$ such that $\text{Hom}_{R}(u_{_{X}},A)$ is surjective for any object $A\in {\mathcal{A}}$; see, for instance, [Reference Auslander and Reiten3]. Dually, a subcategory ${\mathcal{B}}\subseteq \text{mod}R$ is contravariantly finite (or a precovering class) if for any $X\in \text{mod}R$, there is an object $B_{X}\in {\mathcal{B}}$ and a homomorphism $v_{X}:B_{X}\rightarrow X$ such that $\text{Hom}_{R}(B,v_{X})$ is surjective for any object $B\in {\mathcal{B}}$.

A cotorsion pair $({\mathcal{B}},{\mathcal{A}})$ is complete if and only if ${\mathcal{A}}$ is covariantly finite, if and only if ${\mathcal{B}}$ is contravariantly finite; see [Reference Auslander and Reiten3, Proposition 1.9].

Let ${\mathcal{A}}$ be a subcategory of $\text{mod}R$. An $R$-module $T$ is said to be Ext-projective in ${\mathcal{A}}$ if $T\in {\mathcal{A}}\bigcap \text{KerExt}_{R}^{1}(-,{\mathcal{A}})$. Moreover, it is said to be an Ext-projective generator in ${\mathcal{A}}$ if, for any $A\in {\mathcal{A}}$, there exists an exact sequence $0\rightarrow A^{\prime }\rightarrow T_{A}\rightarrow A\rightarrow 0$ with $T_{A}\in \text{add}_{\text{mod}R}T$ and $A^{\prime }\in {\mathcal{A}}$. Dually, an $R$-module $T$ is said to be an Ext-injective cogenerator in ${\mathcal{A}}$ if $T\in {\mathcal{A}}\bigcap \text{KerExt}_{R}^{1}({\mathcal{A}},-)$ and, for any $A\in {\mathcal{A}}$, there exists an exact sequence $0\rightarrow A\rightarrow T_{A}\rightarrow A^{\prime }\rightarrow 0$ with $T_{A}\in \text{add}_{\text{mod}R}T$ and $A^{\prime }\in {\mathcal{A}}$.

Proof. (1) By the assumptions, we can construct the following commutative diagram, where $T_{Z}\in \text{add}_{\text{mod}R}T$ and $Z^{\prime }\in {\mathcal{A}}$:

Since ${\mathcal{A}}$ is closed under extensions and direct summands, we have that $X\in {\mathcal{A}}$ from the middle column.

(2) Dually. ◻

3.2.4

3.2.5

Proof. Since $T$ is an Ext-projective generator in ${\mathcal{A}}$, we see that $T\in \text{KerExt}_{R}^{1}(-,{\mathcal{A}})\bigcap {\mathcal{A}}={\mathcal{B}}\bigcap {\mathcal{A}}$ and that, for any $A\in {\mathcal{A}}$, there is an exact sequence $0\rightarrow A^{\prime }\rightarrow T_{A}\rightarrow A\rightarrow 0$ with $T_{A}\in \text{add}_{\text{mod}R}T$ and $A^{\prime }\in {\mathcal{A}}$. In particular, for any $X\in {\mathcal{A}}\bigcap {\mathcal{B}}$, there is an exact sequence $0\rightarrow X^{\prime }\rightarrow T_{X}\rightarrow X\rightarrow 0$ with $T_{X}\in \text{add}_{\text{mod}R}T$ and $X^{\prime }\in {\mathcal{A}}$, which is clearly split. Hence, $X\in \text{add}_{\text{mod}R}T$. It follows that $\text{add}_{\text{mod}R}T={\mathcal{B}}\bigcap {\mathcal{A}}$. Moreover, by an argument similar to the one used in the proof of [Reference Mantese and Reiten22, Proposition 2.13(b)], we have that $T$ is also an Ext-injective cogenerator in ${\mathcal{B}}$. Note that these facts imply that ${\mathcal{A}}\subseteq _{T}{\mathcal{X}}$ and that ${\mathcal{B}}\subseteq {\mathcal{X}}_{T}$. In particular, $\text{Hom}_{R}(T,-):{\mathcal{A}}_{\longleftarrow }^{\longrightarrow }\text{Hom}_{R}(T,{\mathcal{A}}):-\!\otimes _{S}T$ define an equivalence and $\text{Hom}_{S}(DT,-):{\mathcal{B}}\otimes _{R}DT_{\longleftarrow }^{\longrightarrow }{\mathcal{B}}:-\!\otimes _{R}DT$ define an equivalence, by Proposition 2.1.1.

The above arguments, in particular, show that Lemma 3.2.4 can be applied to ${\mathcal{A}}$ (considering the Wakamatsu-tilting bimodule $_{S}T_{R}$) and ${\mathcal{B}}$ (considering the Wakamatsu-tilting bimodule $_{R}DT_{S}$); thus, we see that $\text{Hom}_{R}(T,{\mathcal{A}})$ is resolving and that ${\mathcal{B}}\otimes _{R}DT$ is coresolving. It is also clear that the bimodules $_{S}T_{R}$ and $_{R}DT_{S}$ represent a counter equivalence between two pairs $({\mathcal{B}},{\mathcal{A}})$ and $(\text{Hom}_{R}(T,{\mathcal{A}}),{\mathcal{B}}\otimes _{R}DT)$, by assumptions. So, it just remains to show that $(\text{Hom}_{R}(T,{\mathcal{A}}),{\mathcal{B}}\otimes _{R}DT)$ is a cotorsion pair.

We divide the remaining proof into three steps.

Step 1. $\text{Ext}_{S}^{i}(\text{Hom}_{R}(T,A),B\otimes _{R}DT)=0$, for any $A\in {\mathcal{A}}$ and any $B\in {\mathcal{B}}$ and for any $i\geqslant 0$.

Note that there is a natural isomorphism

$$\begin{eqnarray}D\text{Hom}_{S}(S,B\otimes _{R}DT)\simeq \text{Hom}_{R}(B,S\otimes _{S}T).\end{eqnarray}$$

It induces a natural isomorphism, for any $S_{i}\in \text{add}_{\text{mod}S}S$,

(¶)

Now take $A\in {\mathcal{A}}$; since $T$ is an Ext-projective generator, there is a long exact sequence

(†)$$\begin{eqnarray}\cdots \rightarrow T_{n}\rightarrow \cdots \rightarrow T_{1}\rightarrow T_{0}\rightarrow A\rightarrow 0,\end{eqnarray}$$

where each $T_{i}\in \text{add}_{\text{mod}R}T$ and each image in ${\mathcal{A}}$. Here we consider the sequence (†) as a (cochain) complex with the term $A$ at the first position.

Since $B\in \text{KerExt}_{R}^{1}(-,{\mathcal{A}})$, we have the following induced exact sequence $\text{Hom}_{R}(B,\dagger )$:

$$\begin{eqnarray}\cdots \rightarrow \text{Hom}_{R}(B,T_{n})\rightarrow \cdots \rightarrow \text{Hom}_{R}(B,T_{1})\rightarrow \text{Hom}_{R}(B,T_{0})\rightarrow \text{Hom}_{R}(B,A)\rightarrow 0.\end{eqnarray}$$

On the other hand, by applying the functor $D\text{Hom}_{S}(\text{Hom}_{R}(T,-),B\otimes _{R}DT)$, we have a complex $D\text{Hom}_{S}(\text{Hom}_{R}(T,\dagger ),B\otimes _{R}DT)$:

$$\begin{eqnarray}\displaystyle & & \displaystyle \cdots \rightarrow D\text{Hom}_{S}(\text{Hom}_{R}(T,T_{n}),B\otimes _{R}DT)\rightarrow \cdots \rightarrow D\text{Hom}_{S}(\text{Hom}_{R}(T,T_{1}),B\otimes _{R}DT)\nonumber\\ \displaystyle & & \displaystyle \quad \rightarrow D\text{Hom}_{S}(\text{Hom}_{R}(T,T_{0}),B\otimes _{R}DT)\rightarrow D\text{Hom}_{S}(\text{Hom}_{R}(T,A),B\otimes _{R}DT)\rightarrow 0.\nonumber\end{eqnarray}$$

Since $\text{Hom}_{R}(T,\dagger )$ is exact, the functor $D\text{Hom}_{S}(\text{Hom}_{R}(T,-),B\otimes _{R}DT)$ is right exact. By the above isomorphism (¶), we obtain the following isomorphisms of complexes:

$$\begin{eqnarray}D\text{Hom}_{S}(\text{Hom}_{R}(T,\dagger ),B\otimes _{R}DT)\simeq \text{Hom}_{R}(B,\text{Hom}_{R}(T,\dagger )\otimes _{S}T)\simeq \text{Hom}_{R}(B,\dagger ).\end{eqnarray}$$

But the later is exact, so we obtain that, for $i\geqslant 1$,

$$\begin{eqnarray}\displaystyle & & \displaystyle \text{Ext}_{S}^{i}(\text{Hom}_{R}(T,A),B\otimes _{R}DT)\simeq \text{H}^{i}(\text{Hom}_{S}(\text{Hom}_{R}(T,\dagger ),B\otimes _{R}DT))\nonumber\\ \displaystyle & & \displaystyle \quad \simeq D\text{H}^{-i}(D\text{Hom}_{S}(\text{Hom}_{R}(T,\dagger ),B\otimes _{R}DT))\simeq D\text{H}^{-i}(\text{Hom}_{R}(B,\dagger ))=0.\nonumber\end{eqnarray}$$

Thus, Step 1 is established. In particular, we obtain that $\text{Hom}_{R}(T,{\mathcal{A}})\subseteq \text{KerExt}_{S}^{1}(-,{\mathcal{B}}\otimes _{R}DT)$ and that ${\mathcal{B}}\otimes _{R}DT\subseteq \text{KerExt}_{S}^{1}(\text{Hom}_{R}(T,{\mathcal{A}}),-)$ due to the arbitrarity of $A\in {\mathcal{A}}$ and $B\in {\mathcal{B}}$.

Step 2. $\text{KerExt}_{S}^{1}(-,{\mathcal{B}}\otimes _{R}DT)\subseteq \text{Hom}_{R}(T,{\mathcal{A}})$.

Take any $Y\in \text{KerExt}_{S}^{1}(-,{\mathcal{B}}\otimes _{R}DT)$ and a projective resolution of $Y$, where we consider as a (cochain) complex with the term $Y$ at the zeroth position:

(♯)$$\begin{eqnarray}\cdots \stackrel{f_{n+1}}{\longrightarrow }S_{n}\stackrel{f_{n}}{\longrightarrow }\cdots \stackrel{f_{3}}{\longrightarrow }S_{2}\stackrel{f_{2}}{\longrightarrow }S_{1}\stackrel{f_{1}}{\longrightarrow }S_{0}\stackrel{f_{0}}{\longrightarrow }Y\rightarrow 0.\end{eqnarray}$$

Note that $DT=R\otimes _{R}DT\in {\mathcal{B}}\otimes _{R}DT$ and that ${\mathcal{B}}\otimes _{R}DT$ is coresolving, so we obtain that

$$\begin{eqnarray}\displaystyle & Y\in \text{KerExt}_{S}^{1}(-,{\mathcal{B}}\otimes _{R}DT)=\text{KerExt}_{S}^{{>}0}(-,{\mathcal{B}}\otimes _{R}DT) & \displaystyle \nonumber\\ \displaystyle & \subseteq \text{KerExt}_{S}^{{>}0}(-,DT)=\text{KerTor}_{{>}0}^{S}(-,T). & \displaystyle \nonumber\end{eqnarray}$$

Then we have an induced exact sequence:

(♯ ⊗_ST)$$\begin{eqnarray}\cdots \rightarrow S_{n}\otimes _{S}T\rightarrow \cdots \rightarrow S_{2}\otimes _{S}T\rightarrow S_{1}\otimes _{S}T\rightarrow S_{0}\otimes _{S}T\rightarrow Y\otimes _{S}T\rightarrow 0.\end{eqnarray}$$

For any $B\in {\mathcal{B}}$, applying the left exact functor $\text{Hom}_{R}(B,-)$, we obtain a complex $\text{Hom}_{R}(B,\sharp \otimes _{S}T)$:

$$\begin{eqnarray}\displaystyle & & \displaystyle \cdots \rightarrow \text{Hom}_{R}(B,S_{n}\otimes _{S}T)\rightarrow \cdots \rightarrow \text{Hom}_{R}(B,S_{2}\otimes _{S}T)\rightarrow \text{Hom}_{R}(B,S_{1}\otimes _{S}T)\nonumber\\ \displaystyle & & \displaystyle \quad \rightarrow \text{Hom}_{R}(B,S_{0}\otimes _{S}T)\rightarrow \text{Hom}_{R}(B,Y\otimes _{S}T)\rightarrow 0.\nonumber\end{eqnarray}$$

Applying the right exact functor $D\text{Hom}_{S}(-,B\otimes _{R}DT)$ to the sequence (♯), we obtain a complex $D\text{Hom}_{S}(\sharp ,B\otimes _{R}DT)$:

$$\begin{eqnarray}\displaystyle & & \displaystyle \cdots \rightarrow D\text{Hom}_{S}(S_{n},B\otimes _{R}DT)\rightarrow \cdots \rightarrow D\text{Hom}_{S}(S_{2},B\otimes _{R}DT)\nonumber\\ \displaystyle & & \displaystyle \quad \rightarrow D\text{Hom}_{S}(S_{1},B\otimes _{R}DT)\rightarrow D\text{Hom}_{S}(S_{0},B\otimes _{R}DT)\rightarrow D\text{Hom}_{S}(Y,B\otimes _{R}DT)\rightarrow 0,\nonumber\end{eqnarray}$$

which is indeed exact since $S_{i},Y\in \text{KerExt}_{S}^{{>}0}(-,{\mathcal{B}}\otimes _{R}DT)$.

By the natural isomorphism (¶) in Step 1 again, we have isomorphisms of truncated complexes

$$\begin{eqnarray}(D\text{Hom}_{S}(\sharp ,B\otimes _{R}DT))^{{<}0}\simeq (\text{Hom}_{R}(B,\sharp \otimes _{S}T))^{{<}0},\end{eqnarray}$$

where $(-)^{{<}0}$ denotes the truncated complex of a complex by replacing the $i$th term with $0$ for all $i\geqslant 0$.

Since $B\in \text{KerExt}_{R}^{1}(-,S_{i}\otimes _{S}T)$ and $\text{Hom}_{R}(B,-)$ is left exact, we obtain that, for $i\geqslant 4$, $\text{Ext}_{R}^{1}(B,Y_{i}\otimes _{S}T)\simeq \text{H}^{-i+2}(\text{Hom}_{R}(B,\sharp \otimes _{S}T))$, where $Y_{i}=\text{Im}f_{i}$. But the latter homology is 0 by the above isomorphism of truncated complexes and by the fact that the complex $D\text{Hom}_{S}(\sharp ,B\otimes _{R}DT)$ is exact. This shows that $Y_{i}\otimes _{S}T\in \text{KerExt}_{R}^{1}({\mathcal{B}},-)={\mathcal{A}}$, for all $i\geqslant 4$.

