I was confused by the sequence of modules in [Yekiutieli and Zhang, Rings with Auslander dualizing complex, p.33, l.-10]. The question is that: why is the second morphism right multiplication of $t$?
Explicitly, assume $A$ is a noetherian connected graded algebra that has a balanced dualizing complex $R$. Assume $t\in A$ is a homogeneous regular normal element of positive degree $l$. Then it gives a graded automorphism $\sigma\colon A\to A$ such that $ta=\sigma(a)t$. Assume $M$ is a $t$-torsionfree graded left-module over $A$. Then one has a short exact sequence $$ 0\to {}^{\sigma^{-1}}M(-l)\xrightarrow{\lambda_t}M\to M/tM \to 0, $$ where $\lambda_t: M\to M$ is the left multiplication of $t$. One has a morphism $$ \lambda_t^*:R\mathrm{Hom}_A(M,R)\to R\mathrm{Hom}_A({}^{\sigma^{-1}}M(-l),R)= R\mathrm{Hom}_A(M,{}^{\sigma} R(l)). $$ Let $\lambda'_t:R\to R$ be the morphism given by left multiplication by $t$. Then $$ \lambda_t^*=(\lambda'_t)_*. $$ We also know that there is an isomorphism $\phi:{}^\sigma R\cong R^{\sigma^{-1}}$ in $D(A^e)$. Let $\rho_t:R\to R$ be the right multiplication by $t$. The question can be reduced to: does $\phi\circ \lambda'_t=\rho_t: R\to R^{\sigma^{-1}}(l)$?
I will be grateful of any comments.
