Proof. For (a), see [Reference CruzCru24c, Proposition 2.4.4.]. For (b), see [Reference CruzCru24c, Theorem 6.0.1, Theorem 6.2.1].

Proof. Assume that $\chi _M$ is surjective. Since $\operatorname {Hom}_A(Q, M)\in B\!\operatorname {-mod} $ , there exists $X\in \operatorname {add} Q$ and a surjective map $\operatorname {Hom}_A(Q, X)\twoheadrightarrow \operatorname {Hom}_A(Q, M)$ , say g. The functor $Q\otimes _B -$ is right exact, so $Q\otimes _B g$ is surjective as well. Define $f:=\chi _M\circ Q\otimes _B g \circ \chi ^{-1}_X\in \operatorname {Hom}_A(X, M)$ . The map f is surjective, and it satisfies $\operatorname {Hom}_A(Q, f)=g$ . So $Df$ is an $(A, R)$ -monomorphism $DM\rightarrow DX$ which remains exact under $\operatorname {Hom}_A(-, DQ)$ . Thus, $DQ\!\operatorname {-domdim}_{(A, R)} DM\geq 1$ . Conversely, assume that ${DQ\!\operatorname {-domdim}_{(A, R)} DM\geq 1}$ . So, there exists $X\in \operatorname {add} DQ$ and an $(A, R)$ -monomorphism $f\colon DM\rightarrow X$ which is also a left $\operatorname {add} DQ$ -approximation. Since $\chi $ is a natural transformation between $Q\otimes _B \operatorname {Hom}_A(Q, -)$ and $\operatorname {id}_{B\!\operatorname {-mod}}$ , the map $Df\circ \chi _{DX}={\chi _{DDM}\circ Q\otimes _B \operatorname {Hom}_A(Q, Df)}$ is surjective. In particular, $\chi _{DDM}$ is surjective. As $DDM\simeq M$ , $\chi _M$ is surjective, and (i) follows.

Now, assume that $DQ\!\operatorname {-domdim}_{(A, R)} DM\geq 2$ . Then, there exists an $(A, R)$ -exact sequence $0\rightarrow DM\xrightarrow {f_0} X_0\xrightarrow {f_1} X_1$ , with $X_0, X_1\in \operatorname {add} DQ$ , which remains exact under $\operatorname {Hom}_A(-, DQ)$ . As $Q\otimes _B -$ is right exact, the following diagram is commutative with exact rows:

(2)

By diagram chasing, $\chi _{DDM}$ is an isomorphism. Since $DDM\simeq M$ , $\chi _M$ is an isomorphism. Conversely, assume that $\chi _M$ is an isomorphism. B is a Noetherian R-algebra, so we can consider a projective B-presentation for $\operatorname {Hom}_A(Q, M)$ of the form for some integers $m, n$ . Since $\operatorname {Hom}_A(Q, -)_{|_{\operatorname {add} T}}$ is full and faithful, there exists $f_1\in \operatorname {Hom}_A(Q^m, Q^n)$ such that $\operatorname {Hom}_A(Q, f_1)=g_1$ . Fix $f_0=\chi _M\circ Q\otimes _B g_0 \circ \chi _{Q^n}^{-1}$ . In particular, $\operatorname {Hom}_A(Q, f_0)=g_0$ . Hence,

(3)

is a commutative diagram. Since the vertical maps are isomorphisms and the upper row is exact, it follows that the bottom row is exact and by construction, and it remains exact under $\operatorname {Hom}_A(Q, -)$ . As $M\in R\!\operatorname {-proj}$ , it is, also, $(A, R)$ -exact. By applying the standard duality D, we obtain that $DQ\!\operatorname {-domdim}_{(A, R)} DM\geq 2$ .