Now consider the exact sequence obtained from (♯ ⊗_ST):

$$\begin{eqnarray}0\rightarrow Y_{4}\otimes _{S}T\rightarrow S_{3}\otimes _{S}T\rightarrow S_{2}\otimes _{S}T\rightarrow S_{1}\otimes _{S}T\rightarrow S_{0}\otimes _{S}T\rightarrow Y\otimes _{S}T\rightarrow 0.\end{eqnarray}$$

As also each $S_{i}\otimes _{S}T\in \text{add}_{\text{mod}R}T\subseteq {\mathcal{A}}$ and ${\mathcal{A}}$ is coresolving, we obtain that each $Y_{i}\otimes _{S}T\in {\mathcal{A}}$, where $Y_{i}=\text{Im}f_{i}$ for $0\leqslant i\leqslant 3$. Thus, the exact sequence $\sharp \otimes _{S}T$ is indeed in ${\mathcal{A}}$. Then we have an induced exact sequence $\text{Hom}_{R}(T,\sharp \otimes _{S}T)$:

$$\begin{eqnarray}\displaystyle & & \displaystyle \cdots \rightarrow \text{Hom}_{R}(T,S_{n}\otimes _{S}T)\rightarrow \cdots \rightarrow \text{Hom}_{R}(T,S_{1}\otimes _{S}T)\nonumber\\ \displaystyle & & \displaystyle \quad \rightarrow \text{Hom}_{R}(T,S_{0}\otimes _{S}T)\rightarrow \text{Hom}_{R}(T,Y\otimes _{S}T)\rightarrow 0.\nonumber\end{eqnarray}$$

Since $S_{i}\simeq \text{Hom}_{R}(T,S_{i}\otimes _{S}T)$ for each $i$, it follows that

$$\begin{eqnarray}Y\simeq \text{Hom}_{R}(T,Y\otimes _{S}T)\in \text{Hom}_{R}(T,{\mathcal{A}}).\end{eqnarray}$$

This shows that $\text{KerExt}_{S}^{1}(-,{\mathcal{B}}\otimes _{R}DT)\subseteq \text{Hom}_{R}(T,{\mathcal{A}})$. Together with Step 1, we obtain that $\text{KerExt}_{S}^{1}(-,{\mathcal{B}}\otimes _{R}DT)=\text{Hom}_{R}(T,{\mathcal{A}})$.

Step 3. $\text{KerExt}_{S}^{1}(\text{Hom}_{R}(T,{\mathcal{A}}),-)\subseteq {\mathcal{B}}\otimes _{R}DT$.

Note that there is an isomorphism

$$\begin{eqnarray}\text{Hom}_{S}(\text{Hom}_{R}(T,A),DS)\simeq D\text{Hom}_{R}(\text{Hom}_{S}(DT,DS),A),\end{eqnarray}$$

for any $A\in \text{mod}R$ and that it induces an isomorphism

$$\begin{eqnarray}\text{Hom}_{S}(\text{Hom}_{R}(T,A),I_{i})\simeq D\text{Hom}_{R}(\text{Hom}_{S}(DT,I_{i}),A),\end{eqnarray}$$

for any $I_{i}\in \text{add}_{\text{mod}S}DS$.

Now take any $X\in \text{KerExt}_{S}^{1}(\text{Hom}_{R}(T,{\mathcal{A}}),-)$ and consider an injective resolution of $X$:

(♮)$$\begin{eqnarray}0\rightarrow X\stackrel{g_{0}}{\longrightarrow }I_{0}\stackrel{g_{1}}{\longrightarrow }I_{1}\stackrel{g_{2}}{\longrightarrow }I_{2}\stackrel{g_{3}}{\longrightarrow }\cdots \stackrel{g_{n}}{\longrightarrow }I_{n}\stackrel{g_{n+1}}{\longrightarrow }\cdots \,;\end{eqnarray}$$

here, we consider (♮) as a (cochain) complex with the term $X$ at the zeroth position. Since $DT\simeq \text{Hom}_{R}(T,DR)\in \text{Hom}_{R}(T,{\mathcal{A}})$ and $\text{Hom}_{R}(T,{\mathcal{A}})$ is resolving, we obtain that

$$\begin{eqnarray}X\in \text{KerExt}_{S}^{1}(\text{Hom}_{R}(T,{\mathcal{A}}),-)=\text{KerExt}_{S}^{{>}0}(\text{Hom}_{R}(T,{\mathcal{A}}),-)\subseteq \text{KerExt}_{S}^{{>}0}(DT,-).\end{eqnarray}$$

Thus, for any $A\in {\mathcal{A}}$, applying the functor $\text{Hom}_{S}(\text{Hom}_{R}(T,A),-)$, we have an induced exact complex $\text{Hom}_{S}(\text{Hom}_{R}(T,A),\natural )$:

$$\begin{eqnarray}\displaystyle & & \displaystyle 0\rightarrow \text{Hom}_{S}(\text{Hom}_{R}(T,A),X)\longrightarrow \,\text{Hom}_{S}(\text{Hom}_{R}(T,A),I_{0})\nonumber\\ \displaystyle & & \displaystyle \quad \longrightarrow \text{Hom}_{S}(\text{Hom}_{R}(T,A),I_{1})\longrightarrow \cdots \longrightarrow \,\text{Hom}_{S}(\text{Hom}_{R}(T,A),I_{n})\longrightarrow \cdots \,.\nonumber\end{eqnarray}$$

On the other hand, by applying the functor $D\text{Hom}_{R}(\text{Hom}_{S}(DT,-),A)$, we also have the following induced complex $D\text{Hom}_{R}(\text{Hom}_{S}(DT,\natural ),A)$:

$$\begin{eqnarray}\displaystyle & & \displaystyle 0\rightarrow D\text{Hom}_{R}(\text{Hom}_{S}(DT,X),A)\longrightarrow D\text{Hom}_{R}(\text{Hom}_{S}(DT,I_{0}),A)\nonumber\\ \displaystyle & & \displaystyle \quad \longrightarrow D\text{Hom}_{R}(\text{Hom}_{S}(DT,I_{1}),A)\longrightarrow \cdots \longrightarrow \,D\text{Hom}_{R}(\text{Hom}_{S}(DT,I_{n}),A)\longrightarrow \cdots \,.\nonumber\end{eqnarray}$$

Since $\text{Hom}_{S}(DT,I_{i})\in \text{add}_{\text{mod}R}\text{Hom}_{S}(DT,DS)=\text{add}_{\text{mod}R}T$ and $A\in \text{KerExt}_{R}^{1}(T,-)$ and since $\text{Hom}_{S}(DT,\natural )$ is exact and $\text{Hom}_{R}(-,A)$ is left exact, we can obtain that, for any $i\geqslant 2$,

$$\begin{eqnarray}\text{Ext}_{R}^{1}(\text{Hom}_{S}(DT,X_{i+1}),A)\simeq \text{H}^{-i}(\text{Hom}_{R}(\text{Hom}_{S}(DT,\natural ),A)),\end{eqnarray}$$

where $X_{i}:=\text{Im}g_{i}$. However, by the above-mentioned isomorphism $(\ast )$ and the fact that $\text{Hom}_{S}(\text{Hom}_{R}(T,A),\natural )$ is exact, we further have that

$$\begin{eqnarray}\displaystyle & & \displaystyle \text{H}^{-i}(\text{Hom}_{R}(\text{Hom}_{S}(DT,\natural ),A))\simeq D\text{H}^{i}(D\text{Hom}_{R}(\text{Hom}_{S}(DT,\natural ),A))\nonumber\\ \displaystyle & & \displaystyle \quad \simeq D\text{H}^{i}(\text{Hom}_{S}(\text{Hom}_{R}(T,A),\natural ))=0.\nonumber\end{eqnarray}$$

It follows that $\text{Hom}_{S}(DT,X_{i+1})\in \text{KerExt}_{R}^{1}(-,{\mathcal{A}})={\mathcal{B}}$ for any $i\geqslant 2$. Since ${\mathcal{B}}$ is resolving and $\text{Hom}_{S}(DT,I_{i})\in \text{add}_{\text{mod}R}T\subseteq {\mathcal{B}}$, we also obtain that each $\text{Hom}_{S}(DT,X_{i})\in {\mathcal{B}}$ for each $0\leqslant i\leqslant 2$, where $X_{0}:=X$ and $X_{i}:=\text{Im}g_{i}$ for $i=1,2$, from the exact sequence

$$\begin{eqnarray}\displaystyle & & \displaystyle 0\rightarrow \text{Hom}_{S}(DT,X)\rightarrow \text{Hom}_{S}(DT,I_{0})\rightarrow \text{Hom}_{S}(DT,I_{2})\nonumber\\ \displaystyle & & \displaystyle \quad \rightarrow \text{Hom}_{S}(DT,I_{3})\rightarrow \text{Hom}_{S}(DT,X_{3})\rightarrow 0.\nonumber\end{eqnarray}$$

The above arguments show that the exact sequence $\text{Hom}_{S}(DT,\natural )$ is indeed in ${\mathcal{B}}$. Then we have an induced exact sequence $\text{Hom}_{S}(DT,\natural )\otimes _{R}DT$ as follows since ${\mathcal{B}}\subseteq \text{KerExt}_{R}^{1}(-,T)=\text{KerTor}_{1}^{R}(-,DT)$:

$$\begin{eqnarray}\displaystyle & & \displaystyle 0\rightarrow \text{Hom}_{S}(DT,X)\otimes _{R}DT\rightarrow \text{Hom}_{S}(DT,I_{0})\otimes _{R}DT\nonumber\\ \displaystyle & & \displaystyle \quad \rightarrow \text{Hom}_{S}(DT,I_{1})\otimes _{R}DT\rightarrow \cdots \rightarrow \text{Hom}_{S}(DT,I_{n})\otimes _{R}DT\rightarrow \cdots \,.\nonumber\end{eqnarray}$$

It follows that

$$\begin{eqnarray}X\simeq \text{Hom}_{S}(DT,X)\otimes _{R}DT\in {\mathcal{B}}\otimes _{R}DT\end{eqnarray}$$

since $I_{i}\simeq \text{Hom}_{S}(DT,I_{i})\otimes _{R}DT$ for each $i$. This shows that

$$\begin{eqnarray}\text{KerExt}_{S}^{1}(\text{Hom}_{R}(T,{\mathcal{A}}),-)\subseteq {\mathcal{B}}\otimes _{R}DT.\end{eqnarray}$$

Together with Step 1, we obtain that $\text{KerExt}_{S}^{1}(\text{Hom}_{R}(T,{\mathcal{A}}),-)={\mathcal{B}}\otimes _{R}DT$.

Altogether, we obtain that $(\text{Hom}_{R}(T,{\mathcal{A}}),{\mathcal{B}}\otimes _{R}DT)$ is a hereditary cotorsion pair.◻

3.2.6

4.1.1 The functor $\text{L}_{T}:{\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)\rightarrow {\mathcal{R}}{\mathcal{C}}(S)$

Let $X=\{X_{i}\}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)$. We define $\text{L}_{T}(X)\in {\mathcal{R}}{\mathcal{C}}(S)$ as follows:

1. (l1) the underlying module $\text{L}_{T}(X)_{i}=\text{Hom}_{R}(T,X_{i-1})\oplus X_{i}\otimes _{R}DT$ and

2. (l2) the structure map $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(\text{L}_{T}(X)):\text{L}_{T}(X)_{i}\otimes _{S}DS\rightarrow \text{L}_{T}(X)_{i-1}$ is given by $\big(\!\begin{smallmatrix}0 & \unicode[STIX]{x1D6FF}_{L_{i}}\\ 0 & 0\end{smallmatrix}\!\big)$, where $\unicode[STIX]{x1D6FF}_{L_{i}}$ is the composition:

   $$\begin{eqnarray}\text{Hom}_{R}(T,X_{i-1})\otimes _{S}DS\stackrel{\simeq }{\longrightarrow }\text{Hom}_{R}(T,X_{i-1})\otimes _{S}T\otimes _{R}DT\stackrel{\unicode[STIX]{x1D716}_{X_{i-1}}^{T}\otimes _{R}DT}{\longrightarrow }X_{i-1}\otimes _{R}DT.\end{eqnarray}$$

From the functor property of $\text{Hom}_{R}(T,-)$ and $-\!\otimes _{R}DT$, one can easily see that $\text{L}_{T}$ is a functor from ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)$ to ${\mathcal{R}}{\mathcal{C}}(S)$.

4.1.2 The functor $-\hat{\otimes }DT:{\mathcal{R}}{\mathcal{C}}(R)\rightarrow {\mathcal{R}}{\mathcal{C}}(S)$

Let $Y=\{Y_{i},\unicode[STIX]{x1D6FF}_{i}^{\otimes }(Y)\}\in {\mathcal{R}}{\mathcal{C}}(R)$. We define $Y\,\hat{\otimes }\,DT\in {\mathcal{R}}{\mathcal{C}}(S)$ by setting

1. (t1) the underlying module is $(Y\,\hat{\otimes }\,DT)_{i}=Y_{i}\otimes _{R}DT$ and

2. (t2) the structure map $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(Y\,\hat{\otimes }\,DT)$ is given by the composition

   $$\begin{eqnarray}\displaystyle & & \displaystyle Y_{i}\otimes _{R}DT\otimes _{S}DS\stackrel{\simeq }{\longrightarrow }Y_{i}\otimes _{R}DT\otimes _{S}T\otimes _{R}DT\nonumber\\ \displaystyle & & \displaystyle \quad \stackrel{\simeq }{\longrightarrow }Y_{i}\otimes _{R}DR\otimes _{R}DT\stackrel{\unicode[STIX]{x1D6FF}_{i}^{\otimes }(Y)\otimes _{R}DT}{\longrightarrow }Y_{i-1}\otimes _{R}DT.\nonumber\end{eqnarray}$$

From the functor property of $-\!\otimes _{R}DT$, one can see that $-\hat{\otimes }DT$ is a functor from ${\mathcal{R}}{\mathcal{C}}(R)$ to ${\mathcal{R}}{\mathcal{C}}(S)$.

