Let $A$ be an algebra over a field k. A module $\_AM$ is called a generator if $\textrm{add}(A) \subseteq \textrm{add}(M)$, a cogenerator if $\textrm{add}\big(D(A)\big) \subseteq \textrm{add}(M)$. $M$ is $n$-rigid if $\textrm{Ext}\_A^i(M,M)=0$ for $1 \leq i \leq n$. 


Now suppose $\_AM$ is a generator-cogenerator which is $n$-rigid and neither projective nor injective. $\Lambda := \textrm{End}\_A(M)$. If $\Lambda$ is at most $(n+m+2)$-Gorenstein, that is the injective dimension $\textrm{inj.dim}(\_{\Lambda} \Lambda)=\textrm{inj.dim} (\Lambda\_{\Lambda}) \leq (n+m+2)$, then for any $\Lambda$-module $N$, how to get that $\Omega^2\_{\Lambda}(N) \cong \textrm{Hom}\_A(M,Y)$ for some $A$-module $Y$(here, $\Omega$ is the syzygy functor)?



How to get that $\Omega^2_{\Lambda}(N) \cong \textrm{Hom}_A(M,Y)$?
This is an application of the Yoneda lemma. There is an exact sequence


$$0\to\Omega^2\_\Lambda(N)\to P\_1\to P\_0\to N\to0$$


with $P\_i$ projective. By definition of $\Lambda$, it follows that $P\_i=\operatorname{Hom}\_A(M,M\_i)$ for some $M\_i\in\operatorname{add}{M}$. (Here I am making the additional assumption that $N$ is finitely presented, which you probably want; if, for example, $A$ and $M$ are finite-dimensional over $k$, then this is automatic.) By Yoneda, the map $P\_1\to P\_0$ is the image under the left exact functor $\operatorname{Hom}\_A(M,-)$ of some $f\colon M\_1\to M\_0$, and so its kernel is $\operatorname{Hom}\_A(M,Y)$, where $Y$ is the kernel of $f$.


Note that this didn't use any of the assumptions on $M$, or the Gorenstein-ness of $\Lambda$.