Proof. We shall prove (i). Assume that $DQ\!\operatorname {-domdim}_{(A, R)} DM\geq n\geq 2$ . By Theorem 3.1.1, $\chi _M$ is an isomorphism. By definition, there exists an $(A, R)$ -exact sequence $ { \delta \colon DM\hookrightarrow X_0\rightarrow X_1\rightarrow \cdots \rightarrow X_{n-1}} $ which remains exact under $\operatorname {Hom}_A(-, DQ)$ with $X_i\in \operatorname {add} DQ$ , $i=0, \ldots , n-1$ . In particular, ${\operatorname {Hom}_A(X_{n-1}, DQ)\rightarrow \cdots \rightarrow \operatorname {Hom}_A(X_0, DQ)\rightarrow \operatorname {Hom}_A(DM, DQ)\rightarrow 0}$ is exact and can be continued to a left projective B-resolution of $\operatorname {Hom}_A(Q, M)$ . Consider the following commutative diagram:

(4)

Observe that the bottom row is exact since the exact sequence $\delta $ is $(A, R)$ -exact. Since all vertical maps are isomorphisms, it follows that the upper row is exact. Thus, $\operatorname {Tor}_i^B(Q, \operatorname {Hom}_A(Q, M))=0$ , $1\leq i\leq n-2$ .

Conversely, assume that $\chi _M$ is an isomorphism and $\operatorname {Tor}_i^B(Q, \operatorname {Hom}_A(Q, M))=0$ for $1\leq i\leq n-2$ . Let $\operatorname {Hom}_A(Q, X_{n-1}) \xrightarrow {g_{n-1}} \cdots \rightarrow \operatorname {Hom}_A(Q, X_0)\xrightarrow {g_0} \operatorname {Hom}_A(Q, M)\rightarrow 0$ be a truncated projective B-resolution of $\operatorname {Hom}_A(Q, M)$ and $X_i\in \operatorname {add}_A Q$ . Furthermore, $\operatorname {Hom}_A(Q, -)_{|_{\operatorname {add} Q}}$ is full and faithful, so each map $g_i$ can be written as $\operatorname {Hom}_A(Q, f_i)$ including $g_0$ since $\chi _M$ is an isomorphism, where $f_i\in \operatorname {Hom}_A(X_i, X_{i-1})$ and $f_0\in \operatorname {Hom}_A(X_0, M).$ So, we have a commutative diagram

(5)

By assumption, $\operatorname {Tor}_i^B(Q, \operatorname {Hom}_A(Q, M))=0$ , $1\leq i\leq n-2$ . So, the upper row is exact. By the exactness and the vertical maps being isomorphisms, the bottom row becomes exact. Since $M\in R\!\operatorname {-proj}$ , it is also $(A, R)$ -exact, and so it remains $(A, R)$ -exact under D. By construction, the bottom row remains exact under $\operatorname {Hom}_A(Q, -)$ ; thus, $DQ\!\operatorname {-domdim}_{(A, R)} DM\geq n\geq 2$ . The statement (ii) is analogous to (i).

Proof. Let S be a commutative R-algebra. Denote by $D_S$ the standard duality with respect to S, $\operatorname {Hom}_S(-, S)$ . The result follows from the following commutative diagram:

(6)

where the map $\theta _{S, M}$ is the isomorphism given in Proposition 2.3 of [Reference CruzCru22b], while $\varphi _S$ is the tensor product of the canonical map given by extension of scalars on $\operatorname {Hom}$ (which is not claimed at the moment to be an isomorphism) with the one providing the isomorphism $S\otimes _R DQ\simeq D_S(S\otimes _RQ)$ .

The implications $(b)\Rightarrow (c)\Rightarrow (d)$ are immediate. Assume that $(a)$ holds. Then, $\chi _{DM}^r$ is surjective. By the commutative diagram, $\chi _{D_SS\otimes _R M}^r$ is surjective, and so $(b)$ follows. Assume that $(d)$ holds. By condition 2, $\varphi _{R(\mathfrak {m})}$ must be an isomorphism for every $\mathfrak {m}\in \operatorname {MaxSpec} R$ . Thus, by the diagram, $\chi ^r_{DM}(\mathfrak {m})$ is surjective for every maximal ideal $\mathfrak {m}$ of R. By Nakayama’s Lemma, $\chi ^r_{DM}$ is surjective and $(a)$ holds.

Proof. By the flatness of S, the vertical maps of the commutative diagram (6) are isomorphisms. So, by Lemma 3.2.1, the implication $(a)\Rightarrow (b)$ is clear for $n=1, 2$ . Again, since S is flat and B is finitely generated projective over R, $S\otimes _R -$ commutes with $\operatorname {Tor}$ functors over B. Therefore, $(b)$ follows by Theorem 3.1.4. Analogously, we obtain $(c)\Rightarrow (a)$ .