4.1.3 The homomorphism $l_{Y}^{X}:\text{Hom}_{{\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)}(X,Y)\rightarrow \text{Hom}_{{\mathcal{R}}{\mathcal{C}}(S)}(X\,\hat{\otimes }\,DT,\text{L}_{T}(Y))$

Recall that we have a forgetful functor from ${\mathcal{R}}{\mathcal{C}}(R)$ to ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)$. For any $X\in {\mathcal{R}}{\mathcal{C}}(R)$ and $Y\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)$, there is a canonical homomorphism

$$\begin{eqnarray}l_{Y}^{X}:\text{Hom}_{{\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)}(X,Y)\longrightarrow \text{Hom}_{{\mathcal{R}}{\mathcal{C}}(S)}(X\,\hat{\otimes }\,DT,\text{L}_{T}(Y))\end{eqnarray}$$

which is functorial in both variables, defined by

$$\begin{eqnarray}l_{Y}^{X}:u=\{u_{i}\}\longmapsto f=\{f_{i}\},\quad \text{with}f_{i}=\left(\begin{array}{@{}cc@{}}-\unicode[STIX]{x1D703}_{l_{i}}, & u_{i}\otimes _{R}DT\end{array}\right),\end{eqnarray}$$

where $\unicode[STIX]{x1D703}_{l_{i}}$ is given by the composition

$$\begin{eqnarray}\displaystyle & & \displaystyle X_{i}\otimes _{R}DT\stackrel{\unicode[STIX]{x1D702}_{X_{i}\otimes _{R}DT}^{T}}{\longrightarrow }\text{Hom}_{R}(T,X_{i}\otimes _{R}DT\otimes _{S}T)\stackrel{\simeq }{\longrightarrow }\text{Hom}_{R}(T,X_{i}\otimes _{R}DR)\nonumber\\ \displaystyle & & \displaystyle \quad \stackrel{\text{Hom}_{R}(T,\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X))}{\longrightarrow }\text{Hom}_{R}(T,X_{i-1})\stackrel{\text{Hom}_{R}(T,u_{i-1})}{\longrightarrow }\text{Hom}_{R}(T,Y_{i-1}).\nonumber\end{eqnarray}$$

Remark.

Using the fact that $_{R}DR_{R}\simeq _{R}(DT\otimes _{S}T)_{R}$ and the adjoint isomorphism

$$\begin{eqnarray}\boldsymbol{\unicode[STIX]{x1D6E4}}^{T}:\text{Hom}_{R}(X_{i}\otimes _{R}DR,Y_{i-1})\simeq \text{Hom}_{S}(X_{i}\otimes _{R}DT,\text{Hom}_{R}(T,Y_{i-1})),\end{eqnarray}$$

one can easily check that $\unicode[STIX]{x1D703}_{l_{i}}$ is just the image of the natural homomorphism $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\cdot u_{i-1}$ under $\unicode[STIX]{x1D6E4}^{T}$, that is, $\unicode[STIX]{x1D703}_{l_{i}}=\boldsymbol{\unicode[STIX]{x1D6E4}}^{T}(\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\cdot u_{i-1})$.

In the following, we simply write $l$ instead of $l_{Y}^{X}$.

It is easy to see that, for any commutative diagram in ${\mathcal{R}}{\mathcal{C}}(R)$

there is an induced commutative diagram in ${\mathcal{R}}{\mathcal{C}}(S)$

4.1.4 A monomorphism $u_{_{X}}:X\rightarrow A_{X}$ in ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{R}})$ with $A_{X}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{A}})$, for $X\in {\mathcal{R}}{\mathcal{C}}(R)$

Let $X=\{X_{i}\}\in {\mathcal{R}}{\mathcal{C}}(R)$. Since $({\mathcal{B}},{\mathcal{A}})$ is a complete cotorsion pair in $\text{mod}R$, there are exact sequences $0\rightarrow X_{i}\stackrel{(u_{_{X}})_{i}}{\longrightarrow }(A_{X})_{i}\stackrel{(\unicode[STIX]{x1D70B}_{_{X}})_{i}}{\longrightarrow }(B_{X})_{i}\rightarrow 0$ with $(A_{X})_{i}\in {\mathcal{A}}$ and $(B_{X})_{i}\in {\mathcal{B}}$, for each $i$. This gives an exact sequence $0\rightarrow X\stackrel{u_{_{X}}}{\longrightarrow }A_{X}\stackrel{\unicode[STIX]{x1D70B}_{_{X}}}{\longrightarrow }B_{X}\rightarrow 0$ in ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)$ with $A_{X}=\{(A_{X})_{i}\}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{A}})$ and $B_{X}=\{(B_{X})_{i}\}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{B}})$.

Now let $Y=\{Y_{i}\}\in {\mathcal{R}}{\mathcal{C}}(R)$ and $h=\{h_{i}\}\in \text{Hom}_{{\mathcal{R}}{\mathcal{C}}(R)}(X,Y)$. Then we have an exact sequence $0\rightarrow Y\stackrel{u_{_{Y}}}{\longrightarrow }A_{Y}\stackrel{\unicode[STIX]{x1D70B}_{_{Y}}}{\longrightarrow }B_{Y}\rightarrow 0$ in ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)$ with $A_{Y}=\{(A_{Y})_{i}\}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{A}})$ and $B_{Y}=\{(B_{Y})_{i}\}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{B}})$, as above. Using that ${\mathcal{B}}=\text{KerExt}_{R}^{1}(-,{\mathcal{A}})$, it is easy to see that there is a homomorphism $h_{A}\in \text{Hom}_{{\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)}(A_{X},A_{Y})$ and further $h_{B}\in \text{Hom}_{{\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)}(B_{X},B_{Y})$ such that the following diagram in ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)$ is commutative with exact rows.

4.1.5 The cokernel $\text{Cok}(l(u_{_{X}}))$

Applying the functor $-\!\otimes _{R}DT$ to the exact sequences

$$\begin{eqnarray}\displaystyle 0\rightarrow X_{i}\stackrel{(u_{_{X}})_{i}}{\longrightarrow }(A_{X})_{i}\stackrel{(\unicode[STIX]{x1D70B}_{_{X}})_{i}}{\longrightarrow }(B_{X})_{i}\rightarrow 0 & & \displaystyle \nonumber\end{eqnarray}$$

above, we obtain induced exact sequences

since $(B_{X})_{i}\in {\mathcal{B}}\subseteq \text{KerExt}_{R}^{1}(-,T)=\text{KerTor}_{1}^{R}(-,DT)$ for each $i$. It follows that, by applying the homomorphism $l$ in 4.1.3 to the homomorphism $u_{_{X}}$ in 4.1.4, there is an induced exact sequence

Remark.

From the definition of $\text{Cok}(l(u_{_{X}}))$, one sees that, for each $i$, $\text{Cok}(l(u_{_{X}}))_{i}$ is given by the pushout

Moreover, for $Y\in {\mathcal{R}}{\mathcal{C}}(R)$ and $h\in \text{Hom}_{{\mathcal{R}}{\mathcal{C}}(R)}(X,Y)$, by applying the homomorphism $l$ to the left square in the commutation diagram in 4.1.4, we obtain the following commutative diagram in ${\mathcal{R}}{\mathcal{C}}(S)$, for some $h_{\text{Cok}}$:

4.1.6 The assignment $\mathbf{S}_{T}:{\mathcal{R}}{\mathcal{C}}(R)\rightarrow \text{}\underline{{\mathcal{R}}{\mathcal{C}}}(S)$ given by $X\longmapsto \text{Cok}(l(u_{_{X}}))$ is a functor

By 4.1.5, it is sufficient to prove that $\mathbf{S}_{T}(h):=h_{\text{Cok}}=0$ in $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(S)$ provided $h=0$. We divide the proof into two steps.

Step 1: Consider each piece in the commutative diagram in 4.1.4. If $h=\{h_{i}\}=0$, then $h_{i}=0$ for each $i$. Thus, we have that $(u_{_{X}})_{i}(h_{A})_{i}=0$ and, consequently, $(h_{A})_{i}=(\unicode[STIX]{x1D70B}_{_{X}})_{i}g_{i}$ for some $g_{i}:(B_{X})_{i}\rightarrow (A_{Y})_{i}$. Since $(B_{X})_{i}\in {\mathcal{B}}\subseteq {\mathcal{X}}_{T}$ for each $i$ and $T$ is an Ext-injective cogenerator in ${\mathcal{B}}$ (see the first part in the proof of Proposition 3.2.5), there are exact sequences $0\rightarrow (B_{X})_{i}\stackrel{b_{i}}{\longrightarrow }T_{(B_{X})_{i}}\rightarrow (B_{X}^{\prime })_{i}\rightarrow 0$ with $T_{(B_{X})_{i}}\in \text{add}_{\text{mod}R}T$ and $(B_{X}^{\prime })_{i}\in {\mathcal{B}}\subseteq \text{KerExt}_{R}^{1}(-,{\mathcal{A}})$. It follows that there exists $t_{i}\in \text{Hom}_{R}(T_{(B_{X})_{i}},(A_{Y})_{i})$ such that $g_{i}=b_{i}t_{i}$. Altogether, we obtain the following commutative diagram:

This induces the following commutative diagram in ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)$, where $T_{B_{X}}=\{T_{(B_{X})_{i}}\}$:

Set $k:=\unicode[STIX]{x1D70B}_{_{X}}b$. Then $\text{L}_{T}(h_{A})=\text{L}_{T}(kt)=\text{L}_{T}(k)\text{L}_{T}(t)$.

Step 2: Consider the commutative diagram in 4.1.5. Since $(u_{_{X}})_{i}(\unicode[STIX]{x1D70B}_{_{X}})_{i}=0$, it holds that $\text{Hom}_{R}(T,(u_{_{X}})_{i})\cdot \text{Hom}_{R}(T,(\unicode[STIX]{x1D70B}_{_{X}})_{i})=0$ and $(u_{_{X}})_{i}\otimes _{R}DT\cdot (\unicode[STIX]{x1D70B}_{_{X}})_{i}\otimes _{R}DT=0$. Then we see that $l(u_{_{X}})\text{L}_{T}(k)=l(u_{_{X}})\text{L}_{T}(\unicode[STIX]{x1D70B}_{_{X}})\text{L}_{T}(b)=0\cdot \text{L}_{T}(b)=0$ by the definition of the functor $\text{L}_{T}$ in 4.1.1 and the morphism $l(u_{_{X}})$ in 4.1.3. Hence, there is some $\unicode[STIX]{x1D703}\in \text{Hom}_{{\mathcal{R}}{\mathcal{C}}(S)}(\text{Cok}(l(u_{_{X}})),\text{L}_{T}(T_{B_{X}}))$ such that $\text{L}_{T}(k)=\unicode[STIX]{x1D70B}_{_{l_{X}}}\unicode[STIX]{x1D703}$. Consequently, we have that $\text{L}_{T}(h_{A})=\text{L}_{T}(k)L(t)=\unicode[STIX]{x1D70B}_{_{l_{X}}}\unicode[STIX]{x1D703}\text{L}_{T}(t)$. Now we obtain that $\unicode[STIX]{x1D70B}_{_{l_{X}}}h_{\text{Cok}}=\text{L}_{T}(h_{A})\unicode[STIX]{x1D70B}_{_{l_{Y}}}=\unicode[STIX]{x1D70B}_{_{l_{X}}}\unicode[STIX]{x1D703}\text{L}_{T}(t)\unicode[STIX]{x1D70B}_{_{l_{Y}}}$. Since $\unicode[STIX]{x1D70B}_{l_{X}}$ is epic, we get that $h_{\text{Cok}}=\unicode[STIX]{x1D703}\text{L}_{T}(t)\unicode[STIX]{x1D70B}_{_{l_{Y}}}$. That is, we have the following commutative diagram:

Note that $\text{L}_{T}(T_{B_{X}})$ is a projective–injective object in ${\mathcal{R}}{\mathcal{C}}(S)$, so $h_{\text{Cok}}=0$ in $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(S)$.

4.1.7 The functor $\mathbf{S}_{T}:\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)\rightarrow \text{}\underline{{\mathcal{R}}{\mathcal{C}}}(S)$

We will show that the functor $\mathbf{S}_{T}$ factors through $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)$.

To see this, it is enough to show that $\mathbf{S}_{T}(X)$ is a projective object in ${\mathcal{R}}{\mathcal{C}}(S)$ whenever $X$ is a projective object in ${\mathcal{R}}{\mathcal{C}}(R)$.

Without loss of generality, we assume that $X=\{X_{i}\}$ is an indecomposable projective object in ${\mathcal{R}}{\mathcal{C}}(R)$. Thus, $X$ is of the form

where $I$ is indecomposable injective and is on the ($k-1$)th position, for some $k$ [Reference Happel18, 2.2 Lemma].

Note that $X_{k}=\text{Hom}_{R}(DR,I)\in \text{add}_{\text{mod}R}R\subseteq {\mathcal{B}}$ and $X_{k-1}=I\in \text{add}_{\text{mod}R}DR\subseteq {\mathcal{A}}$, so, following 4.1.4, we can choose $A_{X}$ to be of the form

where $A_{X_{k}}=T_{k}\in \text{add}_{\text{mod}R}T$. And we have that the homomorphism $u_{_{X}}:X\rightarrow A_{X}$ is of the form

$$\begin{eqnarray}\displaystyle \begin{array}{@{}ccccccccccccccc@{}}X: & & \cdots \, & {\rightsquigarrow} & 0 & {\rightsquigarrow} & 0 & {\rightsquigarrow} & \text{Hom}_{R}(DR,I) & {\rightsquigarrow} & I & {\rightsquigarrow} & 0 & {\rightsquigarrow} & \cdots \\ \downarrow u_{_{X}} & & & & & & & & \downarrow u_{k} & & \downarrow 1 & & & & \\ A_{X}: & & \cdots \, & {\rightsquigarrow} & 0 & {\rightsquigarrow} & 0 & {\rightsquigarrow} & T_{k} & {\rightsquigarrow} & I & {\rightsquigarrow} & 0 & {\rightsquigarrow} & \cdots \end{array} & & \displaystyle \nonumber\end{eqnarray}$$

Then, from the structure of $l(u_{_{X}})$, we can see that $l(u_{_{X}})$ is of the form

$$\begin{eqnarray}\displaystyle \begin{array}{@{}ccccccccccc@{}}X\,\hat{\otimes }\,DT: & & 0 & {\rightsquigarrow} & 0 & {\rightsquigarrow} & \text{Hom}_{R}(DR,I)\otimes _{R}DT & {\rightsquigarrow} & I\otimes _{R}DT & {\rightsquigarrow} & 0\\ \downarrow l(u_{_{X}}) & & & & \downarrow & & \downarrow (-\unicode[STIX]{x1D703}_{l_{k}},u_{k}\otimes _{R}DT) & & \downarrow 1 & & \\ \text{L}_{T}(A_{X}): & & 0 & {\rightsquigarrow} & \text{Hom}_{R}(T,T_{k}) & \stackrel{(0,\unicode[STIX]{x1D6FF}_{L_{k+1}})}{{\rightsquigarrow}} & \text{Hom}_{R}(T,I)\oplus T_{k}\otimes _{R}DT & {\rightsquigarrow} & I\otimes _{R}DT & {\rightsquigarrow} & 0,\end{array} & & \displaystyle \nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D703}_{l_{k}}$ is defined as in 4.1.3 and $\unicode[STIX]{x1D6FF}_{L_{k+1}}$ is defined as in 4.1.1, respectively. One checks that both homomorphisms $\unicode[STIX]{x1D703}_{l_{k}}$ and $\unicode[STIX]{x1D6FF}_{L_{k+1}}$ are, in fact, isomorphisms. So we obtain that $S_{T}(X)=\text{Coker}(l(u_{_{X}}))$ is of the form

$$\begin{eqnarray}\displaystyle \begin{array}{@{}ccccccccccccccc@{}} & & \cdots \, & {\rightsquigarrow} & 0 & {\rightsquigarrow} & \text{Hom}_{R}(T,T_{k}) & \stackrel{\unicode[STIX]{x1D6FF}_{k+1}^{\prime }}{{\rightsquigarrow}} & T_{k}\otimes _{R}DT & {\rightsquigarrow} & 0 & {\rightsquigarrow} & 0 & {\rightsquigarrow} & \cdots \,,\end{array} & & \displaystyle \nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D6FF}_{k+1}^{\prime }$ is the induced isomorphism: $\text{Hom}_{R}(T,T_{k})\otimes _{S}DS\rightarrow T_{k}\otimes _{R}DT$. Since $T_{k}\otimes _{R}DT\in \text{add}_{\text{mod}S}(T\otimes _{R}DT)=\text{add}_{\text{mod}S}DS$, we see that $T_{k}\otimes _{R}DT$ is an injective $S$-module and that $\mathbf{S}_{T}(X)$ is a projective object in ${\mathcal{R}}{\mathcal{C}}(S)$.

