Let $A$ be an algebra over a field k. $D$ is the standard duality functor. A module $_AM$ is called a generator if $add(A) \subseteq add(M)$, a cogenerator if $add(D(A)) \subseteq add(M)$. $M$ is n-rigid if $Ext_A^i(M,M)=0$ for $1 \leq i \leq n$. Now suppose $_AM$ is a generator-cogenerator which is n-rigid and neither projective nor injective. $\Lambda := End_A(M)$.

Suppose there is an long exact sequence of $\Lambda^{op}$-modules $$0 \rightarrow Hom_A(M^-,M) \rightarrow D(E_{n+1}) \rightarrow \cdots \rightarrow D(E_1) \rightarrow D(E_0) \rightarrow D(_{\Lambda} \Lambda) \rightarrow 0$$ such that $D(E_i)$ are projective-injective $\Lambda$-modules for $0 \leq i \leq n+1$. Let $\mathfrak{C}(Y)$ be the smallest triangulated subcategory of $\mathfrak{D}^b(\Lambda^{op})$ which is closed under direct summands and contains $Y$ and all projective-injective modules (here $\mathfrak{D}^b(\Lambda^{op})$ is the bounded derived category of complexes over $\Lambda ^{op}$-modules). Then $\mathfrak{C}(D(\Lambda))= \mathfrak{C}(Hom_A(M^-,M))$. Up to multiplicity and isomorphism, we know all projective-injective modules occur as direct summands of all tilting modules. Then how to get that $D(\Lambda)$ is a tilting module if and only if so is $Hom_A(M^-,M)$?