Proof. Of course, $\overline {k}$ is free over k. Therefore, $\operatorname {Tor}_i^B(\operatorname {Hom}_A(DQ, DM), DQ)=0$ if and only if $\operatorname {Tor}_i^{\overline {k}\otimes _k B}(\operatorname {Hom}_{\overline {k}\otimes _k A}(\overline {k}\otimes _k DQ, \overline {k}\otimes _k DM), \overline {k}\otimes _k DQ)=0$ . By the same reason, $\chi ^r_{DM}$ is surjective (or bijective) if and only if $\chi ^r_{\overline {k}\otimes _k DM}$ is surjective (or bijective).

Proof. Assume that $Q\!\operatorname {-domdim}_{(A, R)} M\geq n\geq 1$ for some $n\in \mathbb {N}$ . Then, there exists an $(A, R)$ -exact sequence $\delta \colon 0\rightarrow M\rightarrow X_1\rightarrow \cdots \rightarrow X_n$ which remains exact under $\operatorname {Hom}_A(-, Q)$ , where $X_i\in \operatorname {add}_A Q$ . The functor $S\otimes _R -$ preserves R-split exact sequences. Hence, $0\rightarrow S\otimes _R M\rightarrow S\otimes _R X_1\rightarrow \cdots \rightarrow S\otimes _R X_n$ is $(S\otimes _R A, S)$ -exact and $S\otimes _R X_i\in \operatorname {add}_{S\otimes _R A}S\otimes _RQ$ . Since $\operatorname {Hom}_A(M, Q)\in R\!\operatorname {-proj}$ , the sequence $\operatorname {Hom}_A(\delta , Q)$ splits over R, and thus, $S\otimes _R \operatorname {Hom}_A(\delta , Q)$ is an exact sequence. Using the commutative diagram

it follows that the bottom row is exact. Hence, $S\otimes _RQ\!\operatorname {-domdim}_{(A, R)} S\otimes _R M\geq n$ .

Proof. By Lemma 3.2.4, $Q(\mathfrak {m})\!\operatorname {-domdim}_{A(\mathfrak {m})} M(\mathfrak {m})\geq Q\!\operatorname {-domdim}_{(A, R)} M$ for every maximal ideal $\mathfrak {m}$ of R. Conversely, assume that $Q(\mathfrak {m})\!\operatorname {-domdim}_{A(\mathfrak {m})} M(\mathfrak {m})\geq n$ for every maximal ideal $\mathfrak {m}$ of R. We want to show that $Q\!\operatorname {-domdim}_{(A, R)} M\geq n$ . If $n=0$ , then there is nothing to show. Using the commutative diagram (6), we obtain that if $n\geq 1$ (resp. $n\geq 2$ ), then $\chi ^r_{DM}(\mathfrak {m})$ is surjective (resp. bijective) for every maximal ideal $\mathfrak {m}$ of R. By Nakayama’s Lemma, $\chi ^r_{DM}$ is surjective, and since $DM\in R\!\operatorname {-proj}$ , $\chi ^r_{DM}$ is bijective in case $n\geq 2$ . So, the inequality holds for $n=1, 2$ . Assume now that $n\geq 3$ . In particular, $Q\!\operatorname {-domdim}_{(A, R)} M\geq 2$ , and therefore, $\operatorname {Hom}_A(DQ, DM)\otimes _B DQ\in R\!\operatorname {-proj}$ . By assumption, $\operatorname {Tor}_i^{B(\mathfrak {m})}(\operatorname {Hom}_{A(\mathfrak {m})}(M(\mathfrak {m}), Q(\mathfrak {m})), D_{(\mathfrak {m})}Q(\mathfrak {m}))=0$ for $1\leq i\leq n-2$ for every maximal ideal $\mathfrak {m}$ of R. Let $Q^{\bullet }$ be a deleted B-projective resolution of $DQ$ . So the chain complex $P^{\bullet }=\operatorname {Hom}_A(M, Q)\otimes _B Q^{\bullet }$ is a projective complex over R since $\operatorname {Hom}_A(M, Q)\in R\!\operatorname {-proj}$ . Consider the Künneth spectral sequence for chain complexes (see, for example, [Reference WeibelWei03, Theorem 5.6.4])

(7) $$ \begin{align} E^2_{i, j}=\operatorname{Tor}_j^R(H_i(\operatorname{Hom}_A(M, Q)\otimes_B Q^{\bullet}), R(\mathfrak{m}))\Rightarrow H_{i+j}(\operatorname{Hom}_A(M, Q)\otimes_B Q^{\bullet}(\mathfrak{m})). \end{align} $$

Since $\operatorname {Hom}_A(M, Q)\otimes _B DQ\in R\!\operatorname {-proj}$ , $\operatorname {Hom}_A(M, Q)\otimes _B Q^{\bullet }(\mathfrak {m})$ becomes a deleted projective $B(\mathfrak {m})$ -resolution of $DQ(\mathfrak {m})$ .