It follows that the functor $\mathbf{S}_{T}$ factors through $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)$. We still denote by $\mathbf{S}_{T}$ the induced functor from $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)$ to $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(S)$.

4.2.1 The functor $\text{R}_{DT}:{\mathcal{R}}{\mathcal{C}}^{\text{tr}}(S)\rightarrow {\mathcal{R}}{\mathcal{C}}(R)$

Dually to 4.1.1, for any $X=\{X_{i}\}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}(S)$, we define $\text{R}_{DT}(X)\in {\mathcal{R}}{\mathcal{C}}(R)$ as follows:

1. (r1) the underlying module is $\text{R}_{DT}(X)_{i}=\text{Hom}_{S}(DT,X_{i})\oplus X_{i+1}\otimes _{S}T$ and

2. (r2) the structure map $\unicode[STIX]{x1D6FF}_{i}^{\text{H}}(\text{R}_{DT}(X))$ is given by $\big(\begin{smallmatrix}0 & \unicode[STIX]{x1D6FF}_{R_{i}}^{\text{H}}0 & 0\end{smallmatrix}\big)$, where $\unicode[STIX]{x1D6FF}_{R_{i}}^{\text{H}}$ is the composition

   $$\begin{eqnarray}\displaystyle & & \displaystyle \text{Hom}_{S}(DT,X_{i})\stackrel{\text{Hom}_{S}(DT,\unicode[STIX]{x1D702}_{X_{i}}^{T})}{\longrightarrow }\text{Hom}_{S}(DT,\text{Hom}_{R}(T,X_{i}\otimes _{S}T))\nonumber\\ \displaystyle & & \displaystyle \quad \simeq \text{Hom}_{R}(DT\otimes _{S}T,X_{i}\otimes _{S}T)\simeq \text{Hom}_{R}(DR,X_{i}\otimes _{S}T).\nonumber\end{eqnarray}$$

Equivalently, the structure map $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(\text{R}_{DT}(X))$ is given by $\big(\!\begin{smallmatrix}0 & \unicode[STIX]{x1D6FF}_{R_{i}}^{\otimes }0 & 0\end{smallmatrix}\!\big)$, where $\unicode[STIX]{x1D6FF}_{R_{i}}^{\otimes }$ is the composition

$$\begin{eqnarray}\text{Hom}_{S}(DT,X_{i})\otimes _{R}DR\simeq \text{Hom}_{S}(DT,X_{i})\otimes _{R}DT\otimes _{S}T\stackrel{\unicode[STIX]{x1D716}_{X_{i}}^{DT}}{\longrightarrow }X_{i}\otimes _{S}T.\end{eqnarray}$$

It is easy to see that $\text{R}_{DT}$ is a functor.

4.2.2 The functor $\text{&#x0124;om}(DT,-):{\mathcal{R}}{\mathcal{C}}(S)\rightarrow {\mathcal{R}}{\mathcal{C}}(R)$

Let $Y=\{Y_{i},\unicode[STIX]{x1D6FF}_{i}^{\text{H}}(Y)\}\in {\mathcal{R}}{\mathcal{C}}(S)$. We define $\text{&#x0124;om}(DT,Y)\in {\mathcal{R}}{\mathcal{C}}(R)$ by setting

1. (h1) the underlying module is $\text{&#x0124;om}(DT,Y)_{i}=\text{Hom}_{S}(DT,Y_{i})$ and

2. (h2) the structure map $\unicode[STIX]{x1D6FF}_{i}^{\text{H}}(\text{&#x0124;om}(DT,Y))$ is given by the composition

   $$\begin{eqnarray}\displaystyle & & \displaystyle \text{Hom}_{S}(DT,Y_{i})\stackrel{\text{Hom}_{S}(DT,\unicode[STIX]{x1D6FF}_{i}^{\text{H}}(Y))}{\longrightarrow }\text{Hom}_{S}(DT,\text{Hom}_{S}(DS,Y_{i-1}))\nonumber\\ \displaystyle & & \displaystyle \quad \simeq \text{Hom}_{R}(DR,\text{Hom}_{S}(DT,Y_{i-1})).\nonumber\end{eqnarray}$$

Then from the functor property of $\text{Hom}_{S}(DT,-)$, one can see that $\text{&#x0124;om}(DT,-)$ is a functor from ${\mathcal{R}}{\mathcal{C}}(S)$ to ${\mathcal{R}}{\mathcal{C}}(R)$.

4.2.3 The homomorphism $r_{Y}^{X}$

Dually to the homomorphism $l_{Y}^{X}$, for any $X\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}(S)$ and $Y\in {\mathcal{R}}{\mathcal{C}}(S)$, we have a canonical homomorphism

$$\begin{eqnarray}\displaystyle r_{Y}^{X}:\text{Hom}_{{\mathcal{R}}{\mathcal{C}}^{\text{tr}}(S)}(X,Y)\rightarrow \text{Hom}_{{\mathcal{R}}{\mathcal{C}}(R)}(\text{R}_{DT}(X),\text{&#x0124;om}(DT,Y)), & & \displaystyle \nonumber\end{eqnarray}$$

which is functorial in both variables, defined by

$$\begin{eqnarray}\displaystyle r_{Y}^{X}:u=\{u_{i}\}\longmapsto f=\{f_{i}\}\quad \text{with }f_{i}=\left(\begin{array}{@{}c@{}}\text{Hom}_{S}(DT,u_{i})\\ -\unicode[STIX]{x1D701}_{r_{i}}\end{array}\right), & & \displaystyle \nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D701}_{r_{i}}:X_{i+1}\otimes _{S}T\rightarrow \text{Hom}_{S}(DT,Y_{i})$ equals to

$$\begin{eqnarray}\displaystyle (u_{i+1}\otimes _{S}T)\cdot (\unicode[STIX]{x1D6FF}_{i+1}^{\text{H}}(Y)\otimes _{S}T)\cdot \unicode[STIX]{x1D716}_{\text{Hom}_{S}(DT,Y_{i})}^{T}, & & \displaystyle \nonumber\end{eqnarray}$$

that is, the composition

$$\begin{eqnarray}\displaystyle & & \displaystyle X_{i+1}\otimes _{S}T\stackrel{u_{i+1}\otimes _{S}T}{\longrightarrow }Y_{i+1}\otimes _{S}T\stackrel{\unicode[STIX]{x1D6FF}_{i+1}^{H}(Y)\otimes _{S}T}{\longrightarrow }\text{Hom}_{S}(DS,Y_{i})\otimes _{S}T\nonumber\\ \displaystyle & & \displaystyle \quad \simeq \text{Hom}_{S}(T\otimes _{R}DT,Y_{i})\otimes _{S}T\nonumber\\ \displaystyle & & \displaystyle \quad \simeq \text{Hom}_{R}(T,\text{Hom}_{S}(DT,Y_{i}))\otimes _{S}T\nonumber\\ \displaystyle & & \displaystyle \quad \stackrel{\unicode[STIX]{x1D716}_{\text{Hom}_{S}(DT,Y_{i})}^{T}}{\longrightarrow }\text{Hom}_{S}(DT,Y_{i}).\nonumber\end{eqnarray}$$

Remark.

Using the fact that $_{S}DS_{S}\simeq _{S}T\otimes _{R}DT_{S}$ and the adjoint isomorphism

$$\begin{eqnarray}\displaystyle \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}:\text{Hom}_{S}(X_{i+1}\otimes _{S}DS,Y_{i})\simeq \text{Hom}_{S}(X_{i+1}\otimes _{S}T,\text{Hom}_{S}(DT,Y_{i})), & & \displaystyle \nonumber\end{eqnarray}$$

one can easily check that $\unicode[STIX]{x1D701}_{r_{i}}$ is the image of the natural homomorphism $(u_{i+1}\otimes _{S}DS)\cdot \unicode[STIX]{x1D6FF}_{i+1}^{\otimes }(Y)$ under $\unicode[STIX]{x1D6E4}^{DT}$, that is, $\unicode[STIX]{x1D701}_{r_{i}}=\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((u_{i+1}\otimes _{S}DS)\cdot \unicode[STIX]{x1D6FF}_{i+1}^{\otimes }(Y))$.

In the following, we simply write $r$ instead of $r_{Y}^{X}$.

4.2.4 An epimorphism $v_{_{Y}}:G_{Y}\rightarrow Y$ in ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{S}})$ with $G_{Y}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{G}})$, for $Y\in {\mathcal{R}}{\mathcal{C}}(S)$

Since $({\mathcal{G}},{\mathcal{K}})$ is a complete hereditary cotorsion pair in $\text{mod}S$, it follows that, for any $Y=\{Y_{i}\}\in {\mathcal{R}}{\mathcal{C}}(S)$, there is an exact sequence $0\rightarrow K_{Y}\stackrel{k_{_{Y}}}{\longrightarrow }G_{Y}\stackrel{v_{_{Y}}}{\longrightarrow }Y\rightarrow 0$ in ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{S}})$ with $K_{Y}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{K}})$ and $G_{Y}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{G}})$.

Moreover, for any $h\in \text{Hom}_{{\mathcal{R}}{\mathcal{C}}(S)}(X,Y)$, there is an induced commutative diagram as follows since $\text{Ext}_{S}^{1}({\mathcal{G}},{\mathcal{K}})=0$:

4.2.5 The kernel $\text{Ker}(r(v_{_{Y}}))$

Applying the functor $\text{&#x0124;om}(DT,-)$ to the bottom exact sequence in the above diagram, we obtain an induced exact sequence

$$\begin{eqnarray}\displaystyle 0\rightarrow \text{&#x0124;om}(DT,K_{Y})\rightarrow \text{&#x0124;om}(DT,G_{Y})\rightarrow \text{&#x0124;om}(DT,Y)\rightarrow 0 & & \displaystyle \nonumber\end{eqnarray}$$

since $(K_{Y})_{i}\in {\mathcal{K}}\subseteq \text{KerExt}_{S}^{1}(DT,-)$ for each $i$. Thus, after applying the homomorphism $r$ in 4.2.3 to the homomorphism $v_{_{Y}}$ in 4.2.4, we obtain the following exact sequence in ${\mathcal{R}}{\mathcal{C}}(R)$:

$$\begin{eqnarray}\displaystyle 0\rightarrow \text{Ker}(r(v_{_{Y}}))\stackrel{\unicode[STIX]{x1D706}_{r_{Y}}}{\longrightarrow }\text{R}_{DT}(G_{Y})\stackrel{r(v_{_{Y}})}{\longrightarrow }\text{&#x0124;om}(DT,Y)\rightarrow 0. & & \displaystyle \nonumber\end{eqnarray}$$

Moreover, for any $h\in \text{Hom}_{{\mathcal{R}}{\mathcal{C}}(S)}(X,Y)$, by applying the homomorphism $r$ to the right part of the commutative diagram in 4.2.4, we obtain the following commutative diagram in ${\mathcal{R}}{\mathcal{C}}(R)$, for some $h_{\text{Ker}}$:

4.2.6 The assignment $\mathbf{Q}_{DT}:{\mathcal{R}}{\mathcal{C}}(S)\rightarrow \text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)$ given by $Y\longmapsto \text{Ker}(r(v_{_{Y}}))$ is a functor

By 4.2.5, it is sufficient to prove that $\mathbf{Q}_{DT}(h):=h_{\text{Ker}}=0$ in $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)$ provided $h=0$. This is also divided into two steps.

Step 1: Consider each piece in the commutative diagram in 4.2.4. If $h=\{h_{i}\}=0$, then $h_{i}=0$ for each $i$. Thus, we have that $(h_{G})_{i}(v_{_{Y}})_{i}=0$ and, consequently, $(h_{G})_{i}=g_{i}(k_{Y})_{i}$ for some $g_{i}:(G_{X})_{i}\rightarrow (K_{Y})_{i}$. Since $(K_{Y})_{i}\in {\mathcal{K}}\subseteq _{DT}{\mathcal{X}}$ for all $i$, there are exact sequences $0\rightarrow (K_{Y}^{\prime })_{i}\rightarrow DT_{(K_{Y})_{i}}\stackrel{b_{i}}{\longrightarrow }(K_{Y})_{i}\rightarrow 0$ with $DT_{(K_{Y})_{i}}\in \text{add}_{\text{mod}S}DT$ and $(K_{Y}^{\prime })_{i}\in {\mathcal{K}}\subseteq \text{KerExt}_{S}^{1}({\mathcal{G}},-)$. It follows that there exists $t_{i}\in \text{Hom}_{R}((G_{X})_{i},DT_{(K_{Y})_{i}})$ such that $g_{i}=t_{i}b_{i}$. Altogether, we obtain the following commutative diagram:

It follows that there is a commutative diagram in ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}(S)$,

where $DT_{K_{Y}}:=\{DT_{(K_{Y})_{i}}\}$.

Set $\unicode[STIX]{x1D6FD}:=bk_{Y}$. Then $\text{R}_{DT}(h_{G})=\text{R}_{DT}(t\unicode[STIX]{x1D6FD})=\text{R}_{DT}(t)\text{R}_{DT}(\unicode[STIX]{x1D6FD})$.

