is a non-zero finitely generated $R$-module such that ${{\mathrm{G}_{C}\text{-}\!\dim}}_R(M) = r < \infty$ and $M\in{{\mathcal{B}}_C}(R)$. Then Lemma \[lem0701x\] implies that ${\mathrm{G}\text{-}\!\dim}_R({\operatorname{Hom}_{R}(C,M)})=r<\infty$. Since $M\neq 0$, we have ${\operatorname{Hom}_{R}(C,M)}\neq 0$, by Fact \[fact110527a\](b), so Theorem \[thm110527a\] provides an $R$-regular sequence $\underline z = z_1, \ldots, z_r$, integers $n_0, n_1, \ldots , n_r  {\geqslant}0$ such that $n{\geqslant}1$, and an exact sequence $$\label{eq110530c}
0 \to {\operatorname{Hom}_{R}(C,M)} \to Z' \to N' \to 0$$ with $Z' = \oplus_{i=0}^r(R/(\underline z_{[i]}))^{n_i}$ and ${\mathrm{G}\text{-}\!\dim}_R(N') {\leqslant}r$.

The condition $M\in{{\mathcal{B}}_C}(R)$ implies that ${\operatorname{Hom}_{R}(C,M)}\in{{\mathcal{A}}_C}(R)$ by Fact \[disc01xy\](c), and Fact \[disc01xy\](a) implies that $Z'\in{{\mathcal{A}}_C}(R)$. Also from Fact \[disc01xy\](a)–(b), we conclude that $N'\in{{\mathcal{A}}_C}(R)$, which implies that ${\operatorname{Tor}^{R}_{1}(C,N')}=0$. Thus, an application of ${C\otimes_{R}-}$ to the sequence  yields the next exact sequence: $$\label{eq110530d}
0 \to {C\otimes_{R}{\operatorname{Hom}_{R}(C,M)}} \to {C\otimes_{R}Z'} \to {C\otimes_{R}N'} \to 0.$$ The assumption $M\in{{\mathcal{B}}_C}(R)$ gives an isomorphism ${C\otimes_{R}{\operatorname{Hom}_{R}(C,M)}}\cong M$. The equality $Z' = \oplus_{i=0}^r(R/(\underline z_{[i]}))^{n_i}$ implies that $Z:={C\otimes_{R}Z'} = \oplus_{i=0}^r(C/(\underline z_{[i]})C)^{n_i}$.

The condition $N'\in{{\mathcal{A}}_C}(R)$ implies that $N:={C\otimes_{R}N'}\in{{\mathcal{B}}_C}(R)$ by Fact \[disc01xy\](c), and $N'\cong{\operatorname{Hom}_{R}(C,N)}$ by the definition of what it means for $N'$ to be in ${{\mathcal{A}}_C}(R)$. Thus, because of Lemma \[lem0701x\], we have $${{\mathrm{G}_{C}\text{-}\!\dim}}_R(N)={\mathrm{G}\text{-}\!\dim}({\operatorname{Hom}_{R}(C,N)})={\mathrm{G}\text{-}\!\dim}_R(N') {\leqslant}r.$$ Thus, the sequence  satisfies the conclusion of Theorem \[thm110527b\].

Proofs of Theorems \[thm110608a\] and \[thm110530b\] {#sec110608a}
====================================================

We continue with a 