We shall proceed by induction on $1\leq i\leq n-2$ to show that $\operatorname {Tor}_i^B(\operatorname {Hom}_A(M, Q), DQ)=0$ . Observe that $\operatorname {Tor}_1^B(\operatorname {Hom}_A(M, Q), DQ)\otimes _R R(\mathfrak {m})=0$ for every $\mathfrak {m}\in \operatorname {MaxSpec} R$ (see for example [Reference CruzCru22b, Lemma A.3]). Hence, $\operatorname {Tor}_1^B(\operatorname {Hom}_A(M, Q), DQ)=0$ . Assume now that $\operatorname {Tor}_i^B(\operatorname {Hom}_A(M, Q), DQ)=0$ for all $1\leq i\leq l$ with $1\leq l\leq n-2$ for some l. Then, $E_{i, j}^2=0$ , for $1\leq i\leq l$ , $j\geq 0$ and $E_{0, j}^2=0$ , $j>0$ . It follows that $\operatorname {Tor}_{l+1}^B(\operatorname {Hom}_A(M, Q), DQ)(\mathfrak {m})=0$ for every $\mathfrak {m}\in \operatorname {MaxSpec} R$ (see [Reference CruzCru22b, Lemma A.4]). Therefore, $\operatorname {Tor}_i^B(\operatorname {Hom}_A(M, Q), DQ)=0$ for $1\leq i\leq n-2$ . By Theorem 3.1.4, the result follows.

Proof. For each $\mathfrak {m}\in \operatorname {MaxSpec} R$ ,

$$ \begin{align*} \operatorname {Tor}_1^{A(\mathfrak {m})}(DQ(\mathfrak {m}), Q(\mathfrak {m}))=\operatorname {Hom}_{R(\mathfrak {m})}(\operatorname {Ext}_{A(\mathfrak {m})}^1(Q(\mathfrak {m}), Q(\mathfrak {m})), R(\mathfrak {m}))=0. \end{align*} $$

Let $Q^{\bullet }$ be a deleted projective A-resolution of Q. Since $Q\in R\!\operatorname {-proj}$ , $Q^{\bullet }(\mathfrak {m})$ is a deleted projective $A(\mathfrak {m})$ -resolution of $Q(\mathfrak {m})$ . Consider the Künneth Spectral sequence with $P=DQ\otimes _A Q^{\bullet }$ , (see, for example, [Reference WeibelWei03, Theorem 5.6.4]) $E^2_{i,j}=\operatorname {Tor}_i^R(\operatorname {Tor}_j^A(DQ, Q), R(\mathfrak {m})) = \operatorname {Tor}_i^R(H_j(DQ\otimes _A Q^{\bullet }), R(\mathfrak {m}))$ . It converges to $H_{i+j}(DQ\otimes _A Q^{\bullet }\otimes _R R(\mathfrak {m}))=\operatorname {Tor}_{i+j}^{A(\mathfrak {m})}(DQ(\mathfrak {m}), Q(\mathfrak {m})).$ Moreover,

$$ \begin{align*} E_{1, 0}^2=\operatorname {Tor}_1^R(DQ\otimes _A Q, R(\mathfrak {m}))=0 \end{align*} $$

for every maximal ideal $\mathfrak {m}$ of R (see for example [Reference CruzCru22b, Lemma A.3]). Hence, $DQ\otimes _A Q\in R\!\operatorname {-proj}$ .