Step 2: Consider the commutative diagram in 4.2.5. Since $(k_{Y})_{i}(v_{_{Y}})_{i}=0$ in 4.2.4, we see that $\text{R}_{DT}(\unicode[STIX]{x1D6FD})r(v_{_{Y}})=0$ by the definitions. Hence, there is some $\unicode[STIX]{x1D703}\in \text{Hom}_{{\mathcal{R}}{\mathcal{C}}(R)}(\text{R}_{DT}(DT_{K_{Y}}),\text{Ker}(r(v_{_{Y}})))$ such that $\text{R}_{DT}(\unicode[STIX]{x1D6FD})=\unicode[STIX]{x1D703}\unicode[STIX]{x1D706}_{r_{Y}}$. Consequently, we have that $\text{R}_{DT}(h_{G})=\text{R}_{DT}(t)\text{R}_{DT}(\unicode[STIX]{x1D6FD})=\text{R}_{DT}(t)\unicode[STIX]{x1D703}\unicode[STIX]{x1D706}_{r_{Y}}$. Now we obtain that $h_{\text{Ker}}\unicode[STIX]{x1D706}_{r_{Y}}=\unicode[STIX]{x1D706}_{r_{X}}\text{R}_{DT}(h_{G})=\unicode[STIX]{x1D706}_{r_{X}}\text{R}_{DT}(t)\unicode[STIX]{x1D703}\unicode[STIX]{x1D706}_{r_{Y}}$. Since $\unicode[STIX]{x1D706}_{r_{Y}}$ is monomorphic, we get that $h_{\text{Ker}}=\unicode[STIX]{x1D706}_{r_{X}}\text{R}_{DT}(t)\unicode[STIX]{x1D703}$, that is, the following diagram is commutative:

Note that $\text{R}_{DT}(DT_{K_{Y}})$ is a projective–injective object in ${\mathcal{R}}{\mathcal{C}}(R)$, so $h_{\text{Ker}}=0$ in $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)$.

4.2.7 The functor $\mathbf{Q}_{DT}:\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(S)\rightarrow \text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)$

We will show that the functor $\mathbf{Q}_{DT}$ factors through $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(S)$.

To see this, it is enough to show that $\mathbf{Q}_{DT}(X)$ is a projective object in ${\mathcal{R}}{\mathcal{C}}(R)$, whenever $X$ is a projective object in ${\mathcal{R}}{\mathcal{C}}(S)$.

Without loss of generality, we assume that $X=\{X_{i}\}$ is an indecomposable projective object in ${\mathcal{R}}{\mathcal{C}}(S)$. Thus, we have that $X$ has the form

$$\begin{eqnarray}\displaystyle \begin{array}{@{}ccccccccccc@{}}\cdots \, & {\rightsquigarrow} & 0 & {\rightsquigarrow} & P & \stackrel{1}{{\rightsquigarrow}} & P\otimes _{S}DS & {\rightsquigarrow} & 0 & {\rightsquigarrow} & \cdots \,,\end{array} & & \displaystyle \nonumber\end{eqnarray}$$

where $P$ is indecomposable projective and is on the ($k$+1)th position, for some $k$; see [Reference Happel18].

Note that $X_{k+1}=P\in \text{add}_{\text{mod}S}S\subseteq {\mathcal{G}}$ and that $X_{k}=P\otimes _{S}DS\in \text{add}_{\text{mod}S}DS$$\subseteq {\mathcal{K}}$, so, following 4.2.4, we can choose $G_{X}$ to be of the form

$$\begin{eqnarray}\displaystyle \begin{array}{@{}ccccccccccc@{}}\cdots \, & {\rightsquigarrow} & 0 & {\rightsquigarrow} & P & {\rightsquigarrow} & DT_{k} & {\rightsquigarrow} & 0 & {\rightsquigarrow} & \cdots \,,\end{array} & & \displaystyle \nonumber\end{eqnarray}$$

where $(G_{X})_{k}=DT_{k}\in \text{add}_{\text{mod}S}DT$. And we have an epimorphism $v_{_{X}}:G_{X}\rightarrow X$ in ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{S}})$ which is of the form

$$\begin{eqnarray}\displaystyle \begin{array}{@{}ccccccccccccccc@{}}G_{X}: & & \cdots \, & {\rightsquigarrow} & 0 & {\rightsquigarrow} & 0 & {\rightsquigarrow} & P & {\rightsquigarrow} & DT_{k} & {\rightsquigarrow} & 0 & {\rightsquigarrow} & \cdots \\ \downarrow v_{_{X}} & & & & & & & & \downarrow 1 & & \downarrow v_{k} & & & & \\ X: & & \cdots \, & {\rightsquigarrow} & 0 & {\rightsquigarrow} & 0 & {\rightsquigarrow} & P & {\rightsquigarrow} & P\otimes _{S}DS & {\rightsquigarrow} & 0 & {\rightsquigarrow} & \cdots \end{array} & & \displaystyle \nonumber\end{eqnarray}$$

Then, from the structure of $r(v_{_{X}})$ in 4.2.3, we can see that $r(v_{_{X}})$ is of the form

where $\unicode[STIX]{x1D701}_{r_{k}}$ is defined as in 4.2.3 and $\unicode[STIX]{x1D6FF}_{R_{k}}$ is defined as in 4.2.1. One checks that both homomorphisms $\unicode[STIX]{x1D701}_{r_{k}}$ and $\unicode[STIX]{x1D6FF}_{R_{k}}$ are, in fact, isomorphisms. So we obtain that $\mathbf{Q}_{DT}(X)=\text{Ker}(r(v_{_{X}}))$ is of the form

$$\begin{eqnarray}\displaystyle \begin{array}{@{}ccccccccccccccc@{}} & & \cdots \, & {\rightsquigarrow} & 0 & {\rightsquigarrow} & 0 & {\rightsquigarrow} & \text{Hom}_{S}(DT,DT_{k}) & \stackrel{\unicode[STIX]{x1D6FF}_{k}^{\prime }}{{\rightsquigarrow}} & DT_{k}\otimes _{S}T & {\rightsquigarrow} & 0 & {\rightsquigarrow} & \cdots \,,\end{array} & & \displaystyle \nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D6FF}_{k}^{\prime }$ is an induced isomorphism: $\text{Hom}_{S}(DT,DT_{k})\rightarrow \text{Hom}_{R}(DR,DT_{k}\otimes _{S}T)$. Since $\text{Hom}_{S}(DT,DT_{k})\in \text{add}_{\text{mod}R}(\text{Hom}_{S}(DT,DT))=\text{add}_{\text{mod}R}R$, we see that $\text{Hom}_{S}(DT,DT_{k})$ is projective and that $\mathbf{Q}_{DT}(X)$ is a projective object in ${\mathcal{R}}{\mathcal{C}}(R)$.

It follows that the functor $\mathbf{Q}_{DT}$ factors through $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(S)$. The induced functor from $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(S)$ to $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)$ is still denoted by $\mathbf{Q}_{DT}$.

4.3.1 Computing the composition $\mathbf{Q}_{DT}\mathbf{S}_{T}$

Take any $X=\{X_{i},\unicode[STIX]{x1D6FF}_{i}^{\otimes }\}\in {\mathcal{R}}{\mathcal{C}}(R)$. From the chosen exact sequence $0\rightarrow X\stackrel{u_{_{X}}}{\longrightarrow }A_{X}\stackrel{\unicode[STIX]{x1D70B}_{_{_{X}}}}{\longrightarrow }B_{X}\rightarrow 0$ in ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)$ with $A_{X}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{A}})$ and $B_{X}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{B}})$, as in 4.1.4, we obtain an exact sequence

$$\begin{eqnarray}\displaystyle 0\rightarrow X\,\hat{\otimes }\,DT\stackrel{l(u_{_{X}})}{\longrightarrow }\text{L}_{T}(A_{X})\stackrel{s}{\longrightarrow }\mathbf{S}_{T}(X)\rightarrow 0 & & \displaystyle \nonumber\end{eqnarray}$$

by the construction of the functor $\mathbf{S}_{T}$ in 4.1.5. Note that, for each $i$, $\mathbf{S}_{T}(X)_{i}$ is given by the pushout diagram

Now we take a projective $R$-module $P_{(A_{X})_{i}}$ such that $P_{(A_{X})_{i}}\stackrel{p_{i}}{\longrightarrow }(A_{X})_{i}\rightarrow 0$ is exact. Then we have a pullback diagram

Since ${\mathcal{B}}$ is closed under kernels of epimorphisms, we see that $\overline{X}_{i}\in {\mathcal{B}}$. By applying the functor $-\!\otimes _{R}DT$, the diagram above induces the following commutative diagram with exact rows since ${\mathcal{B}}\subseteq \text{KerTor}_{1}^{R}(-,DT)$:

Now, one can check that the following diagram is commutative with exact rows, for each $i$, where the lower row is obtained from the first pushout diagram in this section. Here, $t_{i}^{P}=(-(q_{i}\otimes _{R}DT)\cdot \unicode[STIX]{x1D703}_{l_{i}},(\overline{u}_{_{X}})_{i}\otimes _{R}DT)$, $t_{i}=(-\unicode[STIX]{x1D703}_{l_{i}},(u_{_{X}})_{i}\otimes _{R}DT)$, $s_{i}^{P}=\big(\!\begin{smallmatrix}s_{i}^{1}\\ (p_{i}\otimes _{R}DT)\cdot s_{i}^{2}\end{smallmatrix}\!\big)$ and $s_{i}=\big(\begin{smallmatrix}s_{i}^{1}\\ s_{i}^{2}\end{smallmatrix}\big)$.

Denote $\mathfrak{L}_{A_{X}}^{P}:=\{\text{Hom}_{R}(T,(A_{X})_{i-1})\oplus P_{(A_{X})_{i}}\otimes _{R}DT\}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}(S)$. Note that $\overline{X}_{i}\otimes _{R}DT\in {\mathcal{K}}$ and $\text{Hom}_{R}(T,(A_{X})_{i-1})\oplus P_{(A_{X})_{i}}\otimes _{R}DT\in {\mathcal{G}}$, so we have an exact sequence in ${\mathcal{R}}{\mathcal{C}}^{\text{tr}}(S)$ from the first row in the above commutative diagram

$$\begin{eqnarray}\displaystyle 0\rightarrow \overline{X}\otimes _{R}DT\longrightarrow \mathfrak{L}_{A_{X}}^{P}\stackrel{s^{P}}{\longrightarrow }\mathbf{S}_{T}(X)\rightarrow 0 & & \displaystyle \nonumber\end{eqnarray}$$

with $\overline{X}\otimes _{R}DT\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{K}})$ and $\mathfrak{L}_{A_{X}}^{P}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}({\mathcal{G}})$, as in 4.2.4. By applying the homomorphism $r$ in 4.2.3 to the homomorphism $s^{P}:\mathfrak{L}_{A_{X}}^{P}\rightarrow \mathbf{S}_{T}(X)$, we have an exact sequence in ${\mathcal{R}}{\mathcal{C}}(R)$ by the construction of the functor $\mathbf{Q}_{DT}$ in 4.2.5

$$\begin{eqnarray}\displaystyle 0\rightarrow \mathbf{Q}_{DT}\mathbf{S}_{T}(X)\stackrel{\unicode[STIX]{x1D706}}{\longrightarrow }\text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P})\stackrel{r(s^{P})}{\longrightarrow }\text{&#x0124;om}(DT,\mathbf{S}_{T}(X))\rightarrow 0, & & \displaystyle \nonumber\end{eqnarray}$$

where $r(s^{P})$ is defined as in 4.2.3.

4.3.2 The object $X\oplus \text{L}_{R}(P_{A_{X}}^{+})$ in ${\mathcal{R}}{\mathcal{C}}(R)$

Denote $P_{A_{X}}^{+}:=\{P_{(A_{X})_{i+1}}\}$, then $P_{A_{X}}^{+}\in {\mathcal{R}}{\mathcal{C}}^{\text{tr}}(\text{add}_{\text{mod}R}R)$. Applying the functor $\text{L}_{R}$ in the remark in 4.1.1, we obtain that $\text{L}_{R}(P_{A_{X}}^{+})$ is a projective object in ${\mathcal{R}}{\mathcal{C}}(R)$. Hence, the object $X\oplus \text{L}_{R}(P_{A_{X}}^{+})$ is isomorphic to $X$ in $\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)$.

We will prove that $\mathbf{Q}_{DT}\mathbf{S}_{T}(X)\simeq X\oplus \text{L}_{R}(P_{A_{X}}^{+})$ naturally. And then, $\mathbf{Q}_{DT}\mathbf{S}_{T}\simeq 1_{\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)}$.

The general strategy is as follows. First, we construct a natural homomorphism $\unicode[STIX]{x1D709}:X\oplus \text{L}_{R}(P_{A_{X}}^{+})\rightarrow \text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P})$. Second, we show that $\unicode[STIX]{x1D709}\cdot r(s^{P})=0$, that is, the composition of $\unicode[STIX]{x1D709}$ and the homomorphism $r(s^{P}):\text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P})\rightarrow \hat{\text{H}}\text{om}(DT,\mathbf{S}_{T}(X))$ in the exact sequence above is 0. Thus, we obtain a homomorphism $\unicode[STIX]{x1D719}:X\oplus \text{L}_{R}(P_{A_{X}}^{+})\rightarrow \mathbf{Q}_{DT}\mathbf{S}_{T}(X)$. Finally, we prove that $\unicode[STIX]{x1D719}$ is indeed a natural isomorphism.

4.3.3 The homomorphism $\unicode[STIX]{x1D709}:X\oplus \text{L}_{R}(P_{A_{X}}^{+})\rightarrow \text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P})$

Recall from the construction in 4.3.1 that $X=\{X_{i},\unicode[STIX]{x1D6FF}_{i}^{\otimes }\}$ and,

$\text{L}_{R}(P_{A_{X}}^{+})=\{\text{Hom}_{R}(R,P_{(A_{X})_{i}})\oplus P_{(A_{X})_{i+1}}\otimes _{R}DR\}=\{P_{(A_{X})_{i}}\oplus P_{(A_{X})_{i+1}}\otimes _{R}DR\}$, where the structure map $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(\text{L}_{R}(P_{A_{X}}^{+})):(\text{L}_{R}(P_{A_{X}}^{+}))_{i}\otimes DR\rightarrow (\text{L}_{R}(P_{A_{X}}^{+}))_{i-1}$ is given by $\big(\begin{smallmatrix}0 & 1_{P_{(A_{X})_{i}}\otimes _{R}DR}\\ 0 & 0\end{smallmatrix}\!\big)$, and that

$$\begin{eqnarray}\displaystyle \text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P}) & = & \displaystyle \{\text{Hom}_{S}(DT,(\mathfrak{L}_{A_{X}}^{P})_{i})\oplus (\mathfrak{L}_{A_{X}}^{P})_{i+1}\otimes _{S}T\}\nonumber\\ \displaystyle & = & \displaystyle \{\text{Hom}_{S}(DT,\text{Hom}_{R}(T,(A_{X})_{i-1})\oplus P_{(A_{X})_{i}}\otimes _{R}DT)\nonumber\\ \displaystyle & & \displaystyle \oplus \,(\text{Hom}_{R}(T,(A_{X})_{i})\oplus P_{(A_{X})_{i+1}}\otimes _{R}DT)\otimes _{S}T\},\nonumber\end{eqnarray}$$

where the structure map

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{i}^{\otimes }(\text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P})):(\text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P}))_{i}\otimes _{R}DR\rightarrow (\text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P}))_{i-1}\end{eqnarray}$$

is defined in 4.2.1.

