Inverse of a tilting module

Let $k$ be a field, $A$ an associative unital $k$-algebra, $\operatorname{\mathsf{Mod}} A$ the category of left $A$-modules and $D^b(\operatorname{\mathsf{Mod}} A)$ the bounded derived category. Let $A^{\circ}$ be the opposite algebra and $A^e := A \bigotimes_k A^{\circ}$ the enveloping algebra. Let $T$ be a two-sided tilting complex: $T ∈ D^b(\operatorname{\mathsf{Mod}} A^e)$.

How can I understand the structure of $T^{\wedge}:=\mathbb{R}\operatorname{Hom}_A(T,A)$, and why is $T^{\wedge} \bigotimes_{A}^{\mathbb L} T \simeq T \bigotimes_{A}^{\mathbb L} T^{\wedge} \simeq A$?

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Could you supply a definition of tilting complex? – Ben Webster Jan 26 '11 at 16:57
There are different definitions, for example: Let A be a ring. A tilting complex T over A is an object in Kb(P(A)) which satisfies the following conditions: (I) for all i ≠ 0, the set HomDb(A)(T, T[i]) of homomorphisms in Db(A) vanishes, (II) the category add(T) (that is, the full subcategory of all direct sums of direct summands of T inside Kb(P(A))) generates Kb(P(A)) as a triangulated category. By P(A) we denote the additive category of finitely generated projective left A-modules. – Alex Jan 26 '11 at 17:44
And by Kb(P(A)) the homotopy category of complexes of finite length – Alex Jan 26 '11 at 17:46
Are you sure the isomorphisms in your final question are correct? Left tensoring over A by A, oughtn't to do anything at all. – Hugh Thomas Jan 26 '11 at 18:35
Yes, thank you! I've corrected this error. – Alex Jan 26 '11 at 19:01

An object $X$ in $D^b (\operatorname{\mathsf{Mod}} (A \otimes B^{\mathrm{op}}))$ is called a two-sided tilting complex if $$X \otimes^{\mathbb L}_B - \colon D^b (\operatorname{\mathsf{Mod}} B) \rightarrow D^b (\operatorname{\mathsf{Mod}} A)$$ is an equivalence of triangulated categories.
(In the question $A=B$, but I'll ignore that.)
Let $X$ be a two-sided tilting complex. Then by adjointness, $$\mathbb R \operatorname{Hom}_A(X,-) \colon D^b (\operatorname{\mathsf{Mod}} A) \rightarrow D^b (\operatorname{\mathsf{Mod}} B)$$ is also an equivalence of triangulated categories. There are functorial isomorphisms \begin{align} \operatorname{Hom}_{D^b(\operatorname{\mathsf{Mod}} (A \otimes A^{\mathrm{op}}))}&(X \otimes^{\mathbb L}_B \mathbb R \operatorname{Hom}_A(X,A),-)\\ &\simeq \operatorname{Hom}_{D^b(\operatorname{\mathsf{Mod}} (B \otimes A^{\mathrm{op}}))}(\mathbb R \operatorname{Hom}_A(X,A),\mathbb R \operatorname{Hom}_A(X,-))\\ &\simeq \operatorname{Hom}_{D^b(\operatorname{\mathsf{Mod}} (A \otimes A^{\mathrm{op}}))}(A,-), \end{align} so $$X \otimes^{\mathbb L}_B \mathbb R \operatorname{Hom}_A(X,A) \\ \simeq A$$ in $D^b(\operatorname{\mathsf{Mod}} (A \otimes A^{\mathrm{op}}))$ by Yoneda's Lemma.
After first proving the isomorphism $$\mathbb R \operatorname{Hom}_A(X,A) \otimes^{\mathbb L}_A X \simeq \mathbb R \operatorname{Hom}_A(X,X \otimes^{\mathbb L}_B B),$$ a similar adjunction gives $$B \simeq \mathbb R \operatorname{Hom}_A(X,A) \otimes^{\mathbb L}_A X$$ in $D^b(\operatorname{\mathsf{Mod}} (B \otimes B^{\mathrm{op}}))$.