# How to show that $DA\cong D\operatorname{Hom}_{B}(T, T) \cong DT \otimes_{B} T$?

I am reading the book Elements of the Representation Theory of Associative Algebras: Volume 1: Techniques of Representation Theory by Ibrahim Assem, Daniel Simson, Andrzej Skowronski.

Let $A$ be a $K$-algebra and $T_A$ be a tilting module. Let $B=\operatorname{End}_A(T_A)$. Let $D$ be the standard dual. That is $D(M)=\operatorname{Hom}_{K}(M, K)$.

On page 206, line 1, 2 of the proof of (b) of Lemma 3.7. How to show that $DA\cong D\operatorname{Hom}_{B}(T, T) \cong DT \otimes_{B} T$?

I think that $$D\operatorname{Hom}_{B}(T, T) \cong D (D (DT \otimes_{B} T )) \cong DT \otimes_{B} T.$$ Do we have $\operatorname{Hom}_{B}(DT, DT) \cong D\operatorname{Hom}_{B}(T, T)$? If this is true, then we have $D\operatorname{Hom}_{B}(T, T) \cong DT \otimes_{B} T$. How to show that $DA\cong D\operatorname{Hom}_{B}(T, T)$? Thank you very much.

• if $T_A$ is tilting, what can you say about the map $A\rightarrow End_B(T_A)$, $a\mapsto (t\mapsto t\cdot a)$? Note that now $T_A$ is an $B$-$A$-bimodule. – Avitus Jul 6 '13 at 11:50
• Avitus' comment is explained in detail in the proof of Lemma 3.3 (c) of your book. – Matthew Towers Jul 6 '13 at 12:56
• The first isomorphism should be clear, as Avitus and mt_ commented. Note that the Hom is taken over left $B$-module in the RHS of this isomorphism. The second isomorphism is also explained in the first line of the proof of part (b): $Hom_B(X,DT)=Hom_B(X,Hom_k(T,k))=Hom_k(T\otimes_B X, k)=D(T\otimes_BX)$, then you substitute $X$ by $T$. The reason for the second equality (isomorphism) here is tensor-hom adjunction. – Aaron Jul 8 '13 at 22:04
• $Hom_B(DT,DT)\simeq Hom_B(T,T)$ as $k$-vectorspaces, as $D$ is a duality, note that if $T$ is a left $B$-module, then the Hom space in the RHS of this isom is taken over the category of left $B$-module, in contrast to the LHS, where Hom is taken over right $B$-modules. Now the later abelian group is obviously isomorphic to $DHom_B(T,T)$ as $k$-vectorspace. However, if you consider $A$-module structure of the two spaces, they are different, since one is a left module, the other is a right. – Aaron Jul 8 '13 at 22:08
• @Aaron, thank you very much. – LJR Jul 9 '13 at 1:31

The first isomorphism follows from the fact that $A\cong Hom_B(T_B,T_B)$, as Avitus and mt_ commented, it's proved in earlier part of the book.
The second isomorphism is also explained in the first line of the proof of part (b): $Hom_B(X,DT)=Hom_B(X,Hom_k(T,k))=Hom_k(T⊗BX,k)=D(T⊗BX)$, then you substitute $X$ by $T$. The reason for the second equality (isomorphism) here is tensor-hom adjunction.