By the definition, we have that $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(\text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P}))=\bigg(\!\begin{smallmatrix}0 & 0 & \unicode[STIX]{x1D6FE}_{i}^{11} & 0\\ 0 & 0 & 0 & \unicode[STIX]{x1D6FE}_{i}^{22}\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{smallmatrix}\!\bigg)$, where

$\unicode[STIX]{x1D6FE}_{i}^{11}:\text{Hom}_{S}(DT,\text{Hom}_{R}(T,(A_{X})_{i-1}))\otimes _{R}DR\rightarrow \text{Hom}_{R}(T,(A_{X})_{i-1})\otimes _{S}T$ is given by the composition

$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6FE}_{i}^{11}:\text{Hom}_{S}(DT,\text{Hom}_{R}(T,(A_{X})_{i-1}))\otimes _{R}DR\nonumber\\ \displaystyle & & \displaystyle \quad \simeq \text{Hom}_{S}(DT,\text{Hom}_{R}(T,(A_{X})_{i-1}))\otimes _{R}DT\otimes _{S}T\nonumber\\ \displaystyle & & \displaystyle \quad \stackrel{\unicode[STIX]{x1D716}_{\text{Hom}_{R}(T,(A_{X})_{i-1})}^{DT}\otimes _{S}T}{\longrightarrow }\text{Hom}_{R}(T,(A_{X})_{i-1})\otimes _{S}T,\nonumber\end{eqnarray}$$

and $\unicode[STIX]{x1D6FE}_{i}^{22}:\text{Hom}_{S}(DT,P_{(A_{X})_{i}}\otimes _{R}DT)\otimes _{R}DR\rightarrow P_{(A_{X})_{i}}\otimes _{R}DT\otimes _{S}T$ is defined similarly as $\unicode[STIX]{x1D6FE}_{i}^{11}$ by replacing $\text{Hom}_{R}(T,(A_{X})_{i-1})$ with $P_{(A_{X})_{i}}\otimes _{R}DT$.

Let $\unicode[STIX]{x1D709}=\{\unicode[STIX]{x1D709}_{i}\}:X\oplus \text{L}_{R}(P_{A_{X}}^{+})\rightarrow \text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P})$ be a homomorphism. We may assume that $\unicode[STIX]{x1D709}_{i}=(\unicode[STIX]{x1D709}_{i}^{a},\unicode[STIX]{x1D709}_{i}^{b})$, where

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D709}_{i}^{a}:X_{i}\oplus P_{(A_{X})_{i}}\oplus P_{(A_{X})_{i+1}}\otimes _{R}DR\rightarrow \text{Hom}_{S}(DT,(\mathfrak{L}_{A_{X}}^{P})_{i}) & & \displaystyle \nonumber\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D709}_{i}^{b}:X_{i}\oplus P_{(A_{X})_{i}}\oplus P_{(A_{X})_{i+1}}\otimes _{R}DR\rightarrow (\mathfrak{L}_{A_{X}}^{P})_{i+1}\otimes _{S}T. & & \displaystyle \nonumber\end{eqnarray}$$

4.3.3.1 The homomorphism $\unicode[STIX]{x1D709}_{i}^{a}$ in $\text{mod}R$

We set $\unicode[STIX]{x1D709}_{i}^{a}=\bigg(\!\begin{smallmatrix}\unicode[STIX]{x1D709}_{i}^{a_{11}} & \unicode[STIX]{x1D709}_{i}^{a_{12}}\\ \unicode[STIX]{x1D709}_{i}^{a_{21}} & \unicode[STIX]{x1D709}_{i}^{a_{22}}\\ \unicode[STIX]{x1D709}_{i}^{a_{31}} & \unicode[STIX]{x1D709}_{i}^{a_{32}}\end{smallmatrix}\!\bigg)$:

$$\begin{eqnarray}\displaystyle X_{i}\oplus P_{(A_{X})_{i}}\oplus P_{(A_{X})_{i+1}}\otimes _{R}DR\longrightarrow \text{Hom}_{S}(DT,\text{Hom}_{R}(T,(A_{X})_{i-1})\oplus P_{(A_{X})_{i}}\otimes _{R}DT). & & \displaystyle \nonumber\end{eqnarray}$$

Using the isomorphism $_{S}DS_{S}\simeq _{S}T\otimes _{R}DT_{S}$ and the adjoint isomorphism

$$\begin{eqnarray}\displaystyle \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}:\text{Hom}_{S}(-\!\otimes _{R}DT,-)\simeq \text{Hom}_{R}(-,\text{Hom}_{S}(DT,-)), & & \displaystyle \nonumber\end{eqnarray}$$

we define the components of $\unicode[STIX]{x1D709}_{i}^{a}$ as follows.

$\bullet$ The morphism $\unicode[STIX]{x1D709}_{i}^{a_{11}}=\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(\unicode[STIX]{x1D703}_{l_{i}}):$$X_{i}\rightarrow \text{Hom}_{S}(DT,\text{Hom}_{R}(T,(A_{X})_{i-1}))$, where $\unicode[STIX]{x1D703}_{l_{i}}:X_{i}\otimes _{R}DT\rightarrow \text{Hom}_{R}(T,(A_{X})_{i-1})$ is defined in 4.1.3.

In other words, the morphism $\unicode[STIX]{x1D709}_{i}^{a_{11}}$ is given by the composition:

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D702}_{_{X_{i}}}^{DR}\cdot \text{Hom}_{R}(DR,\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\cdot \text{Hom}_{R}(DR,(u_{_{X}})_{i-1}))=\unicode[STIX]{x1D6FF}_{i}^{\text{H}}(X)\cdot \text{Hom}_{R}(DR,(u_{_{X}})_{i-1}) & & \displaystyle \nonumber\end{eqnarray}$$

and some natural isomorphisms

$$\begin{eqnarray}\displaystyle & & \displaystyle X_{i}\stackrel{\unicode[STIX]{x1D702}_{_{X_{i}}}^{DR}}{\longrightarrow }\text{Hom}_{R}(DR,X_{i}\otimes _{R}DR)\stackrel{\text{Hom}_{R}(DR,\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X))}{\longrightarrow }\text{Hom}_{R}(DR,X_{i-1})\nonumber\\ \displaystyle & & \displaystyle \quad \stackrel{\text{Hom}_{R}(DR,(u_{_{X}})_{i-1})}{\longrightarrow }\text{Hom}_{R}(DR,(A_{X})_{i-1})\simeq \text{Hom}_{R}(DT\otimes _{S}T,(A_{X})_{i-1})\nonumber\\ \displaystyle & & \displaystyle \quad \simeq \text{Hom}_{S}(DT,\text{Hom}_{R}(T,(A_{X})_{i-1})).\nonumber\end{eqnarray}$$

$\bullet$ The morphism

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D709}_{i}^{a_{22}} & = & \displaystyle \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(1_{(P_{(A_{X})_{i}}\otimes _{R}DT)})\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D702}_{_{P_{(A_{X})_{i}}}}^{DT}:P_{(A_{X})_{i}}\rightarrow \text{Hom}_{S}(DT,P_{(A_{X})_{i}}\otimes _{R}DT).\nonumber\end{eqnarray}$$

$\bullet$ The remaining morphisms $\unicode[STIX]{x1D709}_{i}^{a_{12}},\unicode[STIX]{x1D709}_{i}^{a_{21}},\unicode[STIX]{x1D709}_{i}^{a_{31}},\unicode[STIX]{x1D709}_{i}^{a_{32}}$ are all $0$.

So we have that $\unicode[STIX]{x1D709}_{i}^{a}=\bigg(\!\begin{smallmatrix}\unicode[STIX]{x1D709}_{i}^{a_{11}} & 0\\ 0 & \unicode[STIX]{x1D709}_{i}^{a_{22}}\\ 0 & 0\end{smallmatrix}\!\bigg)$, where

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D709}_{i}^{a_{11}} & = & \displaystyle \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(\unicode[STIX]{x1D703}_{l_{i}}),\qquad \text{and}\nonumber\\ \displaystyle \unicode[STIX]{x1D709}_{i}^{a_{22}} & = & \displaystyle \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(1_{(P_{(A_{X})_{i}}\otimes _{R}DT)}).\nonumber\end{eqnarray}$$

4.3.3.2 The homomorphism $\unicode[STIX]{x1D709}_{i}^{b}$ in $\text{mod}R$

We set $\unicode[STIX]{x1D709}_{i}^{b}=\bigg(\!\begin{smallmatrix}\unicode[STIX]{x1D709}_{i}^{b_{11}} & \unicode[STIX]{x1D709}_{i}^{b_{12}}\\ \unicode[STIX]{x1D709}_{i}^{b_{21}} & \unicode[STIX]{x1D709}_{i}^{b_{22}}\\ \unicode[STIX]{x1D709}_{i}^{b_{31}} & \unicode[STIX]{x1D709}_{i}^{b_{32}}\end{smallmatrix}\!\bigg)$:

$$\begin{eqnarray}\displaystyle X_{i}\oplus P_{(A_{X})_{i}}\oplus P_{(A_{X})_{i+1}}\otimes _{R}DR\rightarrow (\text{Hom}_{R}(T,(A_{X})_{i})\oplus P_{(A_{X})_{i+1}}\otimes _{R}DT)\otimes _{S}T, & & \displaystyle \nonumber\end{eqnarray}$$

where the components are defined naturally as follows.

• ∙ The morphism $\unicode[STIX]{x1D709}_{i}^{b_{11}}=(u_{_{X}})_{i}\cdot (\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T})^{-1}:X_{i}\rightarrow \text{Hom}_{R}(T,(A_{X})_{i})\otimes _{S}T$ (note that $\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T}$ is an isomorphism since $(A_{X})_{i}\in {\mathcal{A}}$), that is, is given by the composition

  $$\begin{eqnarray}X_{i}\stackrel{(u_{_{X}})_{i}}{\longrightarrow }(A_{X})_{i}\stackrel{(\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T})^{-1}}{\longrightarrow }\text{Hom}_{R}(T,(A_{X})_{i})\otimes _{S}T.\end{eqnarray}$$

• ∙ The morphism $\unicode[STIX]{x1D709}_{i}^{b_{21}}=p_{i}\cdot (\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T})^{-1}:P_{(A_{X})_{i}}\rightarrow \text{Hom}_{R}(T,(A_{X})_{i})\otimes _{S}T$, that is, is given by the composition

  $$\begin{eqnarray}P_{(A_{X})_{i}}\stackrel{p_{i}}{\longrightarrow }(A_{X})_{i}\stackrel{(\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T})^{-1}}{\longrightarrow }\text{Hom}_{R}(T,(A_{X})_{i})\otimes _{S}T.\end{eqnarray}$$

• ∙ The morphism $\unicode[STIX]{x1D709}_{i}^{b_{32}}:P_{(A_{X})_{i+1}}\otimes _{R}DR\rightarrow P_{(A_{X})_{i+1}}\otimes _{R}DT\otimes _{S}T$ is the natural isomorphism given by $_{R}(DT\otimes _{S}T)_{R}\simeq _{R}DR_{R}$.

• ∙ The remaining morphisms $\unicode[STIX]{x1D709}_{i}^{b_{12}},\unicode[STIX]{x1D709}_{i}^{b_{22}},\unicode[STIX]{x1D709}_{i}^{b_{31}}$ are all $0$.

So we have that $\unicode[STIX]{x1D709}_{i}^{b}=\bigg(\!\begin{smallmatrix}\unicode[STIX]{x1D709}_{i}^{b_{11}} & 0\\ \unicode[STIX]{x1D709}_{i}^{b_{21}} & 0\\ 0 & \unicode[STIX]{x1D709}_{i}^{b_{32}}\end{smallmatrix}\!\bigg)$, where $\unicode[STIX]{x1D709}_{i}^{b_{11}}=(u_{_{X}})_{i}\cdot (\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T})^{-1}$, $\unicode[STIX]{x1D709}_{i}^{b_{21}}=p_{i}\cdot (\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T})^{-1}$, and $\unicode[STIX]{x1D709}_{i}^{b_{32}}$ is the natural isomorphism.

4.3.3.3 $\unicode[STIX]{x1D709}$ is a homomorphism in ${\mathcal{R}}{\mathcal{C}}(R)$

We now show that the above-defined morphism $\unicode[STIX]{x1D709}:X\oplus \text{L}_{R}(P_{A_{X}}^{+})\rightarrow \text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P})$ is compatible with structure maps, that is,

$$\begin{eqnarray}\unicode[STIX]{x1D709}_{i}\otimes _{R}DR\cdot \unicode[STIX]{x1D6FF}_{i}^{\otimes }(\text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P}))=\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X\oplus \text{L}_{R}(P_{A_{X}}^{+}))\cdot \unicode[STIX]{x1D709}_{i-1}\end{eqnarray}$$

holds for each $i$.