Proof. By Theorem 3.1.1, $\chi _{X}$ is an isomorphism for every $X\in \mathcal {X}$ . Fix $F_Q=\operatorname {Hom}_A(Q, -)$ and $\mathbb {I}$ its left adjoint. Then, if $F_Qf=0$ for some $f\in \operatorname {Hom}_{\mathcal {X}}(M, N)$ , we obtain $f\circ \chi _M=\chi _N\circ \mathbb {I}F_Qf=0$ . Hence, in such a case, $f=0$ . So, $F_Q$ is faithful. To show fullness, let $g\in \operatorname {Hom}_B(F_QM, F_QN)$ with $M,N\in \mathcal {X}$ . Fixing $h=\chi _N \circ \mathbb {I} g\circ \chi _M^{-1}$ , we get $F_Qh=g$ .

Proof. Denote by $\psi $ the canonical map of algebras $A\rightarrow \operatorname {End}_B(Q)^{op}$ and denote by $\omega _X$ the natural transformation between the identity functor and the double dual $DD$ for $X\in A\!\operatorname {-mod}$ . The equivalence $(i)\Leftrightarrow (ii)$ follows from the following commutative diagram

with horizontal maps being isomorphisms, where $\sigma _Q\colon \operatorname {Hom}_A(Q, DA)\rightarrow DQ$ and $\theta $ are isomorphisms given by Tensor-Hom adjunction.

For each $M, N\in A\!\operatorname {-mod}$ , consider the map $F_{M, N}\colon \operatorname {Hom}_A(M, N)\rightarrow \operatorname {Hom}_B(F_QM, F_QN)$ , given by $f\mapsto F_Qf=\operatorname {Hom}_A(Q, f)$ . Observe that the maps $\psi $ and $F_{DA, DA}$ are equivalent since the isomorphisms $A\simeq \operatorname {Hom}_A(A, A)\simeq \operatorname {Hom}_A(DA, DA)$ and $\operatorname {Hom}_B(Q, Q)\simeq \operatorname {Hom}_B(DQ, DQ)\simeq \operatorname {Hom}_B(F_QDA, F_QDA)$ are functorial (see, for example, [Reference CruzCru22b, Proposition 2.2]). As $F_{M\oplus X, N}$ is equivalent to $F_{M, X}\oplus F_{X, N}$ and $F_{M, X\oplus N}$ is equivalent to $F_{M, X}\oplus F_{M, N}$ for every $M, X, N\in A\!\operatorname {-mod}$ it follows that $(iii)\Leftrightarrow (i)$ .

Proof. Denote by $R(A)$ the Ringel dual $\operatorname {End}_A(T)^{op}$ . Let $P\twoheadrightarrow T$ be a right projective presentation of T over $R(A)$ . Then, $P\in \mathcal {F}(\tilde {\Delta }_{R(A)^{op}})$ . Note that $T\simeq \operatorname {Hom}_A(A, T)\simeq \operatorname {Hom}_{A^{op}}(DT, DA)\in \mathcal {F}(\tilde {\Delta }_{R(A)^{op}})$ . Since $\mathcal {F}(\tilde {\Delta }_{R(A)^{op}})$ is resolving, so the kernel of $P\rightarrow T$ belongs to $\mathcal {F}(\tilde {\Delta }_{R(A)^{op}})$ . Since $\operatorname {Hom}_{A^{op}}(DT, -)$ gives an exact equivalence between $\mathcal {F}(\tilde {\nabla }_{A^{op}})$ and $\mathcal {F}(\tilde {\Delta }_{R(A)^{op}})$ , there exists an exact sequence of right A-modules $ 0\rightarrow K\rightarrow M' \rightarrow DA\rightarrow 0, $ where $M'$ is a (partial) tilting module and $K\in \mathcal {F}(\tilde {\nabla }_{A^{op}})$ . Applying D, we obtain the desired exact sequence. By [Reference RouquierRou08, Proposition 4.26], since $DK\in \mathcal {F}(\tilde {\Delta }_A)$ , there exists an exact sequence $0\rightarrow DK\rightarrow M"\rightarrow K"\rightarrow 0$ , where $M'$ is a (partial) tilting module and $K'\in \mathcal {F}(\tilde {\Delta }_A)$ . Put $Q=DM'\oplus M"$ . Hence, Q is (partial) tilting module, and the $(A, R)$ -exact sequence $0\rightarrow A\rightarrow DM'\rightarrow M"$ remains exact under $\operatorname {Hom}_A(Q, -)$ . This means that $Q\!\operatorname {-domdim} (A, R)\geq 2$ . By Corollary 3.1.6 and Lemma 5.1.2, the second assertion follows.

