Let $A$ be a k-algebra, where k is a fixed field. Let $S$ be a simple, non-injective $A$-module such that $Ext^{i}_{A}(S,S)=0$ for $1 \leq i \leq n$. Let $P(S)$ be the projective cover of $S$, and let $Q$ be the direct sum of all non-isomorphic indecomposable projective $A$-module which are not isomorphic to $P(S)$. $\nu_A$ is the Nakayama functor.

Suppose $$0 \rightarrow S \rightarrow I(S) \rightarrow I_1 \rightarrow \cdots \rightarrow I_n \rightarrow I_{n+1} \rightarrow \cdots$$ is a minimal injective resolution of $S$, where $I(S)$ is the injective cover of $S$. Let $T':= \tau^{-1} \Omega_A ^{-n+1}(S)$, where $\tau^{-1}:=TrD$ is the Auslander-Reiten inverse translation. $\Omega^{-1}$ is the cosyzygy operator. Who can tell me why $T'$ doesn't have projetive direct summand?

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    $\begingroup$ It's just the way $\tau^{-1}$ is defined. $\tau^{-1}(M)$ for an indecomposable module $M$ is either zero or non-projective. $\endgroup$ – Dag Oskar Madsen Sep 24 '16 at 8:11
  • $\begingroup$ @Dag Oskar Madsen Get it, thank you. $\endgroup$ – Xiaosong Peng Sep 26 '16 at 9:30

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