Indeed, by the involved definitions, we have that

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D709}_{i}\otimes _{R}DR\cdot \unicode[STIX]{x1D6FF}_{i}^{\otimes }(\text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P})) & = & \displaystyle \left(\begin{array}{@{}cccc@{}}\unicode[STIX]{x1D709}_{i}^{a_{11}} & 0 & \unicode[STIX]{x1D709}_{i}^{b_{11}} & 0\\ 0 & \unicode[STIX]{x1D709}_{i}^{a_{22}} & \unicode[STIX]{x1D709}_{i}^{b_{21}} & 0\\ 0 & 0 & 0 & \unicode[STIX]{x1D709}_{i}^{b_{32}}\end{array}\right)\otimes _{R}DR\cdot \left(\begin{array}{@{}cccc@{}}0 & 0 & \unicode[STIX]{x1D6FE}_{i}^{11} & 0\\ 0 & 0 & 0 & \unicode[STIX]{x1D6FE}_{i}^{22}\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right)\nonumber\\ \displaystyle & = & \displaystyle \left(\begin{array}{@{}cccc@{}}0 & 0 & \unicode[STIX]{x1D709}_{i}^{a_{11}}\otimes _{R}DR\cdot \unicode[STIX]{x1D6FE}_{i}^{11} & 0\\ 0 & 0 & 0 & \unicode[STIX]{x1D709}_{i}^{a_{22}}\otimes _{R}DR\cdot \unicode[STIX]{x1D6FE}_{i}^{22}\\ 0 & 0 & 0 & 0\end{array}\right)\nonumber\end{eqnarray}$$

and that

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FF}_{i}^{\otimes }(X\oplus \text{L}_{R}(P_{A_{X}}^{+}))\cdot \unicode[STIX]{x1D709}_{i-1} & = & \displaystyle \left(\begin{array}{@{}ccc@{}}\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X) & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{array}\right)\left(\begin{array}{@{}cccc@{}}\unicode[STIX]{x1D709}_{i-1}^{a_{11}} & 0 & \unicode[STIX]{x1D709}_{i-1}^{b_{11}} & 0\\ 0 & \unicode[STIX]{x1D709}_{i-1}^{a_{22}} & \unicode[STIX]{x1D709}_{i-1}^{b_{21}} & 0\\ 0 & 0 & 0 & \unicode[STIX]{x1D709}_{i-1}^{b_{32}}\end{array}\right)\nonumber\\ \displaystyle & = & \displaystyle \left(\begin{array}{@{}cccc@{}}\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\cdot \unicode[STIX]{x1D709}_{i-1}^{a_{11}} & 0 & \unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\cdot \unicode[STIX]{x1D709}_{i-1}^{b_{11}} & 0\\ 0 & 0 & 0 & \unicode[STIX]{x1D709}_{i-1}^{b_{32}}\\ 0 & 0 & 0 & 0\end{array}\right)\nonumber\end{eqnarray}$$

• ∙ $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\cdot \unicode[STIX]{x1D709}_{i-1}^{a_{11}}=0$. In fact, since $\unicode[STIX]{x1D709}_{i-1}^{a_{11}}$ is the composition of the morphism $\unicode[STIX]{x1D6FF}_{i-1}^{\text{H}}(X)\cdot \text{Hom}_{R}(DR,(u_{_{X}})_{i-2})$ and some natural isomorphisms by the construction, we see that $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\cdot \unicode[STIX]{x1D709}_{i-1}^{a_{11}}$ factors through $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\cdot \unicode[STIX]{x1D6FF}_{i-1}^{\text{H}}(X)$. But the latter is 0 as $X$ is a repe-complex. Thus, $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\cdot \unicode[STIX]{x1D709}_{i-1}^{a_{11}}=0$.

• ∙ $\unicode[STIX]{x1D709}_{i}^{a_{11}}\otimes _{R}DR\cdot \unicode[STIX]{x1D6FE}_{i}^{11}=\unicode[STIX]{x1D6FF}_{i}^{\otimes }(X)\cdot \unicode[STIX]{x1D709}_{i-1}^{b_{11}}$. This follows from the involved definitions and the following commutative diagram:

  
  where $\unicode[STIX]{x1D714}:=\unicode[STIX]{x1D716}_{\text{Hom}_{R}(T,(A_{X})_{i-1})}^{DT}\otimes _{S}T$. The last square is commutative since $\unicode[STIX]{x1D716}_{M}^{DT\otimes _{S}T}=\unicode[STIX]{x1D716}_{\text{Hom}_{R}(T,M)}^{DT}\otimes _{S}T\cdot \unicode[STIX]{x1D716}_{M}^{T}$.

• ∙ $\unicode[STIX]{x1D709}_{i}^{a_{22}}\otimes _{R}DR\cdot \unicode[STIX]{x1D6FE}_{i}^{22}=\unicode[STIX]{x1D709}_{i-1}^{b_{32}}$. This follows from the involved definitions and the equality $1_{(P_{(A_{X})_{i}}\otimes _{R}DT)}=\unicode[STIX]{x1D702}_{(P_{(A_{X})_{i}})}^{DT}\otimes _{R}DT\cdot \unicode[STIX]{x1D716}_{(P_{(A_{X})_{i}}\otimes _{R}DT)}^{DT}$.

Then we can easily conclude that the morphism $\unicode[STIX]{x1D709}:X\oplus \text{L}_{R}(P_{A_{X}}^{+})\rightarrow \text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P})$ is, in fact, a homomorphism in ${\mathcal{R}}{\mathcal{C}}(R)$.

4.3.4 The composition $\unicode[STIX]{x1D709}\cdot r(s^{P})=0$, and so $\unicode[STIX]{x1D709}$ factors through a homomorphism $\unicode[STIX]{x1D719}:X\oplus \text{L}_{R}(P_{A_{X}}^{+})\rightarrow \mathbf{Q}_{DT}\mathbf{S}_{T}(X)$

4.3.4.1 The analysis of the homomorphism $s:\text{L}_{T}(A_{X})\rightarrow \mathbf{S}_{T}(X)$ in 4.3.1

Recall from 4.3.1 that $s=\{s_{i}\}:\text{L}_{T}(A_{X})\rightarrow \mathbf{S}_{T}(X)$ is a homomorphism in ${\mathcal{R}}{\mathcal{C}}(S)$ which is the cokernel of the homomorphism $l(u_{_{X}})$. Note that $(\text{L}_{T}(A_{X}))_{i}=\text{Hom}_{R}(T,(A_{X})_{i-1})\oplus (A_{X})_{i}\otimes _{R}DT$, so we write that $s_{i}=\big(\!\begin{smallmatrix}s_{i}^{1}s_{i}^{2}\end{smallmatrix}\!\big)$ as we have done in the last commutative diagram in 4.3.1. The fact that $s$ is a homomorphism in ${\mathcal{R}}{\mathcal{C}}(S)$ implies that there is the following commutative diagram, for each $i$:

By the definition of $\unicode[STIX]{x1D6FF}_{i}^{\otimes }(\text{L}_{T}(A_{X}))$ (see 4.1.1) and the above commutative diagram, we obtain that $(s_{i}^{2}\otimes _{S}DS)\cdot \unicode[STIX]{x1D6FF}_{i}^{\otimes }(\mathbf{S}_{T}(X))=0$ and that $(s_{i}^{1}\otimes _{S}DS)\cdot \unicode[STIX]{x1D6FF}_{i}^{\otimes }(\mathbf{S}_{T}(X))\simeq (\unicode[STIX]{x1D716}_{A_{X_{i-1}}}^{T}\otimes _{R}DT)\cdot s_{i-1}^{2}$.

4.3.4.2 The homomorphism $r(s^{P}):\text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P})\rightarrow \text{&#x0124;om}(DT,\mathbf{S}_{T}(X))$

Recall from 4.3.1 that

$$\begin{eqnarray}s^{P}=\{s_{i}^{P}\}=\left\{\left(\begin{array}{@{}c@{}}s_{i}^{1}\\ (p_{i}\otimes _{R}DT)\cdot s_{i}^{2}\end{array}\right)\right\}:\mathfrak{L}_{A_{X}}^{P}\rightarrow \mathbf{S}_{T}(X).\end{eqnarray}$$

By 4.2.3, we know that

$$\begin{eqnarray}\text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P})=\{\text{Hom}_{S}(DT,(\mathfrak{L}_{A_{X}}^{P})_{i})\oplus (\mathfrak{L}_{A_{X}}^{P})_{i+1}\otimes _{S}T\}\end{eqnarray}$$

and that

$$\begin{eqnarray}r(s^{P})_{i}=\left(\begin{array}{@{}c@{}}\text{Hom}_{S}(DT,s_{i}^{P})-\unicode[STIX]{x1D701}_{r_{i}}\end{array}\right),\end{eqnarray}$$

where $\unicode[STIX]{x1D701}_{r_{i}}=\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((s_{i+1}^{P}\otimes _{S}DS)\cdot \unicode[STIX]{x1D6FF}_{i+1}^{\otimes }(\mathbf{S}_{T}(X)))$.

For convenience, we set $r(s^{P})_{i}=\big(\!\begin{smallmatrix}r_{i}^{1}r_{i}^{2}\end{smallmatrix}\!\big)$, where

$$\begin{eqnarray}r_{i}^{1}=\text{Hom}_{S}(DT,s_{i}^{P}):\text{Hom}_{S}(DT,(\mathfrak{L}_{A_{X}}^{P})_{i})\rightarrow \text{Hom}_{S}(DT,\mathbf{S}_{T}(X)_{i})\end{eqnarray}$$

and

$$\begin{eqnarray}r_{i}^{2}=-\unicode[STIX]{x1D701}_{r_{i}}:(\mathfrak{L}_{A_{X}}^{P})_{i+1}\otimes _{S}T\rightarrow \text{Hom}_{S}(DT,\mathbf{S}_{T}(X)_{i}).\end{eqnarray}$$

4.3.4.3 Checking $\unicode[STIX]{x1D709}\cdot r(s^{P})=0$

To check $\unicode[STIX]{x1D709}\cdot r(s^{P})=0$, we need only to check that $\unicode[STIX]{x1D709}_{i}^{a}r_{i}^{1}+\unicode[STIX]{x1D709}_{i}^{b}r_{i}^{2}=0$ for each $i$, since $\unicode[STIX]{x1D709}_{i}=(\unicode[STIX]{x1D709}_{i}^{a},\unicode[STIX]{x1D709}_{i}^{b})$ and $r(s^{P})_{i}=\big(\begin{smallmatrix}r_{i}^{1}r_{i}^{2}\end{smallmatrix}\big)$. Note that $r_{i}^{1}=\text{Hom}_{S}(DT,s_{i}^{P})$ and $r_{i}^{2}=-\unicode[STIX]{x1D701}_{r_{i}}$, so it is enough to check that $\unicode[STIX]{x1D709}_{i}^{a}\cdot \text{Hom}_{S}(DT,s_{i}^{P})=\unicode[STIX]{x1D709}_{i}^{b}\cdot \unicode[STIX]{x1D701}_{r_{i}}$.

Since $\unicode[STIX]{x1D709}_{i}^{a}=\bigg(\begin{smallmatrix}\unicode[STIX]{x1D709}_{11}^{a} & 0\\ 0 & \unicode[STIX]{x1D709}_{22}^{a}\\ 0 & 0\end{smallmatrix}\bigg)$, $\unicode[STIX]{x1D709}_{i}^{b}=\bigg(\begin{smallmatrix}\unicode[STIX]{x1D709}_{11}^{b} & 0\\ \unicode[STIX]{x1D709}_{21}^{b} & 0\\ 0 & \unicode[STIX]{x1D709}_{32}^{b}\end{smallmatrix}\bigg)$, $\unicode[STIX]{x1D701}_{r_{i}}=\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((s_{i+1}^{P}\otimes _{S}DS)\cdot \unicode[STIX]{x1D6FF}_{i+1}^{\otimes }(\mathbf{S}_{T}(X)))$ and $s_{i}^{P}=\big(\begin{smallmatrix}s_{i}^{1}\\ (p_{i}\otimes _{R}DT)\cdot s_{i}^{2}\end{smallmatrix}\big)$, we just check the following.

1. (1) $\unicode[STIX]{x1D709}_{11}^{a}\cdot \text{Hom}_{S}(DT,s_{i}^{1})=\unicode[STIX]{x1D709}_{11}^{b}\cdot \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((s_{i+1}^{1}\otimes _{S}DS)\cdot \unicode[STIX]{x1D6FF}_{i+1}^{\otimes }(\mathbf{S}_{T}(X)))$.

   By 4.3.3.1, we have that

   $$\begin{eqnarray}\unicode[STIX]{x1D709}_{11}^{a}\cdot \text{Hom}_{S}(DT,s_{i}^{1})=\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(\unicode[STIX]{x1D703}_{l_{i}})\cdot \text{Hom}_{S}(DT,s_{i}^{1})=\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(\unicode[STIX]{x1D703}_{l_{i}}\cdot s_{i}^{1}),\end{eqnarray}$$
   where the later equality uses the naturality of $\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}$. On the other hand, by 4.3.3.2 and 4.3.4.1, we obtain that
   $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D709}_{11}^{b}\cdot \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((s_{i+1}^{1}\otimes _{S}DS)\cdot \unicode[STIX]{x1D6FF}_{i+1}^{\otimes }(\mathbf{S}_{T}(X)))\nonumber\\ \displaystyle & & \displaystyle \quad =((u_{_{X}})_{i}\cdot (\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T})^{-1})\cdot \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T}\otimes _{R}DT)\cdot s_{i}^{2})\nonumber\\ \displaystyle & & \displaystyle \quad =\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((((u_{_{X}})_{i}\cdot (\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T})^{-1})\otimes _{R}DT)\cdot (\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T}\otimes _{R}DT)\cdot s_{i}^{2})\nonumber\\ \displaystyle & & \displaystyle \quad =\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(((u_{_{X}})_{i}\otimes _{R}DT)\cdot s_{i}^{2}).\nonumber\end{eqnarray}$$

   But $((u_{_{X}})_{i}\otimes _{R}DT)\cdot s_{i}^{2}=\unicode[STIX]{x1D703}_{l_{i}}\cdot s_{i}^{1}$ by the pushout diagram on $\mathbf{S}_{T}(X)_{i}$ in 4.3.1. Hence, equality (1) holds.

2. (2) $\unicode[STIX]{x1D709}_{22}^{a}\cdot \text{Hom}_{S}(DT,(p_{i}\otimes _{R}DT)\cdot s_{i}^{2})=\unicode[STIX]{x1D709}_{21}^{b}\cdot \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((s_{i+1}^{1}\otimes _{S}DS)\cdot \unicode[STIX]{x1D6FF}_{i+1}^{\otimes }(\mathbf{S}_{T}(X)))$.

   By 4.3.3.1 and the naturality of $\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}$,

   $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D709}_{22}^{a}\cdot \text{Hom}_{S}(DT,(p_{i}\otimes _{R}DT)\cdot s_{i}^{2})\nonumber\\ \displaystyle & & \displaystyle \quad =\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(1_{P_{(A_{X})_{i}}\otimes _{R}DT})\cdot \text{Hom}_{S}(DT,(p_{i}\otimes _{R}DT)\cdot s_{i}^{2})\nonumber\\ \displaystyle & & \displaystyle \quad =\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(1_{P_{(A_{X})_{i}}\otimes _{R}DT}\cdot ((p_{i}\otimes _{R}DT)\cdot s_{i}^{2}))\nonumber\\ \displaystyle & & \displaystyle \quad =\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((p_{i}\otimes _{R}DT)\cdot s_{i}^{2}).\nonumber\end{eqnarray}$$

   On the other hand, by 4.3.3.2 and 4.3.4.1 and the naturality of $\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}$,

   $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D709}_{21}^{b}\cdot \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((s_{i+1}^{1}\otimes _{S}DS)\cdot \unicode[STIX]{x1D6FF}_{i+1}^{\otimes }(\mathbf{S}_{T}(X)))\nonumber\\ \displaystyle & & \displaystyle \quad =(p_{i}\cdot (\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T})^{-1})\cdot \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T}\otimes _{R}DT)\cdot s_{i}^{2})\nonumber\\ \displaystyle & & \displaystyle \quad =\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(((p_{i}\cdot (\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T})^{-1})\otimes _{R}DT)\cdot (\unicode[STIX]{x1D716}_{(A_{X})_{i}}^{T}\otimes _{R}DT)\cdot s_{i}^{2})\nonumber\\ \displaystyle & & \displaystyle \quad =\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((p_{i}\otimes _{R}DT)\cdot s_{i}^{2}).\nonumber\end{eqnarray}$$

   Hence, equality (2) holds.

