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 $Hom\_A(M^-,M)$?



How to get that one module is tilting iff the other one is?
From the comments, it seems that you know how to prove equivalence for the first two conditions for being a tilting module.


For the last condition you can use $\mathfrak{C}(D(\Lambda))= \mathfrak{C}(\operatorname{Hom}\_A(M^-,M))$, since the final condition for a $\Lambda^{\mathrm{op}}$-module $T$ to be tilting is that $\Lambda^{\mathrm{op}}\in\mathfrak{C}(T)$ and $\operatorname{add}{T}$ contains all projective-injective modules. (This second part is true automatically for both $D(\Lambda)$ and $\operatorname{Hom}\_A(M^-,M)$.)