Proof. Let $X\in A\!\operatorname {-mod}$ such that $\operatorname {Ext}_A^{i>0}(Q, X)=0$ . The desired isomorphism for $i=0$ follows immediately by Tensor-Hom adjunction. To obtain the result for higher values, we will use Theorem 10.49 of [Reference RotmanRot09]. Fix $f=\operatorname {Hom}_A(-, X)$ and $g=Q\otimes _B -$ . So, f is a contravariant left exact and g is covariant. We note that $gP$ is f-acyclic for any $P\in B\!\operatorname {-proj}$ . In fact, $\operatorname {R}^{j>0}f(gP)=\operatorname {Ext}_A^{j>0}(gP, X)=0$ , since $gP=Q\otimes _B P\in \operatorname {add}_A Q$ . So, for each $a\in B\!\operatorname {-mod}$ , there is a spectral sequence $ E_2^{i, j}=(\operatorname {R}^if)(L_j g)(a)$ which converges to $\operatorname {R}^{i+j}(f\circ g)(a).$ By Tensor-Hom adjunction, $f\circ g(N)=\operatorname {Hom}_A(Q\otimes _B N, X)\simeq \operatorname {Hom}_B(N, \operatorname {Hom}_A(Q, X))=\operatorname {Hom}_B(-, F_QX)(N)$ for every $N\in B\!\operatorname {-mod}$ . Hence, the spectral sequence $E_2^{i, j}=\operatorname {Ext}_A^i(\operatorname {Tor}_j^B(Q, a), X)$ converges to $\operatorname {Ext}_B^{i+j}(a, F_QX).$ For each $M\in A\!\operatorname {-mod}$ , fix $a=F_QM$ . By assumption, $\operatorname {Tor}_i^B(Q, F_QM)=0$ for $1\leq i\leq q$ . Hence, $E_2^{i, j}=0$ , $1\leq i\leq q$ . By the dual of Lemma A.4 of [Reference CruzCru22b], the result follows.

Proof. The proof of $(a)$ is analogous to [Reference CruzCru24c, Theorem 6.2.1]. Thanks to $F_Q$ and D being exact on short exact sequences of modules belonging to $\mathcal {F}(\tilde {\nabla })$ , we obtain that we can apply Lemma 3.1.8 to the filtrations by costandard modules. Further, for every $X\in R\!\operatorname {-proj}$ so that $R^t\simeq X\oplus Y$ for some $t\in \mathbb {N}$ ,

$$ \begin{align*} Q\!\operatorname{-codomdim}_{(A, R)} \nabla(\lambda)&=Q\!\operatorname{-codomdim}_{(A, R)} \nabla(\lambda)^t\\&=\inf \{Q\!\operatorname{-codomdim}_{(A, R)} \nabla(\lambda)\otimes_R X, Q\!\operatorname{-codomdim}_{(A, R)} \nabla(\lambda)\otimes_R Y \}. \end{align*} $$

Therefore, $Q\!\operatorname {-codomdim}_{(A, R)} \displaystyle \bigoplus _{\lambda \in \Lambda }\nabla (\lambda )= Q\!\operatorname {-codomdim}_{(A, R)} \mathcal {F}(\tilde {\nabla })$ . Now using the exact sequences (1) together with Lemma 3.1.8 and the reasoning of Theorem [Reference CruzCru24c, Theorem 6.2.1], assertion (a) follows.

By Proposition 2.1.2, $DQ\otimes _A Q\in R\!\operatorname {-proj}$ . Any $(A,R)$ -injective module being projective over R belongs to $Q^\perp $ (see, for example, [Reference CruzCru22b, Proposition 2.10]). By Lemma 5.1.1 and Theorem 3.1.1, $F_Q$ is fully faithful on $\mathcal {F}(\tilde {\Delta })$ . By Theorem 3.1.4, $\operatorname {Tor}_i^B(Q, F_QM)=0$ , $1\leq i\leq n-2$ for every $M\in \mathcal {F}(\tilde {\nabla })$ . By Lemma 5.2.1, $\operatorname {Ext}_B^i(F_QM, F_QX)\simeq \operatorname {Ext}_A^i(\mathbb {I}F_QM, X)$ for $0\leq i\leq n-2$ , where $M, X\in \mathcal {F}(\tilde {\nabla })$ . Since $\chi _M$ is an isomorphism for every $M\in \mathcal {F}(\tilde {\nabla })$ , (b) follows.

By the exactness of $F_Q$ on $\mathcal {F}(\tilde {\nabla })$ and according to Lemma 2.0.1, $F_Q(\nabla (\lambda )\otimes _R X)\simeq F_Q\nabla (\lambda )\otimes _R X$ for every $\lambda \in \Lambda $ and $X\in R\!\operatorname {-proj}$ , and the restriction of the functor $F_Q$ on $\mathcal {F}(\tilde {\nabla })$ has image in $\mathcal {F}(F_Q\tilde {\nabla })$ . By (b), it is enough to prove that for each module M in $\mathcal {F}(F_Q\tilde {\nabla })$ , there exists $N\in \mathcal {F}(\tilde {\nabla })$ so that $F_QN\simeq M$ . By (b), the functor $\mathbb {I}=Q\otimes _B -$ is exact on short exact sequences of modules belonging to $\mathcal {F}(F_Q\tilde {\nabla })$ . Thanks to $Q\otimes _B F_Q\nabla (\lambda )\otimes _R X\simeq \nabla (\lambda )\otimes _R X$ for every $X\in R\!\operatorname {-proj}$ and $\lambda \in \Lambda $ , we obtain that $\mathbb {I}$ sends $\mathcal {F}(F_Q\tilde {\nabla })$ to $\mathcal {F}(\tilde {\nabla })$ . So, (c) follows.

Assume now that $Q\!\operatorname {-codomdim}_{(A, R)} T\geq n\geq 2$ . Fix $B=\operatorname {End}_A(Q)^{op}$ . By Ringel duality (see, for instance, [Reference CruzCru24b, Lemma 7.1]), for each $M\in \mathcal {F}(\tilde {\nabla })$ , we have

(10) $$ \begin{align} \operatorname{Hom}_{R(A)}(\operatorname{Hom}_A(T, Q), \operatorname{Hom}_A(T, M))\simeq \operatorname{Hom}_A(Q, M) \text{ as } (B, R(A))\text{-bimodules.} \end{align} $$

In particular, $\operatorname {Hom}_{R(A)}(\operatorname {Hom}_A(T, Q), R(A))\simeq \operatorname {Hom}_A(Q, T)$ as $(B, R(A))$ -bimodules. By (c), $F_Q$ is fully faithful on $\mathcal {F}(\tilde {\nabla })$ . Hence, $ \operatorname {End}_B(\operatorname {Hom}_A(Q, T))^{op}\simeq \operatorname {End}_A(T)^{op} $ and $\operatorname {End}_{R(A)}(\operatorname {Hom}_A(T, Q))^{op}\simeq B$ . So, $(R(A), \operatorname {Hom}_A(T, Q))$ is a split quasi-hereditary cover of B. By (b) and [Reference CruzCru24b, Lemma 7.1],

(11) $$ \begin{align} \operatorname{Ext}_B^i(\operatorname{Hom}_A(Q, T), &\operatorname{Hom}_{R(A)}(\operatorname{Hom}_A(T, Q), \operatorname{Hom}_A(T, M)))\simeq \operatorname{Ext}_B^i(\operatorname{Hom}_A(Q, T), \operatorname{Hom}_A(Q, M))\nonumber\\ &\simeq \operatorname{Ext}_B^i(F_QT, F_QM), \quad 0\leq i\leq n-2, \forall M\in \mathcal{F}(\tilde{\nabla}), \end{align} $$

and these $\operatorname {Ext}$ groups vanish if $1\leq i\leq n-2$ . By [Reference CruzCru24b, Lemma 7.1] and [Reference CruzCru24c, Proposition 3.0.3, 3.0.4], the first part of (d) follows. Since the maps $\operatorname {Hom}_A(T, DA)\rightarrow \operatorname {Hom}_B(F_QT, F_QDA)$ and ${DT\rightarrow \operatorname {Hom}_B(F_QT, DQ)}$ are equivalent, the second assertion follows by equations (10) and (11), together with Theorem 3.1.4 using $\operatorname {Ext}$ when R is a field (see [Reference Assem, Simson and SkowrońskiASS06, A.4]).