3. (3) $0=\unicode[STIX]{x1D709}_{32}^{b}\cdot \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(((p_{i+1}\otimes _{R}DT)\cdot s_{i+1}^{2})\otimes _{S}DS\cdot \unicode[STIX]{x1D6FF}_{i+1}^{\otimes }(\mathbf{S}_{T}(X)))$.

   In fact, the equality holds by observing that

   $$\begin{eqnarray}\displaystyle & & \displaystyle \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(((p_{i+1}\otimes _{R}DT)\cdot s_{i+1}^{2})\otimes _{S}DS\cdot \unicode[STIX]{x1D6FF}_{i+1}^{\otimes }(\mathbf{S}_{T}(X)))\nonumber\\ \displaystyle & & \displaystyle \quad =\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((p_{i+1}\otimes _{R}DT\otimes _{S}DS)\cdot (s_{i+1}^{2}\otimes _{S}DS)\cdot \unicode[STIX]{x1D6FF}_{i+1}^{\otimes }(\mathbf{S}_{T}(X)))\nonumber\\ \displaystyle & & \displaystyle \quad =0\nonumber\end{eqnarray}$$
   since $(s_{i+1}^{2}\otimes _{S}DS)\cdot \unicode[STIX]{x1D6FF}_{i+1}^{\otimes }(\mathbf{S}_{T}(X))=0$ by 4.3.4.1.

Altogether, we prove that $\unicode[STIX]{x1D709}\cdot r(s^{P})=0$ and, therefore, $\unicode[STIX]{x1D709}$ factors through $\mathbf{Q}_{DT}\mathbf{S}_{T}(X)=\text{Ker}(r(s^{P}))$ by a homomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D719}:X\oplus \text{L}_{R}(P_{A_{X}}^{+})\rightarrow \mathbf{Q}_{DT}\mathbf{S}_{T}(X)\end{eqnarray}$$

in ${\mathcal{R}}{\mathcal{C}}(R)$, that is, $\unicode[STIX]{x1D709}=\unicode[STIX]{x1D719}\cdot \unicode[STIX]{x1D706}$.

4.3.5 The induced homomorphism $\unicode[STIX]{x1D719}:X\oplus \text{L}_{R}(P_{A_{X}}^{+})\rightarrow \mathbf{Q}_{DT}\mathbf{S}_{T}(X)$ is an isomorphism

We now prove that the induced homomorphism $\unicode[STIX]{x1D719}:X\oplus \text{L}_{R}(P_{A_{X}}^{+})\rightarrow \mathbf{Q}_{DT}\mathbf{S}_{T}(X)$ is an isomorphism. Clearly, it is equivalent to show that $\unicode[STIX]{x1D719}_{i}:(X\oplus \text{L}_{R}(P_{A_{X}}^{+}))_{i}\rightarrow \mathbf{Q}_{DT}\mathbf{S}_{T}(X)_{i}$ is an isomorphism, for each $i$.

We will show that there is the following commutative diagram ($\ast$) with exact rows, for each $i$:

where

$$\begin{eqnarray}\displaystyle & C_{1}:=(A_{X})_{i}\oplus P_{(A_{X})_{i+1}}\otimes _{R}DR, & \displaystyle \nonumber\\ \displaystyle & C_{2}:=\text{Hom}_{R}(T,(A_{X})_{i})\otimes _{S}T\oplus P_{(A_{X})_{i+1}}\otimes _{R}DT\otimes _{S}T, & \displaystyle \nonumber\end{eqnarray}$$

$\overline{X}_{i}\in {\mathcal{B}}$ is obtained in 4.3.1, the morphism $a_{i}$ is given in 4.3.5.2, the morphism $\unicode[STIX]{x1D6FD}_{i}=\bigg(\begin{smallmatrix}(u_{_{X}})_{i} & 0\\ p_{i} & 0\\ 0 & 1\end{smallmatrix}\bigg)$ and the morphism $\unicode[STIX]{x1D70E}_{i}$ is the direct sum of two canonical isomorphisms. Note that $\text{L}_{R}(P_{A_{X}}^{+})_{i}=P_{(A_{X})_{i}}\oplus P_{(A_{X})_{i+1}}\otimes _{R}DR$.

Then, since $\overline{X}_{i}\in {\mathcal{B}}$ implies that $\unicode[STIX]{x1D702}_{\overline{X}_{i}}^{DT}$ is an isomorphism, we obtain that $\unicode[STIX]{x1D719}_{i}$ is also an isomorphism from the above commutative diagram.

4.3.5.1 The upper row in the diagram $(\ast )$ is exact

In fact, the pullback of $p_{i}:P_{(A_{X})_{i}}\rightarrow (A_{X})_{i}$ and $(u_{_{X}})_{i}:X_{i}\rightarrow (A_{X})_{i}$ in 4.3.1 gives an exact sequence

$$\begin{eqnarray}0\rightarrow \overline{X}_{i}\stackrel{(-q_{i},(\overline{u}_{X})_{i})}{\longrightarrow }X_{i}\oplus P_{(A_{X})_{i}}\stackrel{}{\longrightarrow _{}}(A_{X})_{i}\rightarrow 0\end{eqnarray}$$

since $p_{i}$ is surjective. The direct sum of the above exact sequence and the trivial exact sequence

$$\begin{eqnarray}0\rightarrow 0\rightarrow P_{(A_{X})_{i+1}}\otimes _{R}DR\stackrel{1}{\longrightarrow }P_{(A_{X})_{i+1}}\otimes _{R}DR\rightarrow 0\end{eqnarray}$$

gives us the exact sequence in the upper row in the diagram $(\ast )$.

4.3.5.2 The bottom row in the diagram $(\ast )$ is exact

Note that we have the following exact sequence in 4.3.1

$$\begin{eqnarray}0\rightarrow \mathbf{Q}_{DT}\mathbf{S}_{T}(X)\stackrel{\unicode[STIX]{x1D706}}{\longrightarrow }\text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P})\stackrel{r(s^{P})}{\longrightarrow }\text{&#x0124;om}(DT,\mathbf{S}_{T}(X))\rightarrow 0\end{eqnarray}$$

and that

$$\begin{eqnarray}\displaystyle \text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P})_{i} & = & \displaystyle \text{Hom}_{S}(DT,(\mathfrak{L}_{A_{X}}^{P})_{i})\oplus (\mathfrak{L}_{A_{X}}^{P})_{i+1}\otimes _{S}T\nonumber\\ \displaystyle & = & \displaystyle \text{Hom}_{S}(DT,\text{Hom}_{R}(T,(A_{X})_{i-1})\oplus P_{(A_{X})_{i}}\otimes _{R}DT)\nonumber\\ \displaystyle & & \displaystyle \oplus \,(\text{Hom}_{R}(T,(A_{X})_{i})\oplus P_{(A_{X})_{i+1}}\otimes _{R}DT)\otimes _{S}T.\nonumber\end{eqnarray}$$

So we have the following pullback diagram:

where

$$\begin{eqnarray}\displaystyle & C_{3}:=(\text{Hom}_{R}(T,(A_{X})_{i})\oplus P_{(A_{X})_{i+1}}\otimes _{R}DT)\otimes _{S}T, & \displaystyle \nonumber\\ \displaystyle & C_{4}:=\text{Hom}_{S}(DT,\text{Hom}_{R}(T,(A_{X})_{i-1})\oplus P_{(A_{X})_{i}}\otimes _{R}DT), & \displaystyle \nonumber\end{eqnarray}$$

$r_{i}^{1},r_{i}^{2}$ and $\unicode[STIX]{x1D706}_{i}^{1},\unicode[STIX]{x1D706}_{i}^{2}$ are the components of the homomorphisms $r(s^{P})_{i}$ and $\unicode[STIX]{x1D706}_{i}$, respectively, and $b_{i}=\text{Hom}_{S}(DT,t_{i})$ with

$$\begin{eqnarray}\displaystyle & t_{i}=(-(q_{i}\otimes _{R}DT)\cdot \unicode[STIX]{x1D703}_{l_{i}},(\overline{u}_{X})_{i}\otimes _{R}DT): & \displaystyle \nonumber\\ \displaystyle & \overline{X}_{i}\otimes _{R}DT\rightarrow \text{Hom}_{R}(T,(A_{X})_{i-1})\oplus P_{(A_{X})_{i}}\otimes _{R}DT & \displaystyle \nonumber\end{eqnarray}$$

is given in 4.3.1.

As the morphism $\unicode[STIX]{x1D706}_{i}:\mathbf{Q}_{DT}\mathbf{S}_{T}(X)_{i}\rightarrow (\text{R}_{DT}(\mathfrak{L}_{A_{X}}^{P}))_{i}$ is injective, we obtain that the morphism $q$ in the left column is an isomorphism. Note that $r_{i}^{1}:=\text{Hom}_{S}(DT,s_{i}^{P})$ is surjective since $\overline{X}_{i}\otimes _{R}DT\in {\mathcal{K}}\subseteq \text{KerExt}_{S}^{{>}0}(DT,-)$ by the construction (see 4.3.1), so we can deduce that the upper row in the above diagram is exact. Thus, we get the bottom exact sequence in the diagram $(\ast )$ by setting $a_{i}=q^{-1}e$.

4.3.5.3 The diagram $(\ast )$ is commutative

At first, it is easy to see that the right part of the diagram $(\ast )$ is commutative from the constructions of the morphisms $\unicode[STIX]{x1D709}$ in 4.3.3 and $\unicode[STIX]{x1D719}$ in 4.3.4, which show that $\unicode[STIX]{x1D6FD}_{i}\unicode[STIX]{x1D70E}_{i}=\unicode[STIX]{x1D709}_{i}^{b}=\unicode[STIX]{x1D719}_{i}\unicode[STIX]{x1D706}_{i}^{2}$.

As for the left part of the diagram $(\ast )$, we first show the following equality of compositions:

(†₁)$$\begin{eqnarray}\unicode[STIX]{x1D702}_{_{\overline{X}_{i}}}^{DT}\cdot a_{i}\cdot \unicode[STIX]{x1D706}_{i}^{1}=(-q_{i},(\overline{u}_{X})_{i},0)\cdot \unicode[STIX]{x1D719}_{i}\cdot \unicode[STIX]{x1D706}_{i}^{1}.\end{eqnarray}$$

Indeed, we have that

and we also have that

Since $\unicode[STIX]{x1D709}_{11}^{a}=\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(\unicode[STIX]{x1D703}_{l_{i}})$ and $\unicode[STIX]{x1D709}_{22}^{a}=\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(1_{(P_{(A_{X})_{i}}\otimes _{R}DT)})$ by the construction in 4.3.3.1, we obtain that

$$\begin{eqnarray}q_{i}\cdot \unicode[STIX]{x1D709}_{11}^{a}=q_{i}\cdot \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(\unicode[STIX]{x1D703}_{l_{i}})=\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(q_{i}\otimes _{R}DT\cdot \unicode[STIX]{x1D703}_{l_{i}})\end{eqnarray}$$

and that

$$\begin{eqnarray}\displaystyle (\overline{u}_{X})_{i}\cdot \unicode[STIX]{x1D709}_{22}^{a} & = & \displaystyle (\overline{u}_{X})_{i}\cdot \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}(1_{(P_{(A_{X})_{i}}\otimes _{R}DT)})\nonumber\\ \displaystyle & = & \displaystyle \boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((\overline{u}_{X})_{i}\otimes _{R}DT\cdot 1_{(P_{(A_{X})_{i}}\otimes _{R}DT)})=\boldsymbol{\unicode[STIX]{x1D6E4}}^{DT}((\overline{u}_{X})_{i}\otimes _{R}DT).\nonumber\end{eqnarray}$$

Hence, we see that the equality (†₁) holds.

Since $a_{i}\cdot \unicode[STIX]{x1D706}_{i}^{2}=0$ and

$$\begin{eqnarray}(-q_{i},(\overline{u}_{X})_{i},0)\cdot \unicode[STIX]{x1D719}_{i}\cdot \unicode[STIX]{x1D706}_{i}^{2}=(-q_{i},(\overline{u}_{X})_{i},0)\cdot \unicode[STIX]{x1D709}_{i}^{b}=0,\end{eqnarray}$$

we also get that

(†₂)$$\begin{eqnarray}\unicode[STIX]{x1D702}_{_{\overline{X}_{i}}}^{DT}\cdot a_{i}\cdot \unicode[STIX]{x1D706}_{i}^{2}=(-q_{i},(\overline{u}_{X})_{i},0)\cdot \unicode[STIX]{x1D719}_{i}\cdot \unicode[STIX]{x1D706}_{i}^{2}.\end{eqnarray}$$

Now, from the property of the pullback in 4.3.5.2, we know that the two equalities (†₁) and (†₂) together imply that

$$\begin{eqnarray}\unicode[STIX]{x1D702}_{_{\overline{X}_{i}}}^{DT}\cdot a_{i}=(-q_{i},(\overline{u}_{X})_{i},0)\cdot \unicode[STIX]{x1D719}_{i}.\end{eqnarray}$$

Thus, the left part of the diagram is also commutative.

4.3.6 The isomorphism $\unicode[STIX]{x1D719}:X\oplus \text{L}_{R}(P_{A_{X}}^{+})\rightarrow \mathbf{Q}_{DT}\mathbf{S}_{T}(X)$ is natural on $X$

For any $X,Y\in {\mathcal{R}}{\mathcal{C}}(R)$ and $h\in \text{Hom}_{{\mathcal{R}}{\mathcal{C}}(R)}(X,Y)$, there is an induced morphism $h_{A}\in \text{Hom}_{{\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)}(A_{X},A_{Y})$, following from the construction in 4.1.4. Moreover, following from the construction in 4.3.1 and the definition of $P_{A_{X}}^{+}$ in 4.3.2, we see that the morphism $h_{A}$ induces a morphism $p_{h_{A}}^{+}\in \text{Hom}_{{\mathcal{R}}{\mathcal{C}}^{\text{tr}}(R)}(P_{A_{X}}^{+},P_{A_{Y}}^{+})$. Then, one can prove that

$$\begin{eqnarray}\left(\begin{array}{@{}cc@{}}h & 0\\ 0 & \text{L}_{R}(p_{h_{A}}^{+})\end{array}\right):X\oplus \text{L}_{R}(P_{A_{X}}^{+})\rightarrow Y\oplus \text{L}_{R}(P_{A_{Y}}^{+})\end{eqnarray}$$

is a morphism in ${\mathcal{R}}{\mathcal{C}}(R)$.

It is not hard to show that the following diagram is commutative:

Thus, the isomorphism $\unicode[STIX]{x1D719}$ is natural on $X$. This means that $\mathbf{Q}_{DT}\mathbf{S}_{T}\simeq 1_{\text{}\underline{{\mathcal{R}}{\mathcal{C}}}(R)}$